Question

Find compound interest on a sum of ₹1000 for 3 years at 6% per annum compunded annually.​

Answer 1

Amount = p(1+r/100)^3
=1000(1+6/100)^3
=1000 * 106/100 * 106/100 * 106/100
=1191.016
Interest=amount-principle
=1191.016-1000
= 191.016

Answer 2

GiveN:-

• Principal = Rs.1000
• Time = 3 years
• Rate = 6% per annum

To FinD:-

Compound Interest.

SolutioN:-

We know that,

$\large{\green{\underline{\boxed{\bf{Amount=P\left(1+\dfrac{R}{100}\right)^n}}}}}$

where,

• P is principal = Rs.1000
• R is rate = 6%
• n is time = 3 years (compounded annually)

Putting the values,

$\large\implies{\sf{Amount=1000\left(1+\dfrac{6}{100}\right)^3}}$

$\large\implies{\sf{Amount=1000\left(1+\dfrac{\cancel{6}}{\cancel{100}}\right)^3}}$

$\large\implies{\sf{Amount=1000\left(1+\dfrac{3}{50}\right)^3}}$

$\large\implies{\sf{Amount=1000\left(\dfrac{50+3}{50}\right)^3}}$

$\large\implies{\sf{Amount=1000\times\left(\dfrac{53}{50}\right)^3}}$

$\large\implies{\sf{Amount=1000\times\dfrac{53}{50}\times\dfrac{53}{50}\times\dfrac{53}{50}}}$

$\large\implies{\sf{Amount=\dfrac{148877000}{125000}}}$

$\large\therefore\boxed{\bf{Amount=Rs.1191.016}}$

Now,

$\large{\green{\underline{\boxed{\bf{Compound\:Interest=Amount-Principal}}}}}$

where,

• Amount = Rs.1191.016
• Principal = Rs.1000

Putting the values,

$\large\implies{\sf{Compound\:Interest=Rs.(1191.016-1000)}}$

$\large\therefore\boxed{\bf{Compound\:Interest=Rs.191.016}}$

★The Compound Interest is Rs.191.016.