Question

Find compound interest 13,500 for 2 years at 10% per annum compounded annually

Answer 1

Given : ↴

$\sf \: Principal = Rs.\: 13500$

$\sf \: Rate : 10 \: \% \:$

$\sf \: Time \: : \: 2 \: years$

To find : ↴

$\sf \: We \: have \: to \: find \: the \: Compound \: Interest.$

So, Let's Start : ↴

$\sf \: To \: find \: the \: Compound \: Interest \: we \: have \:to \: find \: the \: amount.$

$\sf \: Formula \: to \: find \: Amount \: :$ ↴

$\boxed {\mathrm {Amount \: = \: P \bigg(1 + \cfrac{R}{100} \bigg)^{n} }}$

$\sf \: {\mathrm {Amount \: = \: 13500 \bigg(1 + \cfrac{10}{100} \bigg)^{2} }}$

$\sf \: {\mathrm {Amount \: = \: 13500 \bigg(1 + \cfrac{1 \not0}{10 \not0} \bigg)^{2} }}$

$\sf \: {\mathrm {Amount \: = \: 13500 \bigg(1 + \cfrac{1 }{10 } \bigg)^{2} }}$

$\sf \: {\mathrm {Amount \: = \: 13500 \bigg( \cfrac{11 }{10 } \bigg)^{2} }}$

$\sf \: {\mathrm {Amount \: = \: 13500 \: \bigg( \cfrac{11 }{10 } \times \cfrac{11 }{10 } \bigg) }}$

$\sf \: {\mathrm {Amount \: = \: 135 {\not0\not0} \: \bigg( \cfrac{11 }{1 \not0 } \times \cfrac{11 }{1 \not0 } \bigg) }}$

$\sf \: Amount \: = \: 135 \: \times \: 11 \: \times \: 11 \: = \: Rs.\: 16,335.$

$\sf \: So, \: the \: Amount \: is \: Rs. \: 16,335.$

$\sf \: Now, \: We \: to \:will \: find \: Compound \: Interest \: :$

$\sf \: Formula \: to \: find \: Compound \: Interest \: :$ ↴

$\boxed {\mathrm {Amount \: - \: Principal }}$

$\sf \: Compound \: Interest \: : \: 16,335 \: - \: 13,500.$

$\sf \:\: \: 16,335 \: - \: 13,500. \: = Rs. \: 2,835$