Question

The difference between the compound interest compounded annually and the simple interest on a certain sum for 3 years at 10% per annum is ₹ 62 find the sum .​

Answer 1

Answer:

186 is the answer , got it?

Answer 2

The sum is ₹2000

Given : The difference between the compound interest compounded annually and the simple interest on a certain sum for 3 years at 10% per annum is ₹ 62

To find : The sum.

Solution :

We can simply solve this mathematical problem by using the following mathematical process. (our goal is to calculate the sum)

Let, the sum = ₹ x

Now, in case of compound interest we know that,

$A = P \times {(1 - \frac{r}{100})}^{n}$

Where,

• A = Principal amount + Compound interest
• P = Principal amount
• r = rate of interest per annum
• n = number of years

In this case,

• A = ? (unknown quantity)
• P = ₹ x
• r = 10%
• n = 3

So,

$A = x \times {(1 + \frac{10}{100}) }^{ 3}$

$A = x \times {( \frac{11}{10}) }^{3}$

$A = x \times \frac{1331}{1000}$

$A = ₹ \: \: \frac{1331x}{1000}$

Now,

Compound interest :

= (Principal amount + Compound interest) - Principal amount

$= \frac{1331x}{1000} - x$

$= \frac{1331x - 1000x}{1000}$

$=₹ \: \: \frac{331x}{1000}$

Now, in case of simple interest :

$simple \: \: interest = \frac{P \times r \times t}{100}$

In this case,

• P = ₹x
• r = 10%
• t = 3 years

So,

$simple \: \: interest = \frac{x \times 10\times 3}{100} = ₹ \: \: \frac{3x}{10}$

According to the data mentioned in the question,

$\frac{331x}{1000} - \frac{3x}{10} = 62$

$\frac{331x - 300x}{1000} = 62$

$31x = 62 \times 1000$

$31x = 62000$

$x = \frac{62000}{31}$

$x = 2000$

The sum of money = ₹x = ₹2000

Hence, the sum of money is ₹2000