Question

find the CI and interest of Rs 15,000 for 3 year
at 8% per
annum compounded yearly?​

Answer 1

★ Solution :-

First, we can find the amount easily by the given formula.

Amount :

$\sf \longrightarrow P \bigg(1 + \dfrac{R}{100} \bigg)^{T}$

$\sf \longrightarrow 15000 \bigg(1 + \dfrac{8}{100} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg(1 + \dfrac{2}{25} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg( \dfrac{25 + 2}{25} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg( \dfrac{27}{25} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg( \dfrac{{27}^{3}}{{25}^{3}} \bigg)$

$\sf \longrightarrow 15000 \bigg( \dfrac{19683}{15625} \bigg)$

$\sf \longrightarrow 24 \bigg( \dfrac{19683}{25} \bigg)$

$\sf \longrightarrow \dfrac{24 \times 19683}{25} = \dfrac{472392}{25}$

$\sf \longrightarrow \cancel \dfrac{472392}{25} = 18895.68$

Now, we can find the compound interest.

Compound interest :-

$\sf \longrightarrow Amount - Principle$

$\sf \longrightarrow 18895.68 - 15000$

$\sf \longrightarrow Rs.13895.68$

Therefore, the compound interest is ₹13895.68.

Answer 2

Answer:

★ Solution :-

First, we can find the amount easily by the given formula.

Amount :

$\sf \longrightarrow P \bigg(1 + \dfrac{R}{100} \bigg)^{T}$

$\sf \longrightarrow 15000 \bigg(1 + \dfrac{8}{100} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg(1 + \dfrac{2}{25} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg( \dfrac{25 + 2}{25} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg( \dfrac{27}{25} \bigg)^{3}$

$\sf \longrightarrow 15000 \bigg( \dfrac{{27}^{3}}{{25}^{3}} \bigg)$

$\sf \longrightarrow 15000 \bigg( \dfrac{19683}{15625} \bigg)$

$\sf \longrightarrow 24 \bigg( \dfrac{19683}{25} \bigg)$

$\sf \longrightarrow \dfrac{24 \times 19683}{25} = \dfrac{472392}{25}$

$\sf \longrightarrow \cancel \dfrac{472392}{25} = 18895.68$

Now, we can find the compound interest.

Compound interest :-

$\sf \longrightarrow Amount - Principle$

$\sf \longrightarrow 18895.68 - 15000$

$\sf \longrightarrow Rs.13895.68$

Therefore, the compound interest is ₹13895.68.