Question

. If Rs 10000 amounts to Rs 11200 after one year. The rate of compound interest will be​

Answer 1

$\bf{\underline{Given:-}}$

• Principal = Rs.10000
• Amount = Rs.11200
• Time = 1 year

$\bf{\underline{To\:find:-}}$

Rate of the Compound Interest.

$\bf{\underline{Solution:-}}$

Principal = Rs.10000

Amount = Rs.11200

We know,

$\sf{A = P\bigg(1+\dfrac{r}{100}\bigg)^t}$

= $\sf{11200 = 10000\bigg(1 + \dfrac{r}{100}\bigg)^1}$

= $\sf{\dfrac{11200}{10000} = \dfrac{100+r}{100}}$

= $\sf{\dfrac{112}{100} = \dfrac{100+r}{100}}$

= $\sf{\dfrac{112\times100}{100} = 100 + r}$

= $\sf{112 = 100+r}$

= $\sf{r = 112 - 100}$

= $\sf{r = 12\%}$

$\sf{\therefore}$ The rate of Compound Interest will be 12%.

$\bf{\underline{Formula \:of \:CI}}$

$\sf{A = P\bigg(1+\dfrac{r}{100}\bigg)^t}$

$\sf{CI = A - P}$

Where,

• A = Amount
• P = Principal
• t = Time
• r = rate
• CI = Compound Interest.