Question

A man invests ₹3,000 for three years at compound interest. After one year, the money amounts to ₹3,240. Find the rate of interest and the amount (approximated to the nearest rupee) at the end of 3 years.

Answer 1

Let the rate of interest be r.
For the first year, C.I = S.I
Given that the amount at the end of first year = ₹3,240
Hence, I = ₹3,240 - ₹3,000 = ₹ 240

I = $\frac{PTR}{100}$

For T = 1 year and I = ₹ 240,
240 = $\frac{3000\times 1\times R}{100}$
R = $\frac{240\times 100}{3000}$
R = 8%

Principal for the second year = ₹3,240
Hence, I = $\frac{3240\times 1\times 8}{100}$
= ₹ 259.20
Principal for the third year
= ₹3,240 + ₹259.20
= ₹3,499.20
Hence I = $\frac{3499.20\times 1\times 8}{100}$
= ₹ 279.936
Amount = 3,499.20 + 279.936
=₹3,779.136
= ₹3,779 ( approximated to the nearest rupee)