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. }  

\displaystyle \\

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. }  

\displaystyle \\

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. }  

\displaystyle \\

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. }

\displaystyle \\

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. }  

\displaystyle \\

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. }  

\displaystyle \\

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. }  

\displaystyle \\

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. }  

\displaystyle \\

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\%.  

\displaystyle \\

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\%.  

\displaystyle \\

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\%  

\displaystyle \\

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;

  1. The rate of interest per annum
  2. 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. }  

\displaystyle \\

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 }  

\displaystyle \\

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 }  

\displaystyle \\

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. }  

\displaystyle \\

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. }  

\displaystyle \\

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.

  1. What is the share of each out of \text{ Rs. } \displaystyle 44200 ?
  2. 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. }  

\displaystyle \\

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. }  

\displaystyle \\

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\%  

\displaystyle \\

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.?

Answer:

Let the \displaystyle P = x \text{ Rs. }  

Simple Interest

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

Compound Interest

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

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

\displaystyle C.I. = 1.331x-x = 0.331x  

Given

\displaystyle 0.331x-0.3x= 930 \Rightarrow x = 30000 \text{ Rs. }