Question

A sum of money, invested at compound interest, amounts to Rs.19,360 in 2 years and to Rs.23,425.60 in 4 years. Find the rate percent and the original sum of money.
$
  (a){\text{ 10% and Rs}}{\text{. 17,000}} \\
  (b){\text{ 19% and Rs}}{\text{. 25,637}} \\
  (c){\text{ 10% and Rs}}{\text{. 16,000}} \\
  (d){\text{ 19% and Rs}}{\text{. 26,634}} \\
 $

Answer 1

Hint – The principal amount is the original amount that is being invested. Use the direct formula for compound interest. The amount and time are given, use them to obtain the principal value, and then find the rate of interest.

Complete step-by-step answer:

As we know the formula for compound interest which is given as
$ \Rightarrow A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$…………………………. (1)
Where A = total amount received after the compound interest.
             P = Principle amount.
             r = rate of interest.
             t = time in years.
Now it is given that a sum of money, invested at compound interest, amounts to Rs. 19,360 in 2 years.

Therefore A = 19360, t = 2 years

So from equation (1) we have,
 $ \Rightarrow 19360 = P{\left( {1 + \dfrac{r}{{100}}} \right)^2}$………………………………………. (2)

Now again it is given that a sum of money, invested at compound interest, amounts to Rs. 23,425.60 in 4 years.

Therefore A = 23425.60, t = 4 years
So from equation (1) we have,
 $ \Rightarrow 23425.60 = P{\left( {1 + \dfrac{r}{{100}}} \right)^4}$………………………………………. (3)

Now squaring on both sides in equation (2) we have,
$ \Rightarrow {\left( {19360} \right)^2} = {P^2}{\left( {1 + \dfrac{r}{{100}}} \right)^4}$
$ \Rightarrow {\left( {1 + \dfrac{r}{{100}}} \right)^4} = \dfrac{{{{\left( {19360} \right)}^2}}}{{{P^2}}}$

Now substitute this value in equation (3) we have,
$ \Rightarrow 23425.60 = P\dfrac{{{{\left( {19360} \right)}^2}}}{{{P^2}}}$

Now on simplifying we get,
$ \Rightarrow P = \dfrac{{{{\left( {19360} \right)}^2}}}{{23425.60}} = 16000$ Rs.

So the original sum is Rs. 16,000.

Now substitute this value in equation (2) we have,
$ \Rightarrow 19360 = 16000{\left( {1 + \dfrac{r}{{100}}} \right)^2}$
$ \Rightarrow {\left( {1 + \dfrac{r}{{100}}} \right)^2} = \dfrac{{19360}}{{16000}} = 1.21$

Now take square root on both sides we have,
$ \Rightarrow \sqrt {{{\left( {1 + \dfrac{r}{{100}}} \right)}^2}} = \sqrt {1.21} = \sqrt {{{\left( {1.1} \right)}^2}} = 1.1$
$ \Rightarrow \left( {1 + \dfrac{r}{{100}}} \right) = 1.1$
$ \Rightarrow \dfrac{r}{{100}} = 1.1 - 1 = 0.1$
$ \Rightarrow r = 0.1\left( {100} \right) = 10$ %.

So the rate of interest is 10%.

So the original sum and rate of interest is Rs. 16,000 and 10% respectively.

Hence option (C) is correct.

Note – There are two types of interest that are compounded annually and one is a simple interest. The simple interest is based on the principal amount of a loan however a compound interest is based on the principal amount and the interest that accumulates on it in every period.