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CBSE Questions for Class 11 Commerce Applied Mathematics Basics Of Financial Mathematics Quiz 4 - MCQExams.com
CBSE
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.
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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.
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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.
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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.
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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:
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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?
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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.
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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.
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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?
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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
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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.
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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?
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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
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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
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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
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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?
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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 $$25 \%$$?
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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:
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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
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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?
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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?
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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,
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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.?
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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 ?
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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?
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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 ?
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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?
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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
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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 ?
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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 ?
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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
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