Question

[4] iii. A certain sum amounts to 3,47,200 after 3 years and 4,36,800 after 7 years at simple interest. Find the principal and rate of interest per annum.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A certain sum amounts to 3,47,200 after 3 years and 4,36,800 after 7 years at simple interest.

Let assume that

Sum invested be x

and

Simple interest received per annum be y.

We know,

Simple interest received on investment of Rs P at r % per annum is same every year.

So, According to statement,

A certain sum amounts to 3,47,200 after 3 years.

$\rm\implies \:x + 3y = 347200 - - - - (1)$

According to second condition,

A certain sum amounts to 4,36,800 after 7 years.

$\rm\implies \:x + 7y = 436800 - - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm \: \:x + 7y - (x + 3y)= 436800 - 347200$

$\rm \: \:x + 7y - x - 3y= 89600$

$\rm \: \:4y= 89600$

$\rm\implies \:y = 22400$

On substituting the value of y in equation (1), we get

$\rm \: \:x + 3 \times 22400= 347200$

$\rm \: \:x + 67200= 347200$

$\rm \: \:x = 347200 - 67200$

$\rm\implies \:x = 280000$

So, it means

Sum invested, x = 280000

Interest received for one year, y = 22400

Time, n = 1 year

Let assume that rate of interest be r % per annum.

We know that,

Simple interest received on a certain sum of money of P invested at the rate of r % per annum for n years is given by

$\boxed{\sf{  \: \: Simple \: interest = \frac{P \times r \times n}{100} \: }}$

So, on substituting the values, we get

$\rm \: 22400 = \dfrac{280000 \times r \times 1}{100}$

$\rm \: 2800r = 22400$

$\rm \: 28r = 224$

$\rm\implies \:r \: = \: 8 \: \% \: per \: annum$

Hence,

$\rm\implies \:\boxed{\sf{  \: \: Principal \: = \: 280000 \: }}$

and

$\rm\implies \:\boxed{\sf{  \: \: Rate \: of \: interest, \: r \: = \: 8 \: \% \: per \: annum \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

Amount received on a certain sum of money of Rs P invested at the rate of r % per annum for n years is given by

$\boxed{\sf{  \:\rm \: Amount \: = \: P\bigg[\dfrac{100 + rn}{100} \bigg] \: \: }}$

Also,

$\boxed{\sf{  \:\rm \: r \: = \: \frac{Simple \: interest \times 100}{P \times n} \: \: }}$

$\boxed{\sf{  \:\rm \: n\: = \: \frac{Simple \: interest \times 100}{P \times r} \: \: }}$

$\boxed{\sf{  \:\rm \: P\: = \: \frac{Simple \: interest \times 100}{n \times r} \: \: }}$