Question

A man borrow ₹ 1000 at 10% p.a. simple interest for 3 years. He immediately lends this money out at compound interest at the same rate for the same time. What is his gain at the end of 3 years?

Answer 1

SI= 1000x10x3
----------------
100
= 300
A= 1000+300
1300
CI =
A= 1000(1+10)^3
---
100
1000X110X110X110
---- ----- -----
100 100 100
=1331
GAIN= 1331-1300=31

Answer 2

Answer-The gain after 3 years is  Rs31.

Explanation-Given-Principal-1000, Rs 10%, T=3 years

First find SI$=\frac{P\times R\times T}{100}$

$=\frac{10\times 10\times 3}{100}$

$=300$

Amount $=P+SI$

$=1000+300=1300$

Now, Find CI

$A=P(1+\frac{R}{100} )^h\\=1000(1+\frac{10}{100} )^3$

$=1000\times \frac{11}{10}\times \frac{11}{10} \times \frac{11}{10}$

$=1331$

So, the gain after 3 years is $1331-1300$

$=Rs 31$

#SPJ3