Question

If the CI on a sum of money for 3 years at
the rate of 2% per annum is 306.04, then
what will be the SI?​

Answer 1

Compound Interest for 3 years at 2% pa = 306.04

Say Principal = p, we have

(p*(102/100)^3)-p = 306.04

= 1.061208p - p = 306.04

= 0.061208p = 306.04

= p = 306.04/ 0.061208

= p = 5000

SI = 5000*2/100*3= 300.

Answer 2

To Find:-

• Find the Simple Interest.

Given:-

• CI on a sum of money for 3 years at the rate of 2% per annum is 306.04.

Solution:-

$\tt\implies\sf C .I \: = P \: [ \: {(\: 1 \: + \: \dfrac{ r}{100} )}^{t} \: - \: 1 \: ]$

$\tt\implies\sf 306.04 \: = P \: [ \: {(\: 1 \: + \: \dfrac{ 2}{100} )}^{3} \: - \: 1 \: ]$

$\tt\implies\sf 306.04 \: = P \: [ \: {(1.02)}^{3} \: - \: 1 \: ]$

$\tt\implies\sf 306.04 \: = 0.061208P \:$

$\tt\sf \implies {{ \: P \: = \frac{ \: 306.04 \: }{ \: 0.061208 \: } \: } }$

$\tt\sf \implies\boxed {{ \: P \: = \: 5000 \: } }$

So,

$\tt\sf \implies {{ \: S.I \: = \frac{ \: PTR \: }{ \: 100 \: } \: } }$

$\tt\sf \implies {{ \: S.I \: = \frac{ \: 5000 \: \times \: 3 \: \times \: 2 \: }{ \: 100 \: } \: } }$

$\tt\sf \implies {{ \: S.I \: = \frac{ \: 30000 \: }{ \: 100 \: } \: } }$

$\tt\sf \implies\boxed {{ \: S.I \: = \: 300 \: } }$