Question

Find the compound interest on Rs 12600 for 2 years at 5% per annum compounded annually.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Compound\:Interest=1291.5\:rupees}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Principal(p)= 12600 \: rupees \\ \\ \tt: \implies Time(t) = 2 \: years \\ \\ \tt: \implies Rate\%(r) = 5\% \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Compound \: Interest = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies A = p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies A = 12600 \times (1 + \frac{5}{100} )^{2} \\ \\ \tt: \implies A = 12600 \times (1 + 0.05)^{2} \\ \\ \tt: \implies A =12600 \times (1.05)^{2} \\ \\ \tt: \implies A=12600 \times 1.1025 \\ \\ \green{\tt: \implies A =13891.5 \: rupees} \\ \\ \bold{For \: Compound \: Interest} \\ \tt: \implies CI=A- p \\ \\ \tt: \implies CI =13891.5 - 12600 \\ \\ \green{\tt: \implies CI=1291.5 \: rupees}$

Answer 2

Given:-

• Principal, P = ₹12600

• Time, n = 2 years

• Rate, R = 5 %

To find:-

The compound interest.

Solution:-

$\bold{A=P(r+\frac{1}{100})^n }\\\\\bold{\implies A=12600(1+\frac{r}{100})^2 }\\\\\bold{\implies A=12600(1+0.05)^2}\\\\\bold{\implies A=12600(1.05)^2}\\\\\bold{\implies A=12600*1.1025}\\\\\bold{\implies A=Rs.13891.5}$

Compound interest = A - P

$\bold{\implies C.I. = 13891.5-12600}\\\\\boxed{\implies C.I.=Rs.\ 1291.5}$