Question

On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is Rs. 180. Find the sum lent out, if the rate of interest in both the cases is 10% per annum. \[\]

Answer 1

Hint: We use the formula for simple interest $ \text{SI}=\dfrac{PNR}{100} $ where $ P $ is the original sum rent out, $ N $ is the number of simple periods and $ R $ is the rate of simple interest. We use the formula for compound interest $ \text{CI}=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}-P $ where $ n $ the number of compound periods is and $ r $ is the rate of compound interest. According to the question $ \text{SI}-\text{CI}=180 $ and solve for $ P $ . \[\]

Complete step by step answer:
We know from simple interest formula that if original principal sum is $ P $ , the number of simple periods as $ N $ and rate of interest $ R $ , then simple interest is given by
\[SI=\dfrac{PNR}{100}\]
We know from compound interest that if we denote the original principal sum as $ P $ , the rate of interest per time period as $ r $ , the number of compound periods or frequency as $ n $ and the amount accumulated after the time period $ nt $ as $ A $ . The compound interest $ I $ generated after time period $ nt $ is given by
\[I=A-P=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}-P\]
We are given that the rate of interest in both the cases is 10% per annum which means $ r=R=10 $ % per annum but since compound interest is calculated half yearly we have $ r=\dfrac{10}{2}=5 $ % per half year. We also see that compound interest for a year is payable half-yearly and the simple interest is payable for a yearly. So the number of simple periods for 1 year is $ N=1 $ and the number of compound periods for 1 year is 2. We put the values of $ N,R,n,r $ in the simple and compound interest formula to have;
 \[\begin{align}
  & SI=\dfrac{PNR}{100}=\dfrac{P\times 1\times 10}{100}=0.1P \\
 & CI=P{{\left( 1+\dfrac{r}{100} \right)}^{n}}-P=P{{\left( 1+\dfrac{5}{100} \right)}^{2}}-P=P{{\left( 1.05 \right)}^{2}}-P=0.1025 \\
\end{align}\]
We are given that the difference between the compound interest and the simple interest for a year is Rs. 180. So we have;
\[\begin{align}
  & \text{CI}-\text{SI}=180 \\
 & \Rightarrow 0.1025P-0.1P=180 \\
 & \Rightarrow 0.0025P=180 \\
 & \Rightarrow P=\dfrac{180}{0.0025}=72000 \\
\end{align}\]
So the amount of sum lent was 72000 rupees. \[\]

Note:
 We note that in compound interest, the interest is added into the principal and in the next compound period the new principal is the sum of the old principal and accumulated interest. The simple interest does not add the accumulated interest to the principal. For the same time period, the compound interest will be greater than the simple interest. If $ T $ is the total time period and $ t $ is the time period to calculate interest then the number of periods is $ n=\dfrac{T}{t} $ . We can use the algebraic identity $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $ to find $ {{\left( 1.05 \right)}^{2}}={{\left( 1+0.05 \right)}^{2}} $ rather multiplying which will take more time.