Question

A certain sum of money at simple interest amounts to ₹1024 in 4 years and to ₹1136 in 6 years .Find the rate per cent and sum of money

Answer 1

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

We know that

If a certain sum of money Rs P is invested at the rate of R % per annum for T years, earned simple interest SI per annum, then total amount A is given by

$\boxed{ \bf \: A = P + T \times SI}$

Let's solve the problem now!!!

Let

• Sum of money invested be Rs P at the rate of R % per annum.

Let

• Simple Interest received per annum on Rs P at rate of R % per annum be SI.

So,

According to statement,

• A sum of Rs P amounts to Rs 1024 in 4 years.

$\rm :\implies\:1024 = P + 4SI - - - (1)$

Also,

• A sum of Rs P amounts to Rs 1136 in 6 years.

$\rm :\implies\:1136 = P + 6SI - - - (2)$

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

$\rm :\implies\:1136 - 1024 = (P + 6SI) -(P + 4SI)$

$\rm :\implies\:112 = \cancel{P} + 6SI -\cancel{P} - 4SI$

$\rm :\longmapsto\:2SI = 112$

$\rm :\implies\:\boxed{ \bf \: SI = 56}$

Put value of SI = 56 in equation (1), we get

$\rm :\implies\:1024 = P + 4 \times 56$

$\rm :\implies\:1024 = P + 224$

$\rm :\implies\:P = 1024 - 224$

$\rm :\implies\:\boxed{ \bf \: P = 800}$

So,

Sum of money invested, P = Rs 800.

Now,

We have,

• Simple Interest for 1 year = Rs 56

• Sum of money invested, P = Rs 800

We know,

$\boxed{ \bf \: R = \dfrac{SI \times 100}{P \times T}}$

On substituting the values, we get

$\rm :\longmapsto\:R = \dfrac{56 \times 100}{800 \times 1}$

$\bf\implies \:\boxed{ \bf \: R = 7}$

Hence,

Rate of interest = 7 % per annum