Question

Sunitabai borrowed Rs 60,000 at 9 p.c.p.a simple interest for 3 years to have her house repaired. Had she borrowed the same sum for the same period and the same rate at compound interest, how much more interest would she have had to pay?

Answer 1

Principal = Rs 60,000
Rate of interest = 9% p.a.
Number of Years (N) = 3 years
In case of compound interest , Amount$=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}$
$$\begin{gathered} =60000{\left(1+\frac{9}{100}\right)}^3 \\ =60000{\left(\frac{109}{100}\right)}^3 \\ =60000\times \frac{109}{100}\times \frac{109}{100}\times \frac{109}{100} \\ =\mathrm{Rs}77701.74 \end{gathered}$$

The amount is Rs 77701.74
Compound Interest = Amount − Principal = 77701.74 − 60,000 = Rs 17701.74
∴ The compound Interest = Rs 17701.74
Simple Interest$=\frac{\mathrm{Principal}\times \mathrm{Rate}\times \mathrm{Time}}{100}=\frac{60000\times 9\times 3}{100}=\mathrm{Rs}16200$
Difference between compound Interest and simple interest = Compound Interest − Simple Interest
= Rs 17701.74 − Rs 16200 = Rs 1501.74
So, if she had borrowed the same amount for the same period and at the same rate
at compound interest then she had to pay Rs 1501.74 more.