Question

what is the compound interest for 3 year on rupees 40
500 at 6.25 percent per annum
(please tell this question's answer also ​

Answer 1

GIVEN :–

• Principal amount(p) = 40,500

• Rate(r) = 6.25 %

• Time(n) = 3 year

TO FIND :–

• Compound interest = ?

SOLUTION :–

• We know that –

$\\ \longrightarrow \: \large{ \boxed{ \bold{Amount = p \left(1 + \dfrac{r}{100} \right)^{n} } }}$

• Put the values –

$\\ \implies \: { \bold{Amount= (40500) \left(1 + \dfrac{6.25}{100} \right)^{3} }}$

$\\ \implies \: { \bold{Amount= (40500) \left( \dfrac{100 + 6.25}{100} \right)^{3} }}$

$\\ \implies \: { \bold{Amount= (40500) \left( \dfrac{106.25}{100} \right)^{3} }}$

$\\ \implies \: { \bold{Amount= (40500) ( 1.0625)^{3} }}$

$\\ \implies \: { \bold{Amount= (40500) (1.2) }}$

$\\ \implies \large { \boxed { \bold{Amount= 48600}}}$

• We also know that –

$\\ \implies { \bold{Compound \: \: interest = Amount - principal}}$

$\\ \implies { \bold{Compound \: \: interest = 48600 - 40500}}$

$\\ \implies \large { \boxed{ \bold{Compound \: \: interest = 8,100}}}$

Answer 2

AnSwer:

★ Provided :-

• Principal Amount (P) = 40500

• Rate (r) = 6.25 % annum

• Time (T) = 3 year

Now as we know,

$\star \sf{ C.P = P \left( \left(1 + \dfrac{r}{100} \right)^{T} - 1 \right) }$

Now put all the values,

$\implies \sf { C.P = 40500 \left( \left(1 + \dfrac{6.25}{100} \right)^{3} - 1 \right)}$

$\implies \sf { C.P = 40500 \left( \left( \dfrac{100 + 6.25}{100} \right)^{3} - 1 \right)}$

$\implies \sf {C.P = 40500 \left( \left( \dfrac{106.25}{100} \right)^{3} - 1\right)}$

$\implies \sf { C.P = 40500 ( (1.0625)^{3} - 1)}$

$\implies \sf { C.P = 40500 ((1.1995) - 1)}$

$\implies \sf { C.P = 40500 ( 0.1995)}$

$\implies \sf { C.P = 8079.75}$

So, CP = 8,079.75