Kumnud borrowed some amount at simple interest of 10% per annum for 1121\frac{1}{2}121 year. Sneha borrowed the same amount at the same rate on compound interest (compounded semiannually) for the same period. If Sneha paid Rs.61 more than Kumud as interest, then how much money did each of them borrow?

Correct option is D

Given:

Rate of interest, R = 10% per annum

Time period, T =

1 \frac{1}{2}

= 1.5 year

Difference in interest paid by Sneha and Kumud = Rs. 61

Interest for Kumud is simple interest (SI).

Interest for Sneha is compound interest (CI) compounded semi-annually.

Concept Used:

Simple Interest (SI) is given by:

SI = \frac{P \times R \times T}{100} ​

The compound interest formula for semi-annual compounding is:

CI = P \left(1 + \frac{R_{\text{half-year}}}{100}\right)^{2T_{\text{in years}}} - P

Solution:

According to the question;

CI – SI = 61 \\ \ \\\left[P \left(1 + \frac{R_{\text{half-year}}}{100}\right)^{2T_{\text{in years}}} – P\right] - \left[\frac{P \times R \times T}{100} ​\right] = 61 \\ \ \\\text{Substituting values}\\ \ \\\left[P \left(1 + \frac{5}{100}\right)^3 – P \right]-\left[\frac{P \times 10 \times 1.5}{100} \right] = 61 \\ \ \\ \left[P \left( \frac{105}{100}\right)^3 – P \right]-\left[\frac{15P }{100} \right] = 61 \\ \ \\ \left[P \left( \frac{21}{20}\right)^3 – P \right]-\left[\frac{3P }{20} \right] = 61 \\ \ \\ \left[P \left( \frac{9261}{8000}\right) – P \right]-\left[\frac{3P }{20} \right] = 61 \\ \ \\ \left[ \frac{1261P}{8000}\ \right]-\left[\frac{3P }{20} \right] = 61 \\ \ \\ P\left[ \frac{1261 - 1200}{8000}\ \right] = 61 \\ \ \\ P\left[ \frac{61 }{8000}\ \right] = 61 \\ \ \\ \implies P = 8000

Thus, the money borrowed is Rs.8000.

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