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

$P=12000\ Rs.; \ r=5\%; \ n=3 \ years$

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

$Compound \ Interest = 13891.50-12000 = 1891.50 \ Rs.$

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

$P=15000\ Rs.; \ r=8\%; \ n=1 \ year$

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

$P=15000\ Rs.; \ r=10\%; \ n=1 \ year$

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

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

$P=6000\ Rs.; \ r=5\%; \ n=1 \ year$

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

$P=6300\ Rs.; \ r=8\%; \ n=1 \ year$

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

$P=6804\ Rs.; \ r=10\%; \ n=1 \ year$

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

$Compound \ Interest = 7484.40-6000 = 1484.40 \ Rs.$

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

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

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

$\Rightarrow x= 4500 \ Rs.$

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

$A= A \ Rs. ;P=x\ Rs.; \ r=5\%; \ n=2 \ year$

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

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

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

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

$A= A \ Rs. ;P=x\ Rs.; \ r=10\%; \ n=3 \ year$

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

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

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

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

$P=x\ Rs.; \ r=10\%; \ n=1 \ year$

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

$P=1.1x\ Rs.; \ r=12\%; \ n=1 \ year$

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

$\Rightarrow 1.232x=9856 \Rightarrow x=8000 \ Rs.$

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

$P=x\ Rs.; \ r=10\%; \ n=1 \ year$

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

$P=1.1x\ Rs.; \ r=15\%; \ n=1 \ year$

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

$\Rightarrow 1.232x - x=4240 \Rightarrow x=16000 \ Rs.$

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

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

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

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

$P=x\ Rs.; A=1.44x\ Rs.; \ r=r\%; \ n=2 \ year$

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

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

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

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

Given $Interest = 1324 \ Rs.$

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

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

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

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

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

At the end of third year

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

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

$P=7000\ Rs.; \ r=10\%; \ n=n \ years; \ A=9217 \ Rs.$

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

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

$P=4000\ Rs.; \ r=5\%; \ n=n \ years; \ A=4630.50 \ Rs.$

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

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

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

For A

$P=x\ Rs.; \ r=10\%; \ n=3 \ years;$

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

For B

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

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

Given

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

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

$\Rightarrow x= 15730 \ Rs$

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

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

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

For A

$P=x\ Rs.; \ r=5\%; \ n=12 \ years;$

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

For B

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

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

Given

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

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

$\Rightarrow x= 16000 \ Rs$

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

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

1. What is the share of each out of Rs. $44200$?
2. What will each receive, when $16$ years old?

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

For A

$P=x\ Rs.; \ r=10\%; \ n=4 \ years;$

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

For B

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

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

Given

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

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

$\Rightarrow x= 20000 \ Rs$

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

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

$P=22000\ Rs.; \ r=10\%; \ n=1 \ years;$

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

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

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

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

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

$P=x\ Rs.; \ r=r\%; \ n=1 \ year$

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

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

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

Given $A1:A2= 20:21$

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

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

$\Rightarrow r=5\%$

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

Let the $P = x \ Rs.$

Simple Interest

$S.I = x \times$ $\frac{10}{100}$ $\times 3 = 0.3x$

Compound Interest

$P=x\ Rs.; \ r=10\%; \ n=3 \ year$

$A=P \Big(1+$ $\frac{r}{100}$ $\Big)^n = x \Big(1+$ $\frac{10}{100}$ $\Big)^3 = 1.331x \ Rs.$

$C.I. = 1.331x-x = 0.331x$

Given

$0.331x-0.3x= 930 \Rightarrow x = 30000 \ Rs.$