Class 10: Compound Interest – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: A person invests Rs. 10000 for 3 years at a certain rate of interest compounded annually. At the end of one year this sum amounts to Rs. 11200. Calculate:

the rate of interest
the amount at the end of second year
the amount at the end of third year [2006]

Answer:

The rate of interest:

$\displaystyle P=10000 \text{ Rs.; } A= 11200 \text{ Rs.; } r=x\%$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 11200=10000 \Big(1+ \frac{x}{100} \Big)^1 \Rightarrow x=12\%$

Amount at the end of the second year

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 10000 \Big(1+ \frac{12}{100} \Big)^2 = 12544 \text{ Rs. }$

Amount at the end of third year

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 10000 \Big(1+ \frac{12}{100} \Big)^3 = 14049.28 \text{ Rs. }$

$\displaystyle \\$

Question 2: A sum of Rs. 9600 is invested for 3 years at 10% per annum at compound interest.

What is the sum due at the end of the first year?
What is the sum due at the end of the second year?
Find the difference between the answers of the 2) and 1) and find the interest on this sum (difference) for one year
Hence, write down the compound interest for the third year [1996]

Answer:

Sum due at the end of the first year?

$\displaystyle P=9600 \text{ Rs.; } A= ? \text{ Rs.; } r=10\%; n=1$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 9600 \Big(1+ \frac{10}{100} \Big)^1 = 10560 \text{ Rs. }$

Sum due at the end of the second year?

$\displaystyle P=9600 \text{ Rs.; } A= ? \text{ Rs.; } r=10\%; n=2$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 9600 \Big(1+ \frac{10}{100} \Big)^2 = 11616 \text{ Rs. }$

Difference between the answers of the 2) and 1)

$\displaystyle 11616-10560=1056 \text{ Rs. }$

$\displaystyle Interest = 1056 \times \frac{10}{100} \times 1 = 105.6 \text{ Rs. }$

Compound Interest for third year

$\displaystyle =1056+105.6 =1161.60 \text{ Rs. }$

$\displaystyle \\$

Question 3: What sum of money will amount to Rs. 9261 in 3 years at 5% per annum compound interest? [2009]

Answer:

Given

$\displaystyle P=x \text{ Rs.; } A= 9261 \text{ Rs.; } r=5\%; n=3$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 9261 = x \Big(1+ \frac{5}{100} \Big)^3 \Rightarrow x = 8000 \text{ Rs. }$

$\displaystyle \\$

Question 4: In what period of time will Rs. 12000 yield Rs. 3972 as compound interest at 10%, if compounded on a yearly basis. [2011]

Answer:

Given $\displaystyle P=12000 \text{ Rs.; } A= (12000+3972)= 15972 \text{ Rs.; } r=10\%; n=n$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 15972 = 12000 \Big(1+ \frac{10}{100} \Big)^n \Rightarrow n = 3 \text{ years }$

$\displaystyle \\$

Question 5: On what sum of money will the difference between the compound interest and the simple interest for two years be equal to Rs. 25, if the rate of interest charged for both is 5% p.a.? [2012]

Answer

Let the sum be $\displaystyle x \text{ Rs. }$

Simple Interest for 2 years $\displaystyle = x \times \frac{5}{100} \times 2 = 0.1x$

Amount Compound Interest

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = x \Big(1+ \frac{5}{100} \Big)^2 \Rightarrow A = 1.1025x \text{ Rs. }$

Given difference = 25 Rs.

Therefore $\displaystyle C.I. - S.I. = 25 \Rightarrow (1.1025x-x)-(0.1x) = 25 \Rightarrow x = 10000 \text{ Rs. }$

$\displaystyle \\$

Question 6: The simple interest on a sum of money for 2 years at 4% p.a. is Rs. 340. Find

the sum of the money
the compound interest on this sum for one year payable half-yearly at the same rate [2008]

Answer:

Simple Interest

Given: $\displaystyle I = 340 \text{ Rs.; } n=2 \text{ years; } r=4\%; P=x \text{ Rs.; }$

$\displaystyle 340 = x \times \frac{4}{100} \times 2 \Rightarrow x = 4250 \text{ Rs. }$

Compound Interest

$\displaystyle P= 4250 \text{ Rs.; } n=1 \text{ years; } r=4\% \text{ compounded half yearly;}$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A = 4250 \Big(1+ \frac{4}{2 \times 100} \Big)^2 \Rightarrow A = 4421.7 \text{ Rs. }$

$\displaystyle C.I. = 4421.7-4250 = 171.70 \text{ Rs. }$

$\displaystyle \\$

Question 7: Simple interest on a sum of money for 2 years at 4% is Rs. 450. Find compound interest on the same sum and at the same rate for 1 year, if the interest on the same sum and at the same rate for year, if the interest is reckoned half-yearly. [1997]

Answer:

Simple Interest

$\displaystyle P=x \text{ Rs.; } T=2 \text{years; } r=4\%$

$\displaystyle S.I=x \times \frac{4}{100} \times 2 = \frac{2}{25}x$

$\displaystyle 450 = \frac{2}{25}x \Rightarrow x= 5625 \text{ Rs. }$

Compounded Half Yearly

$\displaystyle P=5625 \text{ Rs.; } r=4\%; \text{ Compounded half yearly } n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{1 \times 2} = 5625 \Big(1+ \frac{4}{2 \times 100} \Big)^{1 \times 2} = 5852.25 \text{ Rs. }$

$\displaystyle \text{ Compound Interest }=5852.25-5625= 227.25 \text{ Rs. }$

$\displaystyle \\$

Question 8: Rohit borrows Rs. 86000 from Arun for 2 years at 5% per annum simple interest. He immediately lends his money to Akshay at 5% compounded interest annually for the same period. Calculate Rohits profit at the end of two years. [2010]

Answer:

Simple Interest for 2 years

$\displaystyle S.I. = P \times \frac{r}{100} \times 2 \text{ Rs. }$

$\displaystyle S.I. = 86000 \times \frac{5}{100} \times 2 = 8600 \text{ Rs. }$

Compound Interest for 2 years

$\displaystyle P=8600 \text{ Rs.; } r=5\%; \text{ Compounded yearly } n=2 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^{2} \Rightarrow A= 86000 \Big(1+ \frac{5}{100} \Big)^{2} \Rightarrow A = 9481.50 \text{ Rs. }$

Gain $\displaystyle =(94815-86000)-8600 = 215 \text{ Rs. }$

$\displaystyle \\$

Question 9: Nikita invests Rs.6000 for two years at a certain rate of interest compounded annually, At the end of first year it amounts to Rs.6720. Calculate;

The rate of interest;
The amount at the end of the second year. [2010]

Answer:

Compound Interest for 1 year

$\displaystyle P=6000 \text{ Rs.; } r=x\%; \text{ Compounded yearly } n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^{1} \Rightarrow A= 6000 \Big(1+ \frac{x}{100} \Big)^{1}$

Given $\displaystyle 6000 \Big(1+ \frac{x}{100} \Big)^{1}=6720 \Rightarrow x= 12\%$

Amount at the end of second year

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^{1} \Rightarrow A= 6000 \Big(1+ \frac{12}{100} \Big)^{2} = 7526.40 \text{ Rs. }$

$\displaystyle \\$

Question 10: If the interest is compounded half-yearly, calculate the amount when principal is Rs.7400; the rate of interest is 5% per annum and the duration is one year. [2005]

Answer:

$\displaystyle P=7400 \text{ Rs.; } r=5\%; \text{ Compounded half yearly } n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{2 \times 100} \Big)^{n \times 2} = 12000 \Big(1+ \frac{5}{2 \times 100} \Big)^{1 \times 2} = 7774.63 \text{ Rs. }$

$\displaystyle \\$

Question 11: At what rate per cent will a sum of Rs.4000 yield Rs.1324 as compound interest in 3 years? [2013]

Answer:

$\displaystyle P=4000 \text{ Rs.; } r=x\%; n=3 \text{ year; Interest } =1324 \text{ Rs. }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 4000 \Big(1+ \frac{x}{100} \Big)^3.$

Given $\displaystyle \text{ Interest } = 1324 \text{ Rs. }$

$\displaystyle \Rightarrow 4000 \Big(1+ \frac{x}{100} \Big)^3 - 4000 = 1324 \Rightarrow x= 10\%$

$\displaystyle \\$

Question 12: The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at $\displaystyle 10\%$ for a year is $\displaystyle Rs.15$ . Find the sum of money lent out. [1998]

Answer:

Simple Interest

$\displaystyle P=x \text{ Rs.; } T=1 \text{ Years; } r=10\%$

$\displaystyle S.I=x \times \frac{10}{100} \times 1 = 0.1x$

Compound Interest

$\displaystyle P=x; A= A; r=10\%; n=2 \text{ half } \text{ years }$

$\displaystyle A= x \times \Big(1+ \frac{10}{200} \Big)^2 = x \times \Big( \frac{21}{20} \Big)^2$

$\displaystyle C.I. = x \times \Big( \frac{21}{20} \Big)^2 - x$

Given $\displaystyle C.I. - S.I. = x \times \Big( \frac{21}{20} \Big)^2 - x - 0.1x = 15$

$\displaystyle \Rightarrow x= 6000 \text{ Rs. }$

$\displaystyle \\$

Question 13: The compound interest, calculated yearly, on a certain sum of money for the second year is $\displaystyle Rs. 1320$ and for the third year is $\displaystyle Rs. 1452$ . Calculate the rate of interest and the original sum of money. [2014]

Answer:

Difference between the Compound interest of two successive years.

$\displaystyle = 1452-1320 = Rs. 132$

$\displaystyle \Rightarrow Rs. 132$ is the interest on $\displaystyle Rs. 1320$

$\displaystyle \therefore \text{ Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 132}{1320 \times 1} \% = 10\%$

Let the sum of money $\displaystyle = Rs. 100$

Therefore Interest on it for 1st Year $\displaystyle = 10\% of Rs. 100 = Rs. 10$

$\displaystyle \Rightarrow \text{ Amount in one year } = 100 + 10 = Rs. 110$

$\displaystyle \therefore \text{ Interest on it for } 2^{nd} \text{ Year } = 10\% of 110 = Rs. 11$

$\displaystyle \Rightarrow \text{ Amount in 2nd year } = 110 + 11 = Rs. 121$

When interest of 2nd year $\displaystyle = Rs. 11, sum is Rs. 100$

$\displaystyle \Rightarrow \text{ When interest of 2nd year } = Rs. 1320$ , then

$\displaystyle \text{ sum } = \frac{100}{11} \times 1320 = Rs. 12000$

$\displaystyle \\$

Question 14: Ramesh invests $\displaystyle Rs. 12800$ for three years at the rate of $\displaystyle 10\%$ per annum compound interest. Find;

The sum due to that person at the end of the first year.
The interest he earns for the second year.
The total amount due to him at the end of the third year. [2007]

Answer:

For 1st year: $\displaystyle P = Rs. 12800; R=10\% and T=1 year$

$\displaystyle \text{ Therefore Interest }= \frac{12800 \times 10 \times 1}{100} = Rs. 1280$

$\displaystyle \text{ and, Amount }= 12800+1280 = Rs. 14080$

For 2nd year: $\displaystyle P = Rs. 14080; R=10\% and T=1 year$

$\displaystyle \text{ Therefore Interest } = \frac{14080 \times 10 \times 1}{100} = Rs. 1408$

$\displaystyle \text{ and, Amount } = 14080+1408 = Rs. 15488$

For 3rd year: $\displaystyle P = Rs. 15488; R=10\% and T=1 year$

$\displaystyle \text{ Therefore Interest } = \frac{15488 \times 10 \times 1}{100} = Rs. 1548.80$

$\displaystyle \text{ and, Amount } = 15488+1548.80 = Rs. 17036.80$

$\displaystyle \\$

Question 15: The compound interest, calculated yearly, on a certain sum of money for the second year is $\displaystyle Rs. 880$ and for the third year is $\displaystyle Rs. 968$ . Calculate the rate of interest and the sum of money. [1995]

Answer:

Difference between the Compound interest of two successive years

$\displaystyle = 968-880 = Rs. 88$

$\displaystyle \Rightarrow Rs. 88$ is the interest on $\displaystyle Rs. 880$

$\displaystyle \therefore \text{ Rate of Interest } = \frac{100 \times I}{P \times T} \% = \frac{100 \times 88}{880 \times 1} \% = 10\%$

For 1st year: $\displaystyle P = Rs. x; R=10\% and T=1 year$

$\displaystyle \text{Therefore Interest } = \frac{x \times 10 \times 1}{100} = Rs. 0.1x$

$\displaystyle \text{ and, Amount } = x+ 0.1x = Rs. 1.1x$

For 2nd year: $\displaystyle P = Rs. 1.1x; R=10 \% and T=1 year$

$\displaystyle \text{ Therefore Interest } = \frac{1.1x \times 10 \times 1}{100} = Rs. 0.11x$

Given $\displaystyle 0.11x = 880 \Rightarrow x=Rs. 8000$

$\displaystyle \\$

Question 16: Mr. Kumar borrowed $\displaystyle Rs. 15000$ for two years. The rate of interest for the two successive years are $\displaystyle 8\% \text{ and } 10\%$ respectively. If the repays $\displaystyle Rs. 6200$ at the end of the first year, find the outstanding amount at the end of the second year. [2011]