Question

A sum of money invested at C.I. amounts to Rs. 800 in 3 years and Rs. 840 in 4 years. The rate of interest per annum is:
(a) $ 2\dfrac{1}{2} $ %
(b) 5 %
(c) 4 %
(d) $ 2\dfrac{1}{2} $ %

Answer 1

Hint: Assume the rate as r% and apply the formula for compound interest given as $ A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} $ , where A is the amount, P is the principal, t is the time in years and n is the number of times the interest is received or paid annually. Form two relations using the above formula, first by substituting t = 3 years and second by substituting t = 4 years. Now, divide the two equations and cancel the common terms to get the value of r.

Complete step by step answer:
Here we have been given that the invested money amounts to Rs. 800 in 3 years and Rs. 840 in 4 years. We have to find the rate of interest per annum.
Now, let us assume the rate of interest as r% and the sum of money invested, i.e the principal amount as P. We know that in the compound interest the amount after an interval of ‘t’ years is given as:
 $ A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}} $
Here, A = Amount
P = Principal amount
r = rate %
t = time in years
n = number of times the interest is received or paid annually
In the above question, we have not been provided with any information regarding n, so we will assume that the interest is paid only one time during one year. So, n=1. Now, let us form two relations for the given time intervals.
1. Case (1):- when t = 3 years
Here, in this case we have
A = Rs. 800
Rate = r% = $ \dfrac{r}{100} $
Principal = P
N = 1
t = 3 years
Substituting all the values in the given formula, we get,
 $ \begin{align}
  & \Rightarrow 800=P{{\left( 1+\dfrac{r}{100} \right)}^{1\times 3}} \\
 & \Rightarrow 800=P{{\left( 1+\dfrac{r}{100} \right)}^{3}}\ldots \ldots \ldots \left( i \right) \\
\end{align} $
2. Case (2):- when t = 4 years
Here, in this case we have
A = Rs. 840
Rate = r% = $ \dfrac{r}{100} $
Principal = P
N = 1
t = 4 years
Substituting all the values in the given formula, we get,
 $ \begin{align}
  & \Rightarrow 840=P{{\left( 1+\dfrac{r}{100} \right)}^{1\times 4}} \\
 & \Rightarrow 840=P{{\left( 1+\dfrac{r}{100} \right)}^{4}}\ldots \ldots \ldots \left( ii \right) \\
\end{align} $
Now, dividing equation (ii) by equation (i), we get,
 $ \Rightarrow \dfrac{840}{800}=\dfrac{P{{\left( 1+\dfrac{r}{100} \right)}^{4}}}{P{{\left( 1+\dfrac{r}{100} \right)}^{3}}} $
Cancelling the common factors and terms, we get,
 $ \begin{align}
  & \Rightarrow \dfrac{84}{80}=\left( 1+\dfrac{r}{100} \right) \\
 & \Rightarrow \dfrac{r}{100}=\dfrac{84}{80}-1 \\
 & \Rightarrow \dfrac{r}{100}=\dfrac{84-80}{80} \\
 & \Rightarrow \dfrac{r}{100}=\dfrac{4}{80} \\
 & \Rightarrow \dfrac{r}{100}=\dfrac{1}{20} \\
\end{align} $
By cross multiplying the terms, we get
 $ \begin{align}
  & \Rightarrow r=\dfrac{100}{20} \\
 & \Rightarrow r=5 \\
\end{align} $
Therefore, the rate of interest per annum of r% = 5%.
Hence, option (b) is the correct answer.

Note:
One may note that we can also apply another method to solve the question. What we can do is we will assume the time interval of one year, i.e the time interval between the third and fourth year. Assume Rs. 800 as the principal amount and t = 1 year and apply the formula of simple interest given as $ S.I.=\dfrac{P\times r\times t}{100} $ to get the value of r. This method is applicable because, for the interval of 1 year, the compound interest and simple interest are the same.