Question

A sum of money is invested at compound interest payable annually. The interest in two

successive years is Rs.225 and Rs.240. Find​

Answer 1

Given,

• Interest for first year = Rs 225

• Interest for second year = Rs 240

$\sf{\implies Difference = 240 - 225 }$

$\sf{\implies Difference = Rs \ 15 }$

(i) The rate of interest

We know the equation:

$\boxed{\sf{r=\dfrac{I\times100}{P\times t}}}$

Here,

• r = rate of interest = ?

• I = simple interest = Rs 15

• P = principle amount = Rs 225

• t = time period = 1 year

$\sf{\implies r=\dfrac{15\times100}{225\times 1}}$

$\sf{\implies r=\dfrac{1500}{225}}$

$\sf{\implies r=6.67 \ \%}$

$\boxed{\sf{\therefore Rate \ of \ interest=6.67 \ \%}}$

_________________________

(ii) The original sum

We know the equation:

$\boxed{\sf{S=\dfrac{I\times100}{r\times t}}}$

Here,

• S = original sum = ?

• I = simple interest = Rs 225

• r = rate of interest = 6.67

• t = time period = 1 year

$\sf{\implies S=\dfrac{225\times100}{6.67\times 1}}$

$\sf{\implies S=\dfrac{22500}{6.67}}$

$\sf{\implies S=Rs \ 3375}$

$\boxed{\sf{\therefore Sum=Rs \ 3375}}$

_________________________

(iii) The interest earned in third year

So first we have to find the total amount.

$\sf{\implies Total \ amount = 225 + 240 + 3375}$

$\sf{\implies Total \ amount = Rs \ 3840}$

We know the equation:

$\boxed{\sf{I=\dfrac{P\times r\times t}{100}}}$

Here,

• I = simple interest = ?

• P = principal amount = Rs 3840

• r = rate of interest = 6.67

• t = time period = 1 year

$\sf{\implies I=\dfrac{3840\times6.67\times 1}{100}}$

$\sf{\implies I=\dfrac{25600}{100}}$

$\sf{\implies I=Rs \ 256}$

$\boxed{\sf{\therefore Interest=Rs \ 3375}}$