MCQ Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 4

Calculate the compound interest for the third year on

$Rs.\ 7,500$ invested for

$5$ years at

$10\%$ per annum.

0%
$Rs. 907.50$
0%
$Rs.\ 2482.50$
0%
$Rs.\ 2582.50$
0%
$Rs.\ 982.50$

Explanation

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

$P=Rs.\ 7500, R=10% , T=3$ years

$A=7500\left(1+\dfrac{10}{100}\right)^3$

$A=7500\times \dfrac{110}{100}\times \dfrac{110}{100}\times \dfrac{110}{100}$

$A=\dfrac{75\times 121\times 11}{10}$

$A=Rs. 9982.50$

For

$2$ years

$A=7500\left(1+\dfrac{10}{100}\right)^2$

$A=7500\times \dfrac{110}{100}\times \dfrac{110}{100}$

$A=75\times 121$

$A=Rs. 9075$

Compound interest for third year =

$Rs.\ 9982.50-9075=907.50 Rs.$

Hence, option A.

The value of a machine depreciated by

$10 \%$ per year during the first two years and

$15 \%$ per year during the third year. Express the total depreciation of the machine, as per cent, during the three years.

0%
$43.21 \%$
0%
$31.15 \%$
0%
$23.14 \%$
0%
$18.34 \%$

Explanation

Lets Price of article is

$100$
Dep value after

$1$ year

$=100-\dfrac{100\times10}{100}$ =

$90$

Dep. value after

$2$ year

$=90- \cfrac{90\times 10}{100}= 81$
Dep. value after

$3$ years

$=81-\cfrac{81\times 15}{100}= 68.85$
Total Dep.

$=100-68.85$ =

$31.15 \%$

Calculate the amount and the compound interest on Rs.

$6,000$ in

$3$ years at

$5\%$ per year.

0%
Rs. 6,945.75 and Rs. 945.75
0%
Rs. 6,845.75 and Rs. 945.75
0%
Rs. 6,945.75 and Rs. 945.85
0%
Rs. 6,945.75 and Rs. 945.70

Explanation

C.A =

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

Given

$R=5\%$ and

$P=6000 ,T=3$

So C.A =

$6000\left (1+ \frac{5}{100} \right )^{3}$

So C.A =

$6000\left ( \frac{105}{100} \right )\left ( \frac{105}{100} \right )\left ( \frac{105}{100} \right )$

C.A

$=6945.75$ Rs

C.I

$=6945.75 -6000=945.75$ Rs

A man invests Rs.

$30,800$ at

$5$ % per annum compound interest for

$3$ years. Calculate:

the interest for the first year.

0%
Rs. $1,540$
0%
Rs. $1,627$
0%
Rs. $1,721$
0%
Rs. $1,863$

Explanation

Given,
Principle,

$P=30800$ .
Rate of interest,

$R=5$ %.
Time,

$T=3$ years.

Compound interest is given as,

$CI=P[(1+\frac{R}{100})^T-1]$ ...... (1)

Put

$P=30800$ ,

$R=5$ %, and

$T=1$ in equation (1).

$CI=30800[(1+\frac{5}{100})^1-1]$

$=30800[(\frac{105}{100})-1]$

$=30800(\frac{5}{100})$

$=308\times5$

$\therefore CI=1540$ Rs

The value of an article which was purchased

$2$ years ago, depreciates at

$12$ % per annum. If its present value is Rs .

$9680$ , the price at which it was purchased is:

0%
Rs. $10,000$
0%
Rs. $12,500$
0%
Rs. $14,575$
0%
Rs. $16,250$

Explanation

Given,

$A=$ Rs.

$9680,T=2$ years,

$R=12\%$

$\because A = P\left (1 - \displaystyle \frac{R}{100} \right)$

$9680 = P \left(1 - \displaystyle \frac{12}{100} \right)^{2}$

$9680=P\left(\dfrac{88}{100}\right)^2$

$P = 9680 \times \displaystyle \frac{25}{22} \times \frac{25}{22}$

$=$ Rs.

$12,500$

Mohan lent some amount of money at

$9\%$ simple interest and an equal amount of money at

$10\%$ simple interest each for

$2$ years. If his total interest was Rs.

$760$ , what amount was lent in each case?

0%
Rs. $1700$
0%
Rs. $1800$
0%
Rs. $1900$
0%
Rs. $2000$

Explanation

Let the amount lent in each case be Rs.

$x$

$\Rightarrow \dfrac{x\times9\times2}{100}+\dfrac{x\times10\times2}{100}=760$

$\Rightarrow \dfrac{18x}{100}+\dfrac{20x}{100}=760$

$\Rightarrow\;38x=76000$

$\Rightarrow\;x=2000$

Calculate the amount and the compound interest on Rs.

$16,000$ in

$3$ years, when the rates of the interest for successive years are

$10\%$ ,

$14\%$ and

$15\%$ respectively.

0%
Rs. $23.073.60$ and Rs. $7,073.60$
0%
Rs. $23.073.60$ and Rs. $7,873.60$
0%
Rs. $23.873.60$ and Rs. $7,073.60$
0%
Rs. $28.073.60$ and Rs. $7,073.60$

Explanation

Int-rest amt after one year =

$\dfrac{16000\times 10}{100}= 1600$ Rs

C.A after one year

$=16000+1700=17600$ Rs

Intrest after two year =

$\dfrac{17600\times 14}{100}= 2464$

C.A after two years

$=17600+2464=20064$ Rs

Intrest after thrree year =

$\dfrac{20064\times 15}{100}= 3009.60$

C.A after three years

$=20064+3009.60=23073.60$ Rs

intrest after three year

$=23073.60-16000=7073.60$ Rs

Calculate the amount after five years on

$Rs. \ 10,000$ invested for

$5$ years at

$10 \%$ per annum.

0%
$16105.1$
0%
$15000$
0%
$16288.95$
0%
None of these

Explanation

Given P = Rs.

$10000$ , r

$= 10$ %, n

$= 1$ and nt

$= 5$ years

Formula for Amount with compound interest =

$P { \left( 1+\frac { r }{100} \right) }^{ nt }$

A

$= 10000 { \left( 1+\frac {10}{ 100 } \right) }^{ 5}$

$= 10000 { \left( 1.1 \right) }^{ 5 }$

$= 10000 (1.61051)$

A

$= 16105.1$

Ans Amount after

$5$ years = Rs.

$16105.1$

Arun borrowed a sum of money from Jayant at the rate of

$8\%$ simple interest for the first four years,

$10\%$ p.a. for the next

$6$ years and

$12\%$ p.a. beyond

$10$ years. If he pays a total of Rs.

$12160$ as interest only at the end of

$15$ years, how much money did he borrow?

0%
Rs. $12000$
0%
Rs. $10000$
0%
Rs. $8000$
0%
Rs. $9000$

A person lent a certain sum of money at

$4\%$ simple interest and in

$5$ years, the interest amounted to Rs.

$520$ less than the sum lent. The sum lent was

0%
Rs. $.600$
0%
Rs. $650$
0%
Rs. $700$
0%
Rs. $750$

Explanation

Let the sum lent be

$Rs. x$

The interest in

$5$ years

$=\dfrac{x\times 4\times 5}{100}$

Given:

$\dfrac{x\times4\times5}{100}=x-520$

$\Rightarrow \dfrac{x}{5}=x-520$

$\Rightarrow x-\dfrac{x}{5}=520$

$\Rightarrow \dfrac{4x}{5}=520$

$\Rightarrow 4x=520\times 5$

$\Rightarrow x=\dfrac{520\times 5}{4}$

$\Rightarrow\;x=$ Rs.

$650$

Therefore, the sum lent is Rs.

$650$

Find the compound interest by using simple interest approach for Rs 5000 for 3 years at 8% per annum compounded annnually.

0%
$1298.56$
0%
$1289.00$
0%
$1.200.89$
0%
None of these

What is the compound interest on an amount of Rs.

$4800$ at the rate of

$6$ percent p.a. at the end of

$2$ years?

0%
Rs. $544.96$
0%
Rs. $576$
0%
Rs. $593.28$
0%
Rs. $588$

Explanation

Given,

$P =$

$Rs.$

$4800, r = 6\%$

$p.a.$ ,

$n = 2$

$\displaystyle \therefore C.I.=A-P=P\left ( 1+\frac{r}{100} \right )^{n}-P$

$\displaystyle =4800\left ( 1+\frac{6}{100} \right )^{2}-4800$

$\displaystyle =4800\times \dfrac{106\times 106}{100\times 100}-4800$

$= 5393.28 - 4800 =$

$Rs.$

$593.28$

The compound interest on Rs.

$30,000$ at

$7 \%$ per annum Rs.

$4347$ . The period ( in years) is

0%
$2$
0%
$\displaystyle 2\frac{1}{2}$
0%
$3$
0%
$4$

Explanation

$P=$ Rs.

$30000, r = 7\%$ p.a.

$C.I.$

$=$ Rs.

$4347, n = ?$

$\displaystyle \Rightarrow$ Amount

$=$ Rs.

$30000 +$ Rs.

$4347 =$ Rs.

$34347$

$\displaystyle \therefore 34347=30000\left ( 1+\frac{7}{100} \right )^{n}$

$\Rightarrow \left ( \dfrac{107}{100} \right )^{n}=\dfrac{34347}{30000}=\dfrac{11449}{10000}$

$\displaystyle \Rightarrow \left ( \frac{107}{100} \right )^{n}=\left ( \frac{107}{100} \right )^{2}$

$\Rightarrow n=2$

The principle that amounts to Rs.

$4913$ in

$3$ years at

$\displaystyle 6\frac{1}{4}$ % per anmum compound interest compounded annually is

0%
Rs. $4096$
0%
Rs. $4085$
0%
Rs. $4076$
0%
Rs. $3096$

Explanation

Given,

$\displaystyle A=$ Rs.

$4913,n=3,r=6\dfrac{1}{4}\%=\frac{25}{4}\%,P=?$

$\displaystyle \therefore 4913=P\left ( 1+\frac{25}{400} \right )^{3}$

$\Rightarrow 4913=P\left ( 1+\dfrac{1}{16} \right )^{3}$

$\displaystyle \Rightarrow 4913=P\left ( \frac{17}{16} \right )^{3}$

$\Rightarrow 4913=P\times \dfrac{4913}{4096}$

$\Rightarrow P=$ Rs.

$4096$

The value of a machine depreciates at the rate of

$10\%$ every year. it was purchased

$3$ years ago . If its present value is Rs.

$8748$ , its purchase price was

0%
Rs. $10000$
0%
Rs. $11372$
0%
Rs. $12000$
0%
Rs. $12500$

Explanation

We know that

Final price

$=$ initial price

${ \left( 1+\frac { rate }{ 100 } \right) }^{ time }$

Here the final price

$=$ Rs.

$8784$ , time

$=3$ yrs, rate

$=-10\%$ p.a.

The rate is negative since the price is depriciating.

Let the initial price

$=$ Rs.

$x$ .

$\therefore x\times { \left( 1-\dfrac { 10 }{ 100 } \right) }^{ 3 }=8748$

$\Rightarrow x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } =8748$

$\Rightarrow x=8748\times \dfrac { 10 }{ 9 } \times \dfrac { 10 }{ 9 } \times \dfrac { 10 }{ 9 } =$ Rs.

$12000$

So, the price of the machine

$3$ years back

$=$ Rs.

$12000$ .

Ans- Option C.

A tree grows annually by one-eighth of its height. By how much will it grow after 2 years, if it stands today 64 cm high?

0%
72 cm
0%
74 cm
0%
75 cm
0%
81 cm

Explanation

Present height of tree=64cm
It grow annually by one-eighth of its height
Then it grow in first year=

$\frac{1}{8}\times 64=8 cm$
Then total height=

$64+8=72$
In second year it grow =

$\frac{1}{8}\times 72=9 cm$
then total height after 2 year=

$72+9=81 cm$

The half life of Uranium -

$233$ is

$160000$ years i.e., Uranium

$233$ decays at a constant rate in such a way that it reduces to

$50\%$ in

$160000$ years. in how many years will it reduce to

$25 \%$ ?

0%
$80000$ years
0%
$240000$ years
0%
$320000$ years
0%
$40000$ years

Explanation

The half life of

${ U }_{ 233 }$

$=160000$ yrs

$\therefore$ It becomes

$50\%$ of the original amount in

$160000$ yrs

Now

$25\%=$

$\dfrac { 1 }{ 2 } \times$

$50\%$

i.e Another half life is needed to reduce the original amount into

$25\%$

So, another

$160000$ yrs. will reduce the original amount into

$25\%$

$\therefore$ The required total number of years

$=160000+160000=320000$

i.e

$320000$ yrs

The cost of a vehicle is Rs.

$1,75,000$ . If its value depreciates at the rate of

$20$ % per annum, then the total depreciation after

$3$ years will be:

0%
Rs. $86,400$
0%
Rs. $82,500$
0%
Rs. $84,500$
0%
Rs. $85,400$

Explanation

Value of the vehicle after 3 years

$= 1,75,000 \times \left(1 - \displaystyle \frac{20}{100} \right)^{3}$

$= 1,75,000 \times \displaystyle \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}$
= Rs. 89, 600

$\therefore$ Total depreciation = 1, 75, 000 - 89, 600

$= Rs. 85,400$

A sum of money invested at compound interest amount in

$3$ years to Rs.

$2400$ and in

$4$ years to Rs.

$2520$ . The interest rate per annum is

0%
$5 \%$
0%
$6 \%$
0%
$10 \%$
0%
$12 \%$

Explanation

Let the sum of money invested be Rs.

$x$ and interest rate per annum

$= r\%$
Then

$\displaystyle x\left ( 1+\frac{r}{100} \right )^{3}=$ Rs.

$2400$ ........(i)
and

$\displaystyle x\left ( 1+\frac{r}{100} \right )^{4}=$ Rs.

$2520$ ........(ii)
Dividing equation (ii) by (i), we get

$\displaystyle \left ( 1+\frac{r}{100} \right )=\frac{2520}{2400}$

$\Rightarrow \dfrac{r}{100}=\dfrac{2520-2400}{2400}=\dfrac{120}{2400}$

$\Rightarrow r=\dfrac{1}{20}\times 100=5\%$ p.a.

The population of a town increases annually by

$25 \%$ if the preset population is

$1$ crore, then what was the difference between the population three years ago and two years ago?

0%
$25,00,000$
0%
$12,80,000$
0%
$15,60,000$
0%
None of these

Explanation

Given that,

Population increases annually by

$25\%$

Let population

$3$ years ago be

$x$

$\Rightarrow$ Population two years ago was

$\dfrac{125}{100}x$

Population last year was

$\dfrac{125\times125}{10000}x$

Population this year is

$\dfrac{125\times125\times125}{1000000}x = 10000000$

$\Rightarrow$ Population three years ago was

$x = 5120000$

$\therefore$ Difference between population three years ago and two years ago

$= 25\%$ of

$x$

$= 1280000$

A machine depreciates in value each year at the rate of

$10\%$ of its previous value. However every second year there is some maintenance work, so that in that particular year depreciation is only

$5\%$ of its previous value. If at the end of the fourth year the value of the machine stands at Rs.

$146205$ , then find the value of the machine at the start at the first year?

0%
Rs. $190000$
0%
Rs. $200000$
0%
Rs. $195000$
0%
Rs. $210000$

Explanation

Let the value of the machine, at the onset, be

$x$ when

$x$ is in Rs.

Here we shall apply the rule

Final value

$=$ original value

$\times { \left( 1+rate \right) }^{ time }$ .

Here the rate will be negative since the value is depriciating.

For two yrs. the rate of depriciation is

$10\%$ .

$\therefore$ the value after

$2$ yrs.

$=$

$x\times { \left( 1-\dfrac { 10 }{ 100 } \right) }^{ 2 }=x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 }$ .

As given, the rate of depriciation for next

$2$ yrs

$=5\%$

$\therefore$ The value after next

$2$ yrs

$=$

$x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times { \left( 1-\dfrac { 5 }{ 100 } \right) }^{ 2 }$

$=x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 19 }{ 20 } \times \dfrac { 19 }{ 20 }$

But the final value after

$4$ yrs

$=$ Rs.

$146205$ .

$\therefore \quad x\times \dfrac { 9 }{ 10 } \times \dfrac { 9 }{ 10 } \times \dfrac { 19 }{ 20 } \times \dfrac { 19 }{ 20 } =146205$

$\Rightarrow x=\dfrac { 146205\times 40000 }{ 81\times 19\times 19 } =5\times 40000$

$\Rightarrow x=$ Rs.

$200000$ .

So,

$4$ yrs. back, the value of the machine was Rs.

$200000$ .

The current birth rate per thousand is

$32$ , whereas the corresponding death rate is

$11$ per thousand. The net growth rate in terms of population increase in per cent is given by,

0%
$0.0021 \%$
0%
$0.021 \%$
0%
$2.1 \%$
0%
$21 \%$

Explanation

Given, birth rate per

$1000$ is

$32$ and corresponding death rate is

$11$ per thousand.

therefore, net growth on

$1000=(32-11)=21$

Hence, net growth on

$100=\dfrac {21}{1000}\times 100=2.1\%$

Therefore, net growth rate in terms of population increase in per cent is

$2.1\%$ .

What will be the compound interest on a sum of Rs.25000 after 3 years at the rate of 12 per cent p.a.?

0%
Rs. 10123.20
0%
Rs. 11123.20
0%
Rs. 12123.20
0%
Rs. 13123.20

Explanation

P = Rs. 25000, n = 3 years, r = 12% p.a

$\displaystyle \therefore$ Amount

$\displaystyle =P\left ( 1+\frac{r}{100} \right )^{n}=Rs.25000\times \left ( 1+\frac{12}{100} \right )^{3}$

$\displaystyle =Rs.25000\times \left ( \frac{112}{100} \right )^{3}=Rs.25000\times \frac{28}{25}\times \frac{28}{25}\times \frac{28}{25}=Rs.35123.20$

$\displaystyle \therefore$ Compound interest

$= Rs. (35123.20-25000)=Rs.10123.20$

The population of a town increases by 5% annually. If the population in 2009 is 1,38,915, what was it in 2006 ?

0%
100,000
0%
110,000
0%
120,000
0%
130,000

Explanation

Let the population in 2006 be

$x$ .
Pop. in

$\displaystyle 2009=x\times \left ( 1+\cfrac{5}{100} \right )^{3}$

$\Rightarrow 1,38,915=x\times \left ( \cfrac{21}{20} \right )^{3}$

$\Rightarrow x=\cfrac{138915\times 8000}{9261}=120000$

The population of a town was 1,60,000 three years ago. If it increased by 3%, 2.5% and 5%, respectively, in the last three years, then what is the present population?

0%
166,366
0%
177,366
0%
166,377
0%
177,377

Explanation

Present population

$\displaystyle =1,60,000\left ( 1+\frac{3}{100} \right )\left ( 1+\frac{2.5}{100} \right )\left ( 1+\frac{5}{100} \right )$

$\displaystyle =160000\times \frac{103}{100}\times \frac{102.2}{100}\times \frac{105}{100}=177366$

The population of a village isIf the population increases by 10% in the first year, by 20% in the second year, and due to mass exodus, it decreases by 5% in the third year, what will be its population after 3 years ?

0%
10,540
0%
11,540
0%
12,540
0%
13,540

Explanation

Population after 3 years

$\displaystyle =10000\left ( 1+\frac{10}{100} \right )\left ( 1+\frac{20}{100} \right )\left ( 1-\frac{5}{100} \right )$

$\displaystyle =10000\times \frac{11}{10}\times \frac{6}{5}\times \frac{19}{20}=12540$

On what sum of money lent out at 9% per annum for 6 years does the simple interest amounts to Rs. 810?

0%
Rs. 1000
0%
Rs. 1500
0%
Rs. 1200
0%
None

Explanation

Simple Interest

$=\text{Rs.}\ 810$

$R= 9\%$

$T = 6\ \text{years}$

$P=?$

According to the formula of Simple Interest:

Simple Intrest

$I\, =\, \displaystyle \frac {P\times T\times R}{100}$

$P\, =\, \displaystyle \frac {100 \times I}{T\times R}$

$=\, \displaystyle \frac {100 \times 810}{6 \times 9}$

$=\, \text{Rs.}\, 1500$

A sum of Rs. 12,000 deposited at compound interest doubles after 5 years. After 20 years it will become

0%
Rs. 1,20,000
0%
Rs. 1,92,000
0%
Rs. 1,24,000
0%
Rs. 96,000

Explanation

According to the question
After 5 year

$12000\times (1+\frac{R}{100})^5=24000$

$(1+\frac{R}{100})^5=2$
After 10 year

$(((1+\frac{R}{100})^5)^4)=2^4$

$(1+\frac{R}{100})^{20}=16$

$P(1+\frac{R}{100})^{20}=16\times P=16\times 12000=192000$

The annual increase in the population of a town is 10%. If the present population of the town is 180,000, then what will be its population after two years ?

0%
200,000
0%
210,000
0%
217,800
0%
207,800

Explanation

Population after 1 year

$= 180000 + 180000\times \dfrac{10}{100} = 198000$

Population after 2 years

$= 198000 + 198000\times \dfrac{10}{100}$

$=217800$

A sum of money at compound interest (compounded annually) doubles itself in

$4$ years. In how many years will it amount to eight times of itself ?

0%
12 years
0%
10 years
0%
8 years
0%
16 years

Explanation

Given,

$\displaystyle 2P=P\left ( 1+\frac{R}{100} \right )^{4}\\ \Rightarrow \left ( 1+\dfrac{R}{100} \right )^{4}=2 \ \ \ ......(1)$
Let the time in which it amounts to eight times of itself be

$r$ years
Then,

$\displaystyle 8P=P\left ( 1+\frac{R}{100} \right )^{r}\\ \Rightarrow \left ( 1+\dfrac{R}{100} \right )^{r}=8=2^{3}$

$\left ( 1+\dfrac{R}{100} \right )^{r} =2^3=\left ( \left ( 1+\dfrac{R}{100} \right )^{4} \right )^{3}$ from

$(1)$

$\left ( 1+\dfrac{R}{100} \right )^{r}=\left ( 1+\dfrac{R}{100} \right )^{12}$

Bases are equal,

$\implies r =12$ years

$\displaystyle \therefore r = 12$ years

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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