₹200 was invested for 2 years on 10% compound interest per year. If the rate of interest had been 20%, then how much more would the investor have received as interest for the same period?

Correct option is A

Given:

Principal (P) = ₹200

Time (T) = 2 years

Rate of interest 1 (R₁) = 10%

Rate of interest 2 (R₂) = 20%

Formula Used:
The formula for compound interest:

A = $$P \left(1 + \frac{R}{100}\right)^T$$​

Where:

A is the amount after interest.

P is the principal amount.

R is the rate of interest per year.

T is the time in years.

Compound Interest = A - P

Solution:

At 10% interest:

​$$A_1 = 200 \left(1 + \frac{10}{100}\right)^2 = 200 \times (1.1)^2 = 200 \times 1.21 = 242$$​​

Compound interest at 10% = 242 - 200 = 42

At 20% interest:

​$$A_2 = 200 \left(1 + \frac{20}{100}\right)^2 = 200 \times (1.2)^2 = 200 \times 1.44 = 288$$​​

Compound interest at 20% = 288 - 200 = 88

Additional Interest = 88 - 42 = 46 

Alternate Method:

Formula Used:
a + b + $$\frac {ab}{100}$$ 

Solution:

10 + 10 + $$\frac{100}{100}$$ = 21%

20 + 20 + $$\frac{200}{100}$$ = 44%

Difference = 44% - 21% = 23%

Now,

200 $$\times \frac{23}{100}$$ = 46​