Class 11 Commerce Applied Mathematics - Basics Of Financial Mathematics

During every financial year, the value of a machine depreciates by 10%. Find the original value (cost) of a machine which depreciates by Rs. 2,250 during the second year.

Let the original cost of the machine = Rs. 100

$\therefore$ Depreciation during 1st year

$=$ 10% of Rs. 100

$=$ Rs. 10
value of the machine at the beginning of 2nd year

$=Rs. 100-Rs.10=Rs. 90$

$\therefore$ Depreciation during 2nd year

$= 10% of Rs. 90 = Rs. 9$
Now, when depreciation during 2nd year

$=$ Rs. 9, original cost

$= Rs. 100$

$\Rightarrow$ when depreciation during 2nd year

$=Rs. 2,250,$
original cost

$=Rs. \frac{100}{9}\times 2,250 = RS. 25,000$ Ans.

The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally, how many bacteria will be present at the end of $2^{nd}$ hour $4^{th}$ hour and $n^{th}$ hour?

It is given that the number of bacteria doubles every hour.
Therefore, the number of bacteria after every hour will form a G.P.
with first term

$(a=30)$ and common ratio

$(r=2)$

$\displaystyle \therefore {a} _ {3} = {ar} ^ {2} = (30){(2)} ^ {2} =120$
Therefore, the number of bacteria at the end of 2nd hour will be 480.

$\displaystyle {a} _ {5} = {ar} ^ {4} = (30){(2)} ^ {4} =480$
and

${a} _ {n+1} = {ar} ^ {n} = (30){(2)} ^ {n}$
Thus the number of bacteria at the end of nth hour will be

$\displaystyle (30){(2)} ^ {n}$

A certain sum of money, placed out at compound interest, amounts to Rs.6,272 in 2 years and to Rs. 7,024.64 in 3 years. Find the rate of interest and the sum of money.

$We\quad know\quad that\quad Compound\quad Interest=Amount-Principal$

$Amount\quad for\quad 3rd\quad year-Amount\quad for\quad 2nd\quad year\quad will\quad be\quad the\quad interest\quad for\quad 3rd\quad year$

$Amount\quad for\quad 2nd\quad year\quad will\quad serve\quad as\quad Principal\quad for\quad 3rd\quad year$

$As\quad per\quad problem\quad 7024.64-6272=752.64(compound\quad interest)$

$\Rightarrow P{ \left( 1+\frac { r }{ 100 } \right) }^{ 3 }-P{ \left( 1+\frac { r }{ 100 } \right) }^{ 2 }=752.64$

$\Rightarrow P{ \left( 1+\frac { r }{ 100 } \right) }^{ 2 }\left[ { \left( 1+\frac { r }{ 100 } \right) }-1 \right] =752.64$

$\Rightarrow 6272\left[ \frac { r }{ 100 } \right] =752.64$

$\Rightarrow 6272r=75264$

$r=\frac { 75264 }{ 6272 } =12%$

Let the sum be P

AS PER PROBLEM,

$6272=P{ \left( 1+\frac { 12 }{ 100 } \right) }^{ 2 }$

$6272=P1.2544$

$P=\frac { 6272 }{ 1.2544 } =5000$

Sum of money is 5000 and rate of interest is 12 percent.

Reduction in price of vehicle and machinery with time is called _______

Any vehicle / machine which is in constant use undergoes wear and tear. This wear and tear has an effect on its resale value.

More the use, greater is the wear and tear and hence lower is the resale value.

Such a reduction in the price of the vehicle / machinery with time, is called depriciation.

Simple interest is given by $I=\dfrac{PTR}{100}$ .If p=Rs.900, T=2 years, and $R=5\%$ , find the simple interest.

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What is the rate of interest (%) of Rs.92,800 for $3\dfrac{1}{2}$ years , the interest of which is Rs.16240 ?

We know that:

$SI=\dfrac{PRT}{100}$ .

So,

$R=\dfrac{SI\times 100}{P\times T}$

$=\dfrac{16240\times 100}{92800\times 3.5}=5\%$

The interest is supposed compound unless the contrary is state.
A person borrows $672$ pounds to be repaid in $5$ years by annual instalments of $120$ pounds; find the rate of interest, reckoning simple interest.

Principal

$= 672$

Amount paid

$=6\times120=720$

Interest

$=720-672=48$

Simple Interest

$=\dfrac{PRT}{100}$

$\Rightarrow 48 = \dfrac{672\times R\times5}{100}$

$\Rightarrow R=1.43\text%$

Rate of depreciation is always ..................

Depreciation means a reduction in value of a certain machinery with time.

The older the machine, the higher is its depreciation.

Hence, it is always negative.

A certain sum of money amounts to $Rs. 6300/-$ in $2$ years and $Rs. 7875/-$ in $3$ years $9$ months at simple interest find the rate of interest per annum.

Let, Principle be

$Rs.P$ and Rate of interest be

$R$ % p.a.

Amount

$(A) = P\left(1 + \dfrac{TR}{100}\right)$

Given that,

$6300 = P\left(1 + \dfrac{2R}{100}\right)$ ...............(i)

$7875 = P\left(1 + \dfrac{15R}{400}\right)$ ...............(ii)

$\because T = 3\dfrac{9}{12} = \dfrac{15}{4}$ years

$\dfrac{(ii)}{(i)} \implies \dfrac{(1+\dfrac{15R}{400})}{(1 + \dfrac{2R}{100})} = \dfrac{7875}{6300}$

$\implies \dfrac{400 + 15R}{400 + 8R} = \dfrac{7875\div 1575}{6300\div 1575} = \dfrac{5}{4}$

$\implies 4(400 + 15R) = 5(400 + 8R)$

$\implies 1600 + 60R = 2000 + 40R$

$\implies 60R - 40R = 2000 - 1600$

$\implies 20R = 400$

$\therefore R = 20$ %(Ans)

A certain sum of money double itself & no. of years is equal to rate of interest. Find rate of interest?

$A=2P,r=t$

$A=P+(\cfrac { P\times r\times t }{ 100 })$

$\implies 2P=P+(\cfrac { P\times t\times t }{ 100 })\\\implies r^2=100\\\implies r=10\%$

Hence rate

$=10\%$ per annum.

A girl deposits $300$ in a bank account. The bank offers $7%$ intersection per year.
Assuming the girl does not take any money out of the account, calculate:

Let, A girl deposits in a bank account per mounth.= $300rs$

At one year total deposits money(p) = $12\times 300\,rs=3600\,rs$

Given interest per year= $7%$

Interest per month(R) = $\dfrac{7}{12}$ month.

T=12 month

Now , money after 12 month ?

Interest in 12 month

$=\dfrac{P\times R\times T}{100}$ =$$

$\dfrac{3600\times 12\times \dfrac{7}{12}}{100}=\dfrac{3600\times 7}{100}$

$=252rs$

Hence money will becomes in 12 month $3600+252=3852rs$

Find depreciation at $10\%\ pa$ on furniture costing $Rs\,10,000$ for $9$ month

In accounting terms, depreciation is defined as the reduction of recorded cost of a fixed asset in a systematic manner until the value of the asset becomes zero or negligible.

Depriciation

$=$ fixed price

$\times$ rate

$\times$ time

$(in\ years)$

$=10000\times \dfrac{10}{100}\times \dfrac9{12}$

$=Rs.\ 750$

At what rate percent will Rs. 1600 amounts to Rs. 2000 in 5 years.

r=? A=2000 , P=1600, T=5 YRS

S.I=A-P=400

$S.I = \dfrac{{P \times R \times T}}{{100}}$

$400 = \dfrac{{1600 \times R \times 5}}{{100}}$

$\dfrac{{400}}{{16 \times 5}} = R$

$R = 5\%$

The population of Hyderabad was $68,09,000$ in the year $2011$ . If it increases at the rate of $4.7\%$ per annum. What will be the population at the end of the year $2015$ .

The population of Hydrabad was

$6809000$ in the year

$2011$ .

It increases by

$4.7$ % per annum.

To calculate population after

$4$ years with a given rate of interest.

Using the compounding formula

$Z = A + {\left( {1 + \dfrac{r}{{100}}} \right)^t}$

Where

$Z$ is Population at the end of duration.

$A$ is present population

$t$ is time perior

$r$ is rate of intrest

We know that

$4.7%=0.047$

Substitute the values,

$Z = 6809000 + {\left( {1 + \frac{{4.7}}{{100}}} \right)^4}$

$Z = 6809000{\left( {1.047} \right)^4}$

$Z = 6809000 \times 1.2016741717$

$Z = 8182199$

Michael finished coloring a picture in $\dfrac{7}{{12}}$ hour. Vaibhav finished coloring the same picture in $\dfrac{3}{{4}}$ hour. Who worked longer. By what fraction was it longer

Michael painted

$=\dfrac{7}{12}$ hour

$=\dfrac{7}{12}\times 60=35 minutes$

Vaibhav painted

$=\dfrac{3}{4}$ hour

$=\dfrac{3}{4}\times 60=45minutes$

$\therefore$ Vaibhav worked longer by a fraction.

$\Rightarrow \dfrac{10}{35}$

$\Rightarrow \dfrac{2}{7}$ fraction longer.

Find the rate per cent per annum if :
$Rs\ 600$ yields $Rs\ 238$ interest in $4$ years.

Given that,

$P = Rs. 600$

$R = R$ % p.a

$S.I = Rs. 238$

$T = 4 \ years$

$\because S.I = \dfrac{PTR}{100}$

$\implies 238 = \dfrac{600\times 4 \times R}{100}$

$\implies 238 = 24 R$

$\implies R = \dfrac{238}{24}$

$\implies R = 9.92$

$\therefore Rate = 9.92$ % p.a.

Find the rate per cent per annum if :
$Rs\ 700$ yields $Rs\ 105$ interest in $3$ years.

Given that,

$P = Rs. 700$

$R = R$ % p.a

$S.I = Rs. 105$

$T = 3 \ years$

$\because S.I = \dfrac{PTR}{100}$

$\implies 105 = \dfrac{700\times 3 \times R}{100}$

$\implies 105 = 21 R$

$\implies R = \dfrac{105}{21}$

$\implies R = 5$

$\therefore Rate = 5$ % p.a.

What principal will amount to $Rs \ 15000$ at $10\%$ per annum in $5 \ years$ ?

$A = 15,000$

$r = 10 %$

$t = 5\ years$

$P = ?$

Formula for Amount when simple interest is calculated

$A = P+\dfrac{P \times r\times t}{100} = P(1+\dfrac{r\times t}{100})$

$\implies 15000 = P(1+\dfrac{10 \times 5}{100})$

$15000 = P(\dfrac{3}{2})$

$15000 \times \dfrac{2}{3} = P$

$\therefore \ (P) = 10,000$

At what rate per cent will a sum of money tripe itself in 25 years?.

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Find the rate per cent per annum if :
$Rs\ 350$ yields $Rs\ 105$ interest in $6$ year.

Given that,

Principal

$P = Rs. 350$

Rate

$= R\%$ p.a

S.I.

$= Rs. 105$

Time

$T = 6 \ years$

$\because S.I = \dfrac{PTR}{100}$

$\implies 105 = \dfrac{350\times 6 \times R}{100}$

$\implies 105 = 21 R$

$\implies R = \dfrac{105}{21}$

$\implies R = 5$

$\therefore Rate = 5\%$ p.a.

At what rate per cent per annm will $Rs 3600$ amount to $Rs \ 4734$ in $3\dfrac {1}{2}years?$

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The price of a garment has been reduced by $15\%$ in a sale to $Rs\ 306$ . Find its original price.

Let the original price of the garment be

$Rs. x$ .

Cost of garment after reducing

$15 \% = Rs. 306$

Now according to the question-

$x - 15 \%$ of

$x = 306$

$\Rightarrow x - \left( \cfrac{15}{100} \times x \right) = 306$

$\Rightarrow x - \cfrac{3x}{20} = 306$

$\Rightarrow \cfrac{20x - 3x}{20} = 306$

$\Rightarrow 17x = 6120$

$\Rightarrow x = \cfrac{6120}{17} = 360$

Hence the original price of the garment is

$Rs. 360$ .

Find the principal, when the following are given.
$I=$ Rs. $192$
Rate $=$ $6\%$ per annum
Time $=$ $4$ years

$I=\cfrac{Prt}{100}\\P=\cfrac{100\times I}{r\times t}=\cfrac{100\times 192}{6\times 4}=Rs800$

The population of a city decreased from $25,000$ to $24,500$ . Find the percentage decrease.

A] Population decreased = 25000 - 24500 = 500

% decrease

$= \frac{500}{25000} \times 100$ %

$= \frac{1}{5}$ %

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A sum of money doubles itself in $8$ years. What is the rate of interest?

Let the sum of money that's principal be

$x$ .

Time

$=8$ years

Amount after

$8$ years becomes double=

$2x$

Therefore,

Simple interest=Amount -principal=

$2x-x=x$

Rate per annum

$= \dfrac{{S.I \times 100}}{{P \times T}} = \dfrac{{x \times 100}}{{x \times 8}} = \dfrac{{100}}{8} = \dfrac{{25}}{2} = 12.5\%$

Then,

We get interest of

$8$ years is

$12.5$ %

Find the rate of interest when the following are given.
Principal $=$ Rs. $9600$
Time $=6$ years
$I=$ Rs. $72$

$I=\cfrac{Prt}{100}\\r=\cfrac{100\times I}{P\times t}=\cfrac{100\times 72}{6\times 9600}=\cfrac{1}{8}=0.125\%$

Find the principal, when the following are given.
$I=$ Rs. $20$
Rate $=$ $2\%$ per annum
Time $=$ $2$ years

$I=\cfrac{Prt}{100}\\P=\cfrac{100\times I}{r\times t}=\cfrac{100\times 20}{2\times 2}=Rs500$

At what rate of simple interest will a sum of money double itself in 20 years?

$A = P + SI$

$\Rightarrow 2P = P + SI$ so

$SI = P$

$SI = \frac{PRT}{100}$

$\Rightarrow P = \frac{P\times R\times 20}{100}$

$\Rightarrow R = \frac{100}{20} = 5 \%$

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Find the difference between compound interest and simple interest on Rs. $8000$ in $2$ years an at $5\%$ per annum.

We have,

$P=Rs.8000, T=2\ years, R=5\%$

We know that,

$SI=\dfrac{P\times R\times T}{100}$

or,

$SI=\dfrac{8000\times 5\times 2}{100}$

or,

$SI=Rs.800$

We know that,

$A=CI+P=P\left(1+\dfrac{R}{100}\right)^T$

or,

$CI+8000=8000\left(1+\dfrac{5}{100}\right)^2$

or,

$CI+8000=8000\left(1+\dfrac{1}{20}\right)^2$

or,

$CI+8000=8000\left(\dfrac{21}{20}\right)^2$

or,

$CI+8000=8820$

or,

$CI=Rs.820$

Therefore, the required difference

$CI-SI=820-800$

$=Rs.20$

Certain sum of money deposited at simple interest yield Rs. $4400$ in two years and Rs. $5500$ in a $4$ years. Find the rate of interest.

Simple Interest for the first

$2$ years

$=Rs. 4400$

Simple Interest for

$4$ years

$=Rs.5500$

Net simple interest for third and fourth years

$=Rs(5500-4400)=Rs.1100$

$\Rightarrow 1100=\dfrac{4400\times 2\times R}{100}$

$\Rightarrow R=\dfrac{100\times 1100}{4400\times 2}$

$=\dfrac{25}{2}\%$

$=12\dfrac{1}{2}\%$

A loan of $Rs\ 2500$ was repaid giving $Rs\ 2625$ what is the interest paid ?

We know that,

S.I

$=A-P$

$2625-2500=Rs.125$

So,

$\begin{array}{l} =\dfrac { { 2625-2500 } }{ { 2500 } } \times 100 \\ =\dfrac { { 125 } }{ { 2500 } } \times 100 \end{array}$

$=5$ %

Then,

We get interest is

$5$ %.

The simple interest on a certain sum of money for 3 years at 5% per annum is Rs.1,200. Find the principle amount.

Let Principle amount be

$P$

Interest is

$1200$

Time period is

$3yrs$

Rate is

$5\%$

$1200=\dfrac{P\times 5\times 3}{100}\\P=\dfrac{1200}{15}\times 100=₹8000$

Find the interest on $Rs\ 10,000$ for $3\dfrac {1}{4}years$ at the rate of $3.4\%$ p.a. Also find the amount.

Principal sum

$= Rs. 10000$

$t = 3 \cfrac{1}{4} \text{ years} = 3.25 \text{ years}$

$r = 3.4 \%$

As we know that,

$S.I. = \cfrac{P \times r \times t}{100}$

$\Rightarrow S.I. = \cfrac{10000 \times 3.4 \times 3.25}{100}$

$\Rightarrow S.I. = Rs. 1105$

Therefore,

Amount

$= P + S.I. \\ = 10000 + 1105 = Rs. 11105$

A sum of $Rs. 400$ is lent at the rate of $5\%$ per annum. Find the interest at the end of $2$ years.

$SI=\dfrac{P*R*T}{100}$

$=\dfrac{400*5*2}{100}$

$=Rs. 40$

Interest

$= Rs. 40$

The cost of the car is increased $20\%$ . This year cost is $240000$ . What was the cost is last year.

Let the cost of the car in the last year was is

$x$ .

Given the cost of the car is increased

$20\%$ .

Then according to the problem,

$x\left(1+\dfrac{20}{100}\right)=240,000$

or,

$\dfrac{120x}{100}=240,000$

or,

$x=200,000$ .

So the cost of the car in the last year was

$200,000$ .

If a sum of Rs. $16$ becomes Rs. $81$ in $4$ yrs, then find the rate of interest at compound interest.

Some extra :

$4\sqrt{16}= 4\sqrt{81}$

$4:9$

$2:3$

$3-2=1$

$\frac{1}{2}\times 100=50$ %

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The simple interest on certain principle at the rate of 15% per annum for 5 years is Rs.Find the principle amount ?

$R=15\%\\T=5\\I=9375\\\dfrac{PRT}{100}=9375\\\dfrac{P\times 5\times 15}{100}=9375\\P=\dfrac{3975\times 100}{5\times 15}=5300$

A sum of $Rs. 400$ is lent for $3$ years at the rate of $6\%$ per annum. Find the interest.

$SI=\dfrac{P*R*T}{100}$

$=\dfrac{400*6*3}{100}$

$=Rs. 72$

Interest

$= Rs. 72$

Find the compound interest on ${\rm{Rs}}\,{\rm{3000}}$ at $5\%$ per annum for $2$ years.

1349854_1343741_ans_27984d8803944027be605de519057a60.jpg

At what rate percent per annum will $Rs. 630$ produce an interest of $Rs. 126$ in $4$ years?

$P=Rs. 630, I=Rs. 126, T=4$

$=R=\dfrac {100\times I}{P\times T}=\dfrac {100\times 126}{630\times 4}=\dfrac {100}{20}=5\%$

On what principle will the simple interest be $Rs. 7008$ in $6$ years $3$ month at $5\%$ per year?

Consider principle

$=Rs. P$
Time

$(T)=6$ Years

$3$ month

$=6$ years

$+\dfrac {3}{12}$
Year

$=\dfrac {75}{12}=\dfrac {25}{4}$ year

$=6\dfrac {1}{4}$ years
Rate

$(R)=5\%$
Simple interest

$=Rs. 7008$
Here
Simple interest

$=\dfrac {P\times T\times R}{100}$

$=7008=\dfrac {P\times \dfrac {25}{4}\times 5}{100}\Rightarrow P=\dfrac {7008\times 100\times 4}{25\times 5}$

$=\dfrac {7008\times 16}{5}=\dfrac {112128}{5}=Rs, 22425.60$

$15\%$ increase in price of an article, which is $Rs. 1620$
is the increase of $Rs.$ ________

Suppose, the price of the circle

$=Rs. x$
After

$15\%$ increased in the price,
Now, the price

$=Rs. 1620$

$1620=x+x\times \dfrac {15}{100}$
of

$1620=\dfrac {115x}{100}$
of

$115x=1620\times 100$
of

$x=Rs. 1408$

Find the principle which will amount to $Rs. 4000$ in $4$ years at $6.25\%$ per annum.

Consider principle

$=Rs. P$ , Time

$(T)=4$ years
Rate

$=6\dfrac {1}{4}=\dfrac {25}{4}\%$
Simple interest

$=\dfrac {P\times T\times R}{100}=\dfrac {P\times \dfrac {25}{4}\times 4}{100}=\dfrac {P}{4}$

$=\therefore$ Amount

$=P+\dfrac {P}{4}=\dfrac {5P}{4}=\dfrac {5P}{4}=4000$

$5P=4\times 4000$

$P=Rs. 3200$
Therefore, Principle

$=Rs. 3200$

Increase of a number from $150$ to $162$ is equal to increase of _____ percent.

Given, initial number

$=150$ and Fill number

$=162$
Increased number

$=162-150=12$
Percent increased number

$=\dfrac {12}{150}\times 100=\dfrac {120}{15}=8\%$

A sum of Rs 1,780 becomes Rs 2,316 in 4 years, Find:
The rate of interest.

It is given that

$P = Rs. 1780$

$A = Rs. 2136$

We know that

$S.I = A-P$

Substituting the values

$= 2,136-1780 = Rs. 356$

Also,

$T = 4$ years

Here, Rate

$(R) =(S.I\times 100)/(P\times T)$

Substituting the values

$=(356\times 100)/(1780\times 4)$

$=5$

So we get

$= 5\%$ p.a

The simple interest earned on a certain sum is 30% of the sum. Find the rate of interest.

Consider sum

$P = Rs. 100$

We know that

$SI = 30/100\times 100= Rs. 30$

$T = 5$ years

Here, Rate

$(R)= (S.I\times 100)/(P\times T)$

Substituting the values

$=(30\times 100)/(100\times 5)$

$=6$

So we get

$(R) = 6 \%$

What is the rate of interest, if Rs 3,750 amount to Rs 4,650 in 4 years?

It is given that

$P = Rs. 3,750$

$A = Rs. 4,650$

We know that

$S.I = A-P$

Substituting the values

$= 4,650 - 3,750 = Rs. 900$

$T = 4$ years

Here, Rate

$(R)= (S.I\times100 )/(P\times T)$

Substituting the values

$=(900\times 100 )/(3750\times 4)$

$=6$

So we get

$R= 6 \%$ p.a.

Match each of the entries in Column I with the appropriate entries in Column II :

$\dfrac{3}{5} = \dfrac{3}{5} \times \dfrac{100}{100} = \dfrac{300}{5}\% = 60\%$

ii.

$2.5 = \dfrac{25}{10} \times \dfrac{100}{100} = \dfrac{2500}{10}\% = 250\%$

iii.

$100\%= \dfrac{100}{100} = 1 = 1 \times \dfrac{3}{3} = \dfrac{3}{3} \,i.e. 3 : 3$

iv.

$\dfrac{2}{3} = \dfrac{2}{3} \times \dfrac{100}{100} = \dfrac{200}{3}\% = 66\dfrac{2}{3}\%$

$6\dfrac{1}{4}\% = \dfrac{25}{4}\% = \dfrac{25}{4} \times \dfrac{1}{100} = \dfrac{1}{16}$

vi.

$12.5\% = \dfrac{12.5}{100} = 0.125$

vii. Cost price

$= Rs. 50,$ Loss

$= 6\%$

Selling price

$= \left (\dfrac{100 - 6}{100} \right ) \times 50 = \dfrac{94 \times 50}{100} = Rs.47$

viii. Cost price

$= Rs.50,$ profit

$= Rs.4$

$\therefore$ Selling price

$= Rs.(50 + 4) = Rs.54$

ix. Cost price

$= Rs.40,$ selling price

$= Rs.50$

$\therefore profit = \left (\dfrac{50 - 40}{40} \right ) \times 100 = \dfrac{10}{40} \times 100 = 25\%$

x. Cost price

$= Rs.50,$ selling price

$= Rs.60$

$\therefore profit = \left ( \dfrac{60 - 50}{50} \right ) \times 100 = \dfrac{10}{50} \times 100 = 20\%$

xi.

$Time = 2\,years, Principal = Rs.800, Rate = 10\% \,p.a.$

$I = \dfrac{800 \times 10 \times 2}{100} = Rs.160$

xii.

$Time = 1\,year, Principal = Rs.150, Rate = 6\% \,p.a.$

$I = \dfrac{150 \times 6 \times 1}{100} = Rs.9$

Hence, Amount

$= 150 + 9 = Rs.159.$

The simple interest on a certain sum of money is $\dfrac 38$ of the sum in $64$ years. Find the rate percent charged.

Consider

$P=Rs. 8$

$S.I-Rs. =\dfrac 38\times 8$

$=Rs. 3$

$T=6\dfrac 14$ years

$=\dfrac {25}{4}$ years
Here

$R=\dfrac {100\times I}{P\times T}=\dfrac {100\times 3}{8\times \dfrac {25}{4}}=\dfrac {100\times 3}{8}\times \dfrac {4}{25}=2\times 3$

$=6\%$

Peter invested $240000$ for $2$ years at $10\%$ per annum compounded annually. If $20\%$ of the accrued interest at the end of each year is deducted as income tax, find the amount he received at the end of two years.

$\textbf{Given:}$

Amount invested

$=$ Principal

$= Rs. 2,40,000$

Time

$= 2yrs$

Rate

$=10\% \ p.a$

Interest for 1st year

$= S. I = \dfrac{PRI}{100}$

$S.I = \dfrac{2,40,000\times 10 \times 1}{100}$

$S.I = Rs. 24,000$

Amount at the end of 1st year

$= P + S. I - \dfrac{20}{100} \ of \ S. I.$

$=2,40,000+24000-\dfrac{20}{100}\times 24000$

$=2,40,000 + \dfrac{80}{100}\times 24000$

$=2,40,000+19200$

$=Rs. 2,59,200$

Principal for 2nd year

$=Rs. 2,59,200$

Interest for 2nd year

$=\dfrac{2,59,200\times 10 \times 1}{100}$

$=Rs. 25,920$

Amount at the end of 2nd year

$=P+S.I -20\% \ of \ S.I.$

$=2,59,200+2,5920 - \dfrac{20}{100}\ of \ S.I$

$= 2,59,200 + \dfrac{80}{100} \times 25920$

$= 2,59,200+20736$

$=Rs. 2,79,936$

$\boxed{\text{Hence, the amount he recieved at the the end of 2 years is Rs. 2,79,936}}$

The cost of a scooter depreciates every year by 15% of its value at beginning of the year. If the present cost of the scooter is Rs. 8,000; find its cost (in Rs.) after one year:

Cost after depreciation in 'n' years

$= \text{Original cost} \times {(1-\dfrac {R}{100})}^{n}$

So after one year, cost

$= Rs 8,000 \times {(1-\dfrac {15}{100})}^{1} = 8,000 \times \dfrac {85}{100} = Rs 6800$

A sum of $Rs.8,000$ is invested for $2$ years at $10 \%$ per annum compound interest. Calculate final amount at the end of the second year.

Simple Interest

$I = \dfrac {PNR}{100}$
For the first year,

$P = Rs. 8,000$

$N = 1\ year$

$R = 10$ %
So, Interest at the end of fist year

$I = \dfrac {8,000 \times 1 \times 10 }{100} = Rs\ 800$
For second year,

$P = Rs. 8000 + Rs.800 = Rs.8800$

$N = 1 \ year$

$R = 10$ %

$I = \dfrac {8800 \times 1 \times 10}{100} =Rs.880$
And Amount

$= Rs.8800 + Rs.880 = Rs.9680$

A man borrowed Rs. $20,000$ for $2$ years at $8$ % per year compound interest. Calculate the compound interest of two years.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P =$ Rs.

$20,000$

$N = 1$ year

$R = 8\%$
So, Interest at the end of 1st year is

$I = \dfrac {20,000 \times 1 \times 8 }{100} = Rs 1,600$
So, for second year,

$P =$ Rs.

$20,000 +$ Rs.

$1,600 =$ Rs.

$21,600$

$N = 1$ year

$R = 8\%$

$I = \dfrac {21,600 \times 1 \times 8}{100} =$ Rs.

$1,728$
So, compound interest for two years

$=$ Rs.

$1,600 +$ Rs.

$1,728 =$ Rs.

$3,328$

A man borrowed Rs.20,000 for 2 years at 8 % per year compound interest. Calculate the final amount at the end of the second year.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P = Rs 20,000$

$N = 1 year$

$R = 8$ %
So, Interest at the end of 1st year

$I = \dfrac {20,000 \times 1 \times 8 }{100} = Rs 1,600$
So, for second year,

$P = Rs 20,000 + Rs 1,600 = Rs 21,600$

$N = 1 year$

$R = 8$ %

$I = \dfrac {21,600 \times 1 \times 8}{100} =Rs 1,728$
So, final amount at the end of second year

$= Rs 21,600 + Rs 1,728 = Rs 23,328$

Mohan borrowed $Rs. 16,000$ for $3$ years at $5 \%$ per annum compound interest. Calculate the amount that Mohan will pay at the end of $3$ years to complete the debt.

Given,

Principal

$P=Rs. 16,000$

Rate of interest

$R=5\%$

Time

$T=3$ years

So,

$n=3$

Amount

$A = P\left( 1+\dfrac { R }{ 100 } \right) ^{ n }$

$= Rs .16,000 \times \left( 1+\dfrac { 5 }{ 100 } \right) ^{ 3 }$

$= Rs .16,000 \times \left( 1+\dfrac { 1 }{ 20 } \right) ^{ 3 }$

$= Rs .16,000 \times \left(\dfrac { 21 }{ 20 } \right) ^{ 3 }$

$= Rs .16,000 \times \dfrac { 21 }{ 20 } \times\dfrac{21}{20}\times\dfrac{21}{20}$

$= Rs. 18,522$

Hence, Mohan will pay

$Rs. 18,522$ at the end of

$3$ years to complete the debt.

A man invests Rs. $9{,}600$ at $10 \%$ per annum compound interest for $3$ years. Calculate the amount at the end of the first year.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P =$ Rs.

$9{,}600$

$N = 1$ year

$R = 10 \%$
So, Interest at the end of

$1^{st}$ year

$I = \dfrac {9,600 \times 1 \times 10 }{100} =$ Rs.

$960$
Amount

$= P + I =$ Rs.

$9,600\, +$ Rs.

$960 =$ Rs.

$10,560$

Therefore, the amount at the end of the first year is Rs.

$10,560$ .

A sum of Rs.8,000 is invested for 1 years at 10 % per annum compound interest. Calculate compound interest earned in 2 years.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P = Rs 8,000$

$N = 1 year$

$R = 10$ %
So, Interest at the end of 1st year

$I = \dfrac {8,000 \times 1 \times 10 }{100} = Rs 800$
So, for second year,

$P = Rs 8000 + Rs 800 = Rs 8800$

$N = 1 year$

$R = 10$ %

$I = \dfrac {8800 \times 1 \times 10}{100} =Rs 880$
So compound interest earned in

$2$ years

$= Rs 800 + Rs 880 = Rs 1,680$

Calculate the compound interest for the second year on $Rs. 15,000$ invested for $2$ years at $6\%$ per annum.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P = Rs .15,000$

$N = 1 \ year$

$R = 6\%$
So, Interest at the end of

$1^{st}$ year

$I = \dfrac {15,000 \times 1 \times 6 }{100} = Rs. 900$
Compound interest for the second year is the simple interest calculated on new principal after first year.
And for second year,

$P = Rs. 15,000 + Rs .900 = Rs. 15,900$

$N = 1 \ year$

$R = 6 \%$

$I = \dfrac {15,900 \times 1 \times 6}{100} =Rs. 954$

Rekha borrowed Rs. 40,000 for 3 years at 10 % per annum compound interest. Calculate the interest earned in the second year.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P = Rs 40,000$

$N = 1 year$

$R = 10$ %
So, Interest at the end of 1st year

$I = \dfrac {40,000 \times 1 \times 10 }{100} = Rs 4,000$
And for second year,

$P = Rs 40,000 + Rs 4,000 = Rs 44,000$

$N = 1 year$

$R = 10$ %

$I = \dfrac {44,000 \times 1 \times 10}{100} =Rs 4,400$

A man borrowed $\text{Rs }20,000$ for $2$ years at $8\%$ per year compound interest. Calculate the interest of the second year.

Simple Interest,

$I = \dfrac {P\times N\times R}{100}$
Here for first year,
Principal,

$P =$

$\text{Rs }20,000$
Number of years,

$N = 1$ year
Rate,

$R = 8\%$
So, Interest at the end of 1st year,

$I = \dfrac {20,000 \times 1 \times 8 }{100} =$

$\text{Rs }1,600$
So, for second year,

$P =$

$\text{Rs }20,000 +$

$\text{Rs }1,600 =$

$\text{Rs }21,600$

$N = 1$ year

$R = 8\%$
Therefore,

$I = \dfrac {21,600 \times 1 \times 8}{100} =$

$\text{Rs }1,728$

A man borrowed Rs. $20,000$ for $2$ years at $8\%$ per year compound interest. Calculate the interest of the first year.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P = Rs\ 20,000$

$N = 1 year$

$R = 8$ %
So, Interest at the end of 1st year

$I = \dfrac {20,000 \times 1 \times 8 }{100} = Rs\ 1,600$

A man invests Rs. $5,600$ at $14$ % per annum compound interest for $2$ years. Calculate the interest for the first year.

For the first year

$P = Rs 5,600$

$N = 1 year$

$R = 14$ %

We have

$S.I. = \dfrac { PNR}{100} = \dfrac {5,600 \times 1 \times 14}{100} =Rs 784$

Calculate the compound interest on $\text{Rs. }3,500$ at $10\%$ per annum in $2$ years.

Given:

$P=\text{Rs. }3500$

$R=10\%$

$t=2$ years

$\begin{aligned}{}A &= P{\left( {1 + \frac{R}{{100}}} \right)^n}\\ &= 3500{\left( {1 + \frac{{10}}{{100}}} \right)^2}\\& = 3500{\left( {1.1} \right)^2}\\& = \text{Rs. }4235\end{aligned}$

So, compound interest will be,

$A-P=4235-3500$

$=\text{Rs. }735$

A person invests Rs. $5,000$ for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. $5,600$ . Calculate the rate of interest per year. Find the amount at the end of the second year.

Simple Interest

$, I = \dfrac {PNR}{100}$
Here for first year,

$P =$ Rs.

$5,000$

$N = 1$ year

$R = R\%$
So, Interest at the end of 1st year

$I = A - P = 5,600 - 5,000 =$ Rs.

$600$

$\Rightarrow 600 = \dfrac {5,000 \times 1 \times R }{100}$

$\Rightarrow R = 12\%$
And for the second year,

$P =$ Rs.

$5,600$

$N = 1 year$

$R = 12\%$
So, Interest at the end of 2nd year

$= \dfrac {5,600 \times 1 \times 12 }{100} =$ Rs.

$672$
And Amount at the end of 2nd year

$=$ Rs.

$5,600 +$ Rs.

$672 =$ Rs.

$6,272$

A man invests Rs. $5,600$ at $14$ % per annum compound interest for $2$ years. Calculate the amount at the end of the first year.

For the first year

$P = Rs 5,600$

$N = 1 year$

$R = 14$ %

We have

$S.I. = \dfrac { PNR}{100} = \dfrac {5,600 \times 1 \times 14}{100} =Rs 784$
And Amount at the end of first year

$P + S.I. = Rs 5,600 + Rs 784 = Rs 6,384$

A man invests Rs. $9,600$ at $10 \%$ interest per annum for $3$ years. The interest for the second year is

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P =$ Rs.

$9{,}600$

$N = 1$ year

$R = 10 \%$
So, interest at the end of

$1^{st}$ year

$I = \dfrac {9,600 \times 1 \times 10 }{100} =$ Rs.

$960$

And for second year,

$P =$ Rs.

$9,600\, +$ Rs.

$960 =$ Rs.

$10,560$

$N = 1$ year

$R = 10\%$

$I = \dfrac {10,560 \times 1 \times 10}{100} =$ Rs.

$1,056$

So, interest for the second year is Rs.

$1,056$ .

A sum of Rs. $1,200$ lent out at C.I. amounts to Rs. $1,452$ in $2$ years. Calculate the rate of interest.

For ,
$N = 2 years$
$P = Rs 1,200$

On interest being compounded for 2 years,

$Amount = P{(1 + \dfrac {R}{100})}^{N}$
$1,452 = 1,200 \times {(1 + \dfrac {R}{100})}^{2}$
$1.21 = {(1 + \dfrac {R}{100})}^{2}$
Taking square root on both sides,

$1.1 = 1 + \dfrac {R}{100}$

$=> R = 10$ %

$4,500$ in $2$ years when the rates of interest on successive years are $10$ % and $12$ % respectively.

For the first year, on interest being compounded at

$R = 10$ %, we have

$Amount = P{(1 + \dfrac {R}{100})}^{N} = 4,500 \times {(1 + \dfrac {10}{100})}^{1} = 4,500 \times 1.1 = Rs 4,950$

For the second year,

$P = Rs 4,950$ on interest being compounded at

$R = 12$ %, we have

$Amount = P{(1 + \dfrac {R}{100})}^{N} = Rs 4,950 \times {(1 + \dfrac {12}{100})}^{1} = Rs 4,950 \times 1.12 = Rs 5544$

And

$Total C.I. = A - P = Rs 5,544 - Rs 4,500 = Rs 1,044$

A person invests Rs. $5,000$ for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. $5,600$ . Calculate the rate of interest per year.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P = Rs 5,000$

$N = 1$ year

$R = R$ %
So, Interest at the end of 1st year

$I = A - P = Rs 5,600 - Rs 5,000 = Rs 600$

$\Rightarrow 600 = \dfrac {5,000 \times 1 \times R }{100}$

$\Rightarrow R = 12$ %

The compound interest for the two years on Rs. 8,000/- invested for 3 years at 10% per annum is :

We know, for compound interest

$A=P\left ( 1+\dfrac{R}{100} \right )^{T}$

$=8000\left ( 1+\dfrac{10}{100} \right )^{2}$

$= 8000\times \dfrac{110}{100}\times \dfrac{110}{100}= \,Rs\,9680$

$C.I=A-P=9680-8000=\,Rs\,1680$

A man invests Rs. $30,800$ at $5$ % per annum compound interest for $3$ years. Calculate the interest for the first year.

For the first year

$P = Rs 30,800$

$N = 1 year$

$R = 5$ %

We have

$S.I. = \dfrac { PNR}{100} = \dfrac {30,800 \times 1 \times 5}{100} =Rs 1,540$

Calculate the compound interest for the second year on Rs. $8,000/-$ invested for $3$ years at $10$ % per annnum.

Compound interest for the second year will be the simple interest gained in the second year, with the amount after first year being the principal for the second year.
For the first year
$P = Rs 8,000$
$N = 1 year$
$R = 10$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {8,000 \times 1 \times 10}{100} =Rs 800$
And Amount at the end of first year $P + S.I. = Rs 8,000 + Rs 800 = Rs 8,800$
Now, for the second year
$P = Rs 8,800$
$N = 1 year$
$R = 10$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {8,800 \times 1 \times 10}{100} =Rs 880$
Thus, Compound interest for the second year $= Rs 880$

$A$ borrowed Rs. $2,500$ from $B$ at $12\%$ per annum at compound interest. After $2$ years, $A$ gave Rs. $2,936$ and a watch to $B$ to clear the account. Find the cost of the watch.

We know, for compound interest

$A =P\left \{ 1+\dfrac{R}{100} \right \}^{T}$

Here

$P=2500, R =12, T=2$

$A =2500\left \{ 1+\dfrac{12}{100} \right \}^{2}$

$A =2500\left \{ \dfrac{112}{100} \right \}\left \{ \dfrac{112}{100} \right \} =3136$

As per the question,

$A$ paid Rs.

$2936$ and watch for settlement.

Therefore cost of watch

$= 3136-2936=$ Rs.

$200$

Calculate the difference between the simple interest and the compound interest on Rs. $4,000$ in $2$ years at $8$ % per annum compounded yearly.

${\textbf{Step - 1: Calculating Simple interest}}$

${\text{t = 2 years}}$

${\text{r = 8 % }}$

${\text{P = Rs 4,000}}$

${\text{Simple interest = }}\dfrac{{ptr}}{{100}}$

$= \dfrac{{4000 \times 2 \times 8}}{{100}}$

$= 640$

${\textbf{Step - 2: Calculating Compound interest}}$

${\text{n = 2 years}}$

${\text{r = 8 % }}$

${\text{P = Rs 4,000}}$

${\text{And on interest being compounded for 2 years and R = 8 % }}$

${\text{Amount = P}}{\left( {1 + \dfrac{r}{{100}}} \right)^n}$

$= 4000{\left( {1 + \dfrac{8}{{100}}} \right)^2}$

$= 4000{\left( {\dfrac{{108}}{{100}}} \right)^2}$

$= 4000 \times {1.08^2}$

$= 4665.60$

${\text{So compound interest is 4665}}{\text{.60 - 4000 = 665}}{\text{.60}}$

${\textbf{Step - 3: Finding the difference}}$

${\text{So the difference is 665}}{\text{.60 - 640}}$

${\text{ = 25}}{\text{.60}}$

${\textbf{ Hence the answer is 25}}{\textbf{.60 Rs}}{\text{.}}$

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

Lets some of money is Rs 100 and rate is R
A=

$3\times100=300$
300=

$100\left (1+ \frac{R}{100} \right )^{8}$

$\left (1+ \frac{R}{100} \right )^{8}$ =

$\frac{300}{100}= 3$
second step
A=

$27\times100=2700$
2700=

$100\left (1+ \frac{R}{100} \right )^{T}$

$\left (1+ \frac{R}{100} \right )^{T}$ =

$\frac{2700}{100}= 27= 3 ^{3}$
Then T=

$8\times3$ =24

The population of a place increased to $54000$ in $2003$ at a rate of $5\%$ per annum. Its population in $2005$ will be $P$ . What is the sum of the digits in $P$

Let

$p=$ Initial population

$=54000$ , i.e., in the year

$2003$
Rate of growth

$= 5 \%$

$\therefore$ Population after

$2$ years, i.e., in

$2005$

$P=p{\left(1+\dfrac{R}{100}\right)}^{n}=54000\times {\left(1+\dfrac{5}{100}\right)}^{2}$

$P=54000\times \dfrac{105}{100}\times \dfrac{105}{100}=59535$

$\therefore$ sum of digits in

$P = 5+9+5+3+5 = 27$

Find the compound interest on Rs. 1000 at the rate of 10% per annum for 3 years.

$P =$ Rs.

$1000$ ,

$R=10\%$ , p.a.

$T=3$ years

Interest for the

$1$ st year

$=\cfrac{P\times R\times T}{100}=\cfrac{1000\times 1\times 10}{100}=Rs.100$

Amount after the

$1$ st year

$=$ Principal for the

$2$ nd year

$=1000+100=Rs.1100$

Interest after

$2$ nd year

$=\cfrac{1100\times 1\times 10}{100}=Rs.110$

Amount after the

$2$ nd year =Principal ( for the

$3$ rd year)

$=1100+110=Rs.1210$

Interest for the

$3$ rd year

$=\cfrac{1210\times 1\times 10}{100}=Rs.121$

Amount after the

$3$ rd year

$=1210+121=Rs. 1331$

Compound interest for three years

$= Amount$ (after

$3$ years) - (initial)

$Principal$

$=Rs.1331-Rs.1000=Rs.331$

The population of a town increases every year at the rate of 25 per thousand. The present population of the town isFind Population after 2 years.

The present population,

${P}_{o}=262400$
Rate of increase,

$R= 25$ per thousand

$= \displaystyle\frac{25}{1000}\times 100=2.5\%$ ,
Period,

$n=2$ years
Population after

$2$ years is given by:

$P={P}_{o}\times {(1+\displaystyle\frac{R}{100})}^{n}$

$P = 262400\times {(1+\displaystyle\frac{2.5}{100})}^{2}$

$P = 262400\times \displaystyle\frac{102.5}{100}\times \frac{102.5}{100}$

$P = 275684$

Find the compound interest on Rs.8000 for 3 years at 5% per annum.

Principal for the first year = Rs.8000 Rate = 5% per annum T = 1 year
Interest for the first year =

$\displaystyle \frac{P\times R\times T}{100}=Rs.\left [ \frac{8000\times 5\times 1}{100} \right ]=Rs.400$

$\therefore$ Amount at the end of the first year

$= Rs.(8000 + 400) = Rs.8400$

Now principal for the second year = Rs. 8400
Interest for the second year =

$\displaystyle \frac{P\times R\times T}{100}=Rs.\left [ \frac{8400\times 5\times 1}{100} \right ]=Rs.420$

$\therefore$ Amount at the end of the second year = Rs. (8400 + 420) = Rs.8820
Interest for the third year =

$\displaystyle \frac{P\times R\times T}{100}=Rs.\left [ \frac{8820\times 5\times 1}{100} \right ]=Rs.441$

$\therefore$ Amount at the end of the third year

$= Rs. (8820 + 441) = Rs.9261$
Now we know that total C.I. = Amount - Principal

$= Rs. (9261 - 8000) = Rs.1261$
We can also find the C.I. as follows
Total C.I. = Interest for the first year + Interest for the second year + Interest for third year

$= Rs. (400 + 420 + 441) = Rs. 1261$

In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of $2.5\%$ per hour. Find the bacteria at the end of $2$ hours, if the count was initially $5,06,000$ .

We have, $P_0$ $=$ Original count of bacteria $=506000$ ,
Rate of increase $=R =2.5\%$ per hour,
Time $=2$ hours

$\therefore$ Bacteria count after $2$ hours $=$ $P$

$P = P_0(1 +\dfrac{R}{100})^T$

$=506000\times {\left(1+\dfrac{2.5}{100}\right)}^{2}$

$=506000\times \dfrac{102.5}{100}\times \dfrac{102.5}{100}$

$=531616.25=531616$ (approx)

The value of a car depreciates 20% every year. If after two years the price of a car is Rs.2,08,000, find the original price of the car.

Given:

Let the original price of the car $= Rs.\ x$

Percentage of depreciation per year $= 20\%$

Time period $= 2$ yrs

Price of a car after two years $= Rs.\ 208000$

Amount of depreciation $A=P\bigg( 1-\dfrac{r}{100} \bigg)^n,$

here $A$ is the amount after depreciation, $r$ is the rate of depreciation and $n$ is time.

$\Rightarrow 208000=x\bigg( 1-\dfrac{20}{100} \bigg)^2$

$\Rightarrow 208000=x\bigg( \dfrac{100-20}{100} \bigg)^2$

$\Rightarrow 208000=x\bigg( \dfrac{80}{100} \bigg)^2$

$\Rightarrow 208000=x\bigg( \dfrac{4}{5} \bigg)^2$

$\Rightarrow 208000=x\bigg( \dfrac{16}{25} \bigg)$

$\Rightarrow x=208000\times \dfrac{25}{16}$

$\Rightarrow x=Rs.\ 325000$

$\therefore$ Original price of the car

$=Rs.\ 325000$

A new car costs Rs. $360000$ .Its price depreciates at the rate of $10$ % a year during the first two years and at the rate of $20$ % a year thereafter.What will be the price of the car after $3$ years?

Current price of the car, $P_o = Rs 360000$
Rate of depreciation, 10% for first 2 years, 20% for the third year
Now, Let price after 2 years= $P$

Thus, $P = P_o(1 + \dfrac{R}{100})^T$

$P = 360000(1 + \dfrac{-10}{100})^2$

$P = 360000\times \dfrac{9\times 9}{10 \times 10}$

$P = 291600$

After the third year, let the price be $P'$

$P' = P (1 + \dfrac{R'}{100})^T$

$P' = 291600(1 + \dfrac{-20}{100})^1$

$P' = \dfrac{291600\times 80}{100}$

$P' = 233280$

Thus, the car will cost Rs $233280$ after three years.

The population of a village declines by $5\%$ every year.If the present population of the village is $1444$ , find the population of the village $2$ year back.

Given:-

Present population,

$P = 1444$ ,
Rate

$= 5 \%$ and Time

$= 2$ years

Let the population of the village

$2$ years back be

$P_o.$

Now,

$P = P_o\left (1 - \dfrac{R}{100}\right)^n$

$\Rightarrow 1444 = P_o \left (1 - \dfrac{5}{100}\right)^2$

$\Rightarrow 1444 = P_o \left ( \dfrac{95}{100}\right)^2$

$\Rightarrow P_o = \dfrac{1444 \times 100 \times 100}{95 \times 95}$

$\Rightarrow P_o = 1600$

Hence, the population of the village

$2$ years back is equal to

$1600.$

The current year population of a town is 10% more than that of last year.If the current year population is 9900 find the last year population of the town.

Let the population last year be $x$
Now, Current year population = Last year + 10% of last year
$9900 = x + \dfrac{10}{100} \times x$
$9900 = \dfrac{110}{100} x$
$x = 9000$
Last year the population was $9000$

The population of a city was 200000 in the yearIt increased at the rate of 5% p.a. Find the population at the end of the year 2000.

Initial Population, $P_o = 20000$
Rate of increase $= 5\%$
Let say population after $3$ years at the end of the year $2000= P$

$P = P_o(1 + \dfrac{R}{100})^T$

$P = 200000(1 + \dfrac{5}{100})^3$

$P = \dfrac{200000\times 21\times 441}{20\times 400}$

$P = 231525$

Thus, the population after $3$ years in $2000$ will be $231525.$

The bacteria in a a culture grows by $10$ % in the first hour, decreases by $10$ % in the second hour and again increases by $10$ % in the third hour.If the count of bacteria in the sample is $121000000$ , what will be the count of bacteria after 3 hours?

After 1st Hour bacteria increase by 10%
therefore Count of bacteria after 1st hr

$= 121000000+121000000 \times \dfrac{10}{100} = 133100000$
After 2nd Hour bacteria decrease by 10%
therefore Count of bacteria after 2nd hr

$= 133100000- 133100000 \times \dfrac{10}{100} = 119790000$
After 3rd Hour bacteria increase by 10%
therefore Count bacteria after 3rd hr

$= 119790000 + 119790000 \times \dfrac{10}{100} = 131769000$

The population of a town 3 years ago was 25000.If the population in these three years has increased at the rate of 10%,15% and 8% respectively,find the present population.

Population 3 years ago = 25000
Population increased at the rate of 10%, 15%, 8%
Now, after an increase in 10% of population,
New population = 25000 + 10% of 25000 =

$25000 + \dfrac{10}{100}\times 25000 = 27500$
After an increase in 15% of the population,
New population= 27500 + 15% of 27500 =

$27500 + \dfrac{15}{100}\times 27500 = 31625$
After an increase in 8% of the population,
New population = 31625 + 8% of 31625 =

$31625 + \dfrac{8}{100}\times 31625 = 34155$
Hence, the present population is 34155

A sum of money amount to Rs 1237.50 and Rs 1443.75 in 4 and 6 years respectively at a simple rate of interest Find principle

Principal + interest of 4 years = Rs. 1237.50 Same principal + interest of 6 years = Rs. 1443.75

$\therefore$ 2 years interest on the given principal = Rs. (1443.75 - 1237.50) = Rs. 206.25

$\therefore$ 4 years interest on the given principal =

$\displaystyle Rs.\frac{206.25}{2}\times 4=412.50$

$\therefore$ Principal = Amount of 4 year - Interest of 4 year = Rs. 1237.50 - 412.50 = Rs. 825

Find the compound interest on Rs. 8000 for $\displaystyle 1\frac{1}{2}$ years at 10% per annum, interest being payable half yearly.

Given that, principal amount

$=Rs.\ 8000$ , rate of interest

$= 10\%$ per annum, time

$=1\dfrac{1}{2}$ years.

Also, given that the interest is compounded half yearly.

So, rate of interest

$= 5\%$ per half year

and time

$= \displaystyle 1\frac{1}{2}$ years

$=3$ half years

We know that,

$\text{simple interest}=\dfrac{P\times R\times T}{100}$

So, interest for the first half year

$=Rs.\ \displaystyle \left [ \frac{8000\times 5\times 1}{100} \right]$

$=Rs.\ 400$

Now, amount at the end of the first half year

$= Rs.\ 8000 + Rs.\ 400$

$= Rs. \ 8400$
So, principal for the second half year

$= Rs.\ 8400$

So, interest for the second half year

$=Rs.\ \displaystyle \left [ \frac{8400\times 5\times 1}{100} \right]$

$=Rs.\ 420$

So, amount at the end of the second half year

$= Rs.\ 8400 + Rs.\ 420$

$= Rs.\ 8820$

So, principal for the third half year

$= Rs.\ 8820$

So, interest for the third half year

$=Rs.\ \left[\dfrac{8820\times 5\times 1}{100}\right]$

$=Rs.\ 441$

So, amount at the end of third half year

$=Rs.\ 8820 + Rs.\ 441 = Rs.\ 9261$

$\therefore\$ Compound interest

$= Rs.\ 9261 - Rs.\ 8000 = Rs.\ 1261$

Hence, the compound interest on

$Rs.\ 8000$ at

$10\%$ per annum for

$1\dfrac{1}{2}$ years, compounded half yearly, is

$Rs.\ 1261$ .

In a factory, the production of scooters rose to $48400$ from $40000$ in $2$ years. Find the rate of growth per annum.

Let the rate of growth per annum be

$R$ .

We have, Present production

$=48400$ scooters.

Previous production

$=40000$ scooters

Number of years

$=2$

$\therefore 48400=40000\times {\left(1+\cfrac{R}{100}\right)}^{2}$

$\Rightarrow \cfrac{48400}{40000}={\left(1+\cfrac{R}{100}\right)}^{2}$

$\Rightarrow \cfrac{484}{400}={\left(1+\cfrac{R}{100}\right)}^{2}$

$\Rightarrow \cfrac{121}{100}={\left(1+\cfrac{R}{100}\right)}^{2}$

$\Rightarrow {\left(1+\cfrac{R}{100}\right)}^{2}={\left(\cfrac{11}{10}\right)}^{2}$

$\Rightarrow 1+\cfrac{R}{100}=\cfrac{11}{10}$

$\Rightarrow \cfrac{R}{100}=\cfrac{11}{10}-1$

$\Rightarrow \cfrac{R}{100}=\cfrac{1}{10}$

$\Rightarrow R=\cfrac{100}{10}=10$

Hence, the required rate of growth

$=10\%$ per annum.

Afridi purchased an old scooter for $Rs. 16000$ . If the cost of scooter after $2$ years depreciates to $Rs. 14440$ , find the rate of depreciation.

Current cost of the scooter,

$C_o = 16000$

Cost after two years,

$C = 14440$

Let the rate of depreciation be R, then

$C = C_o\left(1 - \cfrac{R}{100}\right)^T$

$\Rightarrow 14400 = 16000\left(1 - \cfrac{R}{100}\right)^2$

$\Rightarrow \cfrac{14400}{16000} = \left(1 - \cfrac{R}{100}\right)^2$

$\Rightarrow \left(\cfrac{38}{40}\right)^2 = \left(1 - \cfrac{R}{100}\right)^2$

$\Rightarrow \cfrac{38}{40} = 1 - \cfrac{R}{100}$

$\Rightarrow \cfrac{R}{100} = 1 - \cfrac{38}{40}$

$\Rightarrow R = \cfrac{2 \times 100}{40}$

$\Rightarrow R = 5\%$

The present population of a village is $130,000$ . If the population increase each year by $4$ % of what it had been at the beginnning of each year, then the population of the village after four years will be $x$ . Find the sum of the digits of $x$ .

${P}_{0}=130,000$ ,
$R=4$ %,
$n=4$
If the population after 4 years is P,then
$P={P}_{0}\times {(1+\dfrac{R}{100})}^{n}$
$=130000{(1+\dfrac{4}{100})}^{4}$

$=130000\times \displaystyle\dfrac{104}{100}\times \dfrac{104}{100}\times \dfrac{104}{100}\times \dfrac{104}{100}$
= $152082$
Population at the end of 4 years = $152082$

Hence, the sum of digits $= 1+5+2+0+8+2 = 18$

A certain sum of money invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is Rs. 2,Find the sum.

Let the sum of money=

$Rs. x$

Rate

$=10%$ per annum

$Amount=P\left(1+\dfrac{R}{100}\right )^n$

For the first year

$Amount=x(1+\dfrac{10}{100})=\dfrac{11}{10}x$

$Interest=\dfrac{11}{10}x-x=\dfrac{1}{10}x$

For the second year

$P=Rs.\dfrac{11}{10}x$

$Amount=\dfrac{11}{10}x(1+\dfrac{10}{100})=\dfrac{121}{100}x$

For the third year

$P=Rs.\dfrac{121}{100}x$

$Amount=\dfrac{121}{100}(1+\dfrac{10}{100})=\dfrac{1331}{1000}x$

$Interest=\dfrac{1331}{1000}x-\dfrac{121}{100}x=\dfrac{121}{1000}x$

According to the question

The interest for the first year plus interest for the third year=Rs.2652

$\Rightarrow \dfrac{1}{10}x+\dfrac{121}{1000}x=2652$

$\Rightarrow 0.221x=2652$

$\Rightarrow x=\dfrac{2652}{0.221}$

$\Rightarrow x=12000$

What will $Rs 500$ amounts to in $10$ years after its deposit in a bank which pays annual interest rate of $10 \%$ compounded annually?

The amount deposited in bank is

$\text{Rs} 500.$
At the end of first year , amount=

$\displaystyle \text{Rs}\quad 500\cdot\left( 1+\frac { 1 }{ 10 } \right) =\text{Rs}\quad 500 (1.1)$
At the end of second year, amount

$=\text{Rs}\quad 500(1.1) (1.1)$
At the end of third year , amount

$=\text{Rs} \quad 500(1.1) (1.1) (1.1)$ and so on

$\displaystyle\therefore$ Amount at the end of 10 years

$=\text{Rs}\quad 500(1.1)(1.1).... (10 \text{times})= \displaystyle \text{Rs}\quad 500 {(1.1)} ^ {10}$

A manufacturer reckons that the value of the machine, which cost him Rs. $15625$ , will depreciate each year by $20$ %. Find the estimated value at the end of $5$ years.

Cost of machine

$= Rs.15625$
Machine depreciates by

$20$ % every year.
Therefore its value after every year is

$80$ % of the original cost i.e.

$\dfrac{4}{5}$ of the original,
Value at the end of five years

$\displaystyle =15625\times \frac { 4 }{ 5 }\times \frac { 4 }{ 5 }\times \frac { 4 }{ 5 }\times \frac { 4 }{ 5 }\times \frac { 4 }{ 5 }$

$=5\times 1024=5120$
Thus the value of the machine at the end of 5 years is

$Rs.5120.$

A sum of money is borrowed and paid back in two equal annual installments of Rs 882 allowing 5% compound interest. Find the sum.

The sum borrowed= Present Worth of Rs.882 due 1 year hence + Present Worth of Rs.882 due 2 year

$\Rightarrow \frac{882}{(1+\frac{5}{100})^1}+\frac{882}{(1+\frac{5}{100})^2}$

$\Rightarrow \frac{882}{\frac{105}{100}}+\frac{882}{(\frac{105}{100})^2}$

$\Rightarrow \frac{882\times 20}{21}+\frac{882\times 20\times 20}{21\times 21}$

$\Rightarrow 840+800=1640$

Calculate the compound interest accrued on Rs. 16,000 in 3 years, when the rates of interest for successive years are 10%, 12% and 15% respectively.

$When\quad rates\quad of\quad different\quad year\quad is\quad given\quad formula\quad to\quad calculate\quad amount\quad is$

$A=P\left( 1+\dfrac { { R }_{ 1 } }{ 100 } \right) \left( 1+\dfrac { { R }_{ 2 } }{ 100 } \right) \left( 1+\dfrac { { R }_{ 3 } }{ 100 } \right)$

$Given\quad principal\quad is\quad 16000$

$A=16000\left( 1+\dfrac { 10 }{ 100 } \right) \left( 1+\dfrac { 12 }{ 100 } \right) \left( 1+\dfrac { 15 }{ 100 } \right)$

$A=16000\left( 1.1\times 1.12\times 1.15 \right)$

$A=16000\times 1.4168$

$A=22668.8$

$Amount\quad will\quad be\quad 22668.8$

∴ Compound interest accrued will be =22,668.80−Rs.16,000=Rs.6,668.80

On a certain sum and at a certain rate percent, the simple interest for the first year is Rs. 270, and the compound interest for the first two years is Rs. 580.Find the sum and the rate percent.

S.I for

$1$ year

$=$ Rs

$270$

$\therefore$ C.I for

$1$ year

$=$ Rs.

$270$ [For 1st year:C.I=S.I]

C.I for 2 years=Rs.580.50

$\therefore$ C.I for 2nd year=

$580.50-270.00=Rs.310.50$

Difference between the C.I of two successive years

$=310.50-270=Rs. 40.50$

$\therefore$ Rate of interest can be calculated using the Simple Interest formula i.e

$S.I.=\dfrac{P\times R \times T}{100}$

$\Rightarrow R=\dfrac{100\times 40.50}{270\times 1}=15$ %

Now to calculate the sum or Principal

$S.I=\dfrac{P\times R \times T}{100}$

$\therefore P=\dfrac{S.I\times 100}{R\times T}$

$\Rightarrow P=\dfrac{270\times 100}{15\times 1}$

$\Rightarrow P=Rs.1800$

Mrs. Kapoor invested Rs. 6,000 every year at the beginning of the year, at 10% per annum compound interest. Calculate the amount of her total savings :
(i) upto the end of the second year.
(ii) at the beginning of the third year.

(i) For 1st year :
Since, money invested at the beginning of the year

$= Rs. 6,000$

$\Rightarrow$ Principal for 1st year

$= Rs. 6,000$

$\therefore$ Interest

$= \displaystyle Rs. \frac{6,000\times 10\times 1}{100}=Rs. 600$
And, amount

$=Rs. 6,000+Rs. 600=Rs. 6,600$
For 2nd year:
Since,

$Rs. 6,000$ is invested again at the beginning of the second year, therefore, for the second year, principal

$= Rs. 6,600+Rs. 6,000 = Rs. 12,600$
Interest

$=\displaystyle Rs. \frac{12,600\times 10\times 1}{100}=Rs. 1,260$
And, amount

$= Rs. 12,600+Rs. 1,260=Rs. 13,860$

$\therefore$ Amount of her total saving upto the end of the second year

$=Rs. 13,860$
(ii) since,

$Rs. 6,000$ is invested again at the beginning of the third year,

$\therefore$ Amount of her total saving at the beginning of the third year

$=Rs. 13,860+Rs. 6,000=Rs. 19,860$ Ans.

What sum of money will amount to Rs. $9,261$ in $3$ years at $5\%$ per annum compound interest

$A = Rs. 9261 ; n = 3$ years and

$r = 5 \%$

$\because A=P{ \left( 1+\dfrac { r }{ 100 } \right) }^{ n }$

$\displaystyle \therefore \quad 9261=P{ \left( 1+\dfrac { 5 }{ 100 } \right) }^{ 3 }\quad$

$\displaystyle \Rightarrow \quad P=9261\times { \left( \frac { 20 }{ 21 } \right) }^{ 3 }$

$\displaystyle \Rightarrow 9261\times \frac{20}{21}\times \frac{20}{21}\times \frac{20}{21}=Rs.8000$

$\displaystyle \therefore$ Required Sum

$= Rs. 8000$

the amount invested by him.

No. of share=400

N.V of 1 share=20

Premium=Rs.4

$\therefore$ M.V of 1 share

$=20+4=Rs.24$

$\therefore$ Sum invested=No.of shares

$\times$ M.V of 1 share

$\Rightarrow 400\times 24=Rs.9600$

Arif took a loan of $\text{Rs. } 80,000$ from a bank. If the rate of interest is $10\%$ per annum, find the amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is compounded annually.

For the first year,

$P = 80000, R = 10\%,$

$T=1$

$\begin{aligned}{}\text{Interest} &= \dfrac{{P \cdot R \cdot T}}{{100}}\\ &= \frac{{(80000 \cdot 10 \cdot 1)}}{{100}}\\ &= 8000\end{aligned}$

$\begin{aligned}{}\text{Amount}& = \text{Interest} + \text{Principle}\\& = 80000 + 8000\\& = 88000\end{aligned}$

For second-year principle will be the amount of the previous year (since interest is compounded annually).

$P=88000,T= 0.5,R=10\%$ .

$\begin{aligned}{}\text{Interest} &= \dfrac{{P \cdot R \cdot T}}{{100}}\\ &= \frac{{(88000 \cdot 10 \cdot 0.5)}}{{100}}\\ &= 4400\end{aligned}$

$\text{Total amount}=88000+4400$

$=\text{Rs. }92400$

What is the compound interests on Rs. $5000$ for $\dfrac{3}{2}$ years at $4\%$ per annum compounded yearly?

$P=5000$

$rate=4\%$

$time=\dfrac{3}{2}years=1\dfrac{1}{2}\,years=1\,year\,+1\,halfyear$

$A=P\left({1+\dfrac{r}{100}}\right)\left({1+\dfrac{\dfrac{r}{2}}{100}}\right)$

$=5000\left({1+\dfrac{4}{100}}\right)\left({1+\dfrac{\dfrac{4}{2}}{100}}\right)$

$=5000\times 1.04\times1.02$

$=5304$

$\therefore C.I=A-P=5304-5000=Rs\,304$

Find the amount of Rs. $6000$ for $2\dfrac{1}{2}$ years, compounded annually at $5\%$ per annum. Also find the C.I.

We know the formula,

$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given,

$P =$ Rs.

$6000$ ,

$r = 5\%$ ,

$n = 1$ and

$t =$

$2\dfrac{1}{2}$ years
Therefore, we have

$A = 6000\left (1+\dfrac{0.05}{1}\right)^{1\times 2.5}$

$\Rightarrow A = 6000\times 1.05^{2.5}$

$A = 6000\times 1.129726$

$A = 6778.36$
To find interest we use formula

$A = P + I$ , since

$A = 6778.36$ and

$P = 6000$

we have:

$A = P + I$

$\Rightarrow 6778.36 = 6000 + I$

$I = 6778.36 - 6000$

$= 778.36$
Interest,

$I =$ Rs.

$778.36$

Find the compound interest on Rs. $12000$ for $4$ years at $5\%$ per annum.

Given,

$P = 12000, r = 5\%, n = 4$ years

We need to find compound interest.

We know that

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A = 12000\left [\left (1+\dfrac{5}{100}\right)^4\right]$

$= 12000\times \left ( \dfrac{21}{20}\right )^4$

$=12000\times \dfrac{194481}{160000}$

$\Rightarrow A =$ Rs.

$14586.075 \approx 14586$

Thus, compund interest =

$14586\,-\,12000=Rs\,2586$

Find the value of a air conditioner after $2$ years, if its price while new is Rs. $3000$ and it depreciates by $10\%$ per month.

Given,

$P = 3000, r = 10\%, n = 2$ years

$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$

$A = 3000\left [\left (1-\dfrac{10}{100}\right)^2\right]$

$A =$ Rs.

$2430$

The value of air conditioner after

$2$ yeras is Rs.

$2430$ .

Toby invested Rs. $1200$ for $2$ years in a savings account. His account depreciates by $8\%$ per annum compound interest. How much did Toby have in his savings account after $2$ years?

Given,

$P = 1200, r = 8\%, n = 2$ years
We know,

$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A = 1200\left [\left (1-\dfrac{8}{100}\right)^2\right]$

$\Rightarrow A =$ Rs.

$1015.68$

Thus, after

$2$ years Toby have Rs.

$1015.68$ .

The annual growth of capital of a business is $7\%$ . At present, the capital is Rs. $12000$ . What will be the capital after $3$ years?

Given:

$P = 12000, r = 7\%, n = 3$ years

To find: amount

$(A)$

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$A = 12000\left [\left (1+\dfrac{7}{100}\right)^3\right]$

$A =$ Rs.

$14700$

Therefore, capital after

$3$ years is Rs.

$14700$ .

A computer is sold for Rs. $9000$ cash on Rs. $3000$ cash down followed by $8$ equal instalments of Rs. $2000$ each. What is the rate of interest?

Given,

$n = 8, I =$ Rs.

$2000$ ,

$E = 3000 + 6 \times 2000 - 9000 =$ Rs.

$6000$

$R = \dfrac{2400E}{n(n+1)I-2E}$

$\Rightarrow R = \dfrac{2400\times 6000}{8(8+1)2000-2\times 6000}$

$\Rightarrow R = 109\%$

Due to migration to cities, the population of a town decreases every year by $10\%$ . If the present population of the town is $14600$ , find the population two years hence.

Given,

$P = 14600, r = 10\%, n = 2$ years

We know,

$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A=14600\left[\left ( 1-\dfrac {10}{100}\right)^2\right]$

$\Rightarrow A= 11826$

Hence, the population of the town

$2$ years hence is

$11826$

The annual increase in the population of a state is $2\%$ . What was the population of the state two years before if it is $54,500$ now?

Given,

$P = 54500, r = 2\%, n = 2$ years
We know,

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A = 54500\left [\left (1+\dfrac{2}{100}\right)^2\right]$

$\Rightarrow A= 56701.80$
Hence, the population of the town

$2$ years is

$56701.80$ .

The population of a town increases at the rate of $6\%$ every year. The present population of the town is $14000$ . Find the population at the end of $4$ years.

Given:

$P = 14,000, r = 6\%, n = 4$ years

We know that,

$A = P \left (1+\dfrac{r}{100}\right)^n$

$\Rightarrow A = 14000\left (1+\dfrac{6}{100}\right)^4$

$\Rightarrow A= 17,673$

Hence, the population of the city after

$4$ years will be equal to

$17,673.$

In $2001$ , the world's population was $250000$ . With a growth rate of approximately $14\%$ , what was the population in $2011$ ?

Given,

$P = 250000, r = 14\%, n = 10$ years

We need to find amount

$A$ .
We know,

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$A = 250000\left [\left (1+\dfrac{14}{100}\right)^{10}\right]$

$= 926800$
Hence, the population of the city after

$10$ years

$= 926800$ .

At $7$ pm, you count $13500$ bacteria in your petrie dish. If the growth rate is $12\%$ , how many bacteria will there be at one month?

Given,

$P = 13500, r = 12\%, n = \dfrac{1}{12}=0.083$

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$A = 13500\left [\left (1+\dfrac{12}{100}\right)^{0.083}\right]$

$= 13626$
Hence,

$13626$ bacteria will be there at one month.

If population of city is $15000$ and it increase with $9\%$ rate, find population of city after $2$ years.

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

where,

$A=$ amount

$P=$ principle

$r=$ rate of interest

$n=$ number of years

Given:

$P = 15000, r = 9\%, n = 2$ years

$A = 15000\left [\left (1+\dfrac{9}{100}\right)^2\right]$

$= 17821$

Hence, the population of the city after

$2$ years

$= 17821$ .

A value of a machine depreciates every year at the rate of $12\%$ . The present value of a machine is Rs. $5500$ . What was the value of the machine $2$ years before?

Given,

$P = 5500, r = 12\%$ ,

$n = 2$ years
We know

$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$

$A = 5500\left [\left (1-\dfrac{12}{100}\right)^2\right]$

$A =$ Rs.

$4259.20$

Thus, the value of machine

$2$ years before is Rs.

$4259.20$ .

The members of which estate had to pay taxes?

The society of estates was part of the feudal system in French. The members of the third estate were Big businessmen, merchants, court officials, lawyers, peasants, artisans, small peasants, landless labor and servants. They were required to serve the members of the other two estates. The members of the third estate had to pay direct tax to the state known as 'taille'. Indirect taxes were imposed on tobacco, salt and many other everyday items. Thus, the third estate was seething with financial difficulties. There was the rise and emergence of many social groups in France in the eighteenth century.

A certain sum of money placed on compound interest amounts to Rs. 4000 in 3 years and Rs. 5000 in 4 years. Calculate the rate of interest.

Sum of money

$\rightarrow$ II year Rs. 4000

$\rightarrow$ III year Rs. 5000
There is increase of Rs. 1000 in 1 year due to interest

$1000 = \dfrac{4000 \times R \times 1}{100}$

$\therefore R = 25$ % per annum.

$\left [ S.I. = \dfrac{P.R.T}{100}\right]$

A sum of money doubles itself in 4 years placed on compound interest. In what time will it 8 times of itself?

1421536_556606_ans_d3227ed1324d4779b5d5a3c7a7bc251b.PNG

$Rs. 5,000$ is borrowed at $8\%$ for 1 $\dfrac{1}{2}$ years. Calculate the compound interest being compounded semi-annually (half-yearly). What would have been the interest earned if the same amount was borrowed for simple interest at the same rate for the same duration?

Principal

$= Rs. 5,000,$ time

$= 1$

$\dfrac{1}{2}$ years

$= 3$ half years, and

$R = 8\%.$

Interest for the first half year (using simple interest formula)

$= \dfrac{P \times R \times T}{ 100}$

$= \dfrac{5000 \times 1 \times 8}{2 \times 100}$

$= Rs. 200.$

Principal for the second half year

$= Rs. 5,000 + Rs. 200 = Rs. 5,200.$

Interest for the second half year

$=$

$\dfrac{5200 \times 1 \times 8}{2 \times 100} = Rs. 208$ .

Principal for the third half year

$= Rs. 5,200 + Rs. 208 = Rs. 5,408.$

Interest for the third half year

$=$

$\dfrac{5,408 \times 1 \times 8}{2 \times 100} = Rs. 216.32$

Therefore, amount

$= Rs. 5,408 + Rs. 216.32 = Rs. 5,624.32$

Compound interest

$=$ amount - principal

$= Rs. 5,624.32 - Rs. 5,000 = Rs. 624.32$

Simple interest for

$Rs. 5,000$ for

$1$

$\dfrac{1}{2}$ years at the rate of

$8\%$

$= \dfrac{5000 \times 8 \times 3}{100 \times 2} = Rs. 600$

$Rs.\ 20,000$ is lent for $2$ years at $5\%$ compound interest. Calculate the compound interest and the amount after $2$ years if the interest is compounded annually. What would have been the interest earned if the same amount was lent for simple interest at the same rate for the same duration?

Principal for the first year =

$Rs.\ 20,000$ , rate =

$5\%$ , T =

$1$ year.
Interest for the first year

$= \dfrac{20,000 \times 1 \times 5}{100} = Rs. 1,000$ (using simple interest formula

$I = \dfrac{P \times T_1 \times R}{100}$ ).
Principal for the second year =

$Rs.\ 20,000 + 1,000 = Rs.\ 21,000.$
Interest for the second year

$= \dfrac{21,000 \times 5}{100} = Rs. 1,050$
Amount after 2 years = Rs. 21000 + 1050 = Rs. 22,050.
Compound interest = amount - principal = Rs. 22,050 - Rs. 20,000 = Rs. 2,050.
Hence the compound interest after 2 years is Rs. 2,050.
Simple interest on Rs. 20,000 for 2 years at the rate of 5% is

$I = \dfrac{20000 \times 2 \times 5}{100} = Rs. 2,000$ .

The present population of a village is 18,It is estimated that the population of the village grows by 3% per year. Find the population of the village after 4 years.

Given :

$P=18,000, R=3, n=4$

To find

$A$

We know that,

$A=P\left(1+\dfrac{R}{100}\right)^{n}$

$=18000\left(1+\dfrac{3}{100}\right)^{4}$

$=18000\left(\dfrac{103}{100}\right)^{4}$

$=18000\times (1.03)^{4}$

$=20259.15$

$\therefore A=20259$ Rs.

Hence, the population of the village after

$4$ years will be Rs.

$20,259$ .

Prahlad invests a sum of money in a bank and gets $Rs. 3307.5$ and $Rs. 3472.87$ in $2$ -nd and $3$ -rd years respectively. Find the sum he invested.

Interest on

$3307.5$ Rs for one year

$=3472.87-3307.5=165.37$ Rs.

Rate of interest

$=\dfrac{I\times 100}{P\times T}$

$=\dfrac{165.37\times 100}{3307.5\times 1}$

$=4.99=5\%$

Let

$P$ be the sum of money invested

Amount after

$2$ years

$(A)=P\left(1+\dfrac{R}{100}\right)^{n}$

$\implies 3307.5=P\left(1+\dfrac{5}{100}\right)^{2}$

$\implies 3307.5=P\left(\dfrac{21}{20}\right)^{2}$

$\implies P=3000$

Hence, the sum Prahlad invested is Rs.

$3,000$ .

Jeshu purchased a bike by paying Rs. 52,If the value depreciates by 2% every year, find the value of the bike after 3 years.

Given : Original price of bike

$P=52,000$

Rate of depreciation

$=2$ i.e.

$R=2$ and

$n=3$

To find

$A$

We know that,

$A=P\left(1-\dfrac{R}{100}\right)^{n}$

$=52000 \left(1-\dfrac{2}{100}\right)^{3}$

$=52000\times (0.98)^{3}$

$=48941.98=48942$

Hence, the value of the bike after

$3$ years will be Rs.

$48,942$ .

Find the principal, which amounts to Rs. 16,872.96 in 3 years at 4% compounded annually.

Given,

$A = Rs. 16,872.96$ ,

$R = 4\%$ and

$n = 3$ years, we have to find

$P$ . We have,

$A = P \left ( 1 + \dfrac{R}{100} \right )^n$
Hence,

$16,872.96 = P \left ( 1 + \dfrac{4}{100} \right )^3 = P(1.04)^3 \Rightarrow 16,872.96 = P (1.124864)$

$\Rightarrow P = \dfrac{16,872.96}{1.124864} = 15,000$
Hence, the principal is Rs.

$15,000$ .

Anagha invested Rs. 25,000 at 11% p.a. and got Rs. 42,127.50 at the end. Find the period of investment using ready reckoner.

Given

$P = Rs. \ 25,000$ ,

$A = Rs. \ 42,127.50$ and

$R = 11\%$ , we have to find time

$n$ . Using compound interest =

$A - P$ , we get

$Rs. 42,127.50 - 25,000 = Rs. 17,127.50$ as compound interest.
Thus,

$Rs. 17,127.5$ is the compound interest for

$Rs. 25,000$ at

$11\%$ .
Now, we find the compound interest for

$Rs. \ 1$ . For the principal

$Rs. 25,000$ , the compound interest is

$Rs. 17,127.5.$
For the principal Rs. 1, the compound interest is

$\dfrac{17,127.5}{25,000} = 0.6851$ .
Now, refer to the table corresponding to the column with

$11\%$ . Here

$0.6851$ corresponds to

$5$ years.
Hence, the period is

$5$ years.

The value of a new bike is $Rs. 50,000$ . If the value depreciates by $5\%$ every year, find the value of the bike after $2$ years.

Given,

Initial price

$P=Rs. 50,000$

Rate of depreciation

$R=5\%$

Time

$T=2$ years

So,

$n=2$

Amount,

$A = P\left( 1-\dfrac { R }{ 100 } \right) ^{ n }$

$= Rs .50,000 \times \left( 1-\dfrac { 5 }{ 100 } \right) ^{ 2 }$

$= Rs. 50,000 \times \left( 1-\dfrac { 1 }{ 20 } \right) ^{ 2 }$

$= Rs .50,000 \times \left(\dfrac { 19 }{ 20 } \right) ^{ 2 }$

$= Rs .50,000 \times \dfrac { 19 }{ 20 } \times\dfrac{19}{20}$

$= Rs. 45,125$

Hence, the price of the bike after

$2$ years will be

$Rs. 45,125$ .

Calculate the compound interest on a principal 10,000 at 8% per anum for 3 years.

We are given P =

$Rs. \ 10,000$ ,

$R = 8\%$ ,

$n = 3$ years.
at

$8\%$ column with the row

$3$ years. It is

$0.2597$ (by ready recknors) . This means

$Rs.\ 0.2597$ is the interest for

$Rs. \ 1$ for

$3$ years at

$8\%$ per anum.
Therefore compound interest for

$Rs.\ 10,000 = 0.2597 \times 10,000 = Rs. 2,597$ .

What is the result of depreciation of fixed assets?

Depreciation is the permanent and continuous decrease in the book value of a depreciable fixed asset due to use, effluxion of time, obsolescence, expiration of legal rights or any other cause.

Depreciation is a non cash expenditure which is charged against profit. The end result of depreciation of fixed asset is it reduces the net profit of business and provides funds to replace the fixed asset.

Find the compound interest on Rs. 12,000 for 2 years at 4% per anum.

Given:

$P = Rs. 12,000, R = 4\%$ and

$n = 2$ years.
Using the formula,
Total amount after two years =

$A = P \left( 1 + \dfrac{R}{100} \right )^n$ we get

$A = 12000 \left( 1 + \dfrac{4}{100} \right)^2 = 12000 \left( \dfrac{104}{100} \right)^2$

$= 12000 (1.04)^2 = 12,979.2.$ Rs.

Therefore,

$C.I. = A - P = Rs. 12,979.2 - Rs. 12,000 = Rs. 979.2$ .

Fill in the blanks with appropriate words in the following sentences.
"The permanent assets lose a portion of their value every year" it is called __________.

Depreciation

Depreciation is a fall in the value of an asset because of usage or with passage of time or with obsolescence or accident.

Mary invests Rs. 10,000 at compound interest in a financial institution and gets Rs. 11,449 at the end of 2 years. Find the rate of interest using ready reckoner.

Here,

$A = Rs. \ 11,449$ ;

$P = Rs. \ 10,000$ and

$n = \ 2$ years.
Therefore, compound interest =

$Rs.\ 11,449$ -

$Rs. \ 10,000$ =

$Rs \. 1,449$ .
This is the CI for

$Rs. \ 10,000$ . Find the CI for

$Rs. \ 1$ .
For

$Rs.\ 10,000$ principal, compound interest is

$Rs. \ 1,449$ .
For

$Rs.\ 1$ principal, compound interest is

$= \dfrac{1,449}{10,000} = Rs. 0.1449$ .
In the ready reckoner, we find out the figure

$0.1449$ corresponding to

$2$ years. We observe that, the rate of interest is

$7\%$ .
Hence, the rate of interest is

$7\%$ .

The population of a state in 2011 census was 1.2 crore. If it is estimated that the population increases by 2.25% by every year, what will be the population in 2014?

Given present population = 1.2 crore, R = 2.25%, we have to find population after n = 3 years.
Population after 3 years = Present population

$\times \left ( 1 + \dfrac{R}{100} \right )^n$

$= 1.2 \times \left ( 1 + \dfrac{2.25}{100} \right )^3 1.2 (1 + 0.0225 )^3 = 1.2 \times (1.0225)^3 \approx 1.28$ crore (estimated to 2 decimal places.)
Therefore, population after 3 years is 1.28 crore.

Find the rate of interest at which Rs. 5,000 amounts to Rs. 5,304.5 in 2 years, compounded annually.

Given P = Rs. 5,000, A = Rs. 5,304.5 and n = 2 years, we have to find R.
Using the formula,

$A = P \left ( 1 + \dfrac{R}{100} \right )^n$ ,
we get

$5,304.5 = 5,000 \left (1 + \dfrac{R}{100} \right )^2 \Rightarrow \dfrac{5,304.5}{5,000} = \left (1 + \dfrac{R}{100} \right )^2$

$\Rightarrow 1 + \dfrac{R}{100} = \sqrt{\dfrac{5,304.5}{5,000}} = \dfrac{103}{100}$ .
Hence,

$\dfrac{R}{100 } = \dfrac{103}{100} - 1 = \dfrac{3}{100}$ .
This implies that R = 3. Hence, the rate of interest is 3%

The present population of a town is $2,00,000$ . Its population increases by $10$ % in the first year and $15$ % in the second year. Find the population of the town at the end of the two years.

Given :

$P = 2,00,000$
Population after first year

$= 2,00,000{ \left( 1+\dfrac { 10 }{ 100 } \right) }^{ 1 }$

$=2,00,000\left( \dfrac { 110 }{ 100 } \right) =2,20,000$

Principal for the second year

$=2,20,000$
Population at the end of second year

$=2,20,000{ \left[ 1+\dfrac { 15 }{ 100 } \right] }^{ 1 }$

$=2,20,000\times \dfrac { 115 }{ 100 }$

$=2,53,000$

The population of Hyderabad was 68,09,000 in the yearIf it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.

$\textbf{Step 1 : Calculate number of people after 4 years.}$

${\text{Number of people at the end of 2015}=P\left(1+\dfrac{r}{100}\right)^n}$

${\text{where,}}$

${\text{P= population in 2011 = 68,09,000}}$

${\text{r = rate of increase of population = 4.7 %}}$

${\text{n = number of years = 4 years}}$

$\text{Number of people at the end of 2015}=6809000\left(1+\dfrac{4.7}{100}\right)^4$

$=6809000\left(\dfrac{104.7}{100}\right)^4$

$=8182199.42$

$\text{Number of people at the end of 2015} = 8182199.42$

$\textbf{Since, number of people cannot be fraction so the final number of people at the}$

$\textbf{end of 2015 is 8182199.}$

A rubber ball is dropped from a certain height. It is found to rebounce only $90\%$ of its previous height. If it is dropped from the top of a $25\ m$ tall building, to what height would it raise after bounding on the ground two times.

The ball rises to a height of

$90\%$ at the first bounce. So at each bounce, the loss in height is $$10\%$$

So taking

$R = -10\%$ the problem can be solved.

$P = 25$ m and

$n = 2$
The height to which it raises after bouncing two times on the ground

$A = P \left ( 1 + \dfrac{R}{100} \right )^n$

$A = 25 \left( 1 - \dfrac{10}{100} \right )^2$

$= 25 \left ( \dfrac{90}{100} \right )^2$

$= 20.25 m$

Mr. Lalit invested Rs. $5,000$ at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to Rs. $5,325$ . The rate of interest is $R$ . Then $2R =$ (in $\%$ )?

Given: Principal

$=$ Rs.

$5,000$ , Time

$= 2$ years, After one year amount

$=$ Rs.

$5,325$
We know that Amount

$(A) = P + I$
For

$1^{st}$ year:

$5,324 = 5,000 + \dfrac {P\times R\times T}{100}$

$5,325 = \dfrac {5,000 \times R\times T}{100}$

$R = \dfrac {325}{50}$
Rate of interest

$= 6.5\%$ .

$\therefore 2R = 13\%$

Machinery worth $Rs. 10000$ is depreciated by $5\%$ . Find its value after $1$ year.

Initial worth $= 10000,$ Depreciation $= 5$ %

Value after $1$ year = Initial worth – Depreciation $= 10000 – 5$ % of $10000$

Value $=10000-\left( \cfrac { 5 }{ 100 } \right) \times 10000=9500$

the machinery will be worth $Rs.9500$ after a year.

On what sum of money will the difference between the compound interest and simple interest for $2$ years be equal to Rs. $25$ if the rate of interest charged for both is $5\%$ p.a?

Let the principal be Rs.

$P$
Given :

$R =5\%, T=24$ years
C.I. for

$2$ years

$\displaystyle =P\left(1+\frac{5}{100}\right)^2-P$
S.I. for

$2$ years

$=\displaystyle \frac{P\times 5\times 2}{100}=\frac{P}{10}$

$\because$ Difference between C.I. and S.P.

$=$ Rs.

$25$

$\displaystyle P\left(1+\frac{5}{100}\right)^2-P-\frac{P}{10}=25$

$\displaystyle \frac{441P}{400}-\frac{11P}{10}=25$

$\displaystyle \frac{441P-440P}{400}=25$

$P=10,000$
Hence, the principle should be Rs.

$10,000$

Find the population of a city after $2$ years which is at present $12$ lakh, if the rate of increase is $4\%$ .p.a.

Given: Population 12 lakh R=4%

Population after

$2$ years.

$= P{\left(1 + \dfrac {R}{100}\right)}^{N}$

$=1200000{ \left( 1+\dfrac { 4 }{ 100 } \right) }^{ 2 }$

$=1200000{ \left( \dfrac { 26 }{ 25 } \right) }^{ 2 }$

$=1200000\times \dfrac { 26 }{ 25 } \times \dfrac { 26 }{ 25 }$

$=1200000\times \dfrac { 676 }{ 625 }$

$=1297920$ .

At what rate $\%$ p.a. will a sum of Rs. $4000$ yield Rs. $1324$ as compound interest in $3$ years

Given: Principle

$=Rs. 4,000, C.I.=Rs. 1,324,$

Amount

$=P+C.I.$

$=Rs.(4,000+1,324)=Rs. 5,324$

Time

$=3$ years

We know that,

$A=P\left(1+\displaystyle\frac{r}{100}\right)^T$

$5,324=4,000\left(1+\displaystyle\frac{r}{100}\right)^3$

$\displaystyle\frac{5,324}{4,000}=\left(1+\displaystyle\frac{r}{100}\right)^3$

$\displaystyle\frac{1,331}{1,000}=\left(1+\displaystyle\frac{r}{100}\right)^3$

$\left(\displaystyle\frac{11}{10}\right)^3=\left(1+\displaystyle\frac{r}{100}\right)^3$

$\therefore1+\displaystyle\frac{r}{100}=\frac{11}{10}$

$\Rightarrow \displaystyle\frac{r}{100}=\frac{11}{10}-1$

$\Rightarrow \displaystyle\frac{r}{100}=\frac{1}{10}$

$\Rightarrow r=\displaystyle\frac{100}{10}$

$\Rightarrow r=10\%$

A TV was bought at a price of $Rs. 21000$ . After 1 year, the value of the TV was depreciated by $5\%$ . (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.

${\textbf{Step-1: Using relation between cost and depreciation.}}$

${\text{Cost of TV = 21000, Depreciation = 5 %.}}$

${\text{Value after one year = Cost – Depreciation}}$

$=21000- \dfrac{5}{100} \times 21000$

$= 19950.$

${\textbf{Hence, Value of TV after one year is Rs.19950.}}$

The population of a village is $6250$ . It is found that the rate of increase in population is $8\%$ per annum. Find the population after $2$ years.

Here P = 6250 R = 8% T = 2 years
Population after 2 years

$A = P \left ( 1 + \dfrac{R}{100}\right )^n$

$A = 6250 \left( 1 + \dfrac{8}{100} \right )^2$

$= 6250 \left( \dfrac{108}{100} \right )^2$

$= 7290$

The compound interest, calculated yearly on a certain sum of money for the second year is $Rs.\ 1320$ and for the third year is $Rs.\ 1452$ . Calculate the rate of interest ad the original sum of money (in $Rs.$ ).

C.I. for the third year

$=$ Rs.

$1,452$ .
C.I. for the second year

$=$ Rs.

$1,320$

$\therefore$ S.I on Rs.

$1,320$ for one year

$=$ Rs.

$1,452 -$ Rs.

$1,320 =$ Rs.

$132$ .
Rate of interest

$= \dfrac {132\times 100}{1,320} = 10\%$ .
Let the original money be Rs.

$P$ .
Amount after

$2$ year

$-$ amount after one year

$= C.I.$ for second year.

$P\left (1 + \dfrac {10}{100}\right )^{2} - P \left (1 + \dfrac {10}{100}\right ) = 1,320$

$P \left [\left (\dfrac {110}{100}\right )^{2} - \dfrac {110}{100}\right ] = 1,320$

$\Rightarrow P\left [\left (\dfrac {11}{10}\right )^{2} - \dfrac {11}{10}\right ] = 1,320 \Rightarrow P \left (\dfrac {121}{100} - \dfrac {11}{10}\right ) =$ Rs.

$1,320$

$\Rightarrow P\times \dfrac {11}{100} = Rs. 1,320\Rightarrow P = \dfrac {1,320 \times 100}{11} =$ Rs.

$12,000$

$\therefore$ Rate of interest

$= 10\%$
and Original sum of money

$=$ Rs.

$12,000$

Mr. Lalit invested Rs. $5,000$ at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to Rs. $5,325$ . Calculate the amount at the end of second year, to the nearest rupee.

Given: Principal

$=$ Rs.

$5,000$ , Time

$= 2$ years

After one year amount

$=$ Rs.

$5,325$
Amount at the end of

$2^{nd}$ year

$A = P\left (1 + \dfrac {r}{100}\right )^{n}$

$A = 5,000 \left (1 + \dfrac {65}{100}\right )^{2} = 5,000 \left (\dfrac {106.5}{100}\right )^{2}$

$A = 5000\times \dfrac {106.5\times 106.5}{100\times 100}$

$A =$ Rs.

$5671.125$

$A \approx$ Rs.

$5671$

Sangeetha borrowed $8,000$ from Alex for $2$ years at $12\dfrac{1}{2}$ % per annum. What interest did Sangeetha pay to Alex if the interest is compounded annually?

Simple interest on the principal

$P$ at rate of interest

$R$ for

$T$ years is given as,

$I=\dfrac{PRT}{100}$

Interest for first year

$= \cfrac {8000 \times 25 \times 1 }{100 \times 2}$

$= 1000$

Amount after first year

$=1000 + 8000$

$= 9000$

Interest for second year

$= \cfrac {9000 \times 25 \times 1 }{100 \times 2}$

$= 1125$

Ramlal deposited $8,000$ with a finance company for $3$ years at an interest of $15$ % per annum. What is the compound interest that Ramlal gets after $3$ years?

Step 1: Principal for the first year

$=$

$8,000$
Interest for the first year

$= \dfrac{P \times n \times r}{100}$

$= \dfrac{8000 \times 1 \times 15}{100} =$

$1,200$
Amount at the end of first year

$= P + I = 8,000 + 1,200 =$

$9,200$
Step 2: The amount at the end of the first year becomes the Principal for the second year.
Principal for the second year

$=$

$9,200$
Interest for the second year

$= \dfrac{P \times n \times r}{100}$

$= \dfrac{9200 \times 1 \times 15}{100} =$

$1,380$
Amount at the end of second year

$= P + I = 9,200 + 1,380 =$

$10,580$
Step 3: The amount at the end of the second year becomes the Principal for the third year.
Principal for the third year

$=$

$10,580$
Interest for the third year

$= \dfrac{P \times n \times r}{100}$

$= \dfrac{10580 \times 1 \times 15}{100} =$

$1,587$
Amount at the end of third year

$= P + I$

$= 10,580 + 1,587 =$

$12,167$
Hence, the Compound Interest the Ramlal gets after three years is

$A - P = 12,167 - 8,000 =$

$4,167$

Find the difference between simple interest and compound interest on Rs. $2,400$ at $2$ years at $5$ % per annum compounded annually.

$= \cfrac{P \times R \times T}{100}$

$= \cfrac{2400 \times 2 \times 5}{100} = 240$

Compound interest:

Interest for first year

$= \cfrac{ 2400 \times 5 \times 1 }{100} = 120$

Amount after first year

$= 2400+120=2520$

Interest for second year

$= \cfrac{ 2520 \times 5 \times 1 }{100} = 126$

Amount after first year

$= 2520+126=2646$

Hence compound interest

$=2646-2400 = 246$

Difference between CI and SI

$= 246-240=6$

The population of a village has a constant growth of $5$ % every year. If its present population is $1,04,832$ , what was the population two years ago?

Let

$P$ be the population two years ago.

$\therefore$

$P{\left(1+\dfrac{5}{100}\right)}^{2} = 104832$

$P{\left(\dfrac{105}{100}\right)}^{2} = 104832$

$P \times \dfrac{105}{100} \times \dfrac{105}{100} = 104832$

$P = \dfrac{104832 \times 100 \times 100}{105 \times 105}$

$= 95085.71$

$= 95,086$ (rounding off to the nearest whole number)

$\therefore$ Two years ago the population was

$95,086$ .

Find the rate percent per annum, if Rs. $2,000$ amount to Rs. $2,315.25$ in an year and a half, interest being compounded half-yearly.

Amount(A)=

$Rs 2315.25$

Principal

$(P)$ = Rs

$2000$

Time=

$1.5$ yrs

Let rate be '

$R'$

Since

$C . I$ is compounded half yearly

$A=P(1+\cfrac{R}{2\times100})^2\times T$

$2315.25=2000(1+\cfrac{R}{200})^2\times1.5$

$\cfrac{2315.25}{100\times2000}=(1+\cfrac{R}{100})^3$

$\cfrac{9261}{100\times80}=(1+\cfrac{R}{100})^3$

$(\cfrac{21}{20})^3=(1+\cfrac{R}{100})^3$

$\cfrac{21}{20}=1+\cfrac{R}{100}$

$R= (\cfrac{21}{20}-1)\times100$

$R= \cfrac{1}{20}\times100$

$R= 5$

Therefore Rate is

$5$ .

The number of students enrolled in a school is $2000$ . If the enrollment increases by $5$ % every year, how many students will be there after two years?

Let number of students be

$P$

Therefore

$P= 2000$

Rate of increment

$(R) = 5\%$

Time

$(T)= 2$

We need to find amount

$(A)$

The formula is given by

$A= P(1+\dfrac{R} {100})^T$

$= 2000(1+\dfrac{5}{100})^2$

$= 2000 (\dfrac{21}{20})^2$

$=\dfrac{2000\times21\times21}{20\times20}$

$=2205$

Number of students after

$2$ years is

$2205$

At what rate percent compoud interest per annum will Rs. $640$ amount to Rs. $774.40$ in $2$ years?

$A=P(1+\cfrac{r}{100})^n$

$\implies774.40=640(1+\cfrac{r}{100})^2\\ \implies r=10\%$

Gayathri invested a sum of Rs. $12,000$ at $5$ % p.a. at compound interest. She received an amount of $13,230$ after ' $n$ ' years. Find the value of ' $n$ '.

Interest for the

$1$ st year

$=\dfrac{12000\times5\times1}{100}$

$=600$

Amount After the

$1$ st year

$=12000+$ Rs

$600$

$=12600$

Interest for the

$2$ nd year

$=\dfrac{12600\times5\times1}{100}$

$=630$

Amount After the

$2$ nd year

$=12600+$ Rs

$630$

$=13230$

$\therefore$ Required value of

$'n'$

$=2$ years.

The population of a village increases at the rate of 7 $\%$ every year. If the present population is 90,000, what will be the population after 2 years?

Given that, the population of a village increases at a rate of

$7\%$ every year. The present population of the village is

$90,000$ .

To find out: Population of the village after

$2$ years.

We know that, population after '

$n$ ' years

$= P {\left(1+\dfrac{r}{100}\right)}^{n}$

Here,

$P = 90,000$ , rate of increase

$(r) = 7\%$ and number of years

$n = 2$ .

$\therefore\$ The population after two years

$= 90000 {\left(1+\dfrac{7}{100}\right)}^{2}$

$= 90000 {\left(\dfrac{107}{100}\right)}^{2}$

$= 90000 \times \dfrac{107}{100} \times \dfrac{107}{100}$

$= 103041$

Hence, the population of the village after two years will be

$1,03,041$ .

The difference between C.I. and S.I. for $2$ years on a sum of money lent at $5$ % p.a. is Rs. $5$ . Find the sum of money lent.

$T=2$ years,

$R=5\%$ p.a,

$CI-SI=$ Rs

$5$

Let the sum of money lent be

$x$ .

Then,

Simple Interest=Rs

$\dfrac{x\times2\times5}{100}=$ Rs

$\dfrac{x}{10}$

Compound Interest:-

Interest for the

$1$ st year=Rs

$\dfrac{x\times5\times1}{100}=$ Rs

$\dfrac{x}{20}$

Amount after the

$1$ st year

$=Rs\ x+\dfrac{x}{20}=$ Rs

$\dfrac{21x}{20}$

Interest for the

$2$ nd year

$=Rs\dfrac{21x}{20}\times\dfrac{5\times1}{100}$

$=Rs\dfrac{21x}{400}$

Amount after the

$2$ nd year=Rs

$x+\dfrac{21x}{400}+\dfrac{21x}{20}$

=Rs

$\dfrac{441}{400}$

Then,

$\dfrac{ 41x }{ 400 } -\dfrac{ x }{ 10 } =5\\ \Rightarrow \dfrac { 41x-40x }{ 400 } =5\\ \Rightarrow x=Rs\ 2000$

$\therefore$ Required sum of money lent=Rs

$2000$ .

The value of a machine depreciates by $5$ % each year. A man pays $Rs. 30,000$ for the machine. Find its value after three years.

Present value of the machine

$P =$

$30,000$ , Rate of depreciation

$r = 5$ %, Number of years

$n =3$
The value of the machine after '

$n$ ' years

$= P {\left(1 - \dfrac{r}{100}\right)}^{n}$

$\therefore$ The value of the machine after three years

$= 30000 {\left(1 - \dfrac{5}{100}\right)}^{3}$

$= 30000 {\left(\dfrac{95}{100}\right)}^{3}$

$= 30000 \times \dfrac{95}{100} \times \dfrac{95}{100} \times \dfrac{95}{100}$

$= 25721.25$
The value of the machine after three years is

$Rs. 25,721.25$

The population of a village increases by $9$ % every year which is due to the job opportunities available in that village. If the present population of the village is $11,881$ , what was the population two years ago?

Present Population (A)

$=11881$

Let population two years back be

$P$

$Rate (R) =9%$

$Time (T)= 2$ years

$A = P(1+\cfrac{R}{100})^T$

$11881 =P(1+\cfrac{9}{100})^2$

$P=11881\times(\cfrac{100}{109})^2$

$P= 10000$

$\therefore$ Population two years back was

$10,000$

Find the compound interest on Rs. $2,000$ at the rate of interest $5$ % per annum for 3 years.

Given principal

$P=$ Rs.

$2000$ , Rate

$r=5$ % yearly, Time

$n=3$ years

We know that Compound Intererst

$=$ Amount

$-$ Principal

Also we know that Amount can be calculated using the formula :

$A=P{ \left( 1+\cfrac { r }{ 100 } \right) }^{ n }$
Therefore,Compound interest

$=P{ \left( 1+\cfrac { r }{ 100 } \right) }^{ n }-P\quad$

$=2000{ \left( 1+\cfrac { 5 }{ 100 } \right) }^{ 3 }-2000$

$=2000{ \left( 1+\cfrac { 1 }{ 20 } \right) }^{ 3 }-2000\quad \quad$

$=2000{ \left( \cfrac { 21 }{ 20 } \right) }^{ 3 }-2000$

$=2000\times \cfrac { 21 }{ 20 } \times \cfrac { 21 }{ 20 } \times \cfrac { 21 }{ 20 } -2000$

$=\cfrac { 9261 }{ 4 } -2000$

$=2315.25-2000$

$=Rs.315.25$

A motorcycle was bought at Rs. $50,000$ . The value depreciated at the rate of $8$ % per annum. Find the value after one year.

$Cost \space price = Rs 50000= Principal \space value$

$Rate (R) \space of \space depreciation =8\%$

$T=1year$

$\therefore A= P(1-\frac{R}{100})^T since \space value \space depreciates,$

$=> A=50000(1-\dfrac{8}{100})$

$=50000\times\dfrac{23}{25}$

$=46000$

$\therefore value \space of \space motorcycle \space after \space one \space year \space is \space Rs .46000$

If a person joins his work in $2010$ with an annual salary of $Rs. 30,000$ and receives an annual increment of $Rs. 600$ every year, in which year, will his annual salary be $Rs. 39,000$ ?

The person will get

$Rs.30000$ at the end the year

$2010$

And he will get the first increment of

$Rs.600$ in the year

$2011$

So the salary every year will form an A.P.

Salary in the year

$(2009+n)=Rs.(30000+(n-1)\times 600)$

$30000+(n-1)\times 600=39000$

$n-1=15$

$n=16$

Therefore his salary will be incremented to

$Rs.39000$ in the year

$2025$

A company purchases an office copier machine for $Rs. 50,000$ . It is estimated that the copier depreciates in its value at a rate of $15$ % per year. What will be the value of the copier after $15$ years?

$Present \space value(P)=50000Rs.$

$Rate \space of \space depreciation=15\%$

$T=15 years$

Amount after depreciation:

$A=P(1-\frac{R}{100})^T$

$A=50000\left(1-\dfrac{15}{100}\right)^{15}$

$=50000\left(\dfrac{17}{20}\right)^{15}$

$=4367.71$

$\therefore value \space after \space 15years=Rs.4367.71$

The petrol rate is increased by $Rs. 5$ per litre. Now for $Rs. 1320$ , $2$ litres less petrol can be bought as compared to the previous rate. Find the increased price of petrol per litre.

Let the original price of one-litre petrol be Rs.

$x$

In Rs.

$1320$ we obtain

$\cfrac { 1320 }{ x }$ litres of petrol at this rate

If the price is increased by Rs.

$5$ , then the new price of petrol is Rs.

$(x+5)$ per litre.

Then,
In Rs.

$1320$ we obtain

$\cfrac { 1320 }{ x+5 }$ litres of petrol at the new price

According to the given condition, one gets

$2$ litres less petrol on spending Rs.

$1320$

Petrol obtained at old price

$-$ petrol obtained at new price

$=2$

$\therefore$

$\cfrac { 1320 }{ x } -\cfrac { 1320 }{ x+5 } =2$

$\therefore$

$1320(x+5)-1320x=2x(x+5)$

$\therefore$

$1320x+6600-1320x=2{x}^{2}+10x$

$\therefore$

$2{x}^{2}+10x-6600=0$

$\therefore$

${x}^{2}+5x-3300=0$

$\therefore$

${x}^{2}+60x-55x-3300=0$

$\therefore$

$(x+60)(x-55)=0$

$\therefore$

$x=-60$ or

$x=55$

But

$x=-60$ is not possible (price of petrol cannot be negative), so

$x=55$

The old price of petrol is Rs.

$55$ /litre and the increased price is Rs.

$60$ /litre

Formula of compound interest is $C.I. =$ _________

The formula for calculation of compound interest when the principal, rate of interest and the time are known, is

$C.I. = P \left (1 + \dfrac {r}{100}\right )^{n} -P$

The formula for calculation of compound interest when the amount and the principal are known is

$C.I. = A - P$ .

The present value of an oil engine is Rs. $14,580$ . What was worth of the engine $3$ years before if the value depreciates at the rate of $10\%$ every year?

Here,

$Principal (P)=14580$

$Time (T)= 3 years$

$Rate (R) of depreciation = 10\%$

Since it is case of depreciation, value of engine

$3$ would be greater than present value.

let

$Rs\ P$ be the value of engine

$3$ years back.

$Amount (A) = current \space value =Rs14580$

Apply the concept of compound depreciation

$A= P(1-\dfrac{R}{100})^T ( since \space depreciation )$

$14580 = P (1-\dfrac{10}{100})^3$

$14580=P(\dfrac{9}{10})^3$

$P= 20000$

$\therefore$ value of engine

$3$ years back

$= Rs .20000$

Find the amount and compound interest for Rs. $1,200$ for $4$ years at the rate of $5\%$ per annum interest.

The interest of one year

$=\displaystyle\frac{1,200\times 5\times 1}{100}$

$=Rs. 60$
Now, principle for second year

$=1,200+60$

$=Rs. 1,260$
Further, the interest for

$2$ nd year

$=\displaystyle\frac{1,260\times 5\times 1}{100}$

$=Rs.63$
Now, principal for

$3$ rd

$=1260+63$

$=Rs. 1,323$
Further, the interest for

$3$ rd year

$=\displaystyle\frac{1323\times 5\times 1}{100}$

$=Rs. 66.15$
Now, principal for

$4$ th year

$=1,323+66.15$

$=Rs. 1,389.15$
Further, the interest for

$4$ th year

$=\displaystyle\frac{1,389.15\times 5\times 1}{100}$

$=Rs. 69.46$
Principal after four year

$=1,389.15+69.46$

$=Rs. 1,458.61$
Total compound interest for four year

$=1,458.61-1,200$

$=Rs. 258.61$ .

From a village, people started migrating to nearby cities due to unemployment . The population of the village two years ago was $6,000$ . If the rate of migration is $5\%$ , find the current population of the village.

$Principal(P) = population \space 2\ years \space ago \space$

$=6000$

$Rate \space of \space migration \space(R) =5\%$

$Time(T) = 2 yrs$

$A= current \space population$

$Since \space people \space are \space migrating, \space hence \space current \space population \space is \space lesser \space than \space that \space two \space years \space back.$

$Hence\ there \ is\ a\ depreciation$

$\therefore A=P(1-\dfrac{R}{100})^T$

$=6000(1-\dfrac{5}{100})^2$

$=6000\left(\dfrac{19}{20}\right)^2$

$=6000\left(\dfrac{19}{20}\right)\left(\dfrac{19}{20}\right)$

$=5415$

$\therefore Current \space population \space is \space 5415$

Maria invested $8000$ Rs in a business. She would be paid interest at $5 \%$ per annum compounded annually. Find :
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.

The formula for the compound interest,when interest is compound annually

$Amount=P{ \left(1+\dfrac { R }{ 100 } \right) }^{ n }$

where P = Principal amount

R = Rate of interest in percentage

n = Number of years

(i)

Given,

$P=8000 \ Rs.,\quad R=5,\quad n=2$

Amount credited after end of second year

$=8000{ \left(1+\dfrac { 5 }{ 100 } \right) }^{ 2 }= Rs. \ 8820$

(ii)

Given,

$P=8000 \ Rs.,\quad R=5,\quad n=3$

Amount credited after the end of third year

$=8000{ \left(1+\dfrac { 5 }{ 100 } \right) }^{ 3 }= Rs. \ 9261$

So, the interest for the third year = Amount credited after the end of

${ 3 }^{ rd }$ year - Amount credited after the end of

${ 2 }^{ nd }$ year

$= 9261 - 8820 = Rs. \ 441$

In financial year $2014-15$ the gross income of Kuldeep is Rs. $5,80,000$ (Excluding house rent). He deposit Rs. $9,000$ to its General provident fund. He deposit Rs. $60,000$ in his P.P.F. account. Calculate the income tax paid by Kuldeep in his last month. The total income of Rs. $1,50,000$ is free from tax. The rate of income tax are.
Tax Payable Income Tax
(i) $2,00,000$ Rs upto zero
(ii) From Rs. $2,00,001$ to Rs. $5,00,000$ On amount more than Rs. $2,00,000$ $10\%$
In addition to it, educational cess $3\%$ on income tax payable.

Calculation of Taxable income -

Taxable Income = Gross income - Tax Free Income

Taxable income = Rs. 580000

- Tax Free Income = Rs. 150000

Taxable income = Rs. 430000

As tax for income upto Rs. 200000 is Zero % and in slab Rs. 200001 to Rs. 500000 is 10%.

so tax calculated = 0 + Rs. 23000 * 0.10

tax calculated = Rs. 23000

tax payable = tax calculated + educational cess(which is 3% of tax)

tax payable = Rs. 23000 + Rs. 23000*0.03

tax payable = Rs. 23690

What is the Latin word for reduction or decline in the value of fixed asset due to its use?

The word depreciation has been derived from the Latin word "Depretium" which means "decline" or "reduction" in price or value.

What is 'Depreciation'?

In simple words, Depreciation is a gradual reduction in the value of assets.

Answer the following in one sentence:
What is depreciation?

Depreciation means continuous reduction in the value of property or asset due to wear and tear, accident, fall in market price, passage of time etc.

A person has Rs. $30,000$ to invest.He wants to invest some amount in the fixed deposit and remaining in savings account.The interest rates are $7\%$ and $5\% \ p.a$ for the fixed deposit and the saving account respectively. Find how he should divide the total amount $Rs.30000$ in two accounts if he wants to get the annual interest as (i) $Rs.1800$ (ii) $Rs.2000$ .

Let us assume that the person will invest Rs.

$x$ in savings account

Amount invested in the fixed deposit = Rs.

$(30,000-x)$

Given the rate of interest in savings and fixed deposits are

$5\%$ and

$7\%$ p.a. respectively.

Now, total interest got by the person in annum is

$\dfrac{x\times5}{100}+\dfrac{(30,000-x)\times 7}{100}$ .

i) According to the problem,

$\dfrac{x\times5}{100}+\dfrac{(30,000-x)\times 7}{100}=1800$

$\Rightarrow \dfrac{x\times5}{100}+\dfrac{2,10,000}{100} -\dfrac{x\times 7}{100}=1800$

$\Rightarrow \dfrac{x\times5}{100}-\dfrac{x\times 7}{100}=1800-2100$

$\Rightarrow \dfrac{-2x}{100}=-300$

$x=15,000$ .

For the first case, he has to invest Rs.

$15,000$ in each of the accounts.

ii) For the second case

$\dfrac{x\times5}{100}+\dfrac{(30,000-x)\times 7}{100}=2000$

$\Rightarrow \dfrac{x\times5}{100}+\dfrac{2,10,000}{100} -\dfrac{x\times 7}{100}=2000$

$\Rightarrow \dfrac{x\times5}{100}-\dfrac{x\times 7}{100}=2000-2100$

$\Rightarrow \dfrac{-2x}{100}=-100$

$x=5,000$ .

In this case, he has to invest Rs.

$5,000$ in savings and Rs.

$25,000$ in fixed deposit.

Find the amount and the compound interest on 50000 for $1\frac{1}{2}$ years at 8% per annum, the interest is compounded semi-annually.

We have the formula for compound Interest,

$A=P(1+\dfrac{r}{n})^{n.t}$

$A=Total \ amount$

$P=Principal=50000$

$r=annual \ interest \ rate=8$ %

$n=no. \ of \ times \ compounded \ per \ year=2$ , and

$t=time \ in \ years=\dfrac{3}{2}$

Substituting,

$A=50000(1+\dfrac{0.08}{2})^{2\times\dfrac{3}{2}}$

$A=50000(\dfrac{2.08}{2})^{3}$

$A=56243.2$ This is the amount after

$1$ and

$\dfrac{1}{2}$ years

Compound Interest,

$I=A-P=56243.2-50000=6243.5$

Find the amount and the compound interest on 2000 at 10% p.a. for $2\frac{1}{2}$ years, compounded annually.

$P=Rs$

$2000$

$R=10$ %

$T={ 2 }^{ 1/2 }$ years

$=\dfrac { 5 }{ 2 }$ years.

$A=P{ \left( 1+\dfrac { R }{ 100 } \right) }^{ T }$

$=2000{ \left( 1+\dfrac { 10 }{ 100 } \right) }^{ 5/2 }$

$=2000{ \left( \dfrac { 11 }{ 10 } \right) }^{ 5/2 }$

$=2538.1$

$\simeq$

$2538$

$C.I=A-P$

$=2538-2000$

$=538$ .

Hence Amount

$=Rs$

$2538$

$C.I=Rs$

$538$

A sum of $9600$ is invested for $3$ years at $10\%$ per annum at compound interest.
(i) What is the sum due at the end of the first year?
(ii) What is the sum due at the end of the second year?
(iii) Find the compound interest earned in $2$ years.
(iv) Find the difference between the answers in (ii) and (i) and find the interest on this

Interest for the

$1^{st}$ year

$=Rs\cfrac{9600\times 10\times 1}{100} = Rs960$

Amount after the

$1^{st}$ year

$= Rs 9600+Rs 960 = Rs10560$

$(i)$ The sum due at the end of the

$1^{st}$ year

$= Rs 10560$

Interest for the

$2^{nd}$ year

$=Rs\cfrac{10560\times 10\times 1}{100} = Rs1056$

Amount after the

$2^{nd}$ year

$= Rs 10560+Rs 1056 = Rs11616$

$(ii)$ The sum due at the end of the

$2^{nd}$ year

$= Rs 11616$

$(iii)$ Compound Interest earned in

$2$ years

$=Rs 11616-Rs 9600=Rs 2016$

$(iv)$ Difference

$=Rs 11616-Rs 10560=Rs. 1056$

Interest

$=\cfrac{1056\times 3\times 10}{100}=Rs 316.80$

On what sum of money will the compound interest for 2 years at 5 per cent per annum amount to Rs 768.75?

$A=Rs.768.75,r=5\%,n=2$

$\implies A=P(1+\cfrac{r}{100})^n$

$\implies 768.75=P(1+\cfrac{5}{100})^2\\\implies P=697.27\approx 697.5$

Hence, sum

$=Rs.697.5$ (approx).

Monica borrowed a sum of $Rs. 12,500$ to open a store, at the rate of $5.5\%$ per annum for $3$ years. Find the compound interest paid by her after $3 years$ .

Interest for the

$1^{st}$ year

$=Rs\cfrac{12500\times 5.5\times 1}{100} = Rs687.50$

Amount after the

$1^{st}$ year

$= Rs 12500+Rs 687.50 = Rs13187.50$

Interest for the

$2^{nd}$ year

$=Rs\cfrac{13187.50\times 5.5\times 1}{100} = Rs725.3125$

Amount after the

$2^{nd}$ year

$= Rs 13187.50+Rs 725.3125 = Rs13912.8125$

Interest for the

$3^{rd}$ year

$=Rs\cfrac{13912.8125\times 5.5\times 1}{100} = Rs765.20$

Amount after the

$3^{rd}$ year

$= Rs 13912.8125+Rs 765.20 = Rs14678$

$\therefore$ Compound Interest

$=Rs 14678-Rs12500=Rs 2178$

To renovate his shop Mahendra obtained a loan of $Rs. 10000$ from a bank. If the rate of interest is $5\%$ per annum compounded annually, calculate interest annually that he will have to pay for $3$ years.

Interest for the

$1^{st}$ year

$=Rs\cfrac{10000\times 5\times 1}{100} = Rs500$

Amount after the

$1^{st}$ year

$= Rs 10000+Rs 500 = Rs10500$

Interest for the

$2^{nd}$ year

$=Rs\cfrac{10500\times 5\times 1}{100} = Rs525$

Amount after the

$2^{nd}$ year

$= Rs 10500+Rs 525 = Rs11025$

Interest for the

$3^{rd}$ year

$=Rs\cfrac{11025\times 5\times 1}{100} = Rs551.25$

Rohan borrowed a sum of $Rs. 20,000$ to open a book shop at the rate of $4.5\%$ per annum for $3$ years. Find the compound interest paid by him after $3$ years.

Interest for the

$1^{st}$ year

$=Rs\cfrac{20000\times 4.5\times 1}{100} = Rs900$

Amount after the

$1^{st}$ year

$= Rs 20000+Rs 900 = Rs20900$

Interest for the

$2^{nd}$ year

$=Rs\cfrac{20900\times 4.5\times 1}{100} = Rs940.50$

Amount after the

$2^{nd}$ year

$= Rs 20900+Rs 940.50 = Rs21840.50$

Interest for the

$3^{rd}$ year

$=Rs\cfrac{21840.50\times 4.5\times 1}{100} = Rs982.8225$

Amount after the

$3^{rd}$ year

$= Rs 21840.50+Rs 982.8225 = Rs22823.3225$

$\therefore$ Compound Interest

$=22823.3225-Rs20000=Rs 2823.3225$

Find the difference between S.I. and C.I. on $Rs. 25,000$ at $9\%$ per annum for $2$ years.

Simple Interest

$=Rs \cfrac{25000\times 9\times 2}{100}=Rs 4500$

Compound Interest:

Interest for the

$1^{st}$ year

$=Rs\cfrac{25000\times 9\times 1}{100} = Rs2250$

Amount after the

$1^{st}$ year

$= Rs 25000+Rs 2250 = Rs27250$

Interest for the

$2^{nd}$ year

$=Rs\cfrac{27250\times 9\times 1}{100} = Rs 2452.50$

Amount after the

$2^{nd}$ year

$= Rs 27250+Rs 2452.50 = Rs29702.50$

$\therefore$ Compound Interest

$=Rs29702.50-Rs25000=Rs 4702.50$

$\therefore$ Difference between Compound Interest and Simple Interest

$=Rs 4702.50-Rs 4500=Rs 202.50$

What sum of money will amount to Rs. 5,445 in 2 years at 10% per annum compound interest?

Let sum of money be

$'P'$ .

Amount

$'A'=5445$

$T=2$

$R=10$ %

$A=P{ \left( 1+\dfrac { R }{ 100 } \right) }^{ T }$

$\Rightarrow 5445=P{ \left( 1+\dfrac { 10 }{ 100 } \right) }^{ 2 }$

$=P{ \left( \dfrac { 11 }{ 10 } \right) }^{ 2 }$

$\Rightarrow P=\left( 5445 \right) \times { \left( \dfrac { 10 }{ 11 } \right) }^{ 2 }$

$=4500$

Hence, Sum of Money

$=Rs4500$ .

Meera deposited a sum of $Rs.10,400$ in life insurance for $3\%$ compounded annually at $6\frac{1}{4}$ % per annum. What amount will he get on maturity?

$P= Rs. 10,400$

$T=3years$

$R= { 6 }_{ 4 }^{ 1 }%$

$R=\frac { 25 }{ 4 }%$

$A=P(1+\frac { R }{ 100 })^T$

$A=104000(1+\frac { 25 }{ 4*100 })^3$

$A=104000(\frac { 17 }{ 16 })^3$

$A= Rs.12474.4$

Find the amount and the compounded interest on $Rs. 4500$ for $2 years$ at $10\%$ per annum compounded annually.

Amount (Compound Interest with Principle) is

$:A=P\left(1+r\right)^{n}$

Here,

$P=4500\\rate=r=10\%=\dfrac{10}{100}\\n=no \ of \ years=2\ years$

Now,

$A=4500\left(1+\dfrac{10}{100}\right)^{2}=4500(1+0.1)^2=4500(1.1)^2=5445$

So,Compound interest

$=CI=A-P=5445-4500=945$

Three pieces of cakes of weights $4\cfrac { 1 }{ 2 }$ lbs, $6\cfrac { 3 }{ 4 }$ lbs and $7\cfrac { 1 }{ 5 }$ lbs respectively are to be divided into parts of equal weights. Further each must be as heavy as possible. If one such part is served to each guest then what is the maximum number of guests that could be entertained?

$4\cfrac { 1 }{ 2 }$ lbs

$,6\cfrac { 3 }{ 4 }$ lbs

$,7\cfrac { 1 }{ 5 }$ lbs

$\cfrac { 9 }{ 2 }$ lbs

$,\cfrac { 27 }{ 4 }$ lbs

$,\cfrac { 36 }{ 5 }$ lbs

$\cfrac { 4.5 }{ 0.9 }$ lbs

$,\cfrac { 6.75 }{ 0.9 }$ lbs

$,\cfrac { 7.2 }{ 0.9 }$ lbs

$5+7.5+8$

$=20.5$

Find the compound interest when Principal is Rs $10000$ and rate is $10$ % for $2$ years.

$P= Rs10000$

$R=10%$

$T=2$

Interest for the

$1^{st}$ year

$=Rs\cfrac{10000\times 10\times 1}{100} = Rs1000$

Amount after the

$1^{st}$ year

$= Rs 10000+Rs 1000 = Rs11000$

Interest for the

$2^{nd}$ year

$=Rs\cfrac{11000\times 10\times 1}{100} = Rs1100$

Amount after the

$2^{nd}$ year

$= Rs 11000+Rs 1100 = Rs12100$

$\therefore$ Compound Interest

$=Rs 12100-Rs 10000=Rs 2100$

A sum of money becomes $\dfrac {8}{5}$ of itself in $5$ days at certain $5$ rate of simple interest. Find the rate of interest ?

$A=\cfrac{8}{5}P$

$A=P+SI$

$\implies A=P+(\cfrac { P\times r\times t }{ 100 })$

$\implies \cfrac{8}{5}P=P[1+(\cfrac { r\times 5}{ 100 }) ]$

$\implies r=12\%$

A machinery is bought at $Rs. 60,000$ . Its value depreciated at the rate of $7\%\ p.a.$ . Find its value after $2$ years.

Present value

$=Rs.60,000$ .

It is given that the value depreciates

$10\%$ per year.

Value after

$n$ years

$=A\left(1-\cfrac{r}{100}\right)^n$ {

$A$ is the amount of machine}

Value after

$2$ years

$=Rs.60,000\left(1-\cfrac{7}{100}\right)^2=Rs.51,894$

A man borrows Rs. $6000$ at $5\%$ compound interest. If he repays Rs. $1200$ at the end of each year, find the amount outstanding at the beginning of the third year.

$P=6000$ per month ,

$R=5\%$

After

$1$ year

Interest

$=\dfrac{PRT}{100}$

$=\dfrac{6000\times 5\times 1}{100}$

$=300$

New principle will be

$6000+300=6300$ but Rs

$1200$ was repaid

$\implies P^{'}=6300-1200=5100$

After

$2$ years

Interest

$=\dfrac{P^{'}RT}{100}$

$=\dfrac{5100\times 5\times 1}{100}$

$=255$

Principle for

$3^{rd}$ year

$=5100+255=5355$ but again Rs

$1200$ was repaid

$\therefore$ Outstanding

$=5355-1200$

$=4155$

Calculate the amount and compound interest on $Rs. 10,800$ for $3$ years at $12\dfrac{1}{2}\%$ per annum compounded annually.

Here, the principal amount is

$P=10,800$ , time is

$T=3$ and rate of interest is

$12\dfrac {1}{2}$ % and we know that the formula to calculate amount is

$A=P{ \left[ 1+\dfrac { R }{ 100 } \right] }^{ n }$ , therefore, we have:

$A=P{ \left[ 1+\dfrac { R }{ 100 } \right] }^{ n }$

$=10800{ \left[ 1+\dfrac { 12\dfrac { 1 }{ 2 } }{ 100 } \right] }^{ 3 }$

$=10800{ \left[ 1+\dfrac { \dfrac { 25 }{ 2 } }{ 100 } \right] }^{ 3 }$

$=10800{ \left[ 1+\dfrac { 25 }{ 200 } \right] }^{ 3 }$

$=10800{ \left[ \dfrac { 225 }{ 200 } \right] }^{ 3 }$

$=10800\times \dfrac { 225 }{ 200 } \times \dfrac { 225 }{ 200 } \times \dfrac { 225 }{ 200 }$

$=\dfrac { 675\times 9\times 9\times 9 }{ 4\times 8 }$

$=\dfrac { 492075 }{ 32 }$

$=15377.34$

Thus, the amount is Rs.

$15377.34$

Now, to find the compound interest, we subtract the principal amount from the total amount as follows:

$C.I=15377.34-10800=4577.34$

Hence, the amount is Rs.

$15377.34$ and the compound interest is Rs.

$4577.34$ .

A sum of $\text{Rs.}\ 9600$ is invested for $3$ years at $10\%$ per annum at compound interest, then what is the sum due at the end of the first year?

Given :

$P= \text{Rs.}\ 9600$

$R=10%$

$T=1 \ \text{year}$

$A=P \left (1+\dfrac { R }{ 100 } \right )^T$

$A=9600 \left (1+\dfrac { 10 }{ 100 } \right )$

$A= \text{Rs.}\ 9696$

Amount due after first year

$= \text{Rs.}\ 9696$

Raju took a loan from a bank at 10% simple interest per annum after 4 years he repaid rs.Find the sum he borrowed.

Given that,

Raju took a loan at

$10\%$ simple interest per annum for

$4$ years.

He repaid

$Rs.\ 5500$ after

$4$ years.

To find out,

The sum he borrowed.

Let the borrowed sum be

$Rs.\ P$

We know that,

$SI = \dfrac{{P \times R \times R}}{{100}}$

$SI = \dfrac{{P \times 4 \times 10}}{{100}}$

$= \dfrac{4P}{{10}}$

We also know that,

$A = P + SI$

$\therefore \ 5500 = P + \dfrac{4P}{{10}}$

$\Rightarrow 5500 = \dfrac{{14P}}{{10}}$

$\Rightarrow P = \dfrac{5500\times 10}{14}$

$\therefore \ P=Rs.\ 3928.57$

Hence, the sum Raju borrowed was

$Rs.\ 3928.57$ .

The value of a machine depreciates every year by $10\%$ . If its present value is $Rs.90000$ then what will be its value after two years?

Given Present value

$=Rs.90000$ .

Value depriciates

$10\%$ per year.

Value after

$n$ years

$= PV\times \left(1-\dfrac{\% \ depreciation}{100}\right)^n$

Value after two years

$=Rs.90000\left(1-\cfrac{10}{100}\right)^2=Rs.72,900$

Find the compound interest on $Rs. 40,000$ for $1$ year at $10\%$ per annum interest being payable quarterly

$P=40,000\\ T=1\ year\\ R=10\%$

When payable quarterly,

$C.I=P\left( 1+\dfrac { R }{ 4\times 100 } \right) ^{ 4\times T }-P\\ \quad \quad =40000\left( 1+\dfrac { 10 }{ 400 } \right) ^{ 4 }-40000\\ \quad \quad =40000\left( \dfrac { 41 }{ 40 } \right) ^{ 4 }-40000\\ \quad \quad =4152.5\\ \therefore { C.I=Rs.4152.5 }$

A sum of money becomes $\cfrac{8}{5}$ th of itself in $5$ years at certain a rate of simple interest. Find the rate of interest.

Let the Initial principle amount $= P$

After $5$ years the amount received $= \left(\cfrac{8}{5}\right)P$

As the interest applicable is simple interest, the total interest is Final amount received $-$ Initial principle amount.

Interest amount for $5$ years $=\left(\cfrac{8}{5}\right)P - P$

Interest amount for $5$ years $=\left(\cfrac{3}{5}\right)P$

With simple interest, interest amount every year is the same.

Thus interest for one year $=$ Interest for $5$ years $\div 5$

Interest for one year $=\cfrac{\left(\cfrac{3}{5}\right)P}{5}$

Interest for one year $=\cfrac{3}{25}P$

Simple Rate of interest per year $= \cfrac {Amount\ of\ Interest\ received\ per\ year}{Initial\ Principle\ amount}$ $\times 100\%$

$\therefore$ Rate of interest per year $= \left(\left(\cfrac{3}{25}\right)P \div P\right) \times 100 \%$

$\therefore$ simple rate of interest $= \cfrac{3}{25}\times 100 \%$

$\therefore$ Simple Rate of interest $= 12\%$

The simple interest on a certain sum for $3$ years at $10\%$ per annum is $Rs.\ 829.50$ . Find the sum.

The simple interest is given as ${\rm{Rs}}\;829.50$ with the rate of $10\%$ per annum for $3$ years.

$S.I. = \dfrac{{P \times R \times T}}{{100}}$

$829.50 = \dfrac{{P \times 10 \times 3}}{{100}}$

$P = \dfrac{{829.50 \times 100}}{{10 \times 3}}$

$\Rightarrow P = {\rm{Rs}}\;2765$

At what rate percent per annum will $Rs.\ 6300$ yield an interest of $Rs.\ 2100$ in $4$ years?

The principle is given as ${\rm{Rs}}\;6300$ and the simple interest is given as ${\rm{Rs}}\;2100$ with time as $4$ years.

$S.I. = \dfrac{{P \times R \times T}}{{100}}$

$2100 = \dfrac{{6300 \times R \times 4}}{{100}}$

$R = \dfrac{{2100 \times 100}}{{6300 \times 4}}$

$R = 8.34\%$

At what rate percent per annum simple interest will $Rs.\ 66000$ amount to $Rs.\ 72720$ in $2$ years?

Principal,P = Rs. 66,000

Time,T = 2 years

S.I. = 72,720 - 66,000 = Rs. 6,720

Rate,R = ?

$\therefore$ S.I. =

$\dfrac{P\times R\times T}{100}$

$\Rightarrow$ R =

$\dfrac{6720 \times 100}{66,000 \times 2}$

$\Rightarrow$ Rate =

$5.09$ % per annum

You will receive the amount of INR $10,500$ every year for the periods of $6$ years and the discount rate is $14$ %. What will be the present value of this amount?

$SI=10,500,r=14\%,t=6$

$A=\cfrac { P\times r\times t }{ 100 }$

$\implies Rs,10,500=\cfrac { P\times 14\times 6 }{ 100 }\implies P=75000$

Present value

$=Rs.75,000$ .

A farmer, for purchasing seeds and fertilisers, borrowed a loan from a co-operative bank. After two years he paid $Rs\ 5434$ and settled the account. If the rate of simple interest is $2\dfrac{1}{4}\%$ per annum, what sum did he borrow?'

Given that,

Amount

$(A)= Rs. 5434$

Time

$(T)= 2$ years

Rate of interest

$(R) = 2\dfrac{1}{4}$ %

$= \dfrac{9}{4}$ %

Principle

$(P) = ?$

$\because A = P\left(1 + \dfrac{TR}{100}\right)$

$\implies 5434 = P\left(1+ \dfrac{2\times9}{4\times100}\right)$

$\implies 5434 = P\left(1+ \dfrac{9}{200}\right)$

$\implies 5434 = P\left(\dfrac{209}{200}\right)$

$\implies P = 5434\times \dfrac{200}{209} \\\ \ \ \ \ \ \ \ \ = 26\times 200 \\\ \ \ \ \ \ \ \ \ = 5200$

$\therefore$ Sum borrowed

$= Rs.5200$

The population of a city increases by 30% every year. If the present population is 338000, then What was the population of the city two years ago?

Let initial population be

$p$

As population increased by

$30\%$ per year.

So, population after

$2$ years

$=p(1+\cfrac{30}{100})^2$

$=p(1.3)^2$

Given that ,

$338000=1.69p$

$\implies p =\dfrac{338000}{1.69}$

$\implies p=2,00,000$

So, population two years ago

$=2,00,000$

The simple interest on a sum of money for $5$ years is $\dfrac {3}{5}$ of the sum. Find the rate percent per annum.

Let ${\rm{Rs}}\;x$ be the sum. Then according to the question, the simple interest will be $\dfrac{3}{5}x$ .

$S.I. = \dfrac{{P \times R \times T}}{{100}}$

$\dfrac{{3x}}{5} = \dfrac{{x \times R \times 5}}{{100}}$

$R = \dfrac{{3x \times 100}}{{5 \times x \times 5}}$

$R = 12\%$

What rate gives $Rs. 280$ as interest on a sum of $Rs. 56,000$ in $2$ years

${\textbf{Step - 1: State the formula}}{\text{.}}$

${\text{If P is the Principle , R is the rate of interest , T is time period}}{\text{.}}$

${\text{then Interest , I}} = \dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}.$

${\textbf{Step - 2: Calculate the rate of interest}}{\text{.}}$

${\text{Given the interest I}} = {\text{Rs 280}}{\text{.}}$

${\text{P}} = {\text{Rs 56000}}{\text{.}}$

${\text{Time period}} = {\text{2 years}}$

${\text{Suppose the rate is R% }}{\text{.}}$

${\text{So , I}} = \dfrac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}}$

$\Rightarrow {\text{R}} = \dfrac{{100 \times {\text{I}}}}{{{\text{P}} \times {\text{T}}}}$

$\Rightarrow {\text{R}} = \dfrac{{100 \times 280}}{{56000 \times 2}}$

$\Rightarrow {\text{R}} = \dfrac{{140}}{{560}}$

$\Rightarrow {\text{R}} = \dfrac{7}{{28}}$

$\Rightarrow {\text{R}} = \dfrac{1}{4}$

$\Rightarrow {\text{R}} = 0.25$

${\textbf{Hence, The rate of interest is 0}}{\textbf{.25% per annum}}{\text{.}}$

A man invests $Rs. 20,020$ in buying shares of N.V $Rs. 26$ at $10 \%$ premium. The dividend
on the shares is $15 %$ per annum. Calculate:
(a) the
number of shares he buys

(b) the
dividend he receives annually

(c) the rate of interest he gets on his money

1371065_1059665_ans_9161228676674f1aa2680ad7e84f7fcb.JPG

The difference between the simple interest and the compound interest (compounded annually) on $\text{Rs.}\ 1250$ for $2$ years at $8 \%$ per annum will be?

Difference of S.I & C.I for

$2$ years is

$p\left(\dfrac{r}{100}\right)^2$

where

$p$ is the principal amount and

$r$ is rate of interest

Difference

$=p\left(\dfrac{r}{100}\right)^2$

$=1250\left(\dfrac{8}{100}\right)^2$

$=\text{Rs.}\ 8$

Find the rate at which a sum of money will double itself in $2$ years, if the interest is compounded annually.

$\textbf{Step 1: Calculate Rate of interest For Given Sum}$

$\text{Given-If Principal is Rs. P, then Amount is 2P Rs.}$

$\text{If Sum 'P' is invested at an interest of $r\%$ p.a. for $n$ years then the Amount'A' received is-}$

$\Rightarrow A=P\left( 1+\dfrac{r}{100}\right)^n$

$\Rightarrow 2P=P\left( 1+\dfrac{r}{100}\right)^2$

$\textbf{[Since,A=2P & n=2]}$

$\Rightarrow 2=\left( 1+\dfrac{r}{100}\right)^2$

$\Rightarrow 1+\dfrac{r}{100}=\sqrt{2}$

$\Rightarrow \dfrac{r}{100}=1.414-1$

$\left[\boldsymbol{\because \sqrt{2}\cong 1.41.}\right]$

$\Rightarrow \dfrac{r}{100}=0.414$

$\Rightarrow r=41.4\%$

$\textbf{Therefore, Rate of Interest is 41.4 %.}$

A car loses value at a rate of $27$ % each year. How long will it take for its value to halve?

Formula of compounding can be used here.

Let the price of car be

$x$

We know that compound formula

$\Rightarrow S=P(1+\dfrac{R}{100})^n$

$\Rightarrow \dfrac{x}{2}=x(1-\dfrac{27}{100})^n$

Here we used

$'-'$ sign because the amount is decreasing

$\Rightarrow \dfrac{1}{2}=(1-0.27)^n$

$\Rightarrow \dfrac{1}{2}=(0.73)^n$

$\Rightarrow n\approx 2.2years$

The interest on a sum of money at the end of $2\dfrac{1}{2}$ years is $\dfrac{4}{5}$ of the sum.What is the rate percent?

Consider the given data.

$T=2\dfrac{1}{2}=\dfrac{5}{2}\,years,$

$R=?$

Let the sum be $x$

Since, the interest is $\dfrac{4}{5}$ of sum.

So, interest $=\dfrac{4x}{5}$

We know that

$S.I=\dfrac{P\times R\times T}{100}$

$\dfrac{4x}{5}=\dfrac{x\times R\times \dfrac{5}{2}}{100}$

$\dfrac{4}{5}=\dfrac{R\times 5}{2\times 100}$

$R=\dfrac{4\times 2\times 100}{5\times 5}$

$R=\dfrac{4\times 2\times 100}{25}$

$R=32\%$

Shubhalaxmi took a loan of INR 18000 from Surya Finance to purchase a TV set. If the company charges compound interest at $12\%$ per annum during the first year and $12\frac{1}{2}\%$ per annum during the second year, how much will she have to pay after 2 years ?

For

$r=12\%$ ,

$P=Rs.18000$

$A=18000(1+\cfrac{12}{100})^1=Rs.20,160$

For

$r=12.5\%$ ,

$P=Rs.20,160$

$A=20000(1+\cfrac{12.5}{100})^1=Rs.20,160=Rs. 22,680$

Mrs. Sharma deposited $Rs. 60,000$ in a bank for $2\ years.$ If the interest in $8$ % p.a compounded annually, what amount will she receive at the end of $2\ years?$

Let Principal amount

$=P=60,000$

Time period

$=t=2$

Rate of interest

$=r=\dfrac{8}{100}=0.08$

Total amount

$=P(1+r)^{t}$

$=60000(1+0.08)^{2}$

$=Rs.69,984$

Harvinder deposited $Rs.40,000$ in a post officer for a period of three years. The post officer credits the interest yearly in his account at $7\%$ per annum compounded annually. Find the balance in his account after $3$ years.

Given that,

Principle amount

$=Rs.40,000$ , Rate of compound interest

$=7\%$ , Time

$t=3$ years

Total amount after three years

$=p\left(1+\dfrac{r}{100}\right)^t$

$=40,000\left(1+\dfrac{7}{100}\right)^3$

$=40,000\left(1+0.7\right)^3$

$=40,000\times 1.225$

$=Rs.49001.72$

Hence, this is the required solution.

At the rate of $6\%$ p.a. simple interest, a sum of Rs. $2500$ will earn how much interest by the end of $5$ years?

The simple interest rate is

$6 \% \,p.a$

principal amount is

$Rs.\,2500$

After one year, the interest would be

$6 \%$ of

Rs.

$2500=\dfrac{6}{100}\times 2500$

$=Rs. 300$

Similarily, after

$5$ years, interest =

$5\times 150$

$= Rs. 750$

The population of a city in $Jan. 1987$ was $20,000$ . It increased at the rate of $2%$ per annum. Find the population of the city in $jan. 1997$ .

Initial population

$=20,000,r=2\%,t=10$ years.

Final population

$=20,000(1+\cfrac{2}{100})^10=24,380$ .

A sum at simple interest of $13\dfrac{1}{2}\%$ p.a. amounts to Rs. $2502.50$ after $4$ years. Find the sum.

The amount after simple interest is given by

$A = P ( 1+rt)$

Where P is principal amount

$r$ is interest rate in decimal

$t$ is number of years

Given,

$A = Rs. 2502.50 , r= \dfrac{27}{2\times 100}, t=4$

$2502.5= p \left(1+\dfrac{27}{200}\times 4\right)$

$\Rightarrow 2502.5= 1.54 P$

$\Rightarrow P= Rs.1625$

$\therefore$ The sum is

$Rs. \,1625$ .

A car was bought in $2013$ for Rs. $30,00,000$ . If the value of the car depreciates by $5\%$ each year, then find its worth in the year $2016$ .

Rate of depreciation

$=5\%$ .

Initial price

$=Rs.30,00,000$

After

$n$ years, price after depreciation

$=A\left(1-\cfrac{r}{100}\right)^n$ {

$A$ is the amount of Car in

$2013$ }

After

$3$ years in 2016, the final price will be

$=Rs.30,00,000\left(1-\cfrac{5}{100}\right)^3=Rs.25,72,125$

The national wealth of a country increases by $4\%$ of its value at the beginning of every year. Find the national wealth of the country in 1985, if was estimated at INR $3.125 \times {10^{12}}$ in 1983.

The National wealth in 1983 was

$INR \ \ 3.125 \ \times \ 10^{12}$

The national wealth of a country increases by

$4\%$ . ie.,

$4(\dfrac{1}{100}) \ times = \dfrac{1}{25} \ times$

The national wealth of the country in 1984

$= \left (3.125 \ \times 10^{12} \ \times \dfrac{1}{25} \right ) \ + \ 3.125 \ \times 10^{12}$

$=12.5 \ \times 10^{10} \ + \ 3.125 \ \times 10^{12}$

$=0.125 \ \times 10^{12} \ + \ 3.125 \ \times 10^{12}$

$=3.25 \ \times 10^{12}$

The national wealth of the country in 1985

$= \left (3.25 \ \times 10^{12} \ \times \dfrac{1}{25} \right ) \ + \ 3.25 \ \times 10^{12}$

$=13 \ \times 10^{10} \ + \ 3.25 \ \times 10^{12}$

$=0.13 \ \times 10^{12} \ + \ 3.25 \ \times 10^{12}$

$=3.38 \ \times 10^{12}$

A manufactures purchases a second hand machine for $Rs.60000$ and spends some amount towards repairs then its value goes up to $Rs.90000$ . If depreciation is $10\%$ per annum, what will be the value of the machine after two years ?
(a) $Rs.48600$
(b) $Rs.81000$

Value after

$1$ year

$=9000-10\%\times 9000=Rs.81,000$

Value after

$2$ year

$=81,000-10\%\times 81,000=Rs.72,900$

The value of a machine deprecates every year by 10% what will be its value after $2 years$ if its present value is $Rs.50000.$ ?

= 50000× (1-10/100)^2
= 50000×9×9/10×10
=40500

Calculate the difference between the compound interest and the simple interest on $Rs 20000$ in $2$ years at $8\%$ per annum.

In case of compound interest:-

Amount

$\left({A}_{1}\right)=20000{\left(1+\dfrac{8}{100}\right)}^{2}$

$=20000\times {\left(\dfrac{108}{100}\right)}^{2}$

$=2\times 11664$

$=23328$

$\therefore$ compound interest

$\left({I}_{1}\right)=Rs. \left(23328-20000\right)$

$=Rs.3328$

In case of simple interest

$\therefore$ Simple Interest

$\left({I}_{2}\right)=\dfrac{20000\times \left(8\right)\times 2}{100}$

$=Rs. 3200$

$therefore Difference ={I}_{1}-{I}_{2}$

$=Rs.\left(3328-3200\right)$

$=Rs. 128$ Ans.

Find the compound interest when principal is Rs. $2000$ and the rate is $10\%$ per annum compounded annually for $2$ years.

$P = 2000,\,Rate = 10\%$

$and\,T = 2years$

Then

$Compound\,Interest = p\left( {1 + \dfrac{R}{{100}}} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2000{\left( {\dfrac{{11}}{{10}}} \right)^2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2000 \times \dfrac{{121}}{{100}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2420$

Compound Interest $2420$ Rupees

Arif took a loan of $Rs.\ 80,000$ from a bank. If the rate of interest is $10\%$ per amount, find the difference in amount he would be paying after $1\dfrac{1}{2}$ years if the interest is
compounded annually.

$A=P\left[1+\dfrac{R}{100}\right]^{n}$

$P=Rs.80,000$

$R=10 mo$

$n=\dfrac{3}{2}$

$A=80000\times \left[1+\dfrac{10}{100}\right]^{\dfrac{3}{2}}$

$=80000\times \left(\dfrac{11}{10}\right)^{\dfrac{3}{2}}$

$=80000\times (1.1)^{\dfrac{3}{2}}$

$=Rs.92,295.17$

Difference in amount

$=Rs. (92295.1)$

Determine the rate of interest for a sum that becomes $\dfrac{49}{36}$ times of itself in $2$ years compounded annually.

Let the Principal be

$P$

so that Amount after 2 years will be

$A=\dfrac{49}{36}P$

and we know that,

$A=P(1+r)^t$ for C.I. compounded Annually

$\dfrac{49}{36}P=P(1+r)^2\Rightarrow (\dfrac{7}{6})^2=(1+r)^2\Rightarrow \dfrac76=1+r\Rightarrow r=\dfrac16$

A fast blowing town has now $30,000$ cattle, during the last few years, the number of cattle depreciates at the rate of $20$ %.If depreciation continues at this rate what will be the cattle population after $3$ years?

$P=30,000$

$r=-20\%$

$n=3years$

$\left[A=P\left(1+\dfrac{r}{100}\right)^n\right]$

$A_1=30,000\left(1-\dfrac{20}{100}\right)^3$

$=30,000\times\dfrac{80}{100}\times \dfrac{80}{100}\dfrac{80}{100}$

$=15,360$ is present cattle population

Surbhi borrowed a sum of Rs. $12000$ from a finance company to purchase a refrigerator. If the rate of interest is $5\%$ p.a. compounded annually. Calculate the CI that Surbhi has to pay to the company after $3$ yrs.

Principle = Rs 12000

Rate = 5% P.a.

Time = 3 years

$\begin{array}{l} \Rightarrow C.I=P{ \left( { 1+\frac { R }{ { 100 } } } \right) ^{ T } } \\ \Rightarrow C.I=12000{ \left( { 1+\frac { 5 }{ { 100 } } } \right) ^{ 3 } } \\ \Rightarrow C.I=12000{ \left( { \frac { { 21 } }{ { 20 } } } \right) ^{ 3 } } \\ =12000\times \frac { { 21 } }{ { 20 } } \times \frac { { 21 } }{ { 20 } } \times \frac { { 21 } }{ { 20 } } \\ C.I=\frac { 3 }{ 2 } \times { \left( { 21 } \right) ^{ 3 } }=Rs.13891.5 \end{array}$

Rahul invested $\text{Rs. }7,000$ in shares for $2$ years at the rate of $12\%$ per annum compounded annually. What is the compound interest that Rahul gets after $2$ years?

Compound interest

$=$ Amount

$-$ Principal

So, by using the above formula,

$\begin{aligned}{}A& = P{\left( {1 + \frac{r}{{100}}} \right)^2} - P\\ &= 7000{\left( {1 + \frac{{12}}{{100}}} \right)^2} - 7000\\& = 7000\left( {{{1.12}^2} - {1^2}} \right)\\& = 7000\left( {1.12 - 1} \right)\left( {1.12 + 1} \right)\\ &= 7000\left( {0.12} \right)\left( {2.12} \right)\\& = \text{Rs. }1780.8\end{aligned}$

The population of a town $2$ years ago was $2500$ . Due to migration to cities, it decreases every year at the rate $4\%$ per annum. Find its present population.

Present population

$=2500\left(1-\dfrac{4}{100}\right)^{2}$

$=2500\dfrac{96}{100}\times \dfrac{96}{100}$

$=2304$ Ans

In how much time would Rs 5000 amount to Rs 6655 at $10$ % p.a compound interest?

$5000(1+10/100)^n=6655$

$(11/10)^n=6655/5000$

$(11/10)^n=1331/1000$

$(11/10)^n=(11/10)^3$

$n=3$

In how many years will $Rs. 64,000$ earn an interset of $Rs. 4921$ at $5\%$ per annum, with the interset being compound half-day?

As we know that,

$S.I. = \cfrac{P \times r \times t}{100}$

Given:-

$S.I. = Rs. 4921$

$P = Rs. 64000$

$r = 5 \%$

$t = ?$

Therefore,

$4921 = \cfrac{64000 \times 5 \times t}{100}$

$\Rightarrow t = \cfrac{4921}{640 \times 5} \approx 1.5 \text{ years}$

At what rate precent will Rs 2000 amount to Rs 2315.25 in 3 years at compound interest ?

Amount

$=2315.25$

Time

$=3$ Year

Amount

$=P\left(1+\dfrac{r}{100}\right)^{TM}$

$2315.25=2000\left(1+\dfrac{r}{100}\right)^{3}$

$\dfrac{2315.25}{2000}=\left(1+\dfrac{r}{100}\right)^{3}$

$\dfrac{9261}{3000}=\left(1+\dfrac{r}{100}\right)^{3}$

$\dfrac{21}{20}=1+\dfrac{r}{100}$

$100+r=\dfrac{2100}{20}$

$100+r=105$

$r=5\%$

Meena and Eisha both invest their $$6000$ .
Meena invests her $$6000$ at a rate of $1.5%$ per year compound interest.
Eisha invests her $$6000$ in a bank that pays simple interest.
After $8$ years, their investments are worth the same amount.
Calculate the rate of simple interest per year that Eisha received.

Principal amount =

$$ 6000, r_{CI} = 1.5\% ,\, t = 8$

Amount accounted by

$CI = p(1+\dfrac{r}{100})^t$

Amount accounted by SI =

$p+ \dfrac{p\times r \times t}{100}$

$\Rightarrow CI = 6000(1+\dfrac{1.5}{100})^8$

$= 6000(1.015)^8$

$= $ 6758$

since Amount for SI & CI are same :

$\therefore 6758 = 6000 [ 1+\dfrac{r\times 8}{100}]$

$\dfrac{6758}{6000}-1 = \dfrac{r\times 8}{100}$

$1.1264 -1 \Rightarrow 0.1264 = r\times\dfrac{ 8}{100}$

$\therefore r = \dfrac{0.1264 \times 100}{8}$

$r_{SI} = 1.58 \%$

1121133_1140034_ans_e74df048fc7b4e9bbe6f2ac09de486ce.jpeg

Find the amount and the compound interest on Rs 10000 at 5% per annum, for 2 years compounded annually.

$A=P{\left(1+\dfrac{R}{100}\right)}^{n}$

$A=10,000{\left(1+\dfrac{5}{100}\right)}^{2}$

$A=10,000{\left(\dfrac{105}{100}\right)}^{2}$

$A=10,000\times\dfrac{105}{100}\times\dfrac{105}{100}$

$A=105\times 105=Rs.11,025$

$I=A-P=11,025-10,000=Rs.1,025$

A sum of $Rs 8000$ is inverted for $2$ years at $10\%$ per annum compound interest calculate:
interest for the first year.

1188956_1154347_ans_9f543cf63b0840478d29d9664bb9d768.jpg

A sum of Rs. $16000$ earns a compound interest of Rs. $2522$ in $18$ months, when the interest of compounded half-yearly.Find the rate of interest.

Amount at the end of 18 months

$16000+2522=18522$

Now, we know that for CI

$A=P(1+\dfrac{r}{n})^{nt}$

where

$P=16000\\t=18\ months =\dfrac32 \ years$

$n=2$ (as it is compounded half-yearly)

$r=$ rate of interest

$A=$ Amount

So,

$18522=16000(1+\dfrac{r}{2})^{\frac32*2}$

$\Rightarrow \dfrac{18522}{16000}=(1+\dfrac r2)^3$

$\Rightarrow \sqrt[3]{\dfrac{18522}{16000}}=(1+\dfrac r2)$

$\Rightarrow 10.5=1+\dfrac r2$

$\Rightarrow r=19$

So, Rate of interest is

$19\%$

A sum of money invested at compound interest of $8\%$ compounded annually amounted to $\text{Rs. }7290$ in $2$ years. Find the sum invested.

Let

$P$ be the principal.

As we know that, compound interest is given as-

$A = P {\left( 1 + \cfrac{r}{100} \right)}^{n}$

Here,

Amount,

$A = \text{Rs. }7290$

$P =$ Principal

Rate of interest,

$r = 8 \%$

Time period,

$n= 2 \text{ years}$

Therefore,

$\begin{aligned}{}7290 &= P{\left( {1 + \frac{8}{{100}}} \right)^2}\\7290 &= P{\left( {\frac{{27}}{{25}}} \right)^2}\\P& = 7290 \times {\left( {\frac{{25}}{{27}}} \right)^2}\\& =\text{Rs. } 6250\end{aligned}$

Thus the sum invested is

$\text{Rs. }6250$ .

By using formula, find compound Interest and Amount
Sr. No Principal (Rs.) Rate of Interest Time Period (Year)
1 8,000 10% 2
2 6,400 12% 2
3 15,00 10% 2
4 20,000 8% 3
5 50,000 5% 3

1147106_1144029_ans_83ad788b06c54964b61ba80e0b02522a.jpg

A man saved $\text{Rs} \ 66000$ in $20$ years. In each succeeding year after the first year he saved $\text{Rs}\ 200$ more than what he saved in the previous year. How much did he save in the first year?

Let us assume that the man saved Rs

$a$ in the first year.

In each succeeding year, an increment of Rs.

$200$ is made.

So it forms an A.P. whose first term is

$a$

Common difference

$=200$

$n=20$

$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$

We are given,

$S_{20} = 66000$

$S_{20}=\dfrac{20}{2}[2a+19d]=66000$

$2a+19\times 200=6600$

$\Rightarrow a=1400$

The compound interest, calculated yearly. on a certain sum of money for the second year is Rs. $1089$ and for the third year, it is Rs. $1197.90$ . Then, calculate the rate of interest and the sum of money.

2063127_1147477_ans_b8e7d2c7bbb1474aa31e1af451bcfa38.jpg

A sum of money amounts to Rs. 66,836.70 in 2 years at 3 $\%$ per annum compound interest. The sum is

We know that,

In compound interest, final amount

$A=P\left(1+\dfrac R{100}\right)^n$

Where,

$P$ is the principal amount,

$R$ is the rate of interest and

$n$ is the time period.

Here,

$A=66,836.70,\ R=3\%,\ n=2$

Let the principal amount be

$Rs.\ P$ .

Hence,

$66,836.70=P\left(1+\dfrac3{100}\right)^2$

$\Rightarrow 66,836.70=P\left(\dfrac{103}{100}\right)^2$

$\Rightarrow P=\dfrac{66,836.70\times 100\times 100}{103\times 103}$

$=63000$

Hence, the sum of money that amounts to

$Rs.\ 66,836.70$ in

$2$ years at

$3\%$ per annum compound interest is

$Rs.\ 63,000$ .

$Rs 15625$ at $4\%$ per annum for $3$ years compounded annually.

$P=15625\ Rs$

$R=4\%$

$t=3\ years$

$CP=P\left(1+\dfrac{R}{100}\right)^t=15625\left(1+\dfrac{4}{100}\right)^3$

$=15625\times \left(\dfrac{104}{100}\right)^3=17576\ Rs$

Find the amount and the compound interest on $Rs, 12, 800$ for $1$ years at $7\dfrac{1}{2}\%$ per annum, compounded semi-annually.

$P=12,800\ Rs$

Period

$=1$ year

$\therefore$ half yearly periods

$=N=2$

Rate yearly periods

$=\dfrac {15}{2}\% p.a$

$\therefore \$ half yearly Interest

$=\dfrac {1}{2}\times \dfrac {15}{2}=\dfrac {15}{4}\%$

$A=P\left (1+\dfrac {R}{100}\right)^N$

$A=12800 \left (1+\dfrac {15}{4\times 100}\right)^2$

$=12800\times \left (\dfrac {400+15}{400}\right)^2$

$=12800 \times \dfrac {415}{400}\times \dfrac {415}{400}$

$A=13778\ Rs$

$\rightarrow \$ Compound Interest

$=AP$

$=13778-12800$

$=978\ Rs$

Sanjay borrowed $Rs\,12,800$ at the rate of $\dfrac{25}{4}\%$ per annum at simple interest. On the same day, he lent it to Raiesh at the same rate but compounded annually. What does Sanjay gain at the end of $2$ years?

Given:

$P = 12,800$

$R = \dfrac{25}{4}\%$

$T = 2 \,years$

$SI = \dfrac{PRT}{100} = 12800\times \dfrac{25}{4}\times 2\times\dfrac{1}{100}$

$= 1600$

Amount

$= 12800+1600$

$= 14400$

CI for 2 years

$= P(1+\dfrac{R}{100})^{T}$

$= 12800(1+\dfrac{25}{400})^{2}$

$= 12800\times \dfrac{289}{256}$

$= 14450$

Profit

$= 14450-14400$

$= 50 Rs$

Using the formula , calculate the amount and the compound interest :-
$P =Rs. 500,R = 12\%$ per annum, $n = 2$ years

Given

$P=Rs. 500, R=12\%$ and

$n=2$ .

Now total amount when the principle compounded after

$2$ years will be

$A=P(1+\dfrac{r}{100})^n$

$=500\times \left(1+\dfrac{12}{100}\right)^2$

$=500\times \left(\dfrac{112}{100}\right)^2$

$=500\times \left(\dfrac{112}{100}\right)\times \left(\dfrac{112}{100}\right)$

$=627.20$ .

So the amount will be

$Rs. 627.20$ and compound interest will be

$Rs.(627.20-500)=Rs.127.20$ .

Rs. $12000$ is invested for $1 \dfrac{1}{2}$ years at compound interest annually. If Rs. $15972$ is received of at the end, then $R =$ ?

Principal,

$P=Rs. 12000$

Time,

$t=1.5$ years

Rate of interest,

$r=?$

Amount received after

$1.5$ years

$A=Rs. 15972$

For compound interest,

$A=P\left ( 1+\cfrac{r}{100} \right )^{t}$

$15972=12000\left ( 1+\cfrac{r}{100} \right )^{\cfrac{3}{2}}$

$\cfrac{15972}{12000}=\left ( 1+\cfrac{r}{100} \right )^{\cfrac{3}{2}}$

$\cfrac{1331}{1000}=\left ( 1+\cfrac{r}{100} \right )^{\cfrac{3}{2}}$

$1+\cfrac{r}{100}=\left ( \cfrac{1331}{1000} \right )^{\cfrac{2}{3}}$

$1+\cfrac{r}{100}=\left ( \left ( \cfrac{11}{10} \right )^{3} \right )^{\cfrac{2}{3}}$

$1+\cfrac{r}{100}=\left ( \cfrac{11}{10} \right )^{2}$

$1+\cfrac{r}{100}=\cfrac{121}{100}$

$\cfrac{r}{100}=\cfrac{121}{100}-1=\cfrac{21}{100}$

$\Rightarrow r=21$ % per annum

Suhail deposited a sum of Rs. $8000$ at compound interest in the bank at the rate of $18$ % per annum for $3$ years. What interest will he get after the end of $3$ years?

We know that amount

$A$ at the end of

$n$ years at the rate of

$R%$ per annum is given by

$A=P\left(1+{R}/{100}\right)n$

Given

$P=8000$

$R=18%$

$n=3years$

Now,

$A= P\left(1+{R}/{100}\right)n$

$\Rightarrow \ A=8000(1+{18}/{100})3$

$\Rightarrow \ A=8000 \left (\dfrac{100+18}{100}\right)3$

$\Rightarrow \ A=8000 \left (\dfrac{118}{100}\right)3$

$\Rightarrow \ A=8000.\dfrac{254}{100}$

$\Rightarrow \ A=Rs.13,14425$

And

$CI=A-P=13,14425-8000$

$CI=5,14425$

$\therefore$ so, the answer is

$“5,144.25”$ .

Mohit borrowed a certain sum at $5\%$ per annum compound interest and cleared this loan by paying Rs. $12,600$ at the end of the first year and Rs. $17,640$ at the end of the second year. Find the sum borrowed.

Amount with interest paid at the end of first year = Rs. 12600.

$A = P (1+\dfrac{r}{100})^n$

Principal at the starting of the first year

$= 12600 = P (1+\dfrac{5}{100})^1$

$12600 = 105\dfrac{P}{100}$

$P = \dfrac{12600 \times 100}{105}$

$P = Rs 12000.$

Amount with interest paid at the end of second year

$= Rs 17640.$

Principal at the starting of second year

$= 17640 = P (1+\dfrac{5}{100})^1$

$17640 = \dfrac{105P}{100}$

$P = \dfrac{17640 \times 100}{105}$

$P = Rs 16800$

Total sum borrowed

$= 12000 + 16800 = Rs 28800$ .

A certain sum amount $12167$ in $3$ years at $15\%$ p.a compounded annually. Find sum .

Amount (A)

$=12167$
Time (T)

$=3$ yr
Rate (R)

$=15$ % compounded annually

$A=P{ \left( 1+\cfrac { R }{ 100 } \right) }^{ T }$

$12167=P{ \left( 1+\cfrac { 15 }{ 100 } \right) }^{ 3 }$

$12167=P{ \left( \cfrac { 23 }{ 20 } \right) }^{ 3 }$

$P=\cfrac { 12167\times 20\times 20\times 20 }{ 23\times 23\times 23 }$
Sum (P)=Rs.

$8000$

Ranbir borrows Rs. $20,000$ at $12$ % per annum compound interest. If he repays Rs. $8400$ at the end of the first year and Rs. $9680$ at the end of the second year, find the amount of loan outstanding at the beginning of the third year.

Given that Ranbir borrows Rs.20000 at 12% compound interest.

For First year

Interest

$I=\dfrac{20000\times1\times12}{100}=Rs.2400$

Thus amount after one year

$=Rs.20000+Rs.2400=Rs.22400$

$\therefore$ Money repaid

$=Rs.8400$

$\therefore$ Balance

$=Rs.22400-Rs.8400=Rs.14000$

For Second year

Interest

$I=\dfrac{14000\times1\times12}{100}=Rs.1680$

Thus amount after one year

$=Rs.14000+Rs.1680=Rs.15680$

Ranbir paid Rs.9680 at the end of second year.

$\therefore$ Loan outstanding at the beginning of the third year

$=Rs.15680-Rs.9680=Rs.6000$

Find the rate per cent per annum if :
$Rs\ 425$ yields $Rs\ 238$ interest in $8$ years.

Given that,

$P = Rs. 425$

$R = R$ % p.a

$S.I = Rs. 238$

$T = 8 \ years$

$\because S.I = \dfrac{PTR}{100}$

$\implies 238 = \dfrac{425\times 8 \times R}{100}$

$\implies 238 = 34 R$

$\implies R = \dfrac{238}{34}$

$\implies R = 7$

$\therefore Rate = 7$ % p.a.

If Rs. $250$ amounts to Rs. $285$ in $2$ years, find the rate per cent per annum.

Given

$P=Rs.250$

$A=Rs.285$

$T=2\ yr$

We know

Simple Interest

$=A-P$

$=285-250$

$=25$

$SI=\dfrac{PRT}{100}$

$25=\dfrac{250\times R\times 2}{100}$

$R=\dfrac{25\times100}{250\times2}$

$R=5\%$

A person invests Rs 5,000 for two years at a certain rate of interest compounded annually. At the end of one year , this sum amounts Rs 5,Calcuate.
(i)the rate of interest per annum.
(ii)the amount at the end of the second year.

Solution:-

Amount

$\displaystyle = P(1+\frac{r}{100})^{t}$

$\displaystyle \therefore 5600 = 5000(1+\frac{r}{100})^{1}$

$\displaystyle \frac{56}{50} = 1+\frac{r}{100}$

$\displaystyle \Rightarrow \frac{6}{50} = \frac{r}{100}$

$\displaystyle \Rightarrow r = 12$

Amount after two years =

$\displaystyle 5000(1+\frac{12}{100})^{2}$

$\displaystyle = 5000(1+\frac{3}{25})^{2}$

$\displaystyle = \frac{5000\times 28\times 28}{25\times 25}= 6272$

1105786_1183906_ans_b1a74c9952c446449a668e49dcb89ea2.jpg

Find compounded interest paid when a sum of $Rs. 10,000$ is invested for $1$ year and $3$ months at $17/2\%$ per annum compounded annually.

$P = Rs. 10,000$

$R= \dfrac{17}{2}$ % p.a

$T= 1 \ year \ 3 \ months$

Amount at the end of

$1$ year

$= P\left(1 + \dfrac{R}{100}\right)^{n}$

$= 10000\left(1 + \dfrac{17}{200}\right)^{1}$

$= 10000\times \dfrac{217}{200}$

$= Rs.10850$

$S.I$ for

$3$ months

$= \dfrac{10850\times \dfrac{3}{12} \times \dfrac{17}{2}}{100}$

$= \dfrac{10850}{100}\times \dfrac{1}{4} \times \dfrac{17}{2}$

$= Rs.230.56$

$\therefore Amount = 10850 + 230.56 = Rs. 11080.56$

$C.I = A - P = 11080.56 - 10000 = Rs. 1080.56$

The value of machine depreciates 10% every year. If its present value is Rs.38700, what was its value 1 year ago?.

Let us suppose the value of the machine 1 year ago be 'x'

10% of x

$=x\times\dfrac{10}{100}=\dfrac{x}{10}$

As per question

$x-\dfrac{x}{10}=38700$

$\dfrac{9x}{10}=38700$

$9x=387000$

$x=\dfrac{387000}{9}$

$x=43000$

Hence the value of machine 1 year ago will be Rs.43000

A person deposited $Rs\ 400$ for $2$ years , $Rs\ 550$ for $4$ years and $Rs.1200$ for $6$ years at the same rate of simple interest . If he received a total interest of $Rs.1020$ , what is the rate of interest per annum ?

Let the rate of interest be R per cent per annum.

Given, total simple interest is

$1020$

$\therefore \frac{400\times 2\times R}{100}+\frac{500\times 4\times R}{100} + \frac{1200\times 6\times R}{100} = 1020$

$=> 8R+22R+72R = 1020$

$=> 102R = 1020$

$=> R = \frac{1020}{102} = 10\%$

Simple interest on a certain sum is $\dfrac{4}{9}$ of the sum . Find the rate percent and time if both are numerically equal .

Let, principal

$=P$

Rate of interest

$=R$

Time

$=$ Rate of interest

$=R$

Interest

$=\dfrac{4}{9}R$

Simple interest on

$P=\dfrac{Principal\times Rate\times Time}{100}=\dfrac{P\times R\times R}{100}$

$\therefore\dfrac{P\times R^2}{100}=\dfrac{4}{9}P$

$\Rightarrow \dfrac{R^2}{100}=\dfrac{4}{9}$

$\Rightarrow R^2=\dfrac{4\times 100}{9}$

$\Rightarrow R=\dfrac{2\times 10}{3}$

$\Rightarrow R=\dfrac{20}{3}$

$\therefore$ Rate percent

$=6\dfrac{2}{3}\%$

and time

$=6\dfrac{2}{3}$ years.

The simple interest on a sum of money for $3$ years at $\dfrac{20}{3}\%$ per annum is $Rs.8750$ . What will be the compound interest on the same sum at the same rate for the same period, compounded annually?

$SI = \frac{P\times R\times T}{100}$

$8750 = \frac{P\times \frac{20}{3}\times 3}{100}$

$8750 = \frac{20P}{100}$

$P = 43750$

Compound Interest

$A = P(1+\frac{r}{100})^{t}$

$= 43750 (1+\frac{\frac{20}{3}}{100})^{3}$

$= 43750(1+\frac{20}{3}\times \frac{1}{100})^{3}$

$= 43750(1+\frac{1}{15})^{3}$

$= 43750 (\frac{16}{15})^{3}$

$= 43750\times \frac{16\times 16\times 16}{15\times 15\times 15}$

$= \frac{16\times 16\times 16\times 350}{3\times 3\times 3}$

$= \frac{4096\times 350}{27}$

$= \frac{1433600}{27}$

$= 53096.296$

$= 53096$

Compound Interest = A - P

$= 53096 - 43750$

$= 9346$ Answer.

1146061_1191048_ans_7f843fa008a14dc08b85d4134375a021.jpg

On what sum will the compound interest at $5$ % per annum for $2$ years compounded annually be Rs. $164$ ?

Let the sum be Rs.x

We know that

$CI=A-P$

$CI=P\left(1+\dfrac{R}{100}\right)^n-P$

$CI=P\left[\left(1+\dfrac{R}{100}\right)^n-1\right]$

$164=x\left[\left(1+\dfrac{5}{100}\right)^2-1\right]$

$164=x[(1.05)^2-1]$

$x=\dfrac{164}{0.1025}$

$x=1600$

Thus, required sum is Rs.1600.

In a factory,the production of motor bikes was $40,000$ in a particular year,which rose to $48,400$ in two years . Find the rate of growth per annum ,if it was uniform during two years ?

1146086_1193523_ans_16282c023fc34ce8a0619d393352d895.jpg

The difference between compound interest and simple interest on an amount of $Rs.15,000$ for $2$ years is $Rs.96$ . What is the rate of intrest per annum?

$S.I=\dfrac{P\times r\times T}{100}=\dfrac{15000\times r\times 2}{100}=300r$

$C.I=15000\left[{\left({1+\dfrac{r}{100}}\right)}^2-1\right]=15000\left[{1+\dfrac{r^2}{10000}+\dfrac{2r}{100}-1}\right]=1.5r^2+300r$

$C.I-S.I=96$

$\Rightarrow1.5r^2+300r-300r=96$

$\Rightarrow1.5r^2=96$

$\Rightarrow r^2=\dfrac{96}{1.5}=64$

$\therefore r=8\Rightarrow rate=8\%$

If the interest on $1700$ rupees is $340$ rupees for $2$ years the rate of interest must be

$P = Rs 1700$

$SI = 340$

$T = 2 yrs$

$R= ?$

$SI = \dfrac{P\times T\times R}{100}$

$\dfrac{340\times 100}{1700\times 2}= R$

$R = 10\%$

Rucha borrowed Rs. 2.00.000 at 10% rate compound interest for 3 years from a finite. What total sum should she pay to clear the at the end of the period?

$A=P\left(1+\dfrac{x}{100}\right)^t$

A is Amount

P is Principal

r is rate per year(annual rate)

t is time in years

$\therefore A=2,00,000\left(1+\dfrac{10}{100}\right)^3$

$=2,00,000(1.1)^3$

$=2,66,200$ .

1188146_1194990_ans_2ecfc9f83d66411f9ef3be5409ce8763.jpg

Calculate the amount and the compound interest when $Rs16000$ is invested for 3 years at the rate of $5\%$ p.a compounded annually .

$P=16,000$

$r=5\% pa$

$n=3$ year

Compound sum

$=P\left [1+\dfrac {r}{100}\right]^n=16000\left [1+\dfrac {5}{100}\right]^3$

$=16000[1.05]^3=18522$

compound interest

$=[18522-P]=[18522-16000]$

compound interest

$=2522$

The difference between the compound interest and the simple interest on a certain sum of money at $15$ % per annum for three years is Rs $283.50$ . Find the sum.

Given,

$CI – SI = Rs 283.50$

$R = 15\%$

$n = 3 years$

Consider the sum be $RS$ $X$

we know that

$\begin{array}{l} A=P{ \left( { 1+\frac { R }{ { 100 } } } \right) ^{ n } } \\ =x{ \left( { 1+\frac { 5 }{ { 100 } } } \right) ^{ 3 } } \\ =x{ \left( { 1.15 } \right) ^{ 3 } }\, \, ...\left( 1 \right) \end{array}$

Also

$\begin{array}{l} SI=\frac { { PRT } }{ { 100 } } =\frac { { x\left( { 15 } \right) \left( 3 \right) } }{ { 100 } } 0.45X \\ A=SI+P=1.45X\, \, ...\left( 2 \right) \end{array}$

Thus, we have

$\begin{array}{l} x{ \left( { 1.15 } \right) ^{ 3 } }-1.45X=283.50\, \, \, \, \left[ { From\left( 1 \right) and\left( 2 \right) } \right] \\ 1.523X-1.45X=283.50 \\ 0.070875X=283.50 \\ x=\frac { { 283.50 } }{ { 0.070875 } } \\ =4000 \end{array}$

Hence, the sum is $RS4000$

The simple interest on a sum of money is $\frac { 4 }{ 9 }$ of the principal and the no of years is equal to the rate of interest p.a. Find the rate of interest p.a.

Consider the problem

Let

$P$ be the principle and

$r$ be the rate of interest p.a

Then,

$\begin{array}{l} S.I=\frac { 4 }{ 9 } P \\ t=r \end{array}$

Therefore,

$S.I = \frac{{p \times r \times t}}{{100}}$

$\Rightarrow \frac{4}{9}p = \frac{{p \times r \times r}}{{100}}$

$\Rightarrow {r^2} = \frac{{400}}{9}$

$\Rightarrow r = \frac{{20}}{3}$

$\Rightarrow r = 6.67\%$

What sum of money will amount to Rs. $5445$ in $2$ years at $10$ % per annum compound interest?

Let the sum of money be Rs.

$P$

Here,

$r=10\%$ per annum and

$n=2$ years

$A=Rs.5445$

Then

$A=P\left(1+\dfrac{r}{100}\right)^n$

$5445=P\left(1+\dfrac{10}{100}\right)^2$

$5445=P(1.1)^2$

$P=\dfrac{5445}{1.21}$

$P=$ Rs.

$4500$

Hence the sum of money is Rs.

$4500$

Neha and Priya borrowed $Rs\ 5400$ and $Rs\ 6000$ respectively at the same rate of interest for $1\dfrac {3}{5}\ years$ . If Priya paid $Rs\ 240$ more then, Neha, then find the rate of interest.

Let rate of interest be

$r\%$

As simple interest

$= \dfrac{Prt}{100}$

Where,

$p=$ principal,

$r=$ rate of interest

$t=$ period in years

According to the given condition,

$\Rightarrow \dfrac{(5400)r(\dfrac{8}{5})}{100}+240 = \dfrac{(6000)r(\dfrac{8}{5})}{100}$

(Neha's interest

$+240 =$ Priya's int)

$\Rightarrow 1080\times 8r+24000 = 1200\times 8r$

$\Rightarrow 8r(120) = 24000$

$\Rightarrow 8r = 200 \\\Rightarrow r = 25\%$

Find rate of interest if principal in $Rs.\ 1500$ , interest in $Rs.\ 600$ , time in $4\ years$ preparing bill.

Given

$P=Rs.1500 \quad SI=Rs.600 \quad T=4$ yrs

$R=$ ?

We have

$SI=\cfrac { P\times R\times T }{ 100 }$

$\Rightarrow 600=\cfrac { 1500\times R\times 4 }{ 100 }$

$\Rightarrow R=\cfrac { 600\times 100 }{ 1500\times 4 } =10$ %

Find the amount and the compound interest on Rs. $12,000$ in $3$ years at $5$ % interest being compounded annually.

The amount at the end of 'n' years of investing in compound interest.

$A=P\left(1+\dfrac{r}{100}\right)^n$

Here,

$P=12000$

$n=3$

$r=5\%$

$A=12000\left(1+\dfrac{5}{100}\right)^3$

$=Rs.13891.5$

Compound interest

$CI=Amount-Principal$

$=13891.5-12000$

$=Rs.1891.5$

Alok paid Rs 6720 to clear a debt of Rs 6000 to the bank after 1 year 6 months. What is the rate of interest charged by the bank?

Given, Internal

$= 6720- 6000 = 720$

Time = 1 year 6 month

$=1\dfrac {1}{2} year$

Principal

$= Rs. 6000$

$\therefore Rate =\dfrac {Interest \, \times 100}{time \times principal}$

$Rate = \dfrac {720 \times 100}{\dfrac {3}{2}\times 6000} = 8\%$

Three years ago, the population of a town was $50000$ . If the annual increase during three successive years be at the rate of $5$ %, $4$ % and $3$ % respectively. What is the present population?

Population of town

$3$ years ago

$= 50000$

Annual percentage increase in

$3$ years

$= 5\%, 4\%, 3\%$ .resp.

$\therefore$ Present population

$= 50,000\left(1+\dfrac{A_{1}}{100}\right)\left(1+\dfrac{A_{2}}{100}\right)\left(1+\dfrac{A_{3}}{100}\right)$

$\displaystyle = 5000\left(1+\frac{5}{100}\right)\left(1+\frac{4}{100}\right)\left(1+\frac{3}{100}\right)$

$\displaystyle = 5000\times \frac{105}{100}\times \frac{104}{100}\times \frac{103}{100}$

Present population

$= 56238Rs$

The price of an article increased successively over three consecutive weeks. It increased by $10 \%$ in the first week, $20 \%$ in the second week and $25 \%$ in the third week. Find the effective percentage increase in its price after three weeks.

Prince after first week

$=x+10$ % of

$x=1.1x$

Original Price

$=x$

Price after

$2$ weeks

$=1.1x+20$ % of

$1.1x=1.32x$

Price after

$3$ weeks

$=1.32x+30$ % of

$1.32x=1.716x$

So, Total Percentage increase

$=\left( \dfrac { 1.716x-x }{ x } \right) \times 100=71.6$ %

In the year $2000$ , rice formed $20 \%$ of the total food gain production in a country. In the next year, the total food grain production increased by $20%$ and the rice production was $25 \%$ of the total food grain production. What is the increase in the production of rice from $2000$ to $2001$ ?

$2000$ , Rice Production

$=\dfrac { 1 }{ 5 } K=20$ % of

$K=0.2K$

Where

$K=$ Total Production in

$2000$

Production Total in

$2001=K+20$ % of

$K=1.2K$

Production of Rice in

$2001=1.2K\times \dfrac { 1 }{ 4 } =0.3K$

Percentage increment in Rice Production

$\Rightarrow$

$=\left( \dfrac { 0.3K-0.2K }{ K } \right) \times 100=10$ %

Ranchodbhai took a loan of Rs.50,000 at 9% rate of interest for 2 years at simple interest to renovate his bungalow. If in the place of simple interest the compound interest is paid then how much more interest will he have to pay?.

$P=50000,r=9\%,t=2$

Let

$A_s\to$ Amount for SI

$A_c\to$ Amount for CI

$A_c-A_s=P(1+\cfrac{r}{100})^t-P(1+\cfrac{rt}{100})$

$=50000(1+\cfrac{9}{100})^2-50000(1+\cfrac{9\times 2}{100})$

$=59405-59000$

$=Rs.405$

Saleem invests $Euro 20000$ at $4\%$ per annum compound interest for $3$ years. Work out the interest earned at the end of $3$ years.

Given

$P=20000$ Euro

$R=4$ % Pa

$n=3$

$A=P{ \left[ 1+\cfrac { R }{ 100 } \right] }^{ n }$

$=20000{ \left[ 1+\cfrac { 4 }{ 100 } \right] }^{ 3 }$

$=20000{ \left[ \cfrac { 104 }{ 100 } \right] }^{ 3 }$

$=22497.28$

$\Rightarrow I=22497.28-20000$

$=2497.28$ Euro

Mr.Subbarao started work in $2014$ at an annual salary of $Rs.\ 10,000$ and receive an increment of $Rs.\ 750$ each year. In which year did his income reach $Rs.\ 17,500$

This is an AP problem

His salary increases in AP with a C.D

$\therefore 750$ and initial term

$10000/-$

So let the nth term is

$17500$

$17500=10000+(n-1)750$

$7500=750(n-1)$

$10=n-1$

$\therefore n=11$

Rajesh borrowed RS. $5,000$ from a bank. He will have to pay Rs. $10,000$ after $5$ years.
Find the rate of interest, if the interest is payable half yearly.

Given

$P=5000$

$T=5\times 2=10$

$I=10000-5000=5000$

$\Rightarrow 5000=\cfrac { 5000\times R\times 10 }{ 100 }$

$\Rightarrow R=10$ % per

$6$ months

Anand obtained a loan of $Rs. 125000$ from the Allahabad Bank for buying commercial bank charges compound interest at $8\%$ per annum, compounded annually. How much should he pay after $3$ yrs

Principal

$=Rs.125000$

Rate of interest

$=8\%\text{ }p.a$

Time

$=3\text{ }years$

The amount,

$A=P(1+\cfrac{R}{100})^n$

$=125000{(1+\cfrac{8}{100})}^3$

$=125000\times {(1.08)}^3$

$=157464$

$\therefore$ Anand have to pay

$Rs.$

$157464$ after

$3\text{ }years$ to clean the debt.

The population of a town increases by 10.5 $\%$ every year. If its present population is 7,95,600, what was it a year ago.

Given that, the population of a town increases by

$10.5\%$ every year.

Also, the present population is

$7,95,600$ .

To find out: Population of the town a year ago.

Let the population of the town a year ago be

$x$ .

$\therefore \ \text{Population }1\text{ }years\text{ ago}=\cfrac{100}{100+increase \%}\times \text{present population}$

$\Rightarrow x=\cfrac{100}{110.5}\times 795600=720000$

Hence, the population of the town a year ago was

$7,20,000$ .

A trust fund has Rs. 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of
A) Rs. 1800 . B) Rs. 2000

It is given that

$Rs.30,000$ must be invested into two types of bonds with

$5\%$ and

$7\%$ interest rates.

Let

$Rs.x$ be invested in bonds of the first type.

Thus,

$Rs.(30000-x)$ will be invested in the other type.

Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as :

$X=\begin{bmatrix}x&30000-x\end{bmatrix}$

$A)$ Annual interest obtained is

$Rs.1800.$

We know,

$Interest =\dfrac{PTR}{100}$

Here, the time is one year and thus

$T=1$

Hence, the interest obtained after one year can be expressed in matrix representation as -

$\begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}\dfrac{5}{100}\\\dfrac{7}{100}\end{bmatrix}=[1800]$

$\Rightarrow$

$\left[x\times \dfrac{5}{100}+(30000-x)\times \dfrac{7}{100}\right]=[1800]$

$\Rightarrow$

$\dfrac{5x}{100}+\dfrac{7(30000-x)}{100}=1800$

$\Rightarrow$

$5x+210000-7x=180000$

$\Rightarrow$

$-2x=-30000$

$\therefore$

$x=15000$

Amount invested in the first bond

$=x=Rs.15000$

$\Rightarrow$ Amount invested second bond

$=Rx(30000-x)=Rs.(30000-15000)=Rs.15000$

$\therefore$ The trust has to invest

$Rs.15000$ each in both the bonds in order to obtain an annual interest of

$Rs.1800$

$B)$ Annual interest obtained is

$Rs.2000.$

Hence, the interest obtained after one year can be expressed in matrix representation as -

$\begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}\dfrac{5}{100}\\\dfrac{7}{100}\end{bmatrix}=[2000]$

$\Rightarrow$

$\left[x\times \dfrac{5}{100}+(30000-x)\times \dfrac{7}{100}\right]=[2000]$

$\Rightarrow$

$\dfrac{5x}{100}+\dfrac{7(30000-x)}{100}=2000$

$\Rightarrow$

$5x+210000-7x=200000$

$\Rightarrow$

$-2x=-10000$

$\therefore$

$x=5000$

Amount invested in the first bond

$=x=Rs.5000$

$\Rightarrow$ Amount invested second bond

$=Rx(30000-x)=Rs.(30000-5000)=Rs.25000$

$\therefore$ The trust has to invest

$Rs.5000$ in the first bond and

$Rs.25000$ in the second bond in order to obtain an annual interest of

$Rs.2000$

A sum becomes $Rs.$ $4800$ after two years and $Rs.$ $6400$ after four years at compound interest per annum compounded annually. Find the sum.

Let the sum be

$x$ and the rate be

$r$

Given, Amount in 2 years

$=x\left(1+\dfrac{r}{100}\right)^2=4800\quad...(I)$

and the amount in 4 years

$=x\left(1+\dfrac{r}{100}\right)^4=6400\quad...(II)$

Let us divide equation

$(I)$ by equation

$(II)$

$\Rightarrow \dfrac{x\left(1+ \dfrac{r}{100}\right)^2}{x\left( 1+\dfrac{r}{100} \right)^4}=\dfrac{4800}{6400}$

$\Rightarrow \left(1+\dfrac{r}{100}\right)^2=\dfrac{4}{3}$

$\Rightarrow 1+\dfrac{r}{100}=\dfrac{2}{\sqrt{3}}$

$\Rightarrow \dfrac{r}{100}=\dfrac{2}{\sqrt{3}}-1$

from

$(I),$

$\Rightarrow x\left(1+\dfrac{2}{\sqrt{3}}-1\right)^2=4800$

$\Rightarrow x\times \dfrac{4}{3}=4800$

$\therefore\ x=Rs.3600$

Arvind borrows Rs.20,000 at 12% p.a sample interest and immediately lends it to another person at 12 % p.a compound interest. Find his gain at the end at 2 years.

Let

$A_c\to$ Amount after compound interest

$A_s\to$ Amount after simple interest

Now,

$P=20000,R=12\%,t=2$

$-A_s+A_c=-P(1+\cfrac{Rt}{100})+P(1+\cfrac{R}{100})^t$

$=-20000(1+\cfrac{12\times 2}{100})+20000(1+\cfrac{12}{100})^2$

$=-24800-25088$

$=Rs.288$

At what rate per cent compound interest, does a sum of money become $1.44$ times of itself in $2$ years ?

Let the

$P$ is the principle sum of money and

$R$ is rate of interest.

SInce,

$A=1.44\times P, T=2\ years$

We know that

$A=P\left(1+\dfrac{R}{100}\right)^T$

or,

$1.44P=P\left(1+\dfrac{R}{100}\right)^2$

or,

$\left(1+\dfrac{R}{100}\right)^2=1.44$

or,

$1+\dfrac{R}{100}=1.2$

or,

$\dfrac{R}{100}=0.2$

or,

$R=20\%$

Manish borrowed Rs. $75,000$ from his friend and after one year returned Rs. $80,000$ to him. Find the interest.

Manish borrowed

$Rs. 75000$

So the principal is

$Rs. 750000$

and after a year he returns

$Rs 80000$

so the interest that he paid on

$Rs. 75000$ is

$=Rs. \left( 80,000-75,000\right)$

$=Rs. 5000$

Now, the interest

$=\dfrac{Principal \times rate\times time}{100}$

here time

$=$ 1 year

$\therefore 5000=\dfrac{75000\times r\times 1}{100}$

$\Rightarrow \dfrac{100}{15}=r$

$\Rightarrow r=6\dfrac{10}{15}=6\dfrac{2}{3}\%$

[ Note : Since the time period is only one year thereby it does not matter if we have compound interest or simple ineterset as in either of the case the interest will be

$I=\dfrac{Pr}{100}$ ]

Hence, solved.

Shaily got an increase of $10\%$ in her salary. If this salary after increase was $Rs. 72,710$ . Find her salary before increase.

Let Shaily's salary before increment be

$Rs.x$

Increase in the salary

$=10$ %

salary after increase

$=Rs\ 72,710$

According to question,

$\therefore \ x+\dfrac {10}{100}\times x=72710$

$\therefore \ \dfrac {10x+x}{10}=72710$

$\therefore \ \dfrac {11x}{10}=72710$

$\therefore \ x=\dfrac {72710 \times 10}{11}$

$\therefore \ x=6610\times 10$

$\therefore \ x=Rs\ 66100$

Hence, Shaily's salary before increment

$=Rs\ 66100$ .

The simple interest on a certain sum of money at $10\%$ per annum is Rs. $6000$ in $2$ years. Then find :
(i) The sum.
(ii) The amount due at the end of $3$ years and at the same rate of interest compounded annually.

We have,

$SI=Rs.6000, P=?, T=2\ years, R=10\%$

Part

$(i)$ :

We know that:

$SI=\dfrac{P\times R\times T}{100}$

$6000=\dfrac{P\times 10\times 2}{100}$

$6000=\dfrac{P}{5}$

$P=Rs.30,000$

Part

$(ii)$ :

The amount at the end of the

$3rd$ year at the same rate of interest compounded annually.

or,

$A=P\left(1+\dfrac{R}{100}\right)^T$

or,

$A=30000\left(1+\dfrac{10}{100}\right)^3$

or,

$A=30000\left(1+\dfrac{1}{10}\right)^3$

or,

$A=30000\left(\dfrac{11}{10}\right)^3$

or,

$A=Rs.39,930$

In how many years will Rs. $7000$ amount Rs. $9317$ at $10\%$ annum compound interest

We have,

$A=Rs.9317, P=Rs.7000, R=10\%, T=?$

We know that

$A=P\left(1+\dfrac{R}{100}\right)^T$

$9317=7000\left(1+\dfrac{10}{100}\right)^T$

$\left(1+\dfrac{1}{10}\right)^T=1.331$

$\left(1.1\right)^T=1.331$

$\left(1.1\right)^T=(1.1)^3$

$\therefore T=3\ years$

Find out the compound interest of a sum of INR 10,000 for 2 years at a rate 8% per annum , If the interest is added yearly?

P=INR

$10000$

$8$ %

$2$ years

$P[1+R/100]^T$

$10000[1+8/100]^2$

$11664$ INR

CI=A-P

CI=

$11664-10000$

CI=INR

$1664$

Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After $4$ years, his monthly salary was $Rs.4500$ and after $10$ years his monthly salary become $5400$ rupees, then find his original salary and yearly increment.

Let the original salary be

$x$ and yearly increment be

$y$ .

After

$4$ years, his salary will be

$Rs. 4500$

$\therefore x + 4y = 4500$

$\Rightarrow x = 4500 - 4y ..... \left( 1 \right)$

After

$10$ years, his salary becomes

$5400$ .

$\therefore x + 10y = 5400$

$\Rightarrow 4500 - 4y + 10y = 5400 \quad \left( \text{From } \left( 1 \right) \right)$

$\Rightarrow y = 150$

Substituting the value of

$y$ in equation

$\left( 1 \right)$ , we have

$\Rightarrow x = 4500 - 4 \times 150$

$\Rightarrow x = 4500 - 600 = 3900$

Hence, his original salary was

$Rs. 3900$ and increment per year is

$Rs. 150$

Arun took a loan of Rs. $50,000$ for $3$ years and repaid total amount Rs. $68,000$ . Find the rate on interest.

1206506_1242872_ans_8131b5c1b9c940538f7e66931f662edd.jpg

Find the compound interest on $Rs.\ 10000$ for $12$ month at $10\%$ per annum. If the interest is compounded.
Quarterly

$P=$ Rs.

$10000$

$T=12$ months

$=1$ year

$R=10\%$ p.a

Compounded quarterly

$\therefore n=4$

$A=P\left(1+\dfrac{R}{n}\right)^{nT}$

$=10000\left[1+\dfrac{10}{4}\right]^{4\times 1}$

$=10000\times \left(\dfrac{14}{4}\right)^4$

$=10000\times \left(\dfrac{7}{2}\right)^4$

$=\dfrac{10000\times 7\times 7\times 7\times 7}{2\times 2\times 2\times 2}$

$A=$ Rs.

$1500625$

Compound Interest

$=Amount -P$

$=$ Rs.

$(1500625-10000)$

$=$ Rs.

$1490625$ .

1195556_1242306_ans_f13cab610d874408bdace2115bf43ab6.jpg

8000 for 1 year at 9 % per annum compounded half yearly. what will be the amount? (you could use the year by year calculation using SI formula to verify).

After half year, amount

$= \frac{8000 \times \frac{1}{2}\times 9}{100} = Rs.360+8000$

= Rs.8360

After next half year, amount

$= \frac{8360 + \frac{1}{2} \times 9 }{1001}=Rs.376.2+8360$

$= Rs.8736.2$

1167306_1246944_ans_86ddd91c16124f56b75d51a60c39a642.jpg

Divide $Rs\ 3600$ into two parts such that if one part be lent at $9\%$ per annum and the other at $10\%$ per annum the total annual income is $Rs\ 333$ .

1302549_1256778_ans_68c4c68a298342a2ae6c94f1b5792036.jpeg

Find the sum which will amount to $Rs. 700$ in $5$ years at $8\%$ p.a..

Let the sum be

$P$

$S.I=\cfrac{PRT}{100}=\cfrac{P\times 8\times 5}{100}=\cfrac{40P}{100}=\cfrac{2P}{5}$

$A=S.I+P=\cfrac{2P}{5}+P=\cfrac{7P}{5}$

$700=\cfrac{7P}{5}$

So,

$P=\cfrac{700\times 5}{7}=Rs. 500$

At what rate $%$ of $CI$ will $Rs.4000$ amount to $Rs.5324$ in three years?

Amount is Rs.

$5324$

Principle is Rs.

$4000$

So,

$5234=4000\left(1+\dfrac{r}{100}\right)^3$

$\Rightarrow \dfrac{5234}{4000}=\left(\dfrac{100+r}{100}\right)^3$

$\dfrac{11}{10}\times 100=100+r$

$110=100+r$

$r=110-100$

$r=10$ .

1195745_1268164_ans_7c4c18c9c0084528ae5ecaced749a06e.jpg

At what rate will a sum of Rs. $4000$ yield a compound interest of Rs. $410$ in $2$ years?

Principal

$=4000$

Compound interest for two years

$=Rs 410$

$\therefore$ Amount at the end of

$2$ years

$=A=Rs \left( 4000+410\right)$

$=Rs .4410$

$\Rightarrow$ Time

$2$ years

Let the rate of interest be

$r\%$ p.a.

we have,

$\Rightarrow A=P\left( 1+\dfrac{r}{100}\right)^2$

$\Rightarrow 4410=4000\left( 1+\dfrac{r}{100}\right)^2$

$\Rightarrow \dfrac{441}{400}=\left( 1+\dfrac{r}{100}\right)^2$

$\Rightarrow \dfrac{21\times 21}{20\times 20}=\left( 1+\dfrac{r}{100}\right)^2$

$\Rightarrow \sqrt{\dfrac{21\times 21}{20\times 20}}=\dfrac{r}{100}+1$

$\Rightarrow \dfrac{21}{20}-1=\dfrac{r}{100}$

$\Rightarrow \dfrac{1}{20}\times 100=r$

$\therefore r=5\%\;p.a.$

Hence, the answer is

$5\%.$

In what time will $Rs\ 1000$ amount to $Rs\ 1331$ at $10\%$ per annum compound interest?

$A = Rs.1331$

$P = Rs.1000$

$R = 10\%$

$S.I. = A - P$

$= 1331 - 1000$

$= Rs.331$

$S.I. = \dfrac{P \times R \times T}{100}$

$331 = \dfrac{1000 \times 10 \times T}{100}$

$\therefore T = \dfrac{331}{100}$

$T = 3.31$

$T \approx 3$ years

Nikita invests $\text{Rs. }6000$ for two years at a certain rate of interest compounded annually. At the end of the first year, it amounts to $\text{Rs. }6720$ .
Calculate :
(a) The rate of interest.
(b) The amount at the end of the second year.

Let the rate of interest is equal to

$r$ per annum.

a). For one year,

$P=\text{Rs. } 6000$

$A=\text{Rs. } 6720$

$n=1 \text{ years}$

Now, we will substitute the given values in the formula to calculate amount,

$\begin{aligned}{}A &= P{\left( {1 + \frac{r}{{100}}} \right)^n}\\6720& = 6000{\left( {1 + \frac{r}{{100}}} \right)^1}\\\frac{{672}}{{600}} &= 1 + \frac{r}{{100}}\\\frac{r}{{100}} &= \frac{{72}}{{600}}\\r &= 12\end{aligned}$

Hence the rate of interest is equal to

$12\%$ .

b). Amount at the end of the second year will be,

$\begin{aligned}{}A &= P{\left( {1 + \frac{r}{{100}}} \right)^n}\\ &= 6000{\left( {1 + \frac{{12}}{{100}}} \right)^2}\\ &= 6000{\left( {\frac{{28}}{{25}}} \right)^2}\\ &= 7526.4\end{aligned}$

Hence the amount at the end of the second year will be

$\text{Rs. } 7526.4$ .

If the value of car depreciated by $15\%$ annually. What will be its estimated value at the end of $8$ years; if its present (original) value is $5,00,000$ ?

Given,Principal amount

$P=500000, Rate\ r=15\%$
Estimated value at the end of

$8$ years is-

$=5,00,000\left(1-\dfrac{5}{100}\right)^8$

$=5,00,000\times 0.2724$

$=136,245.26$ .

Find the compound interest on $Rs 48000$ for $2$ years compounded annually at $2\dfrac { 1 } { 2 } \%$ per annum.

Given Rate of interest is

$r=2\dfrac{1}{2}\%=\dfrac{5}{2}\%=0.025$

Principal Amount is

$A=48,000$

Time is

$t=2$ years

Interest is compounded yearly once

$\implies n=1$

Amount

$=P(1+\dfrac{r}{n})^{n t}=48000\times (1+0.025)^{2}=50,430$

Interest is

$A-P=50,430-48,000=2430$

What sum will amount to $Rs. 55125$ in two years at $5\% p.a.$ compound interest?

$P =Rs. 55125$

$n = 2\ years$

$a = 5\% \ p.a$

$A=p(1+\dfrac{r}{100})^{n}$

$=55125(1+\dfrac{5}{100})^{2}$

$= 55125 (1.05)^{2}$

$=55125\times 1.1025$

$A=Rs.60775.3125$

At what rate percent per annum will a sum of $Rs 2000$ amount to $Rs 2205$ in $2$ years, compounded annually?

Given Rate of interest is

$r$

Principal Amount is

$A=2,000$

Time is

$t=2$ years

Interest is compounded yearly once

$\implies n=1$

Amount

$=P(1+\dfrac{r}{n})^{n t}=2000\times (1+r)^{2}=2205$

$\implies r=0.05=5\%$

Neeraj lend $\text{Rs. } 65536$ for $2$ years at $12\dfrac{1}{2}\%$ per annum compound annually. How much more could he earn if interest were compounded half-yearly?

To calculate the amount, if interest were compounded annually,

$\begin{aligned}{}A& = P{\left( {1 + \frac{R}{{100}}} \right)^n}\\ &= 65536{\left( {1 + \frac{{25}}{{100 \times 2}}} \right)^2}\\& = 65536{\left( {\frac{9}{8}} \right)^2}\\ &= 1024 \times 81\\ &= \text{Rs. } 82944\end{aligned}$

To calculate amount if interest were compounded half-yearly, rate of interest will become

$\dfrac{25}{4}\%$ and value of

$n$ will be

$4,$

$\begin{aligned}{}A &= P{\left( {1 + \frac{R}{{100}}} \right)^n}\\ &= 65536{\left( {1 + \frac{{25}}{{100 \times 4}}} \right)^2}\\& = 65536{\left( {\frac{{17}}{{16}}} \right)^4}\\ &= 17 \times 17 \times 17 \times 17\\ &=\text{Rs. } 83521\end{aligned}$

Clearly, we can observe that amount compounded half-yearly is more than compounded annually. So, the difference between the two amounts will be the extra amount that Neeraj can earn,

$\begin{aligned}{}\text{Extra Interest}& = 83521 - 82944\\ &= \text{Rs. }577\end{aligned}$

A weaver paid bank Rs. $18,422$ to clear a loan of Rs $15,100$ after $2$ years. At what rate of interest was the loan taken?

1209947_1274228_ans_c74c048d978a4de0b5815989fa1f1d85.jpg

If Meena gives an interest of $Rs\,45$ for one year at $9\%$ rate p.a. What is the sum she has borrowed.

Interest =

$\dfrac{P\times R\times T}{100}$

$Rs. 45=\dfrac{P\times 9\times 1}{100}$

$P=Rs.500$

The value of a property decrease every year at the rate of $5\%$ . If its present value $Rs.\ 4,11,540$ , what was its value three years ago.

A] present value = Rs 411540

rate of depreciation = 5% p.a

t = 3 years

value 3 yrs ago

$=p(1-\frac{R}{100})^{n}$

$\Rightarrow 411540=p(1-\frac{5}{100})^{3}$

$\Rightarrow 411540=p(\frac{95}{100})^{3}$

$\Rightarrow 411540=p(\frac{19}{20})^{3}$

$\Rightarrow 411540=p(\frac{6859}{8000})^{3}$

$\Rightarrow \frac{411540\times 80000}{6859}=Rs 480000$

1195192_1296663_ans_ed16c445b4a24831ad04024c744ac127.jpg

Vasudevan invested $Rs.60,000$ at an interest rate of $12\%$ per annum compounded half yearly. What amount would he get
(i) after $6$ months?
(ii) after $1$ year?

For compounded half - year, we use

$A = P{\left( {1 + \dfrac{r}{2}} \right)^{2n}},$ where

$n = number\,\,of\,\,year$

$\begin{array}{l} \left( i \right) A=60000{ \left( { 1+\dfrac { { 0.12 } }{ 2 } } \right) ^{ 2\times \dfrac { 1 }{ 2 } } } \\ =60000\left( { 1+0.06 } \right) =63,600 \\ \left( { ii } \right) A=60000{ \left( { 1+\dfrac { { 0.12 } }{ 2 } } \right) ^{ 2\times 1 } } \\ =60000{ \left( { 1+0.06 } \right) ^{ 2 } }=67,416 \end{array}$

Hence, this is the answer.

A scooter was bought at $Rs.42,000$ . Its value depreciated at the rate of $8\%$ per annum. Find its value after one year.

Given that,]

A scooter bought

$=42000$

Depreciated rate of scooter

$=8$ %

$= 42000 - 42000 \times \dfrac{8}{{100}}$

$= 42000 - 3360$

$= 38640$

Hence.

We get after one year

$38640$

Aditi got an amount of $Rs.75,000$ at the end of $5$ years on a deposit of $Rs .67,255 ,$ What was the interest that she earned?

$P=67,255$ Amount earned

$=75000$

$\therefore$ Interest earned

$=$ Amount

$-$ Principal

$(P)$

$=$ Rs.

$(75,000-67,255)$

$=$ Rs.

$7,745$ .

Rahul borrowed $Rs\ 20,000$ from Rampal at $18\%$ per annum on simple interest. Find the total amount he has to pay after $2$ years.

Given,

Principal,

$p=Rs.20,000$

Rate of interest,

$r=18\%$

Time,

$t=2\ years$

We know that,

$\begin{array}{l} SI=\dfrac { { p\times r\times t } }{ { 100 } } \\ Now, \\SI =\dfrac { { 20000\times 18\times 2 } }{ { 100 } } \\ =7200 \end{array}$ .

Total amount after

$2$ years is

$=Principal+Interest= Rs.(20,000 + 7200) = Rs.(27,200)$

Calculate the compounded interest on Rs 18000 for $1\dfrac { 1 }{ 2 }$ years at the rate of 10% p.a. compounded semi annually.

We have,

$P=18000$

$Time=t=\dfrac{3}{2}\times 2=3\left( Half\,yearly \right)$

Rate $r=10%\,p.a.$

Then,

$A=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}$

$A=18000{{\left( 1+\dfrac{10}{100} \right)}^{3}}$

$A=18000\times \dfrac{11}{10}\times \dfrac{11}{10}\times \dfrac{11}{10}$

$A=18\times 11\times 11\times 11$

$A=23958$

Hence, this is the answer.

If $A=P\left[1+\dfrac{R.T}{100}\right]$ , find P if $A=Rs\ 3000, R=10\%p.a$ and $T=10$ years

$A = Rs.\,3000$

$R = 10\% \,\,p/a$

$T = 10\,\,years$

$\Rightarrow A = P\left[ {1 + \frac{{R.T}}{{100}}} \right]$

$\Rightarrow 3000 = P\left[ {1 + \frac{{100}}{{100}}} \right]$

$\Rightarrow 3000 = P \left( 2 \right)$

$P = Rs.1500$

Urvashi borrowed a sum of $Rs.\ 5000$ form Debarati at the rate of $2\%$ per year. What is the interest paid by her at the end of $2$ years.

$\begin{array}{l} P=5000 \\ R=2 \\ T=2 \\ I=\dfrac { { P\times R\times T } }{ { 100 } } \\ I=\dfrac { { 5000\times 2\times 2 } }{ { 100 } } \\ I=200 \end{array}$

The S.I on a sum of money is $\dfrac{1}{9}th$ of the principal and the number of years is equal to the rate percent p.a. Find the rate $\%$ .

Let the principal be

$p$

Rate of intersect be

$R$ %

According to the question ,TIme

$=R$

Simple interest

$= \dfrac{p}{9}$

We know that,

$S.I = \dfrac{{P \times R \times T}}{{100}}$

$\Rightarrow \dfrac{P}{9} = \dfrac{{\left( {P \times R \times R} \right)}}{{100}}$

$\Rightarrow \dfrac{P}{9} = \dfrac{{\left( {P \times {R^2}} \right)}}{{100}}$

$\ \Rightarrow \dfrac{1}{9} = \dfrac{{{R^2}}}{{100}}$

$\Rightarrow {R^2} = \dfrac{{100}}{9}$

$\Rightarrow R = \dfrac{{10}}{3}$

Therefore,

rate of interest is

$R = \dfrac{{10}}{3}$ %

If the difference of the compound interest and the simple interest on a sum of money for $2$ years is $1\%$ of the sum of money then find the annual rate of interest.

$CI = P[1+r/100]^n-P$

$SI = pnr/100$

From the problem,

$N = 2$

$CI – SI = P*1/100$

We need to find

$r.$

$CI – SI = P[1+r/100]^n-P-Pnr/100$

$=P[(1+r/100)^n-1-nr/100]$

Substituting the values of

$n$ and

$r,$ we get

$CI – SI = P * 1/100 = P[1+r/100)^2-1-2*r/100]$

Since

$P$ is common on both sides of the

equation, they get cancelled.

Let us consider the expression within the

square brackets.

Expanding and subtracting the remaining

values, we get

$1+2r/100+r^2/10000-1-2r/100 = r^2/10000$

Therefore,

$1/100 = r^2/10000.$

$R^2 = 10000/100 = 100$

Therefore,

$r = 10%$

Thus, the rate of interest is

$10%.$

Mrs Gupta deposited $Rs\ 10,000$ in a bank for $5$ years at the rate of $1.5\%$ per month. How much money did she get at the end of that period?

We have

$P=Rs\,10,000$

$Time=5years$

$Rate=1.5\%pm i.e. 18\%pa$

$\begin{array}{l} Now, \\ S.I=\dfrac { { P\times R\times T } }{ { 100 } } \\ =\dfrac { { 10000\times 18\times 5 } }{ { 100 } } \\ =Rs\, 9000 \\ Hence,\, Total\, amount\, she\, did\, get\, at\, the\, end\, of\, time=10000+9000 \\ =Rs.\, 19000. \end{array}$

What will be the compound interest on $Rs\ 15,625$ for $3$ years at $8\%$ $p.a,$ if the interest is compounded annually ?

$x^{2}+\frac{1}{x^{2}} =(x+\frac{1}{x})^{2}-2 = (x-\frac{1}{x})^{2}+2$

$x^{2}+\frac{1}{x^{2}} = 2+2 = 4$

$[(x+\frac{1}{x})^{2} = x^{2}+\frac{1}{x^{2}}+2; (x-\frac{1}{x})^{2} = x^{2}+\frac{1}{x^{2}}-2]$

1185770_1348081_ans_f26b0b190c634870a3907725c02cf0e5.jpg

Find the sum of money that amounts to $Rs\ 31,740$ in $2$ years if the interest is compounded annually at the rate of $15\%$ per annum.

Given:-

$A = Rs. 31740$

$n = 2 \text{ years}$

$R = 15$

As we know that,

$A = P {\left( 1 + \cfrac{R}{100} \right)}^{n}$

$31740 = P {\left( 1 + \cfrac{15}{100} \right)}^{2}$

$31740 = P {\left( \cfrac{23}{20} \right)}^{2}$

$\Rightarrow P = \cfrac{31740 \times 20 \times 20}{23 \times 23} = Rs. 24000$

Hence the sum of money is

$Rs. 24000$ .

The interest on $Rs. 500$ amounts to $Rs. 60$ in $2\ years$ at the rate per annum. In how much time will $Rs. 800$ amount to $Rs. 1,232$ with the same rate?

We know, simple intrest

$=\dfrac {Principle \times rate \times time}{100}$

$\Rightarrow 60 =\dfrac {500 \times r \times 2}{100}$

$\Rightarrow r=\dfrac {60}{10}=6\%$

Now,

$(1232 -800)= \dfrac {800\times 6 \times t}{100}$

$\Rightarrow 432 = 48\, t$

$\Rightarrow t =9\, years$

The compound interest on $Rs.\ 4000$ at $10\%$ per annum for $2$ years $3$ months, Find compounded annually.

Principal

$\left( P \right) = Rs. 4000$

Rate of Interest

$\left( r \right) = 10 \%$

Time

$= 2$ years and

$3$ months

For first

$2$ years:

$n = 2$ years

Amount on compound interest,

$A = P {\left( 1 + \cfrac{r}{100} \right)}^{n}$

$\Rightarrow A = 4000 {\left( \cfrac{11}{10} \right)}^{2}$

$\Rightarrow A = 4000 \times \cfrac{121}{100}$

$\Rightarrow A = Rs. 4840$

The Amount after two years

$= Rs. 4840$

Now, for last

$3$ months,i.e. for

$\dfrac{1}{4}$ years:

Principal

$= Rs. 4840$

Simple interest

$= \cfrac{PRT}{100} = \cfrac{4840 \times 10 \times \cfrac{1}{4}}{100} = Rs. 121$

Therefore,

Amount after

$2$ years and

$3$ months

$= Rs.(4840 + 121) = Rs. 4961$

Therefore,

$C.I. = A - P$

$\Rightarrow C.I. = Rs.(4961 - 4000) = Rs. 961$

Hence the compound interest is

$Rs. 961$ .

What will be the compound interest on $Rs.\,4000$ at 5% per annum in 2 years?

1349873_1344815_ans_386fad3c4cb24d09868d2f0dbdccd2b5.jpg

Harish deposited $Rs\ 5120$ in a bank paying interest at the rate of $6\dfrac{1}{4}\%$ per annum. Find the compound interest due to him at the end of $2$ years.

As we know that,

$A = P {\left( 1 + \cfrac{R}{100} \right)}^{n}$

Here,

$A =$ Amount

$P =$ Principle

$= Rs. 5120$

$R =$ Rate of interest

$= \cfrac{25}{4} \%$

$n =$ No. of years

$= 2$

Therefore,

$A = 5120 {\left( 1 + \cfrac{\left( \cfrac{25}{4} \right)}{100} \right)}^{2}$

$\Rightarrow A = 5120 {\left( 1 + \cfrac{1}{16} \right)}^{2}$

$\Rightarrow A = 5120 \times \cfrac{17 \times 17}{16 \times 16} \\= 20 \times 289 \\= Rs. 5780$

Therefore,

$I= A - P \\= 5780 - 5120 \\= Rs. 660$

A man paid $8640$ for a loan of $8000$ taken for $320$ days. Find the rate of simple interest charged by the bank .

$Principal = 8000 Rs.$

Money paid by man (Amount)

$= Rs. 8640$

Simple Interest

$= Amount - Principal = 8640 - 8000 = 640\ Rs.$

Time given by

$= 320\ days$

$= \dfrac {320}{365} years = \dfrac {64}{73} years$

$R = \dfrac {SI\times 100}{T\times P}$

$= \dfrac {640\times 100\times 73}{64\times 8000} = 9.12 \%$ per annum

Rate of interest

$= 9.12 \% p.a.$

At what rate percent annum will Rs. 3600 amount to Rs. 4734 in $3\frac{1}{2}$ years?

We have,

$P=Rs. 3600$

$A=Rs.4734$

$T=3\dfrac{1}{2}=\dfrac{7}{2}\ years$

$R=?$

We know that

$A=P+S.I$

$S.I=\dfrac{PRT}{100}$

Therefore,

$A-P=\dfrac{PRT}{100}$

$4734-3600=\dfrac{3600\times R\times \dfrac{7}{2}}{100}$

$1134=36\times R\times \dfrac{7}{2}$

$1134=18\times R\times 7$

$R=9\%$

Hence, this is the answer.

Madhu deposited Rs. $20000$ in a bank at $10\%$ per annum. Find the compound interest after $1\dfrac{1}{2}$ years if the interest is compounded yearly.

Madhu diposited Rs

$20000$ in a bank.

$\text{Rate }=10\%$ per annum compounded annually.

$\text{Time} = \cfrac{3}{2}$ years.

We know that, total amount when compounded yearly is given by

$\text{Total amount} =P\left(1+\cfrac{R}{100}\right)^n$

$\\= 20000\left(1+\cfrac{10}{100}\right)^{\frac{3}{2}}\\= 22000$

$\text{Interest}=\text{Amount}-\text{Principle}=22000-20000=2000$

Rajan makes fixed deposit of Rs. 8000 in a bank for a period of 2 years. If the rate of interest is 10 % p.a. compounded annually, find the amount payable to him by the bank after two years.

The amount of fixed deposit = Rs. 8000

R = 10 % pa and n =2

The amount returned by the bank =

$P[1+\dfrac{r}{100}]^{n}$

$=8000[1+\dfrac{10}{100}]^{2}=8000\times 1.21 = 9680$ Rs.

Govind opened a bank account on 1.4.2006 by depositing Rs.He deposited Rs. 1000 on 11.4.06 and withdrew Rs. 500 on 15.4.Compute the interest paid by the bank for the month of April, if the rate of interest is 4 % pa.

Balance as on 1.4.2006 = Rs. 3000

As on 11.4.2006 = Rs. 4000

As on 15.4.2006 = Rs. 3500

The minimum balance for the month of April = Rs. 3000

The interest paid by the bank for the month of April =

$\dfrac{3000\times 1 \times 4}{1200}=Rs. 10$

S.No. Principal Amount Rate of interest
(i) $Rs. 12,550$ $9 \%$ semi-annually
(ii) $Rs. 1500$ $Rs. 1900$
(iii) $Rs. 8000$ $Rs. 9261$ $5 \% p.a$
(iv)
$RS. 5202$ $2 \% p.a$
(v) $Rs. 625$ $3 \% p.a$
Complete the table given below.

Given principal =12.550 RS Time 2.5 years

rate of interest = 9% semi annually .

now rate is given semi annually .

so we have 5 day years in given time

then

S.Interest =

$\frac{P\times R\times T}{100 }$ =

$\frac{12,550 \times 905.}{100}$

SI =

$\frac{62750\times 9}{100}=$

$\frac{564750}{100}$ = 5647.5

Then Amount = P+SI= 12550+5647.5

Amount =18197.5 RS.

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Which is a better investment $Rs.5000$ for $10$ years at $5\%$ p.a. or $Rs.5500$ for $9$ years at the rate of $6\%$ p.a?

Given

$P_ 1= 5000$ Rs. Time = 1o years Rate = 5% P.a

then

Simple intrest

$=\frac {PRT}{100} =\frac {500 \times 10 \times 5}{100}=2500$ Rs.

and

$P_2= 5500$ Rs. Time = 9 years Reate = 6 %

Simple interest

$=\frac {PRT}{100} =\frac {5500 \times 9 \times 6 }{100}=2970$ Rs.

$\therefore$ Second interest is better as interest is more.

A man borrowed Rs. $8000$ at $10\%$ per annum simple interest and immediately lent the whole sum at $10\%$ per annum compound interest. What does he gain at the end of $2$ years?

C.I

$=$ Rs.

$8000\times {\left(1+\dfrac{10}{100}\right)}^{2}-8000$

$=$ Rs

$8000\times {1.01}^{2}-8000=$ Rs.

$1680$

S.I

$=$ Rs.

$\dfrac{8000\times 10\times 2}{100}=80\times 10\times 2$ Rs.

$1600$

Gain

$=C.I-S.I$

$=1680-1600=$ Rs.

$80$

Find the amount and the compound interest.

Compound interest is given by =

$P\left(1+\dfrac{r}{100}\right)^{n}-P$

(1) C.I. =

$2000(1+0.05)^{2}-2000=2000(0.1025)= Rs. 205 \Rightarrow$ amount

$=2000+205= 2205$

(2) C.I. =

$5000((1+0.08)^{2}-1)=5000\times 0.166=832\Rightarrow$ Amount

$=5000+832=5832$

(3) C.I =

$4000 ((1+0.075)^{2}-1)=4000\times 0.156=622.5 \Rightarrow$ Amount

$= 4000+622.5=Rs. 4622.5$

A man invests Rs. $46875$ at $4\%$ per annum at compound interest for $3$ year. Calculate the interest for the first year.

Interest in

$1^{st}$ year would be the simple interest given by

$=\dfrac{Principal \times rate \times time}{100}$

$=\dfrac{46875\times 4\times 1}{100}$

$=$

$Rs.1875$ .

If a certain sum becomes double in $3$ years at a certain rate of interest compounded annually then in how many years it will become $16$ times.

Let us assume that the initial sum was

$P$ and the rate of interest is

$R\%$ per annum.

It is given that in

$3$ years the sum doubles,

$\begin{aligned}{}A& = P{\left( {1 + \frac{R}{{100}}} \right)^n}\\2P &= P{\left( {1 + \frac{R}{{100}}} \right)^3}\\2 &= {\left( {1 + \frac{R}{{100}}} \right)^3}\quad\quad\dots(i)\end{aligned}$

Let us assume that the sum becomes

$16$ times in

$t$ years. So,

$\begin{aligned}{}A& = P{\left( {1 + \frac{R}{{100}}} \right)^n}\\16P &= P{\left( {1 + \frac{R}{{100}}} \right)^t}\\16 &= {\left( {1 + \frac{R}{{100}}} \right)^t}\quad\quad\dots(ii)\end{aligned}$

Now, if we exponentiate left side of equation

$(i)$ by

$4$ the it becomes equal to left side of the equation

$(ii)$ . So, from this observation we can conclude that,

$t=3\times4$

$=12$

So, in a period of

$12$ years the amount becomes

$16$ times of the original sum.

A sum of money invested at $6$ % p.a. simple interest for a certain period of time yields Rs. $960$ as interest. If this sum has been invested for $5$ years more, it would have yielded Rs. $2160$ as interest. Find the sum.

Let the sum be

$Rs.x$

And we know

$S.I.=\dfrac{PRT}{100}$

Then according to the question,

$\dfrac{x\times{6}\times{T}}{100}=960$ ..............1

Also,

$\dfrac{x\times{6}\times{(T+5)}}{100}=2160$ ....................2

From eq. 1

$T=\dfrac{1600}{x}$

Putting this value in eq 2 we get,

$\dfrac{x\times{6}\times{\left(\dfrac{16000}{x}+5\right)}}{100}=2160$

$\dfrac{16000+5x}{x}\times{x}\times{6}=216000$

$96000+30x=216000$

$30x=120000$

$x=Rs.4000$

Hence sum

$=Rs. 4000$

At what rate percent annum will $Rs.640$ amount to $Rs.774.40$ in $2$ years, interest being compounds annually?

Amount

$=P{\left(1+\dfrac{R}{100}\right)}^{n}$

$\Rightarrow 774.40=640{\left(1+\dfrac{R}{100}\right)}^{2}$

$\Rightarrow \dfrac{77440}{100}=640{\left(1+\dfrac{R}{100}\right)}^{2}$

$\Rightarrow \dfrac{7744}{10}=640{\left(1+\dfrac{R}{100}\right)}^{2}$

$\Rightarrow \dfrac{7744}{10}\div 640={\left(1+\dfrac{R}{100}\right)}^{2}$

$\Rightarrow \dfrac{7744}{6400}={\left(1+\dfrac{R}{100}\right)}^{2}$

$\Rightarrow {\left(\dfrac{88}{80}\right)}^{2}={\left(1+\dfrac{R}{100}\right)}^{2}$

$\Rightarrow 1+\dfrac{R}{100}=\dfrac{88}{80}$

$\Rightarrow \dfrac{R}{100}=\dfrac{88}{80}-1$

$\Rightarrow \dfrac{R}{100}=\dfrac{88-80}{80}$

$\Rightarrow \dfrac{R}{100}=\dfrac{8}{80}$

$\Rightarrow \dfrac{R}{100}=\dfrac{1}{10}$

$\therefore R=10$

Hence the rate of interest is

$10\%$

mr kamat donated Rs $80,000$ to a welfare trust on the agreement with them that the interest on the sum will be used for awarding $5$ scholarship of equal value every value year to physically handicapped children. If the donation earns an interest of $10\%$ per annum, find the value of each scholarship

Amount donated by kamat = Rs. 80,000

Interest earned at the rate of 10% p.a = 10% of 80,000

$=\frac{10}{100}\times 80,000$

= Rs. 8000

Number of scholarship = 5.

$\therefore$ value of each scholarship = Rs.

$\frac{8000}{5}$

= Rs. 1600

$\therefore$ value of each scholarship = Rs. 1600.

1203211_1421358_ans_a1bd4a38cfea424f87b904821e076772.jpg

Find the principal when the following are given for simple interest
$I=Rs\ 192$ Rate= $6%$ per annum Time= $4$ years

As we know that,

$S.I. = \cfrac{P \times R \times t}{100}$

Given:-

$S.I. = Rs. 192$

$R = 6 \%$

$t = 4 \text{ years}$

$P = ?$

Therefore,

$192 = \cfrac{P \times 6 \times 4}{100}$

$\Rightarrow P = \cfrac{19200}{24} = Rs. 800$

Find the simple interest for $3yrs$ if the principle amount is $10,000$ for $5\%$ as rate of interest

Principle amount is

$10,000$

Rate of interest is

$5\%$

Time period is

$3yrs$

Simple interest is given as

$\dfrac{PRT}{100}\\\dfrac{10,000\times 5\times 3}{100}\\1500$

Sandeep borrowed $\text{Rs. } 3000$ for $4$ years and $\text{Rs. } 8000$ for $2 \dfrac{1}{2}$ years from Sonia at the same rate of interest. The total interest earned by Sonia is $\text{Rs. } 2880$ . Find the rate of interest.

Let the rate of interest be equal to

$r$ .

As we know that,

$\text{Simple interest} = \dfrac{\text{Principle} \times \text{rate} \times \text{time}}{100}$

By using the above formula and given information that the total interest is equal to

$\text{Rs. }2880,$

$\begin{aligned}{}\text{Total interest}& = \frac{{3000 \times r \times 4}}{{100}} + \frac{{8000 \times r \times \dfrac{5}{2}}}{{100}}\\2880& = 120r + 200r\\320r& = 2880\\r& = \frac{{2880}}{{320}}\\r &= 9\% \end{aligned}$

Find the present value of annuity immediate of $\text{Rs. } 18,000$ p.a for $3$ years at $9\%$ p.a compounded annually.
[Given $(1.09)^{3}=1.29]$

Total amount

$=P\left(1+\dfrac{r}{100}\right)^t$

$=18000(1+\dfrac{9}{100})^3$

$=18000(1.09)^3$

$=18000\times1.29=\text{Rs }23,220$

Find the rate of interest, when, Principal $=\ Rs.\ 8,250$ , simple interest $=Rs.\ 1,100$ and time $=2$ years.

Principal

$(P)=Rs.8,250$

Simple Interest

$=Rs.1100$

Time

$(T) =2\,years$

Simple interest

$=\dfrac{P\times R\times T}{100}$

$\Rightarrow$

$1100=\dfrac{8250\times R\times 2}{100}$

$\Rightarrow$

$R=\dfrac{1100\times 100}{8250\times 2}$

$\Rightarrow$

$R=6.67\,\%$

What rate gives Rs. $280$ as interest on a sum of Rs. $56000$ in $2$ years?

${\textbf{Step -1: Stating the formula for calculating simple interest}}{\text{.}}$

${\text{If we have a principal amount of }}P{\text{,which is kept for a time period of }}n{\text{ years}}$

${\text{at a rate of }}r% {\text{,then the simple interest can be written as,}}$

$S.I. = \dfrac{{P \times R \times T}}{{100}}$

${\textbf{Step -2: Calculating the rate of interest}}{\text{.}}$

${\text{Let the rate of interest be }}r,{\text{which results to Rs}}{\text{.280 simple interest,}}$

${\text{then we can write,}}$

$280 = \dfrac{{56000 \times r \times 2}}{{100}}$

$r = \dfrac{{280 \times 100}}{{2 \times 56000}}$

$r = \dfrac{1}{4} = 0.25\%$

${\textbf{Hence, the rate of interest is 0}}{\textbf{.25% }}{\text{.}}$

To start a business Shalaka has taken a loan of $Rs\ 8000$ at a rate of $10\frac{1}{2}$ p.c.p.a. After two years how much compound interest will she have to pay?

1226845_1397924_ans_bcb306d8976d4fa3a407682f34344eb8.jpg

If Manohar pays an interest of Rs. 750 for 2 years on a sum of Rs. 4,500 ,find the rate of interest.

1327953_1495034_ans_48fd9aadf3be4c038a39be5fd6d19cd5.JPG

Find the rate of interest when the following are given.
Principal $=Rs\ 350$ Time $=\dfrac{21}{2}$ years $I=Rs\ 140$

As we know that,

$S.I. = \cfrac{P \times R \times t}{100}$

Given:-

$S.I. = Rs. 140$

$R = ?$

$t = \cfrac{21}{2} \text{ years}$

$P = Rs. 350$

Therefore,

$140 = \cfrac{350 \times R \times \cfrac{21}{2}}{100}$

$\Rightarrow R = \cfrac{14000 \times 2}{21 \times 350} = \cfrac{80}{21} \% \approx 3.8 \%$

Find the rate of interest when the following are given.
Principal $=Rs\ 9600$ Time $=6$ years $I=Rs\ 72$

As we know that,

$S.I. = \cfrac{P \times R \times t}{100}$

Given:-

$S.I. = Rs. 72$

$R = ?$

$t = 6 \text{ years}$

$P = Rs. 9600$

Therefore,

$72 = \cfrac{9600 \times R \times 6}{100}$

$\Rightarrow R = \cfrac{7200}{9600 \times 6} = \cfrac{1}{7} \% \approx 0.14 \%$

If a farmer paid $Rs.2025$ as interest over 3 years at $2\%$ per annum to a moneylender, how much loan had the farmer taken?

Hint: With the help Simple interest we can calculate the amount of interest charged on a sum at a given rate for the given period of time

Solution:

$S.I = \frac{PRT}{100}$

$2025 = \frac{P\times 2\times 3}{100}$

$\Rightarrow P =2025 × 100/5$

= 45×100

= 4500

Amit deposits $Rs. 1600$ per month in a bank for $18$ months in a recurring deposit account .if he gets $Rs.31,080$ at the time of maturity ,what is the rate of interest per annum?

$\textbf{Apply suitable formula to find the required answer:}$

$\text{It is given that,}$

$\text{Principal (P) = Rs.1600}$

$\text{Number of months(n) = 18}$

$\text{Maturity value(M) = Rs.31,080}$

$\text{Let, the rate of interest per annum be r % .}$

$\text{We know,}$

$\Rightarrow M = P \times n + \dfrac{P \times n(n + 1)}{24} \times \dfrac{r}{100}$

$\Rightarrow 31080 = 1600 \times 18 + \dfrac{1600\times 18\times 19}{24} \times \dfrac{r}{100}$

$\Rightarrow 31080 = 28800 +228r$

$\Rightarrow 228r=31080 - 28800$

$\Rightarrow 228r=2280$

$\Rightarrow r=10$

$\textbf{Hence, the rate of interest per annum is 10% .}$

Find the principal when the following are given.
$I=Rs\ 20$ Rate= $2%$ per annum Time= $2$ years

As we know that,

$S.I. = \cfrac{P \times R \times t}{100}$

Given:-

$S.I. = Rs. 20$

$R = 2 \%$

$t = 2 \text{ years}$

$P = ?$

Therefore,

$20 = \cfrac{P \times 2 \times 2}{100}$

$\Rightarrow P = \cfrac{2000}{4} = Rs. 500$

The ratio of the amount for two years under compound interest annually and for one year under simple interest is $6:5$ . When the rate of interest is same, then find the value of rate of interest?

Let the principle = P, Rate = x% per annume

$\Rightarrow \frac{P(1+\frac{r}{100})^{2}}{P(1+\frac{r\times 1}{100})}=\frac{6}{5}$

$(1+\frac{r}{100})=\frac{6}{5}$

$\Rightarrow \frac{r}{100}=\frac{6}{5}-1=\frac{1}{5}$

$r=\frac{100}{5}=20$

$\therefore$ rate of interest = 20%

1196951_1508339_ans_982584294b5b43e78b64729479dc320b.PNG

A man starts a business with a capital of Rs $1000000$ . He incurs a loss of $4\%$ during the first year. But he makes a profit of $5\%$ during the second year on his remaining investment. Finally, he makes a profit of $10\%$ on his new capital during the third year. Find his total profit at the end of three years.

Here, initial capital

$P = 1000000$ ,

loss for the first year

$={r}_{1}\% = 4\%$ ,

gain for the second year

$={r}_{2}\% = 5\%$ and

gain for the third year

$= {r}_{3}\% = 10\%$

$Q = P\left(1 - \dfrac{{r}_{1}}{100}\right)\left(1 +\dfrac{{r}_{2}}{100}\right)\left(1 +\dfrac{{r}_{3}}{100}\right)$

$\Rightarrow Q = 1000000\left(1 -\dfrac{4}{100}\right)\left(1 +\dfrac{5}{100}\right)\left(1 +\dfrac{10}{100}\right)$

$\therefore, Q = 1000000\times\dfrac{24}{25}\times \dfrac{21}{20}\times \dfrac{11}{10}$

$\Rightarrow Q = 200 \times 24 \times 21 \times 11$

$\Rightarrow Q =1108800$

$\therefore,$ profit at the end of three years

$=1108800 -1000000=108800$

The compound interest on a certain sum for $2$ years is Rs. $786$ and S.I. is Rs. $750$ . If the sum is invested such that the S.I. is Rs. $1296$ and the number of years is equal to the rate per cent per annum, Find the rate of interest.

Compound interest for 2 years = Rs. 786

Simple interest for 2 years = Rs. 750

Interest for 2 years = I =

$\frac{CI-SI}{n}\times 100=\frac{786-750}{360}\times 100$

$=\frac{36}{36}\times 10=10$ %

P for first year Rs. 3600. Let the time and rate = x

$I= PRT \Rightarrow 1296 = 3600\times x\times x\Rightarrow x^{2}=0.36 \therefore x=0.6$

1245561_1509381_ans_8765f64a8e5d4d4381f3ab43f72e255a.PNG

$P$ and $Q$ invest Rs. $36000$ and Rs. $25000$ respectively at the same rate of interest per year.If at the end of $4$ years, $P$ gets Rs. $3080$ more interest than $Q$ , find the rate of interest.

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An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, find the effective rate of interest.

Let the sum be Rs 100 then

SI for first 6 months = Rs

$\frac{100\times 10\times 1}{100\times 2}= Rs 5$

SI for last 6 months = Rs

$\frac{102\times 10\times 1}{100\times 2}= Rs 5.25$

So smount at the end of 1 year

= Rs

$(100+5+5.25)= Rs 110.25$

Rate =

$110.25-100=10.25$ %

1242654_1508344_ans_7aa39d7ffc30434d91ee4e512da03aed.PNG

The difference between the compound interest and simple interest on a certain sum at $15\%$ per annum for $3$ years is Rs. $283.50$ .Find the sum.

Given

$CI-SI= Rs. 283.50$

$R=15%$

$n=3$ years

Let the sum be

$Rs.x$

We know that :

$A=P(1+\frac{R}{100})^{n}$

$A=2(1+\frac{5}{100})^{3}=x(1.15)^{3}$ _________ (1)

Aslo,

$SI=\frac{PRn}{100}=\frac{x\times 15\times 3}{100}=1.45x$ _______ (2)

Thus, we have :

$x(1.15)^{3}-1.45x=283.50$ (from (1)&(2))

$\Rightarrow 1.523x-1.45x=283.50$

$\Rightarrow 0.070875x=283.50$

$\Rightarrow x=\frac{283}{0.070875}=4,000$

Thus, the sum is

$Rs 4000$

The ratio of difference between compound interest and simple interest for 3 years to the difference between compound interest and simple interest for 2 years is 31 :What is rate of interest per annum?

1242673_1509200_ans_58ffdd9115324645a5388de7eb4f587d.PNG

The simple interest on a sum of money for $2$ years at $8 \%$ per annum is $Rs. 2400$ . What will be the compound interest on that sum at the same rate and for the same period?

$\textbf{Step 1: Apply Simple Interest Formula}$

$\text{SI =}$

$\dfrac{P\times R\times T}{100}$

$2400 = \dfrac{P\times 8\times 2}{100}$

$P = \dfrac{2400\times 100}{16} = Rs. 15000$

$\textbf{Step 2: Apply Amount Formula}$

$A = P\left(1 + \dfrac{R}{100}\right)^T$

$= 1500\left(1 + \dfrac{8}{100}\right)^2$

$= 1500\times\dfrac{108}{100}\times\dfrac{108}{100}$

$= \dfrac{15\times 108\times 108}{10}$

$\text{A = Rs. 17496}$

$\textbf{Step 3: Apply Compound Interest Formula.}$

$\text{CI = A - P}$

$\text{= 17496 - 15000}$

$\text{Rs. 2496}$

$\textbf{Hence, the compound interest is 2496.}$

If the difference between C.I and S.I earned on a certain amount at 10% pa at the end of 3 years is Rs.46.50, find the sum.

Let the sum be

$Rs. x$ .

$S.I.$ =

$\dfrac{x\times 10\times 3}{100} = \dfrac{3x}{10}$

$C.I. =x\left(1 + \dfrac{10}{100}\right)^3 - x = x\left(\dfrac{10 + 1}{10}\right)^3 - x$

$x\left(\dfrac{11^3}{10^3} - 1\right) = x\left(\dfrac{11^3 - 10^3}{10^3}\right)$

$x\left(\dfrac{\left(11 - 10\right)\left(11^2 + 110 + 10^2\right)}{1000}\right)$

$x\left(\dfrac{121 + 110 + 100}{1000}\right) = \dfrac{331}{1000}x$

Given, difference between compund and simple interest

$=46.50$

$\implies \dfrac{331}{1000}x - \dfrac{3x}{10} = 46.50$

$\dfrac{331x - 300x}{1000} = 46.50$

$31x = 46,500$

$x = \dfrac{46500}{31} = Rs. 1500$

$\therefore$ Sum

$= Rs. 1500$ .

A person invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. He received yearly interest of Rs.But if he had interchanged the amounts received, he would have received Rs. 4 more as interest. How much did he invest at different rates?

1313520_1534593_ans_4fd82b7b9102464fa6f882446d207e53.jpeg

A Man lends Rs. 1540 for five years and Rs. 1800 for four years. If he gets Rs. 1788 as interest on both amounts, what is the rate of interest ?

Let the rate of interest peer annum be

$r$ %

Then According to question ,

$\dfrac{r}{100}\times$ (

$1540\times5+1800\times4$ )=

$1788$

Solving we get

$r=12$ %

The monthly salary of a teacher is $\text{Rs }12800$ . If his annual increment is $3\%$ , find his new salary after a year.

Given, monthly salary is

$\text{Rs }12800$ .

Annual increment is

$3\%$

increment in salary is

$3\%$ of

$12800=\dfrac{3}{100}\times 12800=3\times 128=\text{Rs }384$

So new salary after one year is

$12800+384=\text{Rs }13184$

A sum is invested at compound interest compounded yearly. If the interest for two successive years be Rs. 5,700 and Rs. 7,410, calculate the rate of interest.

Rate of interest

$= \dfrac{Difference\ in\ the\ interest\ of\ two\ consecutive\ periods \times 100}{C.I. of preceding year \times 100} \%$

$= \dfrac{(7410 - 5700) \times 100}{5700 \times 1} \%$

$= 30 \%$

A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are Rs 650 and Rs 760.50 ; find the rate of interest.

Difference between the C.I.of two successive half -years

= Rs. 760.50 - Rs. 650 = Rs. 110.50

$\Rightarrow$ Rs. 110.50 is the interest of one half-year on Rs. 650

Rate of interest = Rs.

$\frac{100 \times 1}{P \times T} \% = \frac{100 \times 110.50}{650 \times \frac{1}{2}} \% = 34 \%$

Population of Rajasthan $=570$ lakhs and population of U.P $=1660$ lakhs. Area of Rajasthan $3$ lakh ${km}^{2}$ and area of U.P $=2$ lakh ${km}^{2}$ . How many people are there per ${km}^{2}$ in both states?

(i) People present per

${km}^{2}=\cfrac{Population}{Area}$
In Rajasthan

$=\cfrac{570lakhs}{3lakhs\quad {km}^{2}}=190$ people

${km}^{2}$
In U.P

$=\cfrac{1660lakhs}{2lakhs\quad {km}^{2}}=830$ people

${km}^{2}$
(ii) Rajasthan is less populated.

In how many years will Rs. $8,000$ amount to Rs. $9,261$ at $5\%$ per annum compounded annually?

1303498_1579714_ans_54fee74fb5514c7caf1cc80478d9e7f0.jpg

The population of a city decreased from $25,000$ to $24,500$ . Find the percentage decrease.

The decreased population of a city from

$25,000$ to

$24,500$
Population decreased

$=25,000-24,500=500$
Decreased Percentage

$=\cfrac{Population\quad decreased}{Original\quad population}\times 100$

$=\cfrac{500}{25000}\times 100=2$ %
Hence, the percentage decreased is

$2$ %

What rate gives Rs. $280$ as interest on a sum of Rs. $56,000$ in $2$ years?

${\textbf{Step - 1: Stating the formula for calculating simple interest}}{\text{.}}$

${\text{If we have a principal amount of }}P{\text{,which is kept for a time period of }}n{\text{ years}}$

${\text{at a rate of }}r% {\text{,then the simple interest can be written as,}}$

$S.I. = \dfrac{{P \times R \times T}}{{100}}$

${\textbf{Step - 2: Calculating the rate of interest}}{\text{.}}$

${\text{Let the rate of interest be }}r,{\text{which results to Rs}}{\text{.280 simple interest,}}$

${\text{then we can write,}}$

$280 = \dfrac{{56000 \times r \times 2}}{{100}}$

$r = \dfrac{{280 \times 100}}{{2 \times 56000}}$

$r = \dfrac{1}{4} = 0.25\%$

${\textbf{Hence, the rate of interest is 0}}{\textbf{.25% }}{\text{.}}$

Find the amount of Rs 2400 after 3 years, when the interest is compounded annually at the rate of 20% per annum (correct upto two decial places).

1872861_1673890_ans_d95e243aae96422da6c7fdc147f317c2.jpg

Population of Sundarnagar was $2,35,471$ in the year $1991$ . In the year $2001$ it was found to have increased by $72,958$ . What was the population of the city in $2001$ ?

Population of the city in

$2001=$ Population of the city in

$1991+$ Increased

$=2,35,471+72,958$

Population of the city in

$2001=3,08,429$

Rohit deposited $Rs. 8000$ with a finance company for $3$ years at an interest of $15\%$ per annum. What is the compound interest that Rohit gets after $3$ years?

1872331_1673793_ans_8d51458ce18746ccb9f3919987538fa1.jpeg

The present population of a town isIf it increases at the rate of 5% per annum, what will be its population after 2 years?

Given,

$Present\ Population (P) = 28000$

$Rate\ of\ growth\ of\ population\ (R) = 5\%\ per\ annum$

$Time (t) = 2\ years$

Formula used:

The population after

$t$ years is given by

$=P \left[1+\dfrac{R}{100}\right]^t$

Where

$P=Initial\ population$

$R=Rate$

$t=Time$

Apply the above formula, we get

$=28000 \left[1+\dfrac{5}{100}\right]^2$

$=28000 \left[\dfrac{105}{100}\right]^2$

$=28000 \left[1.05\right]^2$

$=30870$

Hence, The population after

$2$ years is

$30870$ .

Mewa Lal borrowed Rs 20000 from his friend Rooplal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.

${\textbf{Step 1: Calculate Simple Interest}}$

${\text{Principal}} = 20,000$

${\text{Rate}} = 18\% {\text{ per annum}}$

${\text{Time}} = 2{\text{ years}}$

$\therefore \text{S.I. = }\dfrac{{20000 \times 18 \times 2}}{{100}} = 7200$

${\textbf{Step 2: Calculate Amount to be paid by Mewalal to Rooplal}}$

$\text{Amount = P + S.I.}$

$\Rightarrow \text{Amount = }20,000 + 7,200 = 27,200$

${\text{So, Mewalal has to pay Rs 27,200 to Roopalal}}{\text{.}}$

${\textbf{Step 3: Calculate Amount to be paid by Rampal to Mewalal}}$

$\text{Amount = }20,000{\left( {1 + \dfrac{{18}}{{100}}} \right)^2} = 27848$

${\textbf{Step 4: Calculate the profit earned by Mewalal}}$

${\text{Profit Earned}} = {\text{ Amount paid - Amount Received}}$

$\Rightarrow {\text{Profit}} = 27848 - 27200 = 648$

${\textbf{Final Answer: The profit gained by Mewalal is Rs. 648}}$

Find the compound interest when principal = Rs 3000, rate = 5% per anuum and time = 2 years.

1872218_1673785_ans_de1a924f94b84f00a1425cbbf946b645.jpeg

The present population of a town is $25000$ . It grows at $4\%, 5\%$ and $8\%$ during first year, second year and third year respectively. Find its population after $3$ years.

Given,

$Present\ Population=2500$

The growth rate during the first year

$=4\%$

The growth rate during the second year

$=5\%$

The growth rate during the third year

$=8\%$

$Time= 3\ years$

Formula used:
Let the rate of the population be

$x\%,\ y\%$ , and

$z\%$ , respectively.

Therefore, the population after

$3$ years is given by

$=P\left(1+\dfrac{x}{100}\right)\left(1+\dfrac{y}{100}\right)\left(1+\dfrac{z}{100}\right)$

Apply the above formula, we get

$=25000\left(1+\dfrac{4}{100}\right)\left(1+\dfrac{5}{100}\right)\left(1+\dfrac{8}{100}\right)$

$=25000\times\left(\dfrac{104}{100}\right)\times\left(\dfrac{105}{100}\right)\times\left(\dfrac{108}{100}\right)$

$=29484$

Hence, the population after

$3$ years is

$29484$ .

The population of a city isIf the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.

Given,

$Population (P) = 125000$

$Birth\ rate=5.5\%\ per\ annum$

$Death\ rate=3.5\%\ per\ annum$

$Time (t) = 3\ years$

So, the annual growth rate will be

$=Birth\ rate-Death\ rate$

$=5.5-3.5=2\%$

Formula used:

The population after

$t$ years is given by

$=P \left[1+\dfrac{R}{100}\right]^t$

Where

$P=Initial\ population$

$R=Rate$

$t=Time$

Apply the above formula, we get

$=125000 \left[1+\dfrac{2}{100}\right]3$

$=125000 \left[\dfrac{102}{100}\right]^3$

$=125000 \left[1.02\right]^3$

$=132651$

Hence, The population after

$3$ years is

$132651$ .

What will be the compound interest on Rs 4000 in two years when rate of interest is 5% per annum?

1872225_1673788_ans_173c982bd596498d96036ecb9168aa2c.jpeg

Pritam bought a plot of land for RsIts value is increasing by 5% of its previous value after every six months. What will be the value (correct upto two decimal places) of the plot after 2 years ?

Given:

$\mathrm{P}=\operatorname{Rs} 64,000$

$\mathrm{R}=5 \%$ for every six months

$=\dfrac{5}{2}\%$ for a year

Value of the plot after two years

$=\mathrm{P}\left(1+\dfrac{\mathrm{R}}{100}\right)^{\mathrm{n}}$

$=64,000\left(1+\dfrac{5}{200}\right)^{4}$

$=64,000(1.025)^{4}$

$=706,440.25$

Thus, the value of the plot after two years will be

$\mathrm{Rs.}\space 706,440.25$ .

The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 100000?

Value of the machine after two years

$=\mathrm{P}\left(1-\dfrac{\mathrm{R}}{100}\right)^{\mathrm{n}}$

$\Rightarrow 100,000\left(1-\dfrac{10}{100}\right)^{2}$

$=100,000(0.90)^{2}$

$=81,000$

Thus, the value of the machine after two years will be

$\mathrm{Rs.}\space81,000$ .

Depreciation

$=\mathrm{Rs.}\space 100,000-\mathrm{Rs.}\space 81,000$

$=\mathrm{Rs.}\space19000$

Mohan purchased a house for Rs 30000 and its value is depreciating at the rate of 25% per year. Find the value (correct upto two decimal places) of the house after 3 years.

Given:

$\text{P}=\mathrm{Rs.}\space30000, \space \text{n = 3 years}, \space \text{R}=25\%$

Value of the house after three years

$=\mathrm{P}\left(1-\frac{\mathrm{R}}{100}\right)^{\mathrm{n}}$

$\Rightarrow 30,000\left(1-\dfrac{25}{100}\right)^{3}$

$=30,000(0.75)^{3}$

$=12,656.25$

Thus, the value of the house after three years will be

$\mathrm{Rs.}\space12,656.25$ .

Ms. Cherian purchased a boat for RsIf the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.

Price of a boat

$(P)=Rs.16000$

Depreciation rate

$=5\%$ per annum

$n=2$ years.

We know,

$A=P\left(1+\dfrac{R}{100}\right)^n$

Since it is depreciation we use

$A=P\left(1-\dfrac{R}{100}\right)^n$

$\therefore$

$A=16000\left(1-\dfrac{5}{100}\right)^2$

$=16000\left(\dfrac{95}{100}\right)^2$

$=16000(0.95)^2$

$=14440$

$\therefore$ Value of the boat after two years is

$Rs.14440$

The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in 2001

Given :

Rate

$(R)=4\%$

Years

$(n)=1999-2001=2$ years

$P=6760000$

Population of the city in

$2001=P\left(1+\dfrac{R}{100}\right)^n$

$=6760000\left(1+\dfrac{4}{100}\right)^2$

$=6760000(1.04)^2$

$=7311616$

$\therefore$ Population of the city in

$2001$ is

$7311616$

Three years ago, the population of a town wasIf the annual increase during three successive years be at the rate of 4%,5% and 3% respectively, find the present population.

3 years ago the population (P) was 50,000.

The increasing rate is 4 % , 5 % , 3 % , respectively.

The formula for this problem is,

$P \left(1 + \dfrac{r_1}{100} \right) \left(1 + \dfrac{r_2}{100} \right) \left(1 + \dfrac{r_3}{100} \right) \\$

where

$r_1 , r_2 , r_3$ are increasing rate respectively.

$\therefore$ Present population

$= 50000 \left(1 + \dfrac{4}{100} \right) \left(1 + \dfrac{5}{100} \right) \left(1 + \dfrac{3}{100} \right) \\ = 50000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{103}{100} \\ = 56,238$

The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years?

The production of mixi company in

$1996=8000$ mixies

Production growth rate in next

$2$ years

$=15\%$

Decrease rate in

$3^{rd}$ year

$=5\%$

Let the required production be

$x.$

$x=8000\left(1+\dfrac{15}{100}\right)\left(1+\dfrac{15}{100}\right)\left(1-\dfrac{5}{100}\right)$

$x=8000\times \left(1+\dfrac{15}{100}\right)\left(1+\dfrac{15}{100}\right)\left(1-\dfrac{5}{100}\right)$

$x=800\times 1.15\times 1.15\times 0.95$

$\therefore$

$x=10051$

Meera borrowed a sum of $Rs. 1000$ from Sita for two years. If the rate of interest is $10\%$ compounded annually, find the amount in rupees that Meera has to pay back.

1872871_1673896_ans_8fb0c706fbfb4c4b97cb4dc80b0059ca.jpg

The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in 1997

Given :

Rate

$(R)=4\%$

Years

$(n)=-2$ years

$P=6760000$

Population of the city in

$1997=P\left(1+\dfrac{R}{100}\right)^n$

$=6760000\left(1+\dfrac{4}{100}\right)^{(-2)}$

$=6760000(1.04)^{(-2)}$

$=6250000$

$\therefore$ Population of the city in

$1997$ is

$6250000$

The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43740, find its purchase price.

Given:

$\text{P}=\mathrm{Rs.}\space43740, \space \text{n = 3 years ago} \space \text{and}\space \text{R}=10\%$

Purchase price

$=\mathrm{P}\left(1-\frac{\mathrm{R}}{100}\right)^{-\mathrm{n}}$

$= 43,740\left(1-\frac{10}{100}\right)^{-3}$

$=43,740(0.90)^{-3}$

$=60,000$

Thus, the purchase price of the machine was

$\mathrm{Rs.}\space 60,000$ .

Fill in the blanks to make the statement true.
Interest on $Rs.3000$ at $10\%$ per annum for a period of $3$ years is .....

GIven,

$Principal = Rs.3000, Time = 3\,years, R = 10\%\,p.a.$

Now,

$Interest = \dfrac{P \times R \times T}{100}$

$\,T =$ Time period

$\,R\% =$ Rate of interest

$\,P =$ Principal

$\Rightarrow I = \dfrac{3000 \times 10 \times 3}{100}$

$\Rightarrow I = Rs.900$

Interest on

$Rs.3000$ at

$10\%$ per annum for a period of

$3$ years is

$Rs.900$ .

Three years before the population of a village was $10000$ . If at the end of each year, $20\%$ of the people migrated to a nearby town, what is the present population?

1902347_1710776_ans_41f8459953cd42fa9fe37c2f7d7c6a75.jpg

Bachhu Manjhi earns $Rs. 24000$ in $8$ months. At this rate, how much does he earn in one year?

$\begin{aligned}&\text { Money earned within } 8 \text { month }=\text { Rs. } 24000 \\&\text { Money earned within } 1 \text { month }=\frac{R s 24000}{8} \\& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ =\text { Rs } 3000 \\&\text { Money earned within } 12 \text { months = Number of months x Money earned in } 1 \text { month } \\& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ =12 \times \text { Rs } 3000 \\& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ =\text { Rs } 36,000\end{aligned}$

$\therefore \textbf{Bacchu Manjhi earned an amount of Rs 36,000 in 12 months}$

Anita deposits $Rs. 1000$ in a savings bank account. The bank pays interest at the rate of $5\%$ per annum. What amount can Anita get after one year?(in rupees)

Principal amount (P) = Rs 1000

Time period (T) = 1 year

Rate of interest (R) = 5 % p.a.

Simple Interest =

$\frac{P \times R \times T}{100}$

$\frac {1000 \times 5 \times 1}{100}$

= Rs 50

Total amount paid after 1 year = Principal amount + Interest

= Rs 1000 +Rs 50

= Rs 1050

The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9680, for how much was it purchased?

Given:

Present value of refrigerator is

$= Rs. 9680$

Depreciation rate is

$= 12\%$

Let the price of refrigerator 2 years ago be

$= Rs. x$

By using the formula,

$A = P \bigg(1 + \dfrac{R}{100}\bigg)^n$

$9680 = x \bigg(1 – \dfrac{12}{100}\bigg) \bigg(1 – \dfrac{12}{100}\bigg)$

$9680 = x \bigg(\dfrac{88}{100}\bigg) \bigg(\dfrac{88}{100}\bigg)$

$9680 = x (0.88) (0.88)$

$9680 = 0.7744x$

$x =\dfrac{ 9680}{0.7744} = 12500$

$\therefore$ The refrigerator was purchased for

$Rs. 12500$

Find the rate when:
$Principal = Rs. 5200, SI = Rs. 975$ and $time = 2 \dfrac{1}{2} \,years.$

Given:-

$P = Rs. 5200, SI = Rs. 975, t = 2 \dfrac{1}{2} \,years = 5 / 2$
We know that,

$R = (100 \times SI) / (P \times T)$

$= (100 \times 975 ) / (5200 / \times (5 / 2))$

$= (100 \times 975 \times 2) / (5200 \times 5)$

$= (1 \times 975 \times 2) / (52 \times 5)$

$= (195 \times 1) / (26 \times 1)$

$= (195 / 26)$

$= 7.5$

$= 7 \dfrac{1}{2} \% \,p.a.$

Find the rate when:
$Principal = Rs. 3560, amount = Rs. 4521.20$ and $time = 3 \,years.$

Given:-

$P = Rs. 3560 , amount = Rs. 4521.20, t = 3 \,years.$
We know that,

$SI = A - P$

$= 4521.20 - 3560$

$= Rs. 961.2$

$R = (100 \times SI) / (P \times T)$

$= (100 \times 961.20) / (3560 \times 3)$

$= (100 \times 96120) / (8250 \times 3 \times 100)$

$= (32040 / 3560)$

$= 9\% \,p.a.$

The cost of a T.V. set was quoted Rs 17000 at the beginning ofIn the beginning of 2000 the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning ofWhat was the cost of the T.V. set in 2001 ?

Cost of T.V at beginning of

$1999$ is

$= \mathrm{Rs.}\space 17000$

Hiked in price in the year

$2000$ is

$= 5\%$

Depreciation rate in the year

$2001$ is

$= 4\%$

Cost of the T.V.

$=\mathrm{P}\left(1+\dfrac{\mathrm{R}}{100}\right)\left(1-\frac{\mathrm{R}}{100}\right)$

$=17,000\left(1+\dfrac{5}{100}\right)\left(1-\dfrac{4}{100}\right)$

$=17,000(1.05)(0.96)$

$=17,136$

Thus, the cost the TV in 2001 was

$\mathrm{Rs.}\space17,136$ .

In the year $2001$ , the number of malaria patients admitted in the hospitals of a state was $4,375$ . Every year this number decreases by $8 \%$ . Find the number of patients in $2003$ .

As per given question,
Time period

$=2$ years
Suppose the number of patient in

$2003=A$
We know that,

$A=p\left(1-\dfrac{R}{100}\right)^T$
By using above formula,

$=4375 \left(1-\dfrac{8}{100}\right)^2=4375 \times \dfrac{23}{25} \times \dfrac{23}{25}=3703$

The time period after which the interest is added each time to from a new principle is called the ______.

The time period after which the interest is added each time to from a new principle is called the Conversion period

In $2007-08$ , the number of students appeared for Class $X$ examination was $1,05,332$ and in $2008-09$ , the number was $1,16,054$ . If $88,151$ students pass the examination in $277-08$ and $103804$ students in $2880-09$ . What is the increase or decrease in pass $\%$ in Class $X$ result?

Given, In

$2007-08,$
The number of student appeared for Class

$X$
examination

$=1,05,332$
In

$2008-09,$ the number

$=1,16,054$
Passed

$\%$ of student in

$2007-2008$

$=\dfrac{Number\ of\ student\ passed\ in\ 2007-2008}{Number\ of\ student\ appeared\ in\ 2007-2008} \times 100$

$=\dfrac{88151}{105332} \times 100=83.68 \%$

Passed

$\%$ of student in

$2008-2009$

$=\dfrac{Number\ of\ student\ passed\ in\ 2008-2009}{Number\ of\ student\ appeared\ in\ 2008-2009} \times 100$

$=\dfrac{103804}{116054} \times 100=89.44 \%$
So, increased

$\% =89.44-83.68=5.76 \%$

In $1975$ , the consumption of water for human use was about $3850\ cu.km/year$ . It increased to about $6000\ cu.km/year$ in the year $2000$ . Find the per cent increase in the consumption of water from $1975$ to $2000$ . Also, find the annual per cent increase in consumption (assuming water consumption increases uniformly).

As per given question,
Increase in consumption of water in

$1975\ to\ 2000$

$=6000-3850=2150 cu\ km/yr$
In

$\% \dfrac{2150}{3850} \times 100=55.84 \%$
Total increased water consumption in

$25$ years,

$\dfrac{2150}{25}=86 cu\ km/yr$
so, in percentage

$=\dfrac{86}{3850} \times 100=2.23 \%$

Give the principle =40,000, p.a. compounded annually. Find the interest if period is one year.

Given that compound interest is to be calculated for

$1$ year:

$P=40,000, r=8,n=1$

$A=P\left(1+\dfrac{r}{100}\right)^n =40000 \left(1+\dfrac{8}{100}\right)^1 =40000 \times \dfrac{108}{100}=Rs.\ 43200$

Compound Interest

$=A-P$

$=43200-40000=Rs.\ 3200$

Fill in the blanks to make the statement true:
$Interest = \dfrac{P \times R \times T}{100},$ where
$T$ is ...........
$R\%$ is .............. and
$P$ is ......................

$Interest = \dfrac{P \times R \times T}{100},$ then

$\,T =$ Time period

$\,R\% =$ Rate of interest

$\,P =$ Principal

Find the principal when:
final amount is $Rs.2420$ at $4\%$ per annum for $2\dfrac{1}{2}$ years.

$Amount (A) = Rs.2420$

$Rate (R) = 4\%$ p.a.

$Time (T) = 2\dfrac{1}{2} = \dfrac{5}{2}$ years
Let principal

$= Rs.100$
Then

$S.I. = \dfrac{PRT}{100} = \dfrac{100 \times 4 \times 5}{100 \times 2} = Rs.10$
Amount = P + S.I.

$= Rs.100 + Rs.10 = Rs.110$
If amount is

$Rs.110$ then principal

$= Rs.100$
If amount is

$Rs.2420,$ then principal

$= \dfrac{100 \times 2420}{110}$

$= Rs.2200$

Find the rate of interest when:
$Rs.1200$ amounts to $Rs.1320$ in $2$ years

$Principal (P) = Rs.1200$

$Amount (A) = Rs.1320$

$S.I. = A - P = Rs.1320 - Rs.1200 = Rs.120$

$Time (T) = 2$ years

$Rate\% = \dfrac{S.I. \times 100}{P \times T}$

$= \dfrac{120 \times 100}{1200 \times 2} = 5\%$ p.a.

Find the rate of interest when Rs 800 fetches Rs 130 as a simple interest in 2 years 6 months.

Here,

$P = Rs\ 800$

$T = 2$ years

$6$ months

$=2\dfrac{6}{12}$ years

$=2\dfrac{1}{2}$ years

$= \dfrac52$ years

Hence,

$R = \dfrac{I × 100}{P × T}\\= \dfrac{130 × 100}{800 × \cfrac{5}2} \%$ p.a.

On simplification, we get,

$R= \dfrac{130 × 100 × 2}{800 × 5}\%$ p.a.

$= \dfrac{130 × 2}{40}\%$ p.a.

$= 6.5\%$ p.a.

Therefore, the required rate of interest is

$6.5\%$ p.a.

A new computer costs $Rs\ 1,00,000$ . The depreciation of computer is very high as new models with better technology advantages as coming into the market. The depreciation is as high as $50 \%$ every year. How much will the cost of computer be after two years?

As per question,
Suppose the cost of computer after two years

$=x$
We know that,

$A=P\left(1+\dfrac{R}{100}\right)^T$

$=100000\left(1+\dfrac{50}{100}\right)^2=100000 \times \dfrac{1}{2} \times \dfrac{1}{2}=25000$

Find the principal when:
simple interest at $7\dfrac{1}{2}\%$ per annum for $2$ years $4$ months is $Rs.2730.$

$S.I. = Rs.2730$

$Rate = 7\dfrac{1}{2}\% = \dfrac{15}{2}\%$ p.a.
Time (T) = 2 years 4 months

$= 2\dfrac{1}{3} = \dfrac{7}{3}$ years

$\therefore Principal = \dfrac{S.I. \times 100}{R \times T}$

$= \dfrac{2730 \times 100 \times 2 \times 3}{7 \times 15} = Rs.15600$

The population of a town was decreasing every year due to migration, poverty and unemployment. The present population of the town is $6,31,680$ . Last year the migration was $4\ \%$ and the year before last, it was $6\ \%$ . What was the population two years ago?

As per question,
Suppose, two year ago the population

$=P$
We know that,

$A=P\left(1+\dfrac{R}{100}\right) \left(1-\dfrac{R}{100}\right)$
So,

$\Rightarrow 631680 =P\left(1-\dfrac{4}{100}\right) \left(1-\dfrac{6}{100}\right)$

$\Rightarrow 631680 =P \times \dfrac{24}{25} \times \dfrac{47}{50}$

$\Rightarrow P=\dfrac{631680 \times 25 \times 50}{24 \times 47}=700000$

At what rate will $Rs.7200$ fetch a simple interest of $Rs.3024$ in $4$ years ?

Principal (P)

$= Rs.7200$

$S.I. = Rs.3024$

$Time = 4$ years

$\therefore Rate = \dfrac{S.I. \times 100}{P \times T} \\= \dfrac{3024 \times 100}{7200 \times 4}$

$= \dfrac{21}{2} \\= 10.5\%$

Find the rate of interest when:
$Rs.300$ amounts to $Rs.400$ in $2$ years.

$Principal (P) = Rs.300$

$Amount (A) = Rs.400$

$\therefore S.I. = A - P = Rs.400 - Rs.300 = Rs.100$

$Time = 2 \,years$

$\therefore Rate\% = \dfrac{S.I. \times 100}{P \times T} = \dfrac{100 \times 100}{300 \times 2}$

$= \dfrac{50}{3}\% = 16 \dfrac{2}{3}\%$ p.a.

Find the principal when:
final amount is $Rs.4500$ at $20\%$ per annum for $5$ years

$Amount (A) = Rs.4500$

$Rate (R) = 20\%$

$Time (T) = 5 \,years$
Let Principal (P)

$= Rs.100$
Then

$S.I. = \dfrac{PRT}{100}$

$= \dfrac{100 \times 20 \times 5}{100} = Rs.100$
and amount = P + S.I.

$= Rs.100 + Rs.100 = Rs.200$
If amount is

$Rs.200$ then Principal

$= Rs.100$
and if amount is

$Rs.4500$ then principal

$= \dfrac{100}{200} \times 4500$

$= Rs.2250$

Find the principal when:
simple interest at $16\%$ per annum for $2\dfrac{1}{2}$ years is $Rs.3840$

$S.I. = Rs.3840$

$Rate (R) = 16\% \,p.a.$

$Time (T) = 2\dfrac{1}{2} = \dfrac{5}{2} \,years$

$\therefore Principal = \dfrac{S.I. \times 100}{R \times T}$

$= \dfrac{3840 \times 100 \times 2}{16 \times 5} = Rs.9600$

At what rate percent will Rs 2000 amount to Rs 2315.25 in 3 years at compound interest?

Given

Principal (P) = Rs 2000

Amount (A) = Rs 2315.25

Period (n) = 3 years

Let the rate of interest be r% p.a.

A / P = {1 + (r / 100)} $^n$

2315.25 / 2000 = {1 + (r / 100)} $^3$

{1 + (r / 100)} $^3$ = (231525) / (100 × 2000)

On calculating, we get,

{1 + (r / 100)} $^3$ = 9261 / 8000

{1 + (r / 100)} $^3$ = (21 / 20) $^3$

We get,

1 + (r / 100) = 21 / 20

r / 100 = (21 / 20) – 1

r / 100 = 1 / 20

We get,

r = 100 / 20

r = 5

Therefore, rate of interest = 5% p.a.

Calculate the amount and the compound interest on Rs 5000 in 2 years when the rate of interest for successive years is 6% and 8% respectively.

Principal for the first year = Rs 5000

Rate of interest = 6% p.a.

Interest for the first year = Rs (5000 × 6 × 1) / 100

= Rs 50 × 6

= Rs 300

Amount at the end of first year = Rs 5000 + Rs 300

= Rs 5300

Principal for the second year = Rs 5300

Rate of interest = 8% p.a.

Interest for the second year = Rs (5300 × 8 × 1) / 100

= Rs 53 × 8

We get,

= Rs 424

Amount for the second year = Rs 5300 + Rs 424

= Rs 5724

Compound interest for two years = Final amount – (original) Principal

= Rs 5724 – Rs 5000

We get,

= Rs 724

The cost of a refrigerator isIts value depreciates at the rate of 5% ever year. Find the total depreciation in its value at the end of 2 years.

It is given that
Present value (P) = 9000
Rate of depreciation (r) =5% p.a.
Period (n) = 2 years
We know that
Value after 2 years =

$P( 1 - r/100)^{n}$
Substituting the values
=

$9000(1 - 5/100)^{2}$
By further calculation
=

$9000 \times 19/20 \times 19/20$
= 8122.50
So the total depreciation =

$9000- 8122.50$

$= 877.50$

Find the compound interest on Rs. 3125 for 3 years if the rates of interest for the first, second and third year are respectively 4 $\%$ , 5 $\%$ and 6 $\%$ per annum.

Given that,

The rate of interest for

$3$ years on

$Rs.\ 3125$ are respectively

$4\%,\ 5\%$ and

$6\%$ per annum.

To find out,

The compound interest on the sum after

$3$ years.

We can calculate the compound interest by using the concept of simple interest for each year, adding the interest of the previous year to the principal amount of the next.

We know that,

$Interest=\dfrac{P\times R\times T}{100}$

Let

$I_1,\ I_2,\ I_3$ be the interest amounts for the three years respectively.

For the first year, we have:

$P=Rs.\ 3125,\ R=4\%,\ T=1\ year$

Hence,

$I_1=\dfrac{3125\times 4\times 1}{100}$

$=\dfrac{12500}{100}$

$=Rs.\ 125$

$\therefore \ I_1=Rs.\ 125$

For second year,

$P=Rs.\ 3125+125=Rs.\ 3250,\ R=5\%,\ T=1\ year$

Hence,

$I_2=\dfrac{3250\times 5\times 1}{100}$

$=\dfrac{16250}{100}$

$=Rs.\ 162.5$

$\therefore \ I_2=Rs.\ 162.5$

Now, for third year,

$P=Rs.\ 3250+162.5=Rs.\ 3412.5,\ R=6\%,\ T=1\ year$

Hence,

$I_3=\dfrac{3412.5\times 6\times 1}{100}$

$=\dfrac{20475}{100}$

$=Rs.\ 204.75$

$\therefore \ I_3=Rs.\ 204.75$

So, the total interest amount after three years

$=I_1+I_2+I_3$

$=Rs.\ 125+162.5+204.75$

$=Rs.\ 492.25$

Hence, the required compound interest for

$3$ years is

$Rs.\ 492.25$ .

Anil borrowed Rs 18000 from Rakesh at 8% per annum simple interest for 2 years. If Anil had borrowed this sum at 8% per annum compound interest, what extra amount would he has to pay?

Given

Money borrowed (P) = Rs 18000

Rate (R) = 8% p.a.

Time (n) = 2 years

Simple Interest = PRT / 100

= Rs (18000 × 8 × 2) / 100

= Rs 2880

In case of compound interest

A = P {1 + (R / 100)} $^n$

= Rs 18000 {1 + (8 / 100)} $^2$

= Rs 18000 × (27 / 25) $^2$

= Rs 18000 × (27 / 25) × (27 / 25)

We get,

= Rs 104976 / 5

= Rs 20995.20

Hence,

Compound interest = Amount – Principal

= Rs 20995.20 – Rs 18000

= Rs 2995.20

Now,

Difference between compound interest and simple interest

= Rs 2995.20 – Rs 2880

= Rs 115.20

Aryaman invested Rs 10000 in a company, he would be paid interest at 7% per annum compounded annually. Find
(i) the amount received by him at the end of 2 years

Given

Investment to a company $P = Rs 10000$

Rate of interest $R = 7 \%$ p.a.

Period $n = 2$ years

Hence,

Amount $A = P \left[1 + \left(\dfrac{R} {100}\right)\right]^n$

$=$ Rs 10000 {1 + (7 / 100)} $^2$

$=$ Rs 10000 × (107 / 100) × (107 / 100)

On simplification, we get,

$=$ Rs 11449

What sum of money will amount to Rs 9261 in 3 years at 5% per annum compound interest?

Given

Amount (A) = Rs 9261

Rate of interest = 5% p.a.

Time (T) = 3 years

Principal (P) =?

A = P {1 + (r / 100)} $^t$

9261 = P {1 + (5 / 100)} $^3$

We get,

9261 = P (21 / 20) $^3$

P = (9261 × 20 × 20 × 20) / (21 × 21 × 21)

On simplification, we get,

= Rs 8000

Therefore, the sum of money = Rs 8000

The population of a city increase each year by 4% of what it had been at the beginning of two years of each year . If its present population is 6760000, find:
(i) its population 2 years hence
(ii) its population 2 years ago .

It is given that
Present population = 6760000
Increase percent = 4% p.a.
(i) We know that
Population 2 years hence =

$P(1 + r/100)^{n}$
Substituting the values
=

$6760000(1 + 4/100)^{2}$
By further calculation
=

$6760000 \times 26/25 \times 26/25$
=7311616

(ii) We know that A = 6760000
Population 2 years ago P=

$A + (1 + r/100)^{2}$
Substituting the values
=

$6760000 + (1 + 4/100)^{2}$
By further calculation
=

$6760000 + (26/25)^{2}$
=

$6760000 \times 25/26 \times 25/26$
= 6250000

The present population of a town isIf the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years ?

It is given that
Present population (P) = 15625
Rate of increase (r) = 4% p.a.
Period (n) = 3 years
We know that
Population after 3 years =

$P(1 + r/100)^{n}$
Substituting the values
=

$15625( 1 + 4/100)^{3}$
By further calculation
=

$15625 \times 26/25 \times 26/25 \times 26/25$

$= 17576$
So the increase =

$17576 - 15625 = 1951$

The present population of a town isIts population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of 2 years.

We know that
Population after 2 years = Present population

$\times (1 + r/100)^{n}$
Here the present population = 200000
Population after first year =

$200000 \times ( 1 + 10/100)^{1}$
By further calculation
=

$200000 \times 11/10$

$= 220000$
Population after two years =

$220000 \times ( 1 + 15/100)^{1}$
By further calculation
=

$220000 \times 23/20$

$= 253000$

A farmer increases his output of wheat in his farm every year by 8%. This year produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?

It is given that
Present production of wheat = 2187 quintals
Increase in production = 8% p.a.
We know that
Production of wheat 2 years ago =

$A \div (1 + r/100)^{n}$
Substituting the values
=

$2187 \div (27/25)^{2}$
By further calculation
=

$2187 \div (27/25)^{2}$
So we get
=

$2187 \times 25/27 \times 25/27$
=

$1875$ quintals

The value of a property decreases every year at the rate of 5%. If its present value is 411540, what was its value three years ago?

It is given that
Present value of property = 411540
Rate of decrease = 5% p.a.
We know that
Value of property 3 years ago =

$A \div (1 - r/100)^{n}$
Substituting the values
=

$411540 \div (1 - 5/100)^{3}$
By further calculation
=

$411540 \div (19/20)^{3}$
So we get
=

$411540 \times 20/19 \times 20/19 \times 20/19$
= 480000

the cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If the present value of the car is 315600 find:
(i) value of car 2 years ago
(ii) value of cars after 3 years

It is given that
Present value of car = 315600
Rate of depreciation (r) = 20%
(i) We know that
Value of car 2 years ago =

$A \div (1 - r/100)^{n}$
Substituting the values
=

$315600 \div (1 - 20/100)^{2}$
By further calculation
=

$315600 \times 5/4 \times 5/4$
= 493125
(ii) We know that
Value of car after 3 years =

$315600 \times( 1 - 20/100)^{3}$
By further calculation
=

$315600 \times 4/5 \times 4/5 \times 4/5$

$161587.20$

The population of a city isIf the annual birth rate is 9.2% and the anuual death rate is 1.7% find the population after 3 years.

It is given that
Birth rate = 9.2%
Death rate = 1.7%
So the net growth rate = 9.2 - 1.7 = 7.5%
Present population (P) = 320000
Period (N) = 3 years
We know that
Population after 3 years (A) =

$P(1 + r/100)^{n}$
Substituting the values
=

$320000(1 + 7.5/100)^{3}$
By further calculation
=

$320000(1 + 3/40)^{3}$
=

$320000 \times (43/40)^{3}$
So we get
=

$320000 \times 43/40 \times 43/30 \times 43/40$
=397535

The value of a machine worth 500000 is depreciating at the rate of 10% every year . In how many years will its value be reduced to 364500?

It is given that
Present value = 500000
Reduced value = 364500
Rate of depreciation = 10% p.a.
Consider n years as the period
We know that
A/P =

$(1 - r/100)^{n}$
Substituting the values

$364500/500000 = (1 - 10/100)^{n}$
By further calculation

$(9/10)^{n} = 729/1000 = (9/10)^{3}$
So we get
n=3
Therefore, the period in which its value be reduced to 364500 is 3 years.

Ahmed purchased an old scooter forIf the cost of the scooter after 2 years depreciates to 14440, find the rate of depreciation.

It is given that
Present value = 16000
Value after 2 years = 14440
Consider r% p.a. as the rate of depreciation
We know that

$A/P = (1 - r/100)^{n}$
Substituting the values

$14440/16000 = (1 - r/100)^{2}$
By further calculation

$361/400 = (1 - r/100)^{2}$

$(19/20)^{2} = (1 - r/100)^{2}$
We can write it as

$1- r/100 = 19/20$
So we get

$r/100 = 1 - 19/20 = 1/20$
By cross multiplication

$r= 1/20 \times 100 = 5%$
Hence, the rate of depreciation is 5%.

the value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?

It is given that
Original price of the property (P) = 100
Rate of increase (r) = 25% p.a.
Period (n) = 3 years
We know that
Increased value after 3 years =

$P(1 + r/100)^{n}$
Substituting the values
=

$100(1 + 25/100)^{3}$
By further calculation
=

$100 \times 5/4 \times 5/4 \times 5/4$
= 3125/16
Here
Increased value = 3125/16 - 100
Taking LCM
=

$(3125 - 1600)/16$
=

$1525/16$
So the percent increase after 3 years =

$1525/16 = 955/16%$

Afzal purchased an old motorbike forIf the value of the motorbike after 2 years is 14440, find the rate of depreciation.

It is given that
CP of an old motorbike = 16000
Price after 2 years = 14440
Consider r% as the rate of depreciation
We know that
A/P =

$(1 - r/100)^{n}$
Substituting the values

$14440/16000 = (1 - r/100)^{2}$
By further calculation

$361/400 = (1 - r/100)^{2}$

$(19/20)^{2} = (1 - r/100)^{2}$
So we get

$19/20 = 1- r/100$

$r/100 = 1- 19/20 = (20 - 19)/20 = 1/20$
By cross multiplication

$r= 100/20 =5$
Hence, the rate of depreciation is 5%

Dinesh purchased a scooter for $Rs.24000$ . The value of the scooter is depreciating at the rate of $5\%$ per annum. Calculate its value after $3$ years.

It is given that ,
Present value of scooter

$(P) = 24000$
Rate of depreciation

$(r) = 5%$
Period

$(n) = 3$ years
Formula for depriciated amount =

$P(1 - \dfrac{r}{100})^{n}$
Substituting the values,

$= 24000 (1 - \dfrac{5}{100})^{3}$

$=24000 (1 - \dfrac{1}{20})^{3}$

$=24000 \times (\dfrac{19}{20})^3$

$=Rs.20577$

A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-Find the annual rate of growth of production of cars .

It is given that
Production of cars in 2011-2012 = 80000
Production of cars in 2014-2015= 92610
Period (n) = 3 years
Consider r% as the rate of increase
We know that
A/P =

$(1 + r/100)^{n}$
Substituting the values

$92610/80000 = (1 + r/100)^{3}$
By further calculation

$(21/20)^{3} = (1 + r/100)^{3}$
We can write it as

$1 + r/100 = 21/20$

$r/100 = 21/20 - 1= 1/20$
By cross multiplication

$r= 1/20 \times 100 = 5$
Hence, the annual rate of growth of production of cars is 5% p.a.

P and Q invest Rs 36,000 and Rs 25,000 respectively at the same rate of interest per year. If at the end of 4 years , P gets Rs 3,080 more interest than Q , find the rate of interest.

It is given that

$P's$ Investment

$=P_{1} = Rs. 36,000$ and

$Q's$ investment

$=P_{2} = Rs. 25,000$

$T = 4$ years

Let us consider the rate of investment

$R= x \%$

So we get

$P's$ Interest

$(S.I)_1 = (P_1\times R\times T)/100$

Substituting the values

$=(36000\times x\times 4)/100 = Rs. 1440x$

$Q's$ interest

$(S.I)_2=(P_2\times R\times T)/100$

Substituting the values

$=(25000\times x\times 4)/100 = Rs. 1000x$

The difference in their interest is calculated as

$(S.I)_1-(S.I)_2 = 1440x - 1000x = Rs. 440x$

The difference given is

$= Rs. 3080$

So we get

$440x = 3080 \Rightarrow x = 3080/440 \Rightarrow x = 7$

So the rate of interest

$R= 7 \%$ p.a.

The interest paid for the second year.

For first year:

Principal

$= Rs. 8000$

Rate

$= 7\%$

and Time

$= 1$ year

Interest

$= Rs.\ \dfrac{8,000 \times 7 \times 1}{100}$

$= Rs. 560$

Amount

$= Rs. 8,000 + Rs. 560$

$= Rs. 8,560$

Money returned

$= Rs. 3,560$

Balanced money for second year

$= Rs. 8,560 - Rs .3,560$

$= Rs. 5,000$

Find the amount and the compound interest on RS. 120000 in 3 years at 5%: interest compounded annualy

Given; P= Rs 12000; n = 3 years and r = 5%
Amount =

$p\left(1 + \dfrac{r}{100}\right)^n$
=

$1200\left(1 + \dfrac{5}{100}\right)^n$
=

$12000 \left(\dfrac{21}{20}\right)^3$
=

$Rs 13,891.50$

$\therefore C.I = Rs13891.50 - Rs12,000$

$= Rs1,891.50$

At what rate percent per year will a sum double itself in $=6\dfrac 14$ years?

Let

$P=Rs. 100$
Amount

$=2\times Rs. 100=Rs. 100$
Interest

$=A-P$

$=Rs. 200-Rs. 100=Rs. 100$

$T=6\dfrac 14$ years

$=\dfrac {25}{4}$ years

$R=\dfrac {100\times I}{P\times T}=\dfrac {100\times 100}{100\times \dfrac {25}{10}}\% =\dfrac {100\times 100}{100}\times \dfrac {4}{25}=16\%$

What sum of money will amount to Rs.5445 in 2 years at 10% per annum compound interest?

Given Amount = Rs5445; n = 2years and r = 10%

$\therefore A = P\left(1 + \dfrac{r}{100} \right)^n$

$\Rightarrow 5445 =P \left(1 + \dfrac{10}{100} \right)^2$

$\Rightarrow 5445 = P \left(\dfrac{11}{10} \right)^2$

$\Rightarrow 5445 \left(\dfrac{10}{11} \right)^2=P$

$P=Rs 4500$

The rate of depreciation.

Difference between depreciation in value between the first and second years
Rs.4,000 - Rs.3,600 = Rs.400

$\Rightarrow$ Depreciation of one year on Rs.4,000 = Rs. 400

$\Rightarrow$ Rate of depreciation

$= \frac{400}{4000} \times 100 \% = 10 \%$

The original cost of the machine.

Difference between depreciation in value between the first and second years
Rs.4,000 - Rs.3,600 = Rs.400

$\Rightarrow$ Depreciation of one year on Rs.4,000 = Rs. 400

$\Rightarrow$ Rate of depreciation

$= \frac{400}{4000} \times 100 \% = 10 \%$

Let Rs.100 be the original cost of the machine.
Depreciation during 1st year = 10% of Rs.100 = Rs.10
When the values depreciates by Rs.10 during the 1st year. Original cost = Rs.100

$\Rightarrow$ When the depreciation during 1st year = Rs.4,000.
Original cost

$= \frac{100}{10} \times 4000 = 40000$
The original cost of the machine is Rs.40,000.

A sum of money is lent for 5 years at R% simple interest per annum. If the interest earned be one. fourth of the money lent, find the value of R.

Consider the sum

$P = Rs 100$

We know that

$(S.I)= 1/4\times 100= Rs. 25$

$T = 5$ years

Here, Rate

$R= (S.I\times 100)/(P\times T)$

Substituting the values

$=(25\times 100)/(100\times 5)$

$=5$

So we get

$R= 5 \%$

On what sum of money will the compound interest for 2 years for 2 years at 5 per cent per annum amount to Rs 768.75?

Given : C.I = Rs768.75; n = 2 years and r = 5%

$\therefore A = P\left( 1 + \dfrac{r}{100}\right)^n$

$\Rightarrow A = P\left( 1 + \dfrac{5}{100}\right)$

$\Rightarrow A = P\left( \dfrac{21}{20}\right)^2 = \dfrac{441}{400}P$

$\therefore A - P = C.I$

$\Rightarrow \dfrac{441}{400}p - p = Rs 768.75$

$\Rightarrow \dfrac{41}{400}p = Rs 768.75$

$\Rightarrow \dfrac{768.75 \times 400}{41} = Rs 7,500$

$\Rightarrow P = Rs. 7500$

The sum due to Ramesh at the end of the year.

For 1st year
P = Rs 12,800; R = 10% and T = 1 year
Interest = Rs

$\frac{12,800 \times 10 \times 1}{100} =$ Rs 1,280
Amount = Rs 12,800 + Rs 1,280 = Rs 14,080

Rs.6,000 in 3 years at 5% per year.

For 1

$^{st}$ year,
P = Rs 6,000; R = 5% and T = 1 year

$\therefore$ Interest = Rs

$\frac{6000 \times 5 \times 1}{100} =$ Rs 300
And, Amount = Rs (6,000 + 300) = Rs 6,300

For 2

$^{nd}$ year,
P = Rs 6,300; R = 5% and T = 1 year

$\therefore$ Interest = Rs

$\frac{6,300 \times 5 \times 1}{100} =$ Rs 315
And, Amount = Rs (6,300 + 315) = Rs 6,615

For 3

$^{rd}$ year,
P = Rs 6,615; R = 5% and T = 1 year

$\therefore$ Interest = Rs

$\frac{6,615 \times 5 \times 1}{100} =$ Rs 330.75
And, Amount = Rs (6,615 + 330.75) = Rs 6,945.75

CI accured = Final amount - Initial principal
= Rs (6,945.75 - 6,000)
= Rs 945.75

A man borrows Rs.10,000 at 10% compound interest yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year?

For 1st year
P = Rs 10,000; R = 10% and T = 1 year
Interest = Rs

$\frac{10,000 \times 10 \times 1}{100}$ = Rs 1,000
Amount at the end of 1st year = Rs 10,000 + Rs 1,000 = Rs 11,000
Money paid at the end of 1st year = 20% of Rs 11,000 = Rs 2,200

$\therefore$ Principal for 2nd year = Rs 11,000 = Rs 2,200 = Rs 8,800

For 2nd year
P = Rs 8,800; R =10% and T = 1 year
Interest = Rs

$\frac{8,800 \times 10 \times 1}{100} =$ Rs 880
Amount at the end of 2nd year = Rs 8,800 + Rs 880 = Rs 9,680
Money paid at the end of 2nd year = 20% of Rs 9,680 = Rs 1,936

$\therefore$ Principal for 3rd year = Rs 9,680 - Rs 1,936 = Rs 7,744.

Meenal lends Rs.75,000 at C.I. for 3 years. If the rate of interest for the first two years is 15% per year and for the third year it is 16% calculate the sum Meenal will get at the end of the third year.

Interest for the first year

$= \dfrac{P \times R \times T}{100}$

$= \dfrac{75,000 \times 15 \times 1}{100}$

$=$ Rs.

$11,250$
Amount for the first year = Rs.75,000 + Rs. 3,000 = Rs. 86,250
Interest for the second year

$= \dfrac{P \times R \times T}{100}$

$= \dfrac{86,250 \times 15 \times 1}{100}$

$=$ Rs.

$12,937.5$
Amount for the second year = Rs.86,250 + Rs. 12,937 = Rs. 99,187.5
Interest for the third year

$= \dfrac{P \times R \times T}{100}$

$= \dfrac{99,187.5 \times 16 \times 1}{100}$

$=$ Rs.

$15,870$
Amount for the third year = Rs.99,187.5 + Rs. 15,870 = Rs. 1,15,057.5
Hence, the sum Meenal will get at the end of the third year is Rs.1,15,057.5.

Find the sum on which the compound interest for 3 years at 1% per annum amounts to Rs. 1655.

Given : C.I = Rs 1,655; n = 3 years and r = 10%

$\therefore A = P\left( 1 + \dfrac{r}{100}\right)^n$

$\Rightarrow A = P\left( 1 + \dfrac{10}{100}\right)^{3}$

$\Rightarrow A = P\left( \dfrac{11}{10}\right)^2 = \dfrac{1,331}{1,000}P$

$\therefore A - P = C.I$

$\Rightarrow \dfrac{1,331}{1,000}p - p = Rs 1,655$

$\Rightarrow \dfrac{331}{1,000}p = Rs 1,655$

$\Rightarrow \dfrac{1,655 \times 1,000}{331} = Rs 5,000$

$\Rightarrow P = Rs. 5000$

$\text{Rs. }16,000$ is invested at $5\%$ compound interest compounded per annum. Use the table given below to find the amount in $4$ years.

From the table it can be seen that amount after

$4$ years would be

$\text{Rs }19448.1$

Find the compound interest on Rs.4,000 accrued in three years, when the rate of interest is 8% for the first year and 10% per year for second and third year.

Interest for the first year

$= \dfrac{P \times R \times T}{100}$

$= \dfrac{4,000 \times 8 \times 1}{100}$

$=$ Rs.

$320$
Amount for the first year = Rs.4,000 + Rs. 320 = Rs. 4,320
Interest for the second year

$= \dfrac{P \times R \times T}{100}$

$= \dfrac{4,320 \times 10 \times 1}{100}$

$=$ Rs.

$432$
Amount for the second year = Rs.4,320 + Rs. 432 = Rs. 4,752
Interest for the third year

$= \dfrac{P \times R \times T}{100}$

$= \dfrac{4,752 \times 10 \times 1}{100}$

$=$ Rs.

$475.20$
Amount for the third year = Rs.4,752 + Rs. 475.20 = Rs. 5,227.20
So, the compound interest = Rs. 5,227.20 - Rs.4,000 = Rs.1,227.20

At what rate per cent will a sum of Rs. 4000 yield Rs. 1324 as compound interest in 3 years?

Given: P = Rs 4,000, C.I. = Rs 1,324 and n= 3 years

Now, A = P + I

$A = Rs. \left( 4,000 + 1,324\right) = Rs. 5,324$

$A = P \left( 1 + \dfrac{r} {100}\right)^{3}$

$\Rightarrow 5324 = 4000 \left( 1 + \dfrac{r} {100}\right)^{3}$

$\Rightarrow \dfrac{5324} {4000} = \left( 1 + \dfrac{r} {100}\right)^{3}$

$\Rightarrow \dfrac{1331} {1000} = \left( 1 + \dfrac{r} {100}\right)^{3}$

$\Rightarrow \left( 1 + \dfrac{r} {100}\right)^{3} - \dfrac{1331} {1000} - \left(\dfrac{11} {10}\right)^{3}$

$1 + \dfrac{r} {100} = \dfrac{11} {10}$

$\Rightarrow \dfrac{r} {100} = \dfrac{11} {10} - 1 = \dfrac{1} {10}$

$\Rightarrow r = \dfrac{100} {100} = 10$ %

Thus, the rate of interest is 10%.

Its cost at the end of the third year.

Difference between depreciation in value between the first and second years
Rs.4,000 - Rs.3,600 = Rs.400

$\Rightarrow$ Depreciation of one year on Rs.4,000 = Rs. 400

$\Rightarrow$ Rate of depreciation

$= \frac{400}{4000} \times 100 \% = 10 \%$

Let Rs.100 be the original cost of the machine.
Depreciation during 1st year = 10% of Rs.100 = Rs.10
When the values depreciates by Rs.10 during the 1st year. Original cost = Rs.100

$\Rightarrow$ When the depreciation during 1st year = Rs.4,000.
Original cost

$= \frac{100}{10} \times 4000 = 40000$

The original cost of the machine is Rs.40,000.

Total depreciation during all three years
= Depreciation in value during (1

$^{st}$ + 2

$^{nd}$ year + 3

$^{rd}$ year)
= Rs.4,000 + Rs.3,600 + 10% of (Rs.40,000 - Rs.7,600)
= Rs.4,000 + Rs.3,600 + Rs.3,240
= Rs.10,840

The cost of the machine at the end of the third year
= Rs.40,000 - Rs.10,840 = Rs.29,160

Find the sum, invested at 10% compounded annually, on which the interest for the third year exceeds the interest of the first year by Rs.252.

Let the sum of money be Rs 100
Rate of interest = 10% p.a
Interest at the end of 1

$^{st}$ year = 10% of Rs 100 = Rs 10
Amount at the end of 1

$^{st}$ year = Rs 100 + Rs 10 = Rs 110

Interest at the end of 2

$^{nd}$ year = 10% of Rs 110 = Rs 11
Amount at the end of 2

$^{nd}$ year = Rs 110 + Rs 11 = Rs 121

Interest at the end of 3

$^{rd}$ year = 10% of Rs 121 = Rs 12.10

$\therefore$ Difference between interest of 3rd year and 1st year = Rs 12.10 - Rs 10 = Rs 2.10
When difference is Rs 2.10, principal is Rs 100
When difference is Rs.252, principal =

$\frac{100 \times 252}{2.10} =$ Rs 12,000.

At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years?

$Amount = P\left( 1 + \dfrac{r}{100}\right)^{n}$
Let Principal = Rs y
The Amount = Rs. 1.44y
n = 2 years

$A = P\left( 1 + \dfrac{r}{100}\right)^{n}$

$\Rightarrow 1.44 y = y\left( 1 + \dfrac{r}{100}\right)^{2}$

$\Rightarrow \dfrac{1.44y} {y} = \left( 1 + \dfrac{r}{100}\right)^{2}$

$\Rightarrow \dfrac{36} {25} = \left( 1 + \dfrac{r}{100}\right)^{2}$

$\Rightarrow\left(\dfrac{6} {5}\right)^{2} = \left( 1 + \dfrac{r}{100}\right)^{2}$
On comparing,

$\Rightarrow\dfrac{6} {5} = 1 + \dfrac{r}{100}$
On solving, we get

$r = 20$ %

the interest for the first year.

For

$1^{st}$ year
P = Rs.5600
R = 14%
T = 1 year
t

$= \dfrac{5600 \times 14 \times 1}{100} =$ Rs.784

A sum of money, invested at compound interest, amounts to Rs. 19360 in 2 years and to Rs. 23425.60 in 4 years. Find the rate per cent and the original sum of money.

(For 2 years) A = Rs. 19360
T = 2 years
Let P = X

$x \left(1 + \dfrac{R} {100} \right)^{2} = 19360$ ......(1)
A (for 4 years) = Rs. 23425.60

$x \left(1 + \dfrac{R} {100} \right)^{4} = 23425.60$ ......(2)

$(2) \div (1)$

$\left(1 + \dfrac{R} {100} \right)^{2} = \dfrac{2342560} {1936000}$

$\left(1 + \dfrac{R} {100} \right)^{2} = \dfrac{14641} {12100}$

$\left(1 + \dfrac{R} {100} \right)^{2} = \left(\dfrac{121} {110}\right)^{2}$

$1 + \dfrac{R} {100} = \dfrac{121} {110}$

$\dfrac{R} {100} = \dfrac{121} {110} - 1$

$R = 10\%$

From (1)

$x \left(1 + \dfrac{10} {100} \right)^{2} = 19360$

$x = \dfrac{19360 \times 10 \times 10} {11 \times 11}$

$x = Rs. 16000$
Thus, sum = Rs. 16000

A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by Rs 96, find the sum of money.

Let money be Rs 100

For 1

$^{st}$ year

P = Rs.100; R = 8% andT = 1 year

Interest for the first year

$= \dfrac{100 \times 8 \times 1}{100} =$ Rs. 8

Amount = Rs 100 + Rs.8 = Rs.108

Let money be Rs 100

For 2

$^{nd}$ year

P = Rs.108; R = 8% and T = 1 year

Interest for the second year

$= \dfrac{108 \times 8 \times 1}{100} =$ Rs. 8.64

Difference between the interests for the second and first year = Rs 8.64 - Rs 8 = Rs 0.64

Given that interest for the second year exceeds the first year by Rs.96

When difference between the interests is Rs 0.64, principal is Rs 100

When the difference between the interests is Rs 96, principal = Rs

$\dfrac{96 \times 100}{0.64} =$ Rs 15,000

In how many years will Rs. 7000 amount to Rs. 9317 at 10 per cent per annum compound interest?

Given: P = Rs 7,000, A = Rs 9,317 and r = 10%

$A = P \left( 1 + \dfrac{r} {100}\right)^{n}$

$\Rightarrow 9,317 = 7,000 \left( 1 + \dfrac{r} {100}\right)^{n}$

$\Rightarrow \dfrac{9,317} {7,000} =\left(\dfrac{11} {10}\right)^{n}$

$\Rightarrow \dfrac{1,331} {1000} = \left( \dfrac{11} {10}\right)^{n}$

$\Rightarrow \left(\dfrac{11} {10}\right)^{3} = \left( \dfrac{11} {10}\right)^{n}$
On comparing
n = 3 years

A man borrows Rs 6,000 at 5 percent C.I. per annum. If he repays Rs 1,200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Given that the amount borrowed = Rs.6,000
Rate per annum = 5%
Interest on Rs 6,000 =

$\dfrac{5}{100} \times$ Rs. 6000 = Rs. 300
So, amount at the end of the first year
= Rs. 6,000 + Rs. 300
= Rs. 6,300
Amount left to be paid = Rs. 6,300 - Rs. 1,200
= Rs. 5,100
Interest on Rs 5,100 =

$\dfrac{5}{100} \times$ Rs. 5,100 = Rs. 255
So, amount at the end of the second year
= Rs. 5,100 + Rs. 255
= Rs. 5,355
Amount left to be paid = Rs. 5,355 - Rs. 1,200
= Rs. 4,155
Hence, the amount of the loan outstanding at the beginning of the third year is Rs. 4,155.

A man lends Rs.12,500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly; find the difference between C.I of the first year and the compound interest for the third year.

For 1

$^{st}$ year
P = Rs.12500
R = 12%
T = 1 year
t

$= \dfrac{12500 \times 12 \times 1}{100} =$ Rs.1500
A = 12500 + 1500 = Rs.14000
For 2

$^{nd}$ year
P = Rs.1400
R = 15%
T = 1 year
t

$= \dfrac{1400 \times 15 \times 1}{100} =$ Rs.2898
A = 1400 + 2100 = Rs.16100
For 3

$^{rd}$ year
P = Rs.16100
R = 18%
T = 1 year
t

$= \dfrac{16100 \times 18 \times 1}{100} =$ 2898
A = 16100 + 2898 = Rs.3998
Difference between the compound interest of the third year and first year
= Rs.2893 - Rs. 1500
= Rs. 1398

A man borrows Rs.10000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year at 42% of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt ?

The amount borrowed = Rs.10,000

Interest for the first year

$= \frac{P \times R \times T}{100}$

$= \frac{10000 \times 5 \times 1}{100}$

$=$ Rs.500

So, amount at the end of the first year

= Rs.10,000 + Rs.500

= Rs.10,500

The man pays 35% of Rs.10,500 at the end of the first year

$= \frac{35}{100} \times 10,500 =$ Rs.3,675

So, amount left to be paid

= Rs.10,500 - Rs.3,675 = Rs.6,825

Interest for the second year

$= \frac{P \times R \times T}{100}$

$= \frac{6,825 \times 5 \times 1}{100}$

$=$ Rs.341.25

So, amount at the end of the second year

= Rs.6,825 + Rs.341.25

= Rs.7,166.25

The man pays 42% of Rs.7,166.25 at the end of the second year

$= \frac{42}{100} \times 7166.25 =$ Rs.3,009.825

So, amount left to be paid

= Rs.7,166.25 - Rs.3,009.825 = Rs.4,156.425

Interest for the third year

$= \frac{P \times R \times T}{100}$

$= \frac{4,156.425 \times 5 \times 1}{100}$

$=$ Rs.207.82125

So, amount at the end of the second year

= Rs.4,156.425 + Rs.207.82125

= Rs.4,364.24625

Hence, he must pay Rs.4,364.24625 at the end of the third year in order to dear the debt.

A man saves Rs.3000 every year and invests it at the end of the year at 10% compound interest. Calculate the total amount of his savings at the end of the third year.

Savings at the end of every year = Rs 3000
For

$2^{nd}$ year
P = Rs.3000
R = 10%
T = 1 year
t

$= \dfrac{3000 \times 10 \times 1}{100} =$ Rs.300
A = 3000 + 300 = Rs.3300
For third year, saving = 3000
P = 3000 + 3300 = Rs.6300
R = 10%
T = 1 year
t

$= \dfrac{6300 \times 10 \times 1}{100} =$ Rs.630
A = 6300 + 630 = Rs 6930
Amount at the end of 3rd year
= 6930 + 3000
= Rs.9930

A sum of Rs. 44200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10 per cent per annum compound interest, they will receive equal amounts on reaching 16 years of age.
(i) What is the share of each out of Rs. 44,200?
(ii) What will each receive when 16 years old?

(i) Let share of Jhon = Rs y

Share of Smith = Rs (44,200 - y)

Rate of interest = 10%

According to question

Amount of Jhon in 4 years = Amount of Smith in 2 years

$\Rightarrow y \left(1 + \dfrac{10} {100} \right)^{4} - \left(44,200 - y \right) \left(1 + \dfrac{10} {100} \right)^{2}$

$\Rightarrow y \left(1 + \dfrac{10} {100} \right)^{2} = \left(44,200 - y \right)$

$\Rightarrow y \left( \dfrac{11} {10} \right)^{2} = \left(44,200 - y \right)$

$\Rightarrow 121y = 100 \left(44,200 - y \right)$

$\Rightarrow 121y = 100 \times 44,200 - 100y$

$\Rightarrow 121y + 100y = 100 \times 44,200$

$\Rightarrow 221y = 100 \times 44,200$

$\Rightarrow y = \dfrac{100 \times 44,200} {221} = Rs 20,000$

$\therefore$ Share of John = Rs 20,000

Share of Smith = Rs 44,200 - Rs 20,000

= Rs 24,200

(ii) Amount that each will receive

$= 20,000 \left(1 + \dfrac{10} {100} \right)^{4}$

$= 20,000 \left( \dfrac{11} {10} \right)^{4}$

= Rs 29,282

A manufacturer estimates that his machine depreciates by 15% of its value at the beginning of the year. Find the original value (cost) of the machine, if it depreciates by Rs. 5,355 during second year.

Let the original cost of the machine = Rs 100

$\therefore$ Depreciation during the 1sy year = 15% of Rs.100 = Rs.15

Value of the machine at the beginning of the 2nd year

= Rs.100 - Rs 15

= Rs. 85

$\therefore$ Depreciation during the 2nd year = 15% of Rs.85 = Rs.12.75

Now, when depreciation during 2nd year = Rs. 1275, original cost = Rs.100

$\Rightarrow$ when depreciation during 2nd year = Rs.5,355

Original cost = Rs.

$\dfrac{100}{12.75} \times 5,355 =$ Rs. 42,000

Hence, original cost of the machine is Rs.42,000.

The value of an article decreased for two years at the rate of 10% per year and then in the third year it increased by 10%. Find the original value of the article, if its value at the end of 3 years is Rs. 40095.

${\textbf{Step -1: Writing mathematical equation according to question}}$

${\text{Let the initial value of the article be Rs}}{\text{. x,}}$

${\text{In the first year, value decreased by 10% }}$

${\text{Value after first year, }}{{\text{x}}_{\text{1}}}{\text{ = x - (10% of x)}}$

$\Rightarrow {\text{ }}{{\text{x}}_{\text{1}}}{\text{ = x - }}\dfrac{{\text{x}}}{{{\text{10}}}}{\text{ = }}\dfrac{{{\text{9x}}}}{{{\text{10}}}}$

${\text{In the second year, value decreased by further 10% }}$

${\text{Value after second year, }}{{\text{x}}_{\text{2}}}{\text{ = }}{{\text{x}}_{\text{1}}}{\text{ - (10% of }}{{\text{x}}_{\text{1}}}{\text{)}}$

$\Rightarrow {\text{ }}{{\text{x}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{9x}}}}{{{\text{10}}}}{\text{ - }}\left( {\dfrac{{{\text{9x}}}}{{{\text{10}}}}} \right)\left( {\dfrac{{\text{1}}}{{{\text{10}}}}} \right){\text{ = }}\dfrac{{{\text{81x}}}}{{{\text{100}}}}$

${\text{In the third year, value increased by 10% }}$

${\text{Value after third year, }}{{\text{x}}_{\text{3}}}{\text{ = }}{{\text{x}}_{\text{2}}}{\text{ + (10% of }}{{\text{x}}_{\text{2}}}{\text{)}}$

$\Rightarrow {\text{ }}{{\text{x}}_{\text{3}}}{\text{ = }}\dfrac{{{\text{81x}}}}{{{\text{100}}}}{\text{ + }}\left( {\dfrac{{{\text{81x}}}}{{{\text{100}}}}} \right)\left( {\dfrac{{\text{1}}}{{{\text{10}}}}} \right){\text{ = }}\dfrac{{{\text{891x}}}}{{{\text{1000}}}}$

${\textbf{Step -2: Calculating initial value of article}}$

${\text{Given, value after third year = 40095}}$

$\Rightarrow {\text{ }}\dfrac{{{\text{891x}}}}{{{\text{1000}}}}{\text{ = 40095}}$

$\Rightarrow {\text{ x = 45,000}}$

${\textbf{Final Answer: The initial value of article is Rs}}{\textbf{. 45,000}}$

The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at Rs. 44000 at the beginning of 2008, find its value:
(i) At the end of 2009.
(ii) At the beginning of 2007.

Cost of machine in 2008 = Rs 44,000

Depreciation rate = 12%

(i)

$\therefore$ Cost of machine at the end of 2009

$= P \left(1 - \frac{r} {100} \right)^{ n }$

$= 44,000 \left(1 - \frac{12} {100} \right)^{ 2 }$

$= 44,000 \times \left(\frac{88} {100} \right)^{ 2 } = Rs 34,073.60$

(ii)

$\therefore$ Cost of machine at the beginning of 2007 (P)

$= P \left(1 - \frac{r} {100} \right)^{ n }$

$= 44,000 \left(1 - \frac{12} {100} \right)^{1 }$

$= 44,000 \times \left(\frac{88} {100} \right)^{ 1 }$

$\Rightarrow P = \frac{44,000 \times 100} {88} = Rs 50,000$

During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates by Rs.2,640 during the second financial year of its purchase.

Let original value of machine = Rs 100

For 1st year

P = Rs 100; R = 12% and T = 1 year

Depreciation in 1st year = Rs

$\dfrac{100 \times 12 \times 1}{100} =$ Rs 12

Value at the end of 1st year = Rs 100 - Rs 12 = Rs 88

For 2nd year

P = Rs 88; R = 12% and T = 1 year

Depreciation in 2nd year = Rs

$\dfrac{88 \times 12 \times 1}{100} =$ Rs 10.56

When depreciation in 2nd year is Rs.2,640, original cost =

$\dfrac{100 \times 2640}{10.56}$

= Rs. 25,000

The rate of interest charged

Let

$x$ % be the rate of interest charged

For 1st year

P = Rs 12,000; R =

$x$ % and T = 1 year

$\Rightarrow$ Interest (I) = Rs

$\dfrac{12000 \times x \times 1}{100} = 120x$

For 2nd year

After a year, Ashok paid back Rs.4,000.

P = Rs.12000 + Rs.120

$x -$ Rs.4000 = Rs.8000 + 120

$x$

$\Rightarrow$ Interest (I) = Rs

$\dfrac{(8000 + 120x) \times x \times 1}{100} = (80x + 1.20x^2)$

The compound interest for the second year is Rs.920

Rs.

$(80x + 1.20x^2) =$ Rs.920

$\Rightarrow 1.20x^2 + 80x - 920 = 0$

$\Rightarrow 3x^2 + 200x - 2300 = 0$

$\Rightarrow 3x^2 + 230x - 30x - 2300 = 0$

$\Rightarrow x(3x + 230) - 10(3x + 230) = 0$

$\Rightarrow (3x + 230) (x - 10) = 0$

$\Rightarrow x = -230/3$ or

$x = 10$

As rate of interest cannot be negative so x = 10

Therefore the rate of interest charged is 10%.

The interest for the first year.

For 1st year

P = Rs.13,500; R = 16% and T = 1 year

Interest = Rs

$\dfrac{13,500 \times 16 \times 1}{100} =$ Rs.2,160

According to a census taken towards the end of the year 2009, the population of a rural town was found to beThe census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this reach 74088?

Population in 2009 (P) = 64,000

Let after n years its population be 74,088(A)

Growth rate = 5% per annum

$\therefore A = P\left(1 + \frac{r} {100} \right)^{n}$

$\Rightarrow 74,088 = 64,000 \left(1 + \frac{5} {100} \right)^{n}$

$\Rightarrow \frac{74,088} {64,000} = \left( \frac{21} {20} \right)^{n}$

$\Rightarrow \frac{9,261} {8,000} = \left( \frac{21} {20} \right)^{n}$

On camparing, we get

n = 3 years

Saurabh invests Rs.48,000 for 7 years at 10% per annum compound interest. After a year, he paid back Rs.4,If compound interest for the second year be Rs.920, find:
(i) The rate of interest charged.
(ii) The amount of debt at the end of the second year.

For 1st year

P = Rs.48,000; R = 10% and T = 1 year

Interest = Rs

$\dfrac{48,000 \times 10 \times 1}{100} =$ Rs.4,800

Amount = Rs.48,000 + Rs. 4,800 = Rs. 52,800

For 2nd year

P = Rs.52,800; R = 10% and T = 1 year

Interest = Rs

$\dfrac{52,800 \times 10 \times 1}{100} =$ Rs.5,280

Amount = Rs.52,800 + Rs 5,280 = Rs.58,080

For 3rd year

P = Rs.58,080; R = 10% and T = 1 year

Interest = Rs

$\dfrac{58,080 \times 10 \times 1}{100} =$ Rs.5,808

A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20 %. Find the estimated value at the end of 5 years.

Cost of the machine at the end of year

$15625 \left ( 1 - \dfrac{20}{100} \right )$

$15625 \left ( \dfrac{80}{100} \right )$

$15625 \times \dfrac{4}{5}$
Rs 5120.

Second year

Total interest obtained in the first year = Rs.1500

Interest for the second year - Total interest obtained in the first year

= Rs.1,725 - Rs.1,500

= Rs.225

Rate of interest for the second year

$= \dfrac{225}{1,500} \times 100 = 15 \%$

So rate of interest for the second year is 15%

A man invests Rs. $9{,}600$ at $10 \%$ per annum compound interest for $3$ years. Find the interest for the third year.

Simple Interest

$I = \dfrac {PNR}{100}$
Here for first year,

$P =$ Rs.

$9{,}600$ ,

$N = 1$ year,

$R = 10 \%$
So, Interest at the end of 1st year

$I = \dfrac {9,600 \times 1 \times 10 }{100} =$ Rs.

$960$

And for second year,

$P =$ Rs.

$9,600\, +$ Rs.

$960 =$ Rs.

$10,560$

$N = 1$ year,

$R = 10\%$

$I = \dfrac {10,560 \times 1 \times 10}{100} =$ Rs.

$1,056$
And for third year,

$P =$ Rs.

$10,560\, +$ Rs.

$1,056 =$ Rs.

$11,616$

$N = 1$ year,

$R = 10\%$

$I = \dfrac {11,616 \times 1 \times 10}{100} =$ Rs.

$1,161.60$

Therefore, the interest for the third years is Rs.

$1,161.60$ .

In how many years will the compound interest on Rs. $32000$ at the rate of $5\%$ per annum, be Rs. $5044$ ?

Given, $P = Rs. 32000$
$C.I. = Rs. 5044$

$A = P + C.I. =32000 + 5044 =Rs. 37044$

Where, $P=$ principal

$C.I=$ compound interest

$A=$ amount

$R=$ rate of interest

On interest being compounded for $N$ years,

$\Rightarrow A = P{(1 + \dfrac {R}{100})}^{N}$

$\Rightarrow 37044 = 32000 \times {(1 + \dfrac {5}{100})}^{N}$

$\Rightarrow \dfrac {37044}{32000} = {(\dfrac {21}{20})}^{N}$

$\Rightarrow \dfrac {9261}{8000} = {(\dfrac {21}{20})}^{N}$

$\Rightarrow \dfrac {{21}^{3}}{{20}^{3}} = {(\dfrac {21}{20})}^{N}$

$\Rightarrow N = 3$ years

A man invests Rs. $30,800$ at $5$ % per annum compound interest for $3$ years. Calculate the amount standing to his credit at the end of the second year.

For the first year

$P = Rs 30,800$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {30,800 \times 1 \times 5}{100} =Rs 1,540$
And Amount at the end of first year $P + S.I. = Rs 30,800 + Rs 1,540 = Rs 32,340$
Now, for the second year
$P = Rs 32,340$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {32,340 \times 1 \times 5}{100} =Rs 1617$
And Amount at the end of second year $P + S.I. = Rs 32,340 + Rs 1617 = Rs 33,957$

A sum of money, invested at compound interest, amounts to Rs. $19,360$ in $2$ years and to Rs. $23,425.60$ in $4$ years. Find the original sum of money.

On interest being compounded for

$2$ years,

$Amount = P{(1 + \frac {R}{100})}^{N}$

$\implies 19,360 = P{(1 + \frac {R}{100})}^{2}$ --- (1)

On interest being compounded for

$4$ years,

$Amount = P{(1 + \frac {R}{100})}^{N}$

$\implies 23,425.60 = P{(1 + \frac {R}{100})}^{4}$ --- (2)

Dividing, eqn (2) by eqn (1)

$1.21 = {(1 + \frac {R}{100})}^{2}$

Taking square root on both sides,

$1.1 = 1 + \frac {R}{100}$

$\implies 0.1 = \frac {R}{100}$

$\implies R = 10$ %

Putting the value of

$R$ in eqn

$(1),$ we get

$\implies 19,360 = P{(1 + \frac {10}{100})}^{2}$

$\implies19,360 = P \times {(1.1})^{2}$

$\implies P = \frac {19,360}{1.21}$

$\implies P = Rs. 16,000$

Calculate the compound interest for the third year on Rs. $7,500$ invested for $5$ years at $10$ % per annum.

For the first year
$P = Rs 7,500$
$N = 1 year$
$R = 10$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac { 7,500 \times 1 \times 10}{100} =Rs 750$
And Amount at the end of first year $P + S.I. = Rs 7,500 + Rs 750 = Rs 8,250$
Now, for the second year
$P = Rs 8,250$
$N = 1 year$
$R = 10$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac { 8,250 \times 1 \times 10}{100} =Rs 825$
And Amount at the end of second year $P + S.I. = Rs 8,250 + Rs 825 = Rs 9,075$
Also, for the third year
$P = Rs 9,075$
$N = 1 year$
$R = 10$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac { 9,075 \times 1 \times 10}{100} =Rs 907.50$
Hence, $C.I.$ for the third year $= Rs 907.50$

What sum of money will amount to Rs. $15,972$ in three years at $10$ % per annum compounded yearly?

On interest being compounded for $3$ years and $R = 10$ $\%$
Amount $= P{(1 + \dfrac {R}{100})}^{N}$

$\Rightarrow P=\dfrac{15972 \times 1000}{1331}$
$\Rightarrow P =$ Rs. $12,000$

Therefore, the sum of money is Rs. $12000$ .

A man invests Rs. $5,600$ at $14$ % per annum compound interest for $2$ years. Calculate the interest for the second year, correct to the nearest rupee.

For the first year:

$P =$ Rs.

$5,600$ ,

$N = 1$ year and

$R = 14 \%$
We have

$\text{S.I. } = \dfrac { \text{PNR}}{100} = \dfrac {5,600 \times 1 \times 14}{100} =$ Rs.

$784$
And Amount at the end of first year

$\text{P} + \text{S.I.} =$ Rs.

$5,600 +$ Rs.

$784 =$ Rs.

$6,384$
Now, for the second year :

$P =$ Rs.

$6,384$ ,

$N = 1$ year and

$R = 14 \%$
We have

$\text{S.I.} = \dfrac { \text{PNR}}{100} = \dfrac {6,384 \times 1 \times 14}{100} =$ Rs.

$893.76$
To the nearest rupee, it is Rs.

$894$ .

A sum of money lent out at C. I. at a certain rate per annum becomes three times of itself in $8$ years. Find in how many years will be money become twenty-seven times of itself at the same rate of interest p.a.

On interest being compounded for $N$ years
$Amount = P{(1 + \dfrac {R}{100})}^{N}$

On interest being compounded for 8 years,
$=> 3P = P{(1 + \dfrac {R}{100})}^{8}$

$=> 3= {(1 + \dfrac {R}{100})}^{8}$ --- (1)

ATQ,
$=> 27P = P{(1 + \dfrac {R}{100})}^{n}$
$=> 27 = {(1 + \dfrac {R}{100})}^{n}$
$=> {3}^{3} = {(1 + \dfrac {R}{100})}^{n}$ --- (2)

Putting the value of $3$ from (1) in (2), we get
$=> {( {(1 + \dfrac {R}{100})}^{8}) }^{3} = {(1 + \dfrac {R}{100})}^{n}$
${(1 + \dfrac {R}{100})}^{24} = {(1 + \dfrac {R}{100})}^{n}$

$=> n = 24$ years

What principal will amount to Rs. $9,744$ in two years, if the rates of interest for successive years are $16$ % and $20$ % respectively?

For the first year, on interest being compounded at $R = 16$ $\%$ , we have

Amount $= P {\left (1 + \dfrac {R}{100}\right)}^{N}$

$= P \times {\left (1 + \dfrac {16}{100}\right)}^{1}$

$= P \times 1.16$

$=$ Rs $1.16P$
For the second year, $P =$ Rs. $1.16P$ on interest being compounded at $R = 20$ $\%$ , we have
Amount $= P{\left (1 + \dfrac {R}{100}\right)}^{N}$

$=$ Rs. $1.16P \times {\left (1 + \dfrac {20}{100}\right)}^{1}$

$=$ Rs. $1.16P \times 1.20$

$=$ Rs. $1.392P$
Given, $1.392P = 9,744$
$\Rightarrow P =$ Rs. $7,000$

In what time will Rs, $1,500$ yield Rs. $496.50$ as compound interest at $10$ % per year compounded annually?

$P = Rs 1,500$
$C.I. = Rs 496.50$
$A = P + C.I. = 1,500 + 496.50 =Rs 1996.50$

On interest being compounded for N years,
$Amount = P{(1 + \dfrac {R}{100})}^{N}$
$1996.50 = 1,500 \times {(1 + \dfrac {10}{100})}^{N}$
$\dfrac { 1996.50 }{1,500} = {(\dfrac {11}{10})}^{N}$
$\dfrac {1331}{1000} = {(\dfrac {11}{10})}^{N}$
$\dfrac {{11}^{3}}{{10}^{3}} = {(\dfrac {11}{10})}^{N}$

$=> N = 3$ years

A man invests Rs. $30,800$ at $5$ % per annum compound interest for $3$ years. Calculate the interest for the third year.

For the first year
$P = Rs. 30,800$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {30,800 \times 1 \times 5}{100} =Rs. 1,540$
And Amount at the end of first year $P + S.I. = Rs. 30,800 + Rs. 1,540 = Rs. 32,340$
Now, for the second year
$P = Rs. 32,340$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {32,340 \times 1 \times 5}{100} =Rs. 1617$
And Amount at the end of second year $P + S.I. = Rs. 32,340 + Rs. 1617 = Rs. 33,957$
Also, for the third year
$P = Rs. 33,957$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {33,957 \times 1 \times 5}{100} =Rs. 1697.85$

The production of the factory increases every year by $5\%$ . At present, the production is $160000$ tons. After how many years will the production reach $194481$ tons?

Present production :

$P=160000, R=5$ %

$P'=$ Production after

$T$ years

$=194481$ , So

$T=?$

$P'=P\left( 1+\cfrac { R }{ 100 } \right) ^{ T }\\ =\cfrac { 194481 }{ 160000 } =\left( \cfrac { 21 }{ 20 } \right) ^{ T }\\ =\left( \cfrac { 21 }{ 20 } \right) ^{ 4 }=\left( \cfrac { 21 }{ 20 } \right) ^{ T }$

Hence,

$T=4$ years is the correct answer.

The value of a car depreciates by $15\%$ each year. At the end of $2014$ the value of the car was Rs. $125000$ . Find out the value of car at the end of $2011$ .

Given,

$P = 125000, r = 15\%, n = 3$ years

We know

$A = P\left [\left (1-\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A = 125000\left [\left (1-\dfrac{15}{100}\right)^3\right]$

$\Rightarrow A =$ Rs.

$76765.62$

Thus, the value of car at the end of

$2011$ is Rs.

$76765.62$ .

A city has $20000$ residents. Its population grows at the rate of $10\%$ per annum, what will be its total population after $5$ years?

Given,

$P = 14600, r = 10\%, n = 2$ years
We know,

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A = 20000\left [\left (1+\dfrac{10}{100}\right)^5\right]$

$\Rightarrow A= 32210$
Hence, the population of the city after

$5$ years is

$32210$ .

The value of an estate in January 1905 started gradually declining in such a way that at the end of each year it was worth only x times its value at the beginning of the year. What was its worth in end December 1910 ?
(1) It was worth Rs. 10109 in the end of December 1906
(2) It was worth Rs. 12345 in the beginning of January 1905.

Let rate of declining per annum be

$y$ .

Now,

According to given information, from January

$1905$ to December

$1905$ , time

$=2$ years

Therefore,

$12345-\cfrac{12345\times y\times 2}{100}=10109$

By solving we get

$y=9.05\%$

Now,

Value in December

$1905 =12345-\cfrac{12345\times 9.05\times 6}{100}=$ Rs.

$5641.665$

What will a deposit of Rs. $5,500$ at $8\%$ compounded quarterly interest be worth, if left in the bank for $10$ years?

We know the formula,

$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,

$A =$ total amount

$P =$ principal or amount of money deposited,

$r =$ annual interest rate

$n =$ number of times compounded per year

$t =$ time in years
Given:

$P =$ Rs.

$5500, r = 8\%, n = 4$ and

$t = 10$ years

$A = 5500\left (1+\dfrac{0.08}{4}\right)^{4.10}$

$A = 5500 \times 1.02^40$

$A = 5500 \times 2.20804$

$A =$ Rs.

$12144.22$

A man borrows Rs. $18,000$ at $5$ % per annum compound interest. If he repays Rs. $6,000$ at the end of the first year and Rs. $8,000$ at the end of the second year; how much he should pay at the end of the third year in order to clear the account? Answer correctly to the nearest rupee.

For the first year
$P = Rs 18,000$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {18,000 \times 1 \times 5}{100} =Rs 900$
And Amount at the end of first year $P + S.I. = Rs 18,000 + Rs 900 = Rs 18,900$
Now, for the second year
$P = Rs 18,900 - Rs 6,000 = Rs 12,900$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {12,900 \times 1 \times 5}{100} =Rs 645$
And Amount at the end of second year $P + S.I. = Rs 12,900 + Rs 645 = Rs 13,545$
Also, for the third year
$P = Rs 13,545 - Rs 8,000 = Rs 5,545$
$N = 1 year$
$R = 5$ %
We have $S.I. = \dfrac { PNR}{100} = \dfrac {5,545 \times 1 \times 5}{100} =Rs 277.25$
And Amount at the end of third year $P + S.I. = Rs 5,545 + Rs 277.25 = Rs 5,822.25$
Thus, at the end of the third year, amount to be paid $= Rs 5,822.25$ or approximately $Rs 5,822$

Out of the world population of approximately $6.6$ billion, $1.2$ billion people live in the richer countries of Europe, North America, Japan and Oceania and is growing at the rate of $0.25\%$ per year, while other $5.4$ billion people live in the less developed countries and is growing at the rate of $1.5\%$ . What will be the world population in billions in $5$ years if we assume that these rates of increase will stay constant for the next $5$ years. (round answer to $3$ significant digits)

Given:

$1.2$ billion live in rich countries and their growth rate is

$0.25\%$ every year.

After one year:

${ P }_{ R }=1.2+\cfrac { 0.25 }{ 100 } \times 1.2\\ =1.2\left( 1+\cfrac { 0.25 }{ 100 } \right)$

After

$2$ years:

${ P }_{ R }=1.2\left( 1+\cfrac { 0.25 }{ 100 } \right) +\cfrac { 0.25 }{ 100 } \left( 1.2\left( 1+\cfrac { 0.25 }{ 100 } \right) \right) \\ =1.2\left( 1+\cfrac { 0.25 }{ 100 } \right) ^{ 2 }$

Continuing for

$5$ years we get

${ P }_{ R }=1.2\left( 1+\cfrac { 0.25 }{ 100 } \right) ^{ 5 }$

Similarly for developing countries:

${ P }_{ L }=5.4\left( 1+\cfrac { 1.5 }{ 100 } \right) ^{ 5 }$

Total

$P={ P }_{ R }+{ P }_{ L }=5.4\left( 1+\cfrac { 1.5 }{ 100 } \right) ^{ 5 }+1.2\left( 1+\cfrac { 0.25 }{ 100 } \right) ^{ 5 }$

$=7.03$ billion

Hence, the correct answer is

$7.03$ billion.

The botanist would like to increase the number of plants that the environment can support so that the population of the species will increase more rapidly. If the botanists goal is that the number of plants will increase from $3000$ this year to $3360$ next year, how many plants must the modified environment support?

Substituting

$N_\text{thisyear} = 3000$
and

$N_\text{nextyear} = 3360$ in the given equation, we get

$3360 = 3000 + 0.2(3000)\left(1-\cfrac{3000}{K}\right)$

$\Rightarrow 360 = 600\left(1-\cfrac{3000}{K}\right)$

$\Rightarrow \cfrac {360}{600} = 1-\cfrac {3000}{K}$

$\Rightarrow \cfrac {3}{5} = 1-\cfrac {3000}{K}$

$\Rightarrow \cfrac {3000}{K} = \cfrac {2}{5}$

$\Rightarrow K = 7500$

If the current population of Chennai city is $3888180$ and the average rate of population growth in Bangalore city is $10\%$ p.a., what can city planners expect the population of Chennai city to be in $10$ years?

Given,

$P = 3888180, r = 10\%, n = 10$
We know,

$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$

$\Rightarrow A = 3888180\left [\left (1+\dfrac{10}{100}\right)^{10}\right]$

$\Rightarrow A= 10084772$
Hence, the population of Chennai city in

$10$ years is

$10084772$ .

According to the formula, what will be the number of plants two years from now if $K = 4000$ (Round your answer to the nearest whole number.)

Substituting

$K = 4000$

$N_\text{thisyear} = 3000$

We get the number of plants in the first year as

$N_\text{first year} = 3000 + 0.2(3000)\left(1-\dfrac{3000}{4000}\right) \\\ \ \ \ \ \ \ \ \ \ \ \ \ = 3000 + 150 \\\ \ \ \ \ \ \ \ \ \ \ \ \ = 3150$

Now,

$K = 4000$

$N_\text{this year} = 3150$
We get the number of plants in the second year as

$N_\text{second year} = 3150 + 0.2(3150)\left(1-\dfrac{3150}{4000}\right) \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3283.875 \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3284$

A high school had $1200$ students enrolled in $2003$ and $1500$ students in $2006$ . If the student population $P$ ; grows as a linear function of time $t$ , where $t$ is the number of years after $2003$ .
$(a)$ How many students will be enrolled in the school in $2010$ ?
$(b)$ Find a linear function that relates the student population to the time $t$ .

Given: Population is a linear function of

$t$ .

$\Rightarrow P=kt+C$ , where

$k$ and

$C$ are constants.

$2003$ , given that

$P=1200,t=0$

since

$t$ is time in years from

$2003$

$\Rightarrow 1200=k(0)+C\\ \therefore C=1200$

$2006$

$P=1500,t=2006-2003=3$

$\\ \Rightarrow 1500=k(3)+1200\\ \therefore k=100$

(A) In

$2010$

$t=2010-2003=7$

$P=100\times 7+1200=1900$

(B) The linear function is

$P=100t+1200$

Find the amount and the compound interest on $Rs. 6500$ for $2$ years, compounded annually, the rate of interest being $5\%$ per annum during the first year and $6\%$ per annum during the second year.

$\textbf{Step - 1 : Calculate the amount after 2nd year}$

$\text{Amount}=P(1+\dfrac{r_1}{100})(1+\dfrac{r_2}{100})$

$\text{where, P = principal with which one began}$

$\text{r1 = rate of interest for first year}$

$\text{r2 = rate of interest for second year}$

$\text{Amount}=6500(1+\dfrac{5}{100})(1+\dfrac{6}{100})$

$\text{Amount =Rs } 7234.5$

$\textbf{Step - 2 : Calculate the interest earned}$

$\text{Interest=Amount- Principal}$

$\text{Interest}=7234.5-6500$

$\text{Interest =Rs }734.5$

$\textbf{Hence, the interest earned is Rs 734.5}$

A town's population increased by $1200$ people, and then this new population decreased $11$ %. The town now had $32$ less people than it did before $1200$ increase. Find the original population.

Let original population $= x,$ given it increased by $1200.$

New population $= x + 1200.$ This new population decreased by $11$ %

Final population $=$ New population $– 11$ % of New population

$=\left( x+1200 \right) \times \left( 1-\cfrac { 11 }{ 100 } \right)$

Final Population

$=\left( x+1200 \right) \times \left( 1-\cfrac { 89 }{ 100 } \right)$

Given Final population is $32$ less than original population.

$\Rightarrow \left( x+1200 \right) \times \left( 1-\cfrac { 89 }{ 100 } \right) =x-32\\ 89x+106800=100x-3200\\ 11x=110000\\ x=10000$

$\therefore$ Original population $=10000$

Saloni invest 60% of her money at the rate of 18% per annum for 4years. The remaining sum was invested at 14% per annum for 5 years. Find the sum of money she invested each time, if the difference between the simple interests earned is Rs.1520.

Let sum of money

$=x$

money invested at

$18\%$ is

$60\%$ of x

$\Rightarrow \dfrac{6x}{10}$

Interest

$=\dfrac{PRT}{100}=\dfrac{\dfrac{6x}{10}\times 18\times 4}{100}$

$=\dfrac{6\times 18\times 4x}{1000}=\dfrac{432x}{1000}$

Remaining sum

$=\dfrac{4x}{10}$

Interest

$=\dfrac{PRT}{100}=\dfrac{\dfrac{4x}{10}\times 14\times 5}{100}$

$=\dfrac{(4\times 14\times 5)}{100}x=\dfrac{280x}{1000}$

$\Rightarrow$ Difference of interest

$=1520$

$\dfrac{432x-280x}{1000}=1520$

$152x=1520000$

$\Rightarrow x=10000$ Rs.

1185266_1207859_ans_cf16dd2a523b49a4bbe715bac2d4a8dc.jpg

Two friends meena and tina invested equals sums of money for 2 years at 16.66% per annume, Meena invested at C.I. where as Tina invested at S.I. Meena gets rs 23 more money than what Tina got. How much money did each invest?

Let

$P$ be the principal invested by Meena and Tina

Given Meena invests

$CI$ and invested at

$SI$ at

$16.66\%$ Rate of interest for

$2$ years.

$\therefore$ Let

$A$ be the amount tina gets at the end of

$2$ years

then Meena gets

$A+23$

$A=P+\dfrac{P+R}{100}$

$A+23=P\left(1+\dfrac{R}{100}\right)^T$

$P\left(1+\dfrac{R}{100}\right)^2$

$P\left(1+\dfrac{50}{3\times 100}\right)^2=P\left(1+\dfrac{2\times 50}{3\times 100}\right)+23$

$P\left(1+\dfrac{1}{6}\right)^2=P\left(1+\dfrac{1}{3}\right)+23$

$P\left(\dfrac{7}{6}\right)^2=P\left(\dfrac{4}{3}\right)+23$

$\dfrac{49P}{36}-\dfrac{4P}{3}=23$

$\dfrac{49P-48P}{36}=23$

$P=23\times 36=828$

$\therefore$ The amount they invested is

$828$ each

The difference of the compound interest and simple interest on a certain sum money for 3 years at the rate of 10% p.a. is Rs.93. Find the sum.

Let principle amount

$=x$

Simple interest

$=\dfrac{P\times R\times T}{100}$

$P=$ Principle

$=x$

$R=$ Rate

$=10\%$ Per annum

$T=$ Time

$=3$ years

$\Rightarrow S.I=\dfrac{x\times 10\times 3}{100}=\dfrac{3x}{10}$

Compound interest

$=$ Amount

$-$ Principle

Amount

$=P\left(1+\dfrac{R}{100}\right)^T$

$=x\left(1=\dfrac{10}{100}\right)^3$

$=x\left(\dfrac{11}{10}\right)^3$

$\Rightarrow C.I=\dfrac{1331x}{1000}-x$

$=\dfrac{1331x-1000x}{1000}$

$=\dfrac{331x}{1000}$

Difference of C.I and S.I

$=93$

$\Rightarrow \dfrac{331x}{1000}-\dfrac{3x}{10}=93$

$\Rightarrow \dfrac{31x}{1000}=93$

$\Rightarrow x=\dfrac{93\times 1000}{31}=3000$ Rs.

$\Rightarrow$ Sum

$=3000$ Rs.

1212711_1246016_ans_02f909cc7b8f49b995c3e0fe7cc714a3.jpg

The population of a town was $59000$ . In one year it was increased by $4536$ due to new births. However, $9218$ persons died or left the town during the year. What was the population at the end of the year?

Population of the town $= 59000$

Increase in the population $= 4536$

Decrease in the population $= 9218$

New population $= 59000 + 4536 – 9218 = 54318$

Hence the population at the end of the year is $54318$

In how much time would Rs. $5000$ amount to Rs. $6655$ at $10\%$ per annum compound interest?

If an amount 'P' is invested for 'n' years at

$r\%$ p.a. compound interest the amount obtained is

$\Rightarrow A=P\left( 1+\dfrac{r}{100}\right)^n$

$\Rightarrow P=Rs.5000$

$A=Rs.6655$

$r=10\%p.a.$

$\therefore 6655=5000\left( 1+\dfrac{10}{100}\right)^n$

$\Rightarrow \dfrac{6655}{5000}=\left( \dfrac{11}{10}\right)^n$

$\Rightarrow \dfrac{1331\times 5}{1000\times 5}=\left( \dfrac{11}{10}\right)^n$

$\Rightarrow \dfrac{1331}{1000}=\left( \dfrac{11}{10}\right)^n$

$\Rightarrow \dfrac{11\times 11\times 11}{10\times 10\times 10}=\left( \dfrac{11}{10}\right)^n$

$\Rightarrow \dfrac{11^3}{10^3}=\left(\dfrac{11}{10}\right)^n$

$\Rightarrow \left( \dfrac{11}{10}\right)^3=\left(\dfrac{11}{10}\right)^n$

$\therefore n=3\;years.$

Hence, the answer is

$3.$

The population of Andhra Pradesh $5.4\times { 10 }^{ 7 }$ in $1988$ . If the population was growing at a constant rate of $2.4\%$ p.a, what will be the population in $2008$ ?

The population of a particular year is given by

$\Rightarrow P_n=P_0\left(1+\dfrac{r}{100}\right)^n$

where

$P_0=$ Population in 1988

$\Rightarrow P_n=$ Population in n years after 1988

$r=$ Population growth rate

$n=$ Number of years after 1988

$\Rightarrow P_n=4.5\times 10^7$

$n=20$

$r=2.4\%$

$\Rightarrow P_{2008}=5.4\times 10^7\left(1+\dfrac{2.4}{100}\right)^{20}$

$=5.4\times 10^7\times 1.60694$

$=8.677\times 10^7.$

Hence, the answer is

$8.677\times 10^7.$

Anuj and Rajesh each lent the same sum of
money for $2$ years at $8$ % simple interest and compound interest respectively.
Rajest received $Rs. 64$ more than Anuj. Find the money lent by each and interest
received.

Let lmt by ball

$=x \ Rs.\ /person$

Interest earned by any

$=S.I=\dfrac {P\times R \times I}{100}$

$\dfrac {x\times 8 \times 2}{100}=\dfrac {16x}{100}Rs$

Interest earned by Rajesh

$=Amount -Principle$

$=x\left (1+\dfrac {R}{100}\right)^T-x$

$=x\left (1+\dfrac {8}{100}\right)^2 -x$

$=x \left [ \left(\dfrac {108}{100}\right)^2 -1 \right]$

$=x(0.1664)$

Difference of interest

$=0.1664x -0.16x$

$=0.0064x =64$

$\Rightarrow \dfrac {64x}{1000}=64$

$\Rightarrow \$ Interest required by Anej

$=1600\ Rs$

$\Rightarrow \$ Interest required by Rajesh

$=1664\ Rs$

Rajasthan $=570$ lakhs and population of $UP=160$ lakhs. Area of Rajasthan $=3$ lakh $km^{2}$ and area of $UP=2$ lakh $km^{2}$ .
Which State is less populated?

Given population of Rajasthan

$=570lakh=3{km}^{2}$ area

Population of UP

$=160lakh=2{km}^{2}$ area

Population of Rajasthan in

$1{km}^{2}$ area

$=\cfrac{570}{3}=190lakh$

Population of UP in

$1{km}^{2}$ area

$=\cfrac{160}{2}=80lakh$

$\therefore$ we observe that population per

${km}^{2}$ area is lesser for UP.

$\therefore$ UP is less populated

Find the difference between the simple interest and compound interest on $Rs\ 4800$ for $2$ years at $5\%$ per annum, compound interest being reckoned annually.

Simple Interest:

$P=Rs\ 4800$

$r=5\% p.a$

$t=2\ years$

$\therefore SI=\dfrac{Ptr}{100}=Rs.\dfrac{4800\times 2\times 5}{100}=Rs\ 480$

Compound interest:

Case I

$P=Rs\ 4800$

$r=5\% p.a.$

$t=1\ year$

Interest

$=Rs.\dfrac{4800\times 5\times 1}{100}=Rs\ 240$

$\therefore A=Rs\ (4800+240)=Rs\ 5040$

Case II

$P=Rs\ 5040$

$r=5\% p.a.$

$t=1\ year$

Interest

$=Rs.\dfrac{5040\times 5\times 1}{100}=Rs\ 252.$

$\therefore A=Rs\ (5040+252)=Rs\ 5292$ .

$\therefore$ Compound Interest

$=Rs(5292-4800)=Rs\ 492$

$\therefore$ Difference between compound Interest and Simple Interest

$=Rs\ (492-480)=Rs.12$

Sohan deposited a sum of $rs \quad 45000$ in bank for 5 years.He also invested $rs \quad 20000$ for the same period but at a rate $2 %$ higher than the first rate.He got a total interest of $rs \quad 39000$ from both his deposites at the end of 5 years.Find the rate of simple interest.

2039340_1263880_ans_c4b7319df9f94c94b9e994ea25addb07.JPG

Calculate the amount and compound interest on $Rs.\ 10,800$ for $3$ year at $12\dfrac {1}{2}$ % per annum compounded annually.

$A=P\left(1+\dfrac{R}{100}\right)^{t}$

$P=10800 ; R = 12\dfrac{1}{2}\, \% ; t = 3$ years

$A = 10800\left(1+\dfrac{25}{200}\right)^{3}$

$= 10800 \left(1+\dfrac{1}{8}\right)^{3}$

$= 10800 \left(\dfrac{9}{8}\right)^{3}$

$A = 15377.34$

Interest = A - P

$= 15377.34-10800$

$= 4577.34$

An amount of money invested trebled in $6$ years. Find the rate of interest earned.

1974133_1367702_ans_e1ef9a11144043d38eba259fb10b83a7.jpg

Mr Jane donates $Rs 1\ lakh$ to a school and the interest on it is to be used for awarding five scholarships of equal value. If the value of each scholarship is $Rs\ 1,500$ , find the rate of interest.

$\Pr incipal\,\,\left( P \right) = Mr.\,\,Jane\,\,donates$

$P = Rs.100000$

$Each\,\,scholarship\,\,value\, = Rs.\,1500$

$Value\,\,of\,\,5\,\,scholarship\,\, = \,5 \times 1500$

$Rs = \,7500$

$Interest\,\,(i)\, = value\,\,of\,\,5\,\,scholarships$

$I = \,Rs.\,\,7500$

$Rate\,\,of\,\,\operatorname{int} erest\,\,per\,\,annum\,\, = \,\,R\%$

$Time\,\,(T) = 1\,\,year$

$we\,\,know\,\,that$

$I = \left( {PTR} \right)/100$

$R = \left( {1 \times 100} \right)/\left( {P \times T} \right)$

$R = \frac{{\left( {7500 \times 100} \right)}}{{\left( {100000 \times 1} \right)}}$

$R = \frac{{75}}{{10}}$

$R = 7.5\%$

$Rate\,\,of\,\,\operatorname{int} erest\,\,Rs.\,\,7.5\% \,\,$

Complete the following life table:-
$x$ $l_x$ $d_x$ $q_x$ $p_x$ $L_x$
$4$ $9100$ $60$ $?$ $?$ $?$
$5$ $?$ $45$ $?$ $?$

$d_x=l_{x}-l_{x+1}$

$d_5=l_4-l_5\implies 9100-l_5=45\implies l_5=9055$

$d_x=l_x q_x\implies q_4=\dfrac{d_4}{l_4}=\dfrac{60}{9100}=\dfrac{3}{455}=0.0066$

$q_5=\dfrac{d_5}{l_5}=\dfrac{45}{9055}=0.005$

$p_{x}=\dfrac{l_{x+1}}{l_{x}}\implies p_{4}=0.995,p_5=0$

$L_x=\log p_{x}\implies L_{4}=\log p_{4}=log(0.995)=-0.0022$

$x$

$l_x$

$d_x$

$q_x$

$p_x$

$L_x$

$4$

$9100$

$60$

$0.0066$

$0.995$

$-0.0022$

$5$

$9055$

$45$

$0.005$

$0$

What and borrowed a certain sum for two years at simple interest from Bhim. Mahesh lent this sum to Vishnu at the same rate for two years of compound interest. At the end of two years, Mahesh received Rs.410 as compound interest but paid Rs.400 as simple interest. Find the sum and rate of interest.

2042735_1532740_ans_77a363e048604fc1bf79afff8138cf13.png

Rs. 9000 becomes Rs. 18000 at simple interest in 8 years. Find the rate per cent per annum.

1896595_1444126_ans_8352165a32a249b2b0030e228f4d2999.jpg

I borrowed Rs. $12,000$ from Suresh at $6\%$ per annum simple interest for $2$ years. Had I borrowed this sum at $6\%$ per annum compound interest, what extra amount would i have to pay?

For SI, principal

$=$ Rs.

$12,000$

Time

$=2$ yrs

Rate

$=6\%$ p.a.

$=\dfrac{PRT}{100}=$ Rs.

$\dfrac{12000\times 12\times 6}{100}=$ Rs.

$1440$

For CI, principal

$=$ Rs.

$12,000$ , Time

$=2$ yrs

Rate

$=6\%$ p.a

Amount

$=p\left(1+\dfrac{R}{100}\right)^n$

$=12,000\left(1+\dfrac{6}{100}\right)^2$

$=12,000\times \dfrac{53}{50}\times \dfrac{53}{50}$

$=$ Rs.

$\dfrac{48\times 53\times 53}{10}=$ Rs.

$13483.20$

$\therefore C.I. =$ Rs.

$13483.20-$ Rs.

$12,000=$ Rs.

$1483.20$

Excess amount

$=$ Rs.

$1483.20-$ Rs.

$1440=$ Rs.

$43.20$ .

The population of a town inereases at the proportional to the population existing at that time. The population of the town 5,00,000 in the year 1980 and 8,00,000 in the year 1990 . What will be the population of the town in the year 2010?

2040899_1453371_ans_deb740df592d495ab4d40b06c9096a15.jpg

In what time will a sum of Rs. $3,750$ at $20\%$ per annum compounded annually amount to Rs. $6,480$ ?

1297872_1579718_ans_fecb9984b01a4500b16a3e87340e8f25.jpg

In a factory the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.

2039827_1673996_ans_12fa5598072c4559af5f717b0b8a5925.jpg

The population of a town increases at the rate of 50 per thousand. Its population after 2 years will beFind its present population.

1875016_1674005_ans_422cae6249b3452a84d4b2cc21a288fa.jpeg

A certain sum amounts to Rs. 5292 in two years and Rs. 5556.60 in three years, interest being compounded annually. Find the rate of interest and the original sum.

2044617_1566868_ans_eebcd9d4d8634e18a556a526785cb742.jpg

If the amount on the principal of Rs. 6000 is written as $6000(1+\cfrac 4 {100})^2$ and compound interest payable half yearly; then what will be the the rate of interest p.a and time in years?

2044083_1561693_ans_a7d7fcda5896411c81347e84a9388da9.jpg

The population of a certain city was 72000 on the last day of the yearDuring next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000?

1876504_1674014_ans_784387654ad142598a4e14bfc47f66e9.jpeg

There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago?

2039826_1673994_ans_610de14f181c4dd58dd5b5237d2d653e.png

A bacteria population increases exponentially at a rate of r% each day.After 32 days, the population has increased by 309% .Find value of r

2043846_1538419_ans_333f2e98f69f44b0b052d74c534e6da1.png

A manufacturer reckons that the value of a machine which costs him Rs. $156250$ , will depreciate each year by $20\%$ . Find the estimated value at the end of $5$ years.

1902364_1710778_ans_7e74525d09964a47b0f3b7db8efbdb1b.jpg

The value of a machine costing Rs. $80000$ depreciates at the rate of $15\%$ per annum. What will be the worth of this machine after $3$ years?

1902344_1710775_ans_126f6afd1a1940bba99dc405a7c88e49.jpg

The population of a town decreased by 12% during 1998 and then increased by 8% duringFind the population of the town, at the beginning of 1998, if at the end of 1999 its population was 285120.

${\textbf{Step - 1: Framing the equation to find the initial population}}{\text{.}}$

${\text{Let the population before the beginning of 1998 be }}P.$

${\text{Also we know that the population at the end of 1999 = }}285120(A).$

${\text{So we know that population increases or decreases in a compounded manner, so we can write}}$

$A = P{\left( {1 \pm \dfrac{{{r_1}}}{{100}}} \right)^{{n_1}}}{\left( {1 \pm \dfrac{{{r_2}}}{{100}}} \right)^{{n_2}}}$

$\Rightarrow 285120 = P\left( {1 - \dfrac{{12}}{{100}}} \right)\left( {1 + \dfrac{8}{{100}}} \right)$

${\textbf{Step - 2: Solving the equation to find }}P.$

${\text{285120 = }}P \cdot \dfrac{{88}}{{100}} \cdot \dfrac{{108}}{{100}}$

$\Rightarrow P = \dfrac{{285120 \cdot 100 \cdot 100}}{{88 \cdot 108}} = 300000.$

${\textbf{Therefore, the population at the beginning of the year 1998 was 300000}}{\text{.}}$

The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago.

1876542_1674025_ans_059a742cee2d48e7ab23b2252cde5237.jpeg

Tax Payable Income	Tax
(i) $2,00,000$ Rs upto	zero
(ii) From Rs. $2,00,001$ to Rs. $5,00,000$	On amount more than Rs. $2,00,000$ $10\%$

Sr. No	Principal (Rs.)	Rate of Interest	Time Period (Year)
1	8,000	10%	2
2	6,400	12%	2
3	15,00	10%	2
4	20,000	8%	3
5	50,000	5%	3

S.No.	Principal	Amount	Rate of interest
(i)	$Rs. 12,550$		$9 \%$ semi-annually
(ii)	$Rs. 1500$	$Rs. 1900$
(iii)	$Rs. 8000$	$Rs. 9261$	$5 \% p.a$
(iv)		$RS. 5202$	$2 \% p.a$
(v)	$Rs. 625$		$3 \% p.a$

Basics Of Financial Mathematics - Class 11 Commerce Applied Mathematics - Extra Questions

During every financial year, the value of a machine depreciates by 10%. Find the original value (cost) of a machine which depreciates by Rs. 2,250 during the second year.

The number of bacteria in a certain culture doubles every hour. If there were 3030 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd2^{nd} hour 4th4^{th} hour and nthn^{th} hour?

A certain sum of money, placed out at compound interest, amounts to Rs.6,272 in 2 years and to Rs. 7,024.64 in 3 years. Find the rate of interest and the sum of money.

Reduction in price of vehicle and machinery with time is called _______

Simple interest is given by I=\dfrac{PTR}{100} I=\dfrac{PTR}{100}.If p=Rs.900, T=2 years, and R=5\%R=5\%, find the simple interest.

What is the rate of interest (%) of Rs.92,800 for 3\dfrac{1}{2}3\dfrac{1}{2} years , the interest of which is Rs.16240 ?

The interest is supposed compound unless the contrary is state.A person borrows 672672 pounds to be repaid in 55 years by annual instalments of 120120 pounds; find the rate of interest, reckoning simple interest.

Rate of depreciation is always ..................

A certain sum of money amounts to Rs. 6300/-Rs. 6300/- in 22 years and Rs. 7875/-Rs. 7875/- in 33 years 99 months at simple interest find the rate of interest per annum.

A certain sum of money double itself & no. of years is equal to rate of interest. Find rate of interest?

A girl deposits 300300 in a bank account. The bank offers 7%7% intersection per year.Assuming the girl does not take any money out of the account, calculate:

Find depreciation at 10\%\ pa10\%\ pa on furniture costingRs\,10,000Rs\,10,000 for 99 month

At what rate percent will Rs. 1600 amounts to Rs. 2000 in 5 years.

The population of Hyderabad was 68,09,00068,09,000 in the year 20112011. If it increases at the rate of 4.7\%4.7\% per annum. What will be the population at the end of the year 20152015.

Michael finished coloring a picture in \dfrac{7}{{12}}\dfrac{7}{{12}} hour. Vaibhav finished coloring the same picture in \dfrac{3}{{4}}\dfrac{3}{{4}} hour. Who worked longer. By what fraction was it longer

Find the rate per cent per annum if :Rs\ 600Rs\ 600 yields Rs\ 238Rs\ 238 interest in 44 years.

Find the rate per cent per annum if :Rs\ 700Rs\ 700 yields Rs\ 105Rs\ 105 interest in 33 years.

What principal will amount to Rs \ 15000Rs \ 15000 at 10\%10\% per annum in 5 \ years5 \ years?

At what rate per cent will a sum of money tripe itself in 25 years?.

Find the rate per cent per annum if :Rs\ 350Rs\ 350 yields Rs\ 105Rs\ 105 interest in 66 year.

At what rate per cent per annm will Rs 3600Rs 3600 amount to Rs \ 4734Rs \ 4734 in 3\dfrac {1}{2}years?3\dfrac {1}{2}years?

The price of a garment has been reduced by 15\%15\% in a sale to Rs\ 306Rs\ 306. Find its original price.

Find the principal, when the following are given.I=I= Rs. 192192Rate == 6\%6\% per annum Time == 44 years

The population of a city decreased from 25,00025,000 to 24,50024,500. Find the percentage decrease.

A sum of money doubles itself in 88 years. What is the rate of interest?

Find the rate of interest when the following are given.Principal == Rs. 96009600Time =6=6 yearsI=I=Rs. 7272

Find the principal, when the following are given.I=I= Rs. 2020Rate == 2\%2\% per annumTime == 22 years

At what rate of simple interest will a sum of money double itself in 20 years?

Find the difference between compound interest and simple interest on Rs.80008000 in 22 years an at 5\%5\% per annum.

Certain sum of money deposited at simple interest yield Rs. 4400 4400 in two years and Rs. 5500 5500 in a 44 years. Find the rate of interest.

A loan of Rs\ 2500Rs\ 2500 was repaid giving Rs\ 2625Rs\ 2625 what is the interest paid ?

The simple interest on a certain sum of money for 3 years at 5% per annum is Rs.1,200. Find the principle amount.

Find the interest on Rs\ 10,000Rs\ 10,000 for 3\dfrac {1}{4}years3\dfrac {1}{4}years at the rate of 3.4\%3.4\% p.a. Also find the amount.

A sum of Rs. 400Rs. 400 is lent at the rate of 5\%5\% per annum. Find the interest at the end of 22 years.

The cost of the car is increased 20\%20\%. This year cost is 240000240000. What was the cost is last year.

If a sum of Rs. 1616 becomes Rs. 8181 in 44 yrs, then find the rate of interest at compound interest.

The simple interest on certain principle at the rate of 15% per annum for 5 years is Rs.Find the principle amount ?

A sum of Rs. 400Rs. 400 is lent for 33 years at the rate of 6\%6\% per annum. Find the interest.

Find the compound interest on {\rm{Rs}}\,{\rm{3000}}{\rm{Rs}}\,{\rm{3000}} at 5\% 5\% per annum for 22 years.

At what rate percent per annum will Rs. 630Rs. 630 produce an interest of Rs. 126Rs. 126 in 44 years?

On what principle will the simple interest be Rs. 7008Rs. 7008 in 66 years 33 month at 5\%5\% per year?

15\%15\% increase in price of an article, which is Rs. 1620Rs. 1620is the increase of Rs.Rs. ________

Find the principle which will amount to Rs. 4000Rs. 4000 in 44 years at 6.25\%6.25\% per annum.

Increase of a number from 150150 to 162162 is equal to increase of _____ percent.

A sum of Rs 1,780 becomes Rs 2,316 in 4 years, Find: The rate of interest.

The simple interest earned on a certain sum is 30% of the sum. Find the rate of interest.

What is the rate of interest, if Rs 3,750 amount to Rs 4,650 in 4 years?

Match each of the entries in Column I with the appropriate entries in Column II :

The simple interest on a certain sum of money is \dfrac 38\dfrac 38 of the sum in 6464 years. Find the rate percent charged.

Peter invested 240000240000 for 22 years at 10\%10\% per annum compounded annually. If 20\%20\% of the accrued interest at the end of each year is deducted as income tax, find the amount he received at the end of two years.

The cost of a scooter depreciates every year by 15% of its value at beginning of the year. If the present cost of the scooter is Rs. 8,000; find its cost (in Rs.) after one year:

A sum of Rs.8,000Rs.8,000 is invested for 22 years at 10 \%10 \% per annum compound interest. Calculate final amount at the end of the second year.

A man borrowed Rs.20,00020,000 for 22 years at 88 % per year compound interest. Calculate the compound interest of two years.

A man borrowed Rs.20,00020,000 for 22 years at 88 % per year compound interest. Calculate the final amount at the end of the second year.

Mohan borrowed Rs. 16,000Rs. 16,000 for 33 years at 5 \%5 \% per annum compound interest. Calculate the amount that Mohan will pay at the end of 33 years to complete the debt.

A man invests Rs. 9{,}6009{,}600 at 10 \%10 \% per annum compound interest for 33 years. Calculate the amount at the end of the first year.

A sum of Rs.8,000 is invested for 1 years at 10 % per annum compound interest. Calculate compound interest earned in 2 years.

Calculate the compound interest for the second year on Rs. 15,000Rs. 15,000 invested for 22 years at 6\%6\% per annum.

Rekha borrowed Rs. 40,000 for 3 years at 10 % per annum compound interest. Calculate the interest earned in the second year.

A man borrowed \text{Rs }20,000\text{Rs }20,000 for 22 years at 8\%8\% per year compound interest. Calculate the interest of the second year.

A man borrowed Rs. 20,00020,000 for 22 years at 8\%8\% per year compound interest. Calculate the interest of the first year.

A man invests Rs. 5,6005,600 at 1414% per annum compound interest for 22 years. Calculate the interest for the first year.

Calculate the compound interest on \text{Rs. }3,500\text{Rs. }3,500 at 10\%10\% per annum in 22 years.

A person invests Rs. 5,0005,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,6005,600. Calculate the rate of interest per year. Find the amount at the end of the second year.

A man invests Rs. 5,6005,600 at 1414% per annum compound interest for 22 years. Calculate the amount at the end of the first year.

A man invests Rs. 9,6009,600 at 10 \%10 \% interest per annum for 33 years. The interest for the second year is

A sum of Rs. 1,2001,200 lent out at C.I. amounts to Rs. 1,4521,452 in 22 years. Calculate the rate of interest.

Calculate the amount and the compound interest on Rs 4,5004,500 in 22 years when the rates of interest on successive years are 1010% and 1212% respectively.

A person invests Rs. 5,0005,000 for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. 5,6005,600. Calculate the rate of interest per year.

The compound interest for the two years on Rs. 8,000/- invested for 3 years at 10% per annum is :

A man invests Rs. 30,80030,800 at 55% per annum compound interest for 33 years. Calculate the interest for the first year.

Calculate the compound interest for the second year on Rs. 8,000/-8,000/- invested for 33 years at 1010% per annnum.

AA borrowed Rs. 2,5002,500 from BB at 12\%12\% per annum at compound interest. After 22 years, AA gave Rs. 2,9362,936 and a watch to BB to clear the account. Find the cost of the watch.

Calculate the difference between the simple interest and the compound interest on Rs. 4,0004,000 in 22 years at 88% per annum compounded yearly.

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

The population of a place increased to 5400054000 in 20032003 at a rate of 5\%5\% per annum. Its population in 20052005 will be PP. What is the sum of the digits in PP

Find the compound interest on Rs. 1000 at the rate of 10% per annum for 3 years.

The population of a town increases every year at the rate of 25 per thousand. The present population of the town isFind Population after 2 years.

Find the compound interest on Rs.8000 for 3 years at 5% per annum.

The number of bacteria in a certain culture doubles every hour. If there were $30$ bacteria present in the culture originally, how many bacteria will be present at the end of $2^{nd}$ hour $4^{th}$ hour and $n^{th}$ hour?

Simple interest is given by $I=\dfrac{PTR}{100}$ .If p=Rs.900, T=2 years, and $R=5\%$ , find the simple interest.

What is the rate of interest (%) of Rs.92,800 for $3\dfrac{1}{2}$ years , the interest of which is Rs.16240 ?

The interest is supposed compound unless the contrary is state.
A person borrows $672$ pounds to be repaid in $5$ years by annual instalments of $120$ pounds; find the rate of interest, reckoning simple interest.

A certain sum of money amounts to $Rs. 6300/-$ in $2$ years and $Rs. 7875/-$ in $3$ years $9$ months at simple interest find the rate of interest per annum.

A girl deposits $300$ in a bank account. The bank offers $7%$ intersection per year.
Assuming the girl does not take any money out of the account, calculate:

Find depreciation at $10\%\ pa$ on furniture costing $Rs\,10,000$ for $9$ month

The population of Hyderabad was $68,09,000$ in the year $2011$ . If it increases at the rate of $4.7\%$ per annum. What will be the population at the end of the year $2015$ .

Michael finished coloring a picture in $\dfrac{7}{{12}}$ hour. Vaibhav finished coloring the same picture in $\dfrac{3}{{4}}$ hour. Who worked longer. By what fraction was it longer

Find the rate per cent per annum if :
$Rs\ 600$ yields $Rs\ 238$ interest in $4$ years.

Find the rate per cent per annum if :
$Rs\ 700$ yields $Rs\ 105$ interest in $3$ years.

What principal will amount to $Rs \ 15000$ at $10\%$ per annum in $5 \ years$ ?

Find the rate per cent per annum if :
$Rs\ 350$ yields $Rs\ 105$ interest in $6$ year.

At what rate per cent per annm will $Rs 3600$ amount to $Rs \ 4734$ in $3\dfrac {1}{2}years?$

The price of a garment has been reduced by $15\%$ in a sale to $Rs\ 306$ . Find its original price.

Find the principal, when the following are given.
$I=$ Rs. $192$
Rate $=$ $6\%$ per annum
Time $=$ $4$ years

The population of a city decreased from $25,000$ to $24,500$ . Find the percentage decrease.

A sum of money doubles itself in $8$ years. What is the rate of interest?

Find the rate of interest when the following are given.
Principal $=$ Rs. $9600$
Time $=6$ years
$I=$ Rs. $72$

Find the principal, when the following are given.
$I=$ Rs. $20$
Rate $=$ $2\%$ per annum
Time $=$ $2$ years

Find the difference between compound interest and simple interest on Rs. $8000$ in $2$ years an at $5\%$ per annum.

Certain sum of money deposited at simple interest yield Rs. $4400$ in two years and Rs. $5500$ in a $4$ years. Find the rate of interest.

A loan of $Rs\ 2500$ was repaid giving $Rs\ 2625$ what is the interest paid ?

Find the interest on $Rs\ 10,000$ for $3\dfrac {1}{4}years$ at the rate of $3.4\%$ p.a. Also find the amount.

A sum of $Rs. 400$ is lent at the rate of $5\%$ per annum. Find the interest at the end of $2$ years.

The cost of the car is increased $20\%$ . This year cost is $240000$ . What was the cost is last year.

If a sum of Rs. $16$ becomes Rs. $81$ in $4$ yrs, then find the rate of interest at compound interest.

A sum of $Rs. 400$ is lent for $3$ years at the rate of $6\%$ per annum. Find the interest.

Find the compound interest on ${\rm{Rs}}\,{\rm{3000}}$ at $5\%$ per annum for $2$ years.

At what rate percent per annum will $Rs. 630$ produce an interest of $Rs. 126$ in $4$ years?

On what principle will the simple interest be $Rs. 7008$ in $6$ years $3$ month at $5\%$ per year?

$15\%$ increase in price of an article, which is $Rs. 1620$
is the increase of $Rs.$ ________

Find the principle which will amount to $Rs. 4000$ in $4$ years at $6.25\%$ per annum.

Increase of a number from $150$ to $162$ is equal to increase of _____ percent.

A sum of Rs 1,780 becomes Rs 2,316 in 4 years, Find:
The rate of interest.

The simple interest on a certain sum of money is $\dfrac 38$ of the sum in $64$ years. Find the rate percent charged.

Peter invested $240000$ for $2$ years at $10\%$ per annum compounded annually. If $20\%$ of the accrued interest at the end of each year is deducted as income tax, find the amount he received at the end of two years.

A sum of $Rs.8,000$ is invested for $2$ years at $10 \%$ per annum compound interest. Calculate final amount at the end of the second year.

A man borrowed Rs. $20,000$ for $2$ years at $8$ % per year compound interest. Calculate the compound interest of two years.

A man borrowed Rs.20,000 for 2 years at 8 % per year compound interest. Calculate the final amount at the end of the second year.

Mohan borrowed $Rs. 16,000$ for $3$ years at $5 \%$ per annum compound interest. Calculate the amount that Mohan will pay at the end of $3$ years to complete the debt.

A man invests Rs. $9{,}600$ at $10 \%$ per annum compound interest for $3$ years. Calculate the amount at the end of the first year.

Calculate the compound interest for the second year on $Rs. 15,000$ invested for $2$ years at $6\%$ per annum.

A man borrowed $\text{Rs }20,000$ for $2$ years at $8\%$ per year compound interest. Calculate the interest of the second year.

A man borrowed Rs. $20,000$ for $2$ years at $8\%$ per year compound interest. Calculate the interest of the first year.

A man invests Rs. $5,600$ at $14$ % per annum compound interest for $2$ years. Calculate the interest for the first year.

Calculate the compound interest on $\text{Rs. }3,500$ at $10\%$ per annum in $2$ years.

A person invests Rs. $5,000$ for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. $5,600$ . Calculate the rate of interest per year. Find the amount at the end of the second year.

A man invests Rs. $5,600$ at $14$ % per annum compound interest for $2$ years. Calculate the amount at the end of the first year.

A man invests Rs. $9,600$ at $10 \%$ interest per annum for $3$ years. The interest for the second year is

A sum of Rs. $1,200$ lent out at C.I. amounts to Rs. $1,452$ in $2$ years. Calculate the rate of interest.

$4,500$ in $2$ years when the rates of interest on successive years are $10$ % and $12$ % respectively.

A person invests Rs. $5,000$ for two years at a certain rate of interest compounded annually. At the end of one year, this sum amounts to Rs. $5,600$ . Calculate the rate of interest per year.

A man invests Rs. $30,800$ at $5$ % per annum compound interest for $3$ years. Calculate the interest for the first year.

Calculate the compound interest for the second year on Rs. $8,000/-$ invested for $3$ years at $10$ % per annnum.

$A$ borrowed Rs. $2,500$ from $B$ at $12\%$ per annum at compound interest. After $2$ years, $A$ gave Rs. $2,936$ and a watch to $B$ to clear the account. Find the cost of the watch.

Calculate the difference between the simple interest and the compound interest on Rs. $4,000$ in $2$ years at $8$ % per annum compounded yearly.

The population of a place increased to $54000$ in $2003$ at a rate of $5\%$ per annum. Its population in $2005$ will be $P$ . What is the sum of the digits in $P$

In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of $2.5\%$ per hour. Find the bacteria at the end of $2$ hours, if the count was initially $5,06,000$ .

A new car costs Rs. $360000$ .Its price depreciates at the rate of $10$ % a year during the first two years and at the rate of $20$ % a year thereafter.What will be the price of the car after $3$ years?

The population of a village declines by $5\%$ every year.If the present population of the village is $1444$ , find the population of the village $2$ year back.

The bacteria in a a culture grows by $10$ % in the first hour, decreases by $10$ % in the second hour and again increases by $10$ % in the third hour.If the count of bacteria in the sample is $121000000$ , what will be the count of bacteria after 3 hours?

Find the compound interest on Rs. 8000 for $\displaystyle 1\frac{1}{2}$ years at 10% per annum, interest being payable half yearly.

In a factory, the production of scooters rose to $48400$ from $40000$ in $2$ years. Find the rate of growth per annum.

Afridi purchased an old scooter for $Rs. 16000$ . If the cost of scooter after $2$ years depreciates to $Rs. 14440$ , find the rate of depreciation.

The present population of a village is $130,000$ . If the population increase each year by $4$ % of what it had been at the beginnning of each year, then the population of the village after four years will be $x$ . Find the sum of the digits of $x$ .

What will $Rs 500$ amounts to in $10$ years after its deposit in a bank which pays annual interest rate of $10 \%$ compounded annually?

A manufacturer reckons that the value of the machine, which cost him Rs. $15625$ , will depreciate each year by $20$ %. Find the estimated value at the end of $5$ years.

Mrs. Kapoor invested Rs. 6,000 every year at the beginning of the year, at 10% per annum compound interest. Calculate the amount of her total savings :
(i) upto the end of the second year.
(ii) at the beginning of the third year.

What sum of money will amount to Rs. $9,261$ in $3$ years at $5\%$ per annum compound interest

Arif took a loan of $\text{Rs. } 80,000$ from a bank. If the rate of interest is $10\%$ per annum, find the amounts he would be paying after $1\dfrac{1}{2}$ years if the interest is compounded annually.

What is the compound interests on Rs. $5000$ for $\dfrac{3}{2}$ years at $4\%$ per annum compounded yearly?

Find the amount of Rs. $6000$ for $2\dfrac{1}{2}$ years, compounded annually at $5\%$ per annum. Also find the C.I.

Find the compound interest on Rs. $12000$ for $4$ years at $5\%$ per annum.

Find the value of a air conditioner after $2$ years, if its price while new is Rs. $3000$ and it depreciates by $10\%$ per month.

Toby invested Rs. $1200$ for $2$ years in a savings account. His account depreciates by $8\%$ per annum compound interest. How much did Toby have in his savings account after $2$ years?

The annual growth of capital of a business is $7\%$ . At present, the capital is Rs. $12000$ . What will be the capital after $3$ years?

A computer is sold for Rs. $9000$ cash on Rs. $3000$ cash down followed by $8$ equal instalments of Rs. $2000$ each. What is the rate of interest?

Due to migration to cities, the population of a town decreases every year by $10\%$ . If the present population of the town is $14600$ , find the population two years hence.

The annual increase in the population of a state is $2\%$ . What was the population of the state two years before if it is $54,500$ now?