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

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

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

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

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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.

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

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

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

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

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Question 3: Calculate the compound interest accrued on Rs.6000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.

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

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

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

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

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

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

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

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

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

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

$\Rightarrow x= 4500 \ Rs.$

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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?

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

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

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

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

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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.

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

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

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

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

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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?

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

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

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

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

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

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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.

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

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

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

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

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

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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?

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

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

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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 2years?

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

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

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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]

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

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

Given $Interest = 1324 \ Rs.$

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

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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.

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

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

At the end of third year

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

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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?

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

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

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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.

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

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

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

For B

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

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

Given

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

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

$\Rightarrow x= 15730 \ Rs$

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

For B

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

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

Given

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

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

$\Rightarrow x= 16000 \ Rs$

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

For B

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

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

Given

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

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

$\Rightarrow x= 20000 \ Rs$

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.

year 2011

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

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

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

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

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

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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.

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

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

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

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

Given $A1:A2= 20:21$

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

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

$\Rightarrow r=5\%$

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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

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

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

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

Given

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

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