Class 10: Compound Interest – Exercise 2(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the amount and the compound interest on Rs. $\displaystyle 12000$ in $\displaystyle 3$ years at $\displaystyle 5\%$ ; interest being compounded annually.

Answer:

$\displaystyle P=12000 \text{ Rs.; } r=5\%; n=3 \text{ years }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 12000 \Big(1+ \frac{5}{100} \Big)^3 = 13891.50 \text{ Rs. }$

$\displaystyle \text{Compound Interest } = 13891.50-12000 = 1891.50 \text{ Rs. }$

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Question 2: Calculate the amount, if Rs. $\displaystyle 15000$ is lent at compound interest for $\displaystyle 2$ years and the rates for the successive years are $\displaystyle 8\%$ p.a. and $\displaystyle 10\%$ p.a. respectively.

Answer:

$\displaystyle P=15000 \text{ Rs.; } r=8\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 15000 \Big(1+ \frac{8}{100} \Big)^1 = 16200 \text{ Rs. }$

$\displaystyle P=15000 \text{ Rs.; } r=10\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 15000 \Big(1+ \frac{10}{100} \Big)^1 = 17820 \text{ Rs. }$

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Question 3: Calculate the compound interest accrued on Rs. $\displaystyle 9000$ in $\displaystyle 3$ years, compounded yearly, if the rates for the successive years are $\displaystyle 5\%, 8\%$ and $\displaystyle 10\%$ respectively.

Answer:

$\displaystyle P=6000 \text{ Rs.; } r=5\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 6000 \Big(1+ \frac{5}{100} \Big)^1 = 6300 \text{ Rs. }$

$\displaystyle P=6300 \text{ Rs.; } r=8\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 6300 \Big(1+ \frac{8}{100} \Big)^1 = 6804 \text{ Rs. }$

$\displaystyle P=6804 \text{ Rs.; } r=10\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 6804 \Big(1+ \frac{10}{100} \Big)^1 = 7484.40 \text{ Rs. }$

$\displaystyle \text{Compound Interest } = 7484.40-6000 = 1484.40 \text{ Rs. }$

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Question 4: What sum of money will amount to Rs. $\displaystyle 5445$ in $\displaystyle 2$ years at $\displaystyle 10\%$ per annum compound interest?

Answer:

$\displaystyle A= 5445 Rs. ;P=x \text{ Rs.; } r=10\%; n=2 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 5445= x \Big(1+ \frac{10}{100} \Big)^2 \text{ Rs. }$

$\displaystyle \Rightarrow x= 4500 \text{ Rs. }$

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Question 5: On what sum of money will be compound interest for $\displaystyle 2$ years at $\displaystyle 5$ per cent per annum amount to Rs $\displaystyle 768.75$ ?

Answer:

$\displaystyle A= A Rs. ;P=x \text{ Rs.; } r=5\%; n=2 \text{ year }$

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

$\displaystyle \Rightarrow C.I.=A-P=768.75$

$\displaystyle \Rightarrow C.I.=x \Big(1+ \frac{5}{100} \Big)^2-x=768.75 \Rightarrow x=7500 \text{ Rs. }$

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Question 6: Find the sum on which the compound interest for $\displaystyle 3$ years at $\displaystyle 10\%$ per annum amounts to Rs. $\displaystyle 1655$ .

Answer:

$\displaystyle A= A Rs. ;P=x \text{ Rs.; } r=10\%; n=3 \text{ year }$

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

$\displaystyle \Rightarrow C.I.=A-P=1655$

$\displaystyle \Rightarrow C.I.=x \Big(1+ \frac{10}{100} \Big)^2-x=1655 \Rightarrow x=5000 \text{ Rs. }$

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Question 7: What principal will amount to Rs. $\displaystyle 9856$ in two years, if the rates of interest for Question successive years are $\displaystyle 10\%$ and $\displaystyle 12\%$ respectively?

Answer:

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

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = x \Big(1+ \frac{10}{100} \Big)^1 = 1.1x \text{ Rs. }$

$\displaystyle P=1.1x \text{ Rs.; } r=12\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 1.1x \Big(1+ \frac{12}{100} \Big)^1 = 1.232x \text{ Rs. }$

$\displaystyle \Rightarrow 1.232x=9856 \Rightarrow x=8000 \text{ Rs. }$

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Question 8: On a certain sum, the compound interest in $\displaystyle 2$ years amounts to Rs. $\displaystyle 4240$ . If the rates of interest for successive year are $\displaystyle 10\%$ and $\displaystyle 15\%$ respectively, find the sum.

Answer:

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

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = x \Big(1+ \frac{10}{100} \Big)^1 = 1.1x \text{ Rs. }$

$\displaystyle P=1.1x \text{ Rs.; } r=15\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n = 1.1x \Big(1+ \frac{15}{100} \Big)^1 = 1.265x \text{ Rs. }$

$\displaystyle \Rightarrow 1.232x - x=4240 \Rightarrow x=16000 \text{ Rs. }$

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Question 9: At what rate per cent per annum will Rs. $\displaystyle 6000$ amount to Rs. $\displaystyle 6615$ in $\displaystyle 2$ years when interest is compounded annually?

Answer:

$\displaystyle P=6000 \text{ Rs.; } A=6615 \text{ Rs.; } r=x\%; n=2 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 6615= 6000 \Big(1+ \frac{x}{100} \Big)^2 \Rightarrow x= 5\%.$

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Question 10: At what rate per cent compound interest, does a sum of money become $\displaystyle 1.44$ times of itself in $\displaystyle 2$ years?

Answer:

$\displaystyle P=x \text{ Rs.; } A=1.44x \text{ Rs.; } r=r\%; n=2 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 1.44x= x \Big(1+ \frac{r}{100} \Big)^2 \Rightarrow r= 20\%.$

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

Answer:

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

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

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

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

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Question 12: A person invests Rs. $\displaystyle 5000$ for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to Rs. $\displaystyle 6272$ . Calculate;

The rate of interest per annum
The amount at the end of the third year.

Answer:

$\displaystyle P=5000 \text{ Rs.; } r=x\%; n=2 \text{ years; } A=6272 \text{ Rs. }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 6272= 5000 \Big(1+ \frac{x}{100} \Big)^2 \Rightarrow x = 12\%$

At the end of third year

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= 5000 \Big(1+ \frac{12}{100} \Big)^3 = 7024.64 \text{ Rs. }$

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Question 13: In how many years will Rs. $\displaystyle 7000$ amount to Rs. $\displaystyle 9317$ at $\displaystyle 10\%$ per cent per annum compound interest?

Answer:

$\displaystyle P=7000 \text{ Rs.; } r=10\%; n=n \text{ years; } A=9217 \text{ Rs. }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 9317= 7000 \Big(1+ \frac{10}{100} \Big)^n \Rightarrow n = 3 \text{ years }$

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Question 14: Find the time, in years, in which Rs. $\displaystyle 4000$ will produce Rs. $\displaystyle 630.50$ as compound interest at $\displaystyle 5\%$ p.a. interest being compounded annually.

Answer:

$\displaystyle P=4000 \text{ Rs.; } r=5\%; n=n \text{ years; } A=4630.50 \text{ Rs. }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 4630.50= 4000 \Big(1+ \frac{5}{100} \Big)^n \Rightarrow n = 3 \text{ years }$

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Question 15: Divide Rs. $\displaystyle 28,730$ between $\displaystyle A$ and $\displaystyle B$ so that when their shares are lent out at $\displaystyle 10\%$ compound interest compounded per year, the amount that $\displaystyle A$ receives in $\displaystyle 3$ years is the same as what $\displaystyle B$ receives in $\displaystyle 5$ years.

Answer:

Let the share of $\displaystyle A = x Rs$ . Therefore share of $\displaystyle B = (28730-x) Rs$ .

For A

$\displaystyle P=x \text{ Rs.; } r=10\%; n=3 \text{ years; }$

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

For B

$\displaystyle P=(28730-x) \text{ Rs.; } r=10\%; n=5 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= (28730-x) \Big(1+ \frac{10}{100} \Big)^5$

Given

$\displaystyle x \Big(1+ \frac{10}{100} \Big)^3 = (28730-x) \Big(1+ \frac{10}{100} \Big)^5$

$\displaystyle x = (28730-x)(1.1)^2$

$\displaystyle \Rightarrow x= 15730 Rs$

Therefore B’s share = $\displaystyle (28730-15730)= 13000 \text{ Rs. }$

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Question 16: A sum of Rs. $\displaystyle 34522$ is divided between $\displaystyle A$ and $\displaystyle B$ , $\displaystyle 18$ years and $\displaystyle 21$ years old respectively in such a way that if their shares be invested at $\displaystyle 5\%$ per annum compound interest, both will receive equal money at the age of $\displaystyle 30$ years. Find the shares of each out of Rs. $\displaystyle 34522$ .

Answer:

Let the share of $\displaystyle A = x$ Rs. Therefore share of $\displaystyle B = (34522-x)$ Rs.

For A

$\displaystyle P=x \text{ Rs.; } r=5\%; n=12 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= x \Big(1+ \frac{5}{100} \Big)^{12}$

For B

$\displaystyle P=(34522-x) \text{ Rs.; } r=10\%; n=9 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= (34522-x) \Big(1+ \frac{5}{100} \Big)^9$

Given

$\displaystyle x \Big(1+ \frac{5}{100} \Big)^{12} = (34522-x) \Big(1+ \frac{5}{100} \Big)^9$

$\displaystyle x(1.05)^3 = (34522-x)$

$\displaystyle \Rightarrow x= 16000 Rs$

Therefore B’s share = $\displaystyle (28730-15730)= 13000 \text{ Rs. }$

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Question 17: A sum of Rs. $\displaystyle 44200$ is divided between $\displaystyle A$ and $\displaystyle B$ , $\displaystyle 12$ years and $\displaystyle 14$ years old respectively, in such a way that if their portions be invested at $\displaystyle 10\%$ per annum compound interest, they will receive equal amounts on reaching $\displaystyle 16$ years of age.

What is the share of each out of \text{ Rs. } $\displaystyle 44200$ ?
What will each receive, when $\displaystyle 16$ years old?

Answer:

Let the share of $\displaystyle A = x$ Rs. Therefore share of $\displaystyle B = (44200-x)$ Rs.

For A

$\displaystyle P=x \text{ Rs.; } r=10\%; n=4 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= x \Big(1+ \frac{10}{100} \Big)^4$

For B

$\displaystyle P=(44200-x) \text{ Rs.; } r=10\%; n=2 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= (44200-x) \Big(1+ \frac{10}{100} \Big)^2$

Given

$\displaystyle x \Big(1+ \frac{10}{100} \Big)^4 = (44200-x) \Big(1+ \frac{10}{100} \Big)^2$

$\displaystyle x(1.1)^2 = (44200-x)$

$\displaystyle \Rightarrow x= 20000 \text{ Rs. }$

Therefore B’s share = $\displaystyle (44200-20000)= 24200 \text{ Rs. }$

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Question 18: At the beginning of the year 2011, a man had Rs. $\displaystyle 22000$ in his bank account. He saved some money by the end of this year and deposited it in the bank. The bank pays $\displaystyle 10\%$ per annum compound interest and at the end of the year 2012, he had Rs. $\displaystyle 39820$ in his bank account. Find, what amount of money at the end of the year 2011.

Answer:

$\displaystyle P=22000 \text{ Rs.; } r=10\%; n=1 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A= 22000 \Big(1+ \frac{10}{100} \Big)^1 = 24200 \text{ Rs. }$

Lets us say he saves and deposits $\displaystyle x \text{ Rs. }$ at the end of year 2011.

$\displaystyle P=(24200+x) \text{ Rs.; } A=39820 \text{ Rs.; } r=10\%; n=1 \text{ years; }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow 39820= (24200+x) \Big(1+ \frac{10}{100} \Big)^1 \Rightarrow x = 12000 \text{ Rs. }$

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Question 19: If the amounts of two consecutive years on a sum of money are in the ratio $\displaystyle 20:21$ , find the rate of interest.

Answer:

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

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A1= x \Big(1+ \frac{r}{100} \Big)$

$\displaystyle P=x \Big(1+ \frac{r}{100} \Big) \text{ Rs.; } r=r\%; n=1 \text{ year }$

$\displaystyle A=P \Big(1+ \frac{r}{100} \Big)^n \Rightarrow A2= x(1+ \frac{r}{100} \Big) \Big(1+ \frac{r}{100} \Big)^1$

Given $\displaystyle A1:A2= 20:21$

$\displaystyle x \Big(1+ \frac{r}{100} \Big) : x \Big(1+ \frac{r}{100} \Big)^2 = 20: 21$

$\displaystyle \frac{1}{x(1+\frac{r}{100})} = \frac{20}{21}$

$\displaystyle \Rightarrow r=5\%$

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Question 20: On what sum of money will the difference between the compound interest and simple interest for $\displaystyle 3$ years be equal to Rs. $\displaystyle 930$ , if the rate of interest charged for both is $\displaystyle 10\%$ p.a.?