Question

A certain sum of money amounts to Rs 4838.4 at 2 years the compound interest.If the rates of successive years are 8% p.a. and 12% p.a. respectively , find the sum​

Answer 1

Given :

• Amount = Rs. 4838.4
• Time = 2 years
• Rate of interest ($R_{1}$) = 8 % p.a.
• Rate of interest ($R_{2}$) = 12 % p.a.

To Find :

The sum of money or the principal.

Solution :

According to the Question , there the rates are given for the two successive years.. i.e,

For 1st year the rate is 8 % p.a. and for the second year the rate is 12 % p.a.

So we will take time in the amount formula.

Hence, we will use the formula for different Rate of interest :-

$\underline{\boxed{\bf{A = P\bigg(1 + \dfrac{R_{1}}{100}\bigg)\bigg(1 + \dfrac{R_{1}}{100}\bigg)\bigg(1 + \dfrac{R_{2}}{100}\bigg)....\bigg(1 + \dfrac{R_{n}}{100}\bigg)}}}$

Where :-

• A = Amount
• P = Principal
• R = Rate of interest

Now , using the formula and substituting the values in it , we get :

$:\implies \bf{A = P\bigg(1 + \dfrac{R_{1}}{100}\bigg)\bigg(1 + \dfrac{R_{2}}{100}\bigg)}$

$:\implies \bf{4838.4 = P\bigg(1 + \dfrac{8}{100}\bigg)\bigg(1 + \dfrac{12}{100}\bigg)}$

$:\implies \bf{4838.4 = P\bigg(\dfrac{100 + 8}{100}\bigg)\bigg(\dfrac{100 + 12}{100}\bigg)}$

$:\implies \bf{4838.4 = P\bigg(\dfrac{108}{100}\bigg)\bigg(\dfrac{112}{100}\bigg)}$

$:\implies \bf{\dfrac{4838.4}{P} = \bigg(\dfrac{108}{100}\bigg)\bigg(\dfrac{112}{100}\bigg)}$

$:\implies \bf{\dfrac{1}{P} = \dfrac{\bigg(\dfrac{108}{100}\bigg)\bigg(\dfrac{112}{100}\bigg)}{4838.4}}$

$:\implies \bf{\dfrac{1}{P} = \dfrac{\dfrac{108}{100} \times \dfrac{112}{100}}{4838.4}}$

$:\implies \bf{\dfrac{1}{P} = \dfrac{\dfrac{54}{50} \times \dfrac{56}{50}}{4838.4}}$

$:\implies \bf{\dfrac{1}{P} = \dfrac{\dfrac{27}{25} \times \dfrac{28}{25}}{4838.4}}$

$:\implies \bf{\dfrac{1}{P} = \dfrac{\dfrac{756}{625}}{4838.4}}$

$:\implies \bf{\dfrac{1}{P} = \dfrac{1.2}{4838.4}}$

$:\implies \bf{P = \dfrac{4838.4}{1.2}}$

$:\implies \bf{P = 3998.68}$

$\therefore \bf{Principal = 3998.68}$

Hence , the sum of money is Rs. 3998.68.