Question

find the compound interest on 12000 for 3 years at 10% per annum compounded annually

Answer 1

T=3
R=10%
P=12000

For annual calculations, R and T remains the same.

1st year:
R=10%
P=12000
I=P*R/100
ie
I=12000*10/100=1200
Hence P becomes= 12000+1200=13200

2nd year:
R=10%
P=13200
I=13200*10/100=1320
Hence P becomes=1320+13200=14520

3rd year:
R=10%
P=14520
I=14520*10/100=1452
Hence P becomes=1452+14520=15972

ANS:15972

Answer 2

$\huge\bold\red{Question}$

Find the compound interest on 12000 for 3 years at 10% per annum compounded annually

$\huge\bold\green{Solution}$

We know that the amount A at the end of n years at the rate of R % per annum when the interest is compounded annually is given by $\bold{A=P {(1 + \frac{R}{100}})^{n} }$

Here $\bold{ P = ₹12000, R = 10\%\: per\:annum}$ $\bold{and\: n = 3}$

∴ Amount after 3 years

= $\bold{₹ 12000\times {(1 + \frac{10}{100}})^{3} }$

$\bold{=₹12000\times {( \frac{11}{100}})^{3}}$

=$\bold{12000\times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}}$

$\bold{=₹(12×11×11×11)}$

=$\bold{₹15972}$

Now Compound interest = A - P Compound interest = $\bold{₹ 15972 - ₹12000}$

= $\bold{₹ 3972}$