Question

The compound interest on rs. 30,000 at 7% per annum is rs. 4347. The period(in years) is?

Answer 1

The period is one year as it is given as per annum

Answer 2

$\large\underline\pink{Given:-}$

• Principal, p = 30,000
• Compound Interest, C.I = 4347
• Amount, A = C.I + p = 30,000 + 4347 = 34347
• Rate, r = 7 %

$\large\underline\pink{To find:-}$

• Fine the time period if the interest is compound annually ....?

$\large\underline\pink{Solutions:-}$

• Let the time be n year

$\: \: \: \: \: \leadsto \: \: A \: \: = \: \: p \: {({1} \: + \: \frac{r}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {34347} \: \: = \: \: {30000} \: {({1} \: + \: \frac{7}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: \: = \: \: {3000} \: {(\frac{{100} \: + \: {7}}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: \: = \: \: {3000} \: {(\frac{{100} \: + \: {7}}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {5712} \: \: = \: \: {3000} \: {(\frac{107}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: \frac{34347}{3000} \: \: = \: \: {(\frac{107}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {(\frac{107}{100})}^{2} \: \: = \: \: {(\frac{107}{100})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {(\cancel{\frac{107}{100}})}^{2} \: \: = \: \: {(\cancel{\frac{107}{100}})}^{n}$

$\: \: \: \: \: \: \: \leadsto \: \: {2} \: \: = \: \: {n}$

$\: \: \: \: \: \: \: \leadsto \: \: {n} \: \: = \: \: {2}$

Hence, the time period is 2 year.