Question

if principal is 3200,rate is 10% and time is 3years. Find amount and compound interest? ​

Answer 1

Answer:

$2 {59}^{6y \sqrt[25]{.} } 362 \sqrt[2]{2 |2.6 \leqslant 36 = 62 \beta e log(\pi \sec( \cot( \gamma \pi\%e052) ) ) | }$

A+b = a-B

Answer 2

Step-by-step explanation:

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

• Principal = Rs. 3200

• Rate = 10%

• Time = 3 years

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The Amount

• And the Compound Interest.

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

As we know that:-

The formula for finding Amount is:-

$\boxed{ \sf \leadsto Amount = P \bigg(1 + \frac{r}{100} \bigg)^{n} }$

Here:-

• $\sf P = Principal = Rs. 3200$

• $\sf r = Rate = 10\%$

• $\sf n = Time = 3\: years$

Substituting the values:-

$\sf \implies 3200 \bigg(1 + \dfrac{10}{100} \bigg) ^{3}$

$\sf \implies 3200 \bigg( \dfrac{100 + 10}{100} \bigg) ^{3}$

$\sf \implies 3200 \bigg( \dfrac{110}{100} \bigg)^{3}$

$\sf \implies 3200 \bigg( \dfrac{11}{10} \bigg)^{3}$

$\sf \implies 3200 \bigg( \dfrac{ {(11)}^{3} }{ {(10)}^{3} } \bigg)$

$\sf \implies 3200 \times \dfrac{ 1331 }{ 1000 }$

$\sf \implies \dfrac{4,259,200}{ 1000 }$

$\sf \implies4259.2$

$\boxed{ \pink{\sf \therefore Amount = Rs.4259.2}}$

As we know that:-

$\boxed{ \sf \leadsto Compound \: Interest = Amount - Principal}$

${ \sf \implies 4259.2 - 3200}$

${ \sf \implies 1059.2}$

$\boxed{ \purple{\sf \therefore Compound \: Interest = Rs.1059.2}}$