Question

The differences between the compound
interest and the simple interest for 3 years
at 5% p.a. on a certain sum of money was
Rs.610. Find the sum.​

Answer 1

Answer:-

Let the sum be P.

Given:

Time period (T) = 3 years.

Rate of interest (R) = 5%

Difference between CI and SI = Rs. 610.

We know that;

$\sf \: Simple \: Interest \: (SI) = \frac{PTR}{100} \\ \\ \sf \: Compound \: Interest (CI) = P \bigg(1 + \frac{R}{100} \bigg) ^{T} - P$

Hence;

$\implies \sf \: P \bigg(1 + \frac{5}{100} \bigg) ^{3} - P - \frac{P \times 3 \times 5}{100} = 610 \\ \\ \\ \implies \sf \:P \bigg( \frac{100 + 5}{100} \bigg) ^{3} - \frac{100P}{100} - \frac{15P}{100} = 610 \\ \\ \\ \implies \sf \: P \bigg( \frac{105 \times 105 \times 105}{1000000} \bigg) - \frac{100P }{100} - \frac{15P}{100} = 610 \\ \\ \\ \implies \sf \: \frac{1157625P}{1000000} - \frac{100P}{100} - \frac{15P}{100} = 610\\ \\ \\ \implies \sf \: \frac{1157625P -1000000P - 150000P}{1000000} = 610 \\ \\ \\ \implies \sf \:7625P = 610 \times 1000000 \\ \\ \\ \implies \sf \:P = \frac{610 \times 1000000}{7625} \\ \\ \\ \implies \boxed{\sf \:P = Rs. \:80000}$

∴ The sum is Rs. 80,000.

Answer 2

Given :-

Rate = 5 %

Time = 3 years

Difference = 610

To Find :-

Sum

Solution :-

For Simple interest

$\sf SI = \dfrac{PRT}{100}$

$\sf \dfrac{P\times 5\times 3}{100}$

$\sf \dfrac{15P}{100}$

$\sf 0.15P$

For CI

$\sf CI = P\bigg(1+\dfrac{r}{100}\bigg)^n - P$

$\sf P\bigg(1+ \dfrac{5}{100}\bigg)^3 - P$

$\sf P\bigg(\dfrac{100+5}{100}\bigg)^3 - P$

$\sf P\bigg(\dfrac{105}{100}\bigg)^3 - P$

$\sf P(1.05)^3-P$

$\sf 1.157625P - P = 0.157625P$

$\sf 610 = 0.157625P - 0.15P$

$\sf 610= 0.007625P$

$\sf \dfrac{610}{0.007625} = P$

$\sf 80,000 = P$