X invested ₹m for simple interest at 10% rate of interest per annum for 4 years, and Y invested ₹2,000 more than X for compound interest at 20% rate of interest per annum for 2 years. The ratio of the interest obtained by X and Y is 10 : 33, respectively. Find the investment of X.

Correct option is A

Given:

X invested ₹m for simple interest at 10% rate of interest per annum for 4 years.

Y invested ₹2000 more than X for compound interest at 20% rate of interest per annum for 2 years.

The ratio of the interest obtained by X and Y is 10:33.

Formula Used:

Simple Interest Formula:

SI $$= \frac{P \times R \times T}{100}$$​

Compound Interest Formula:

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

Where:

P is the principal,

R is the rate of interest,

T is the time in years.

Solution:
Let X's investment be ₹m.

X's simple interest:

​$$\text{SI}_X = \frac{m \times 10 \times 4}{100} = \frac{40m}{100}$$​ = 0.4m

Y's investment is ₹(m + 2000). Y's compound interest:

​$$\text{CI}_Y = (m + 2000) \times \left(1 + \frac{20}{100}\right)^2 - (m + 2000) \\ \ \\\text{CI}_Y = (m + 2000) \times (1.2)^2 - (m + 2000) \\ \ \\\text{CI}_Y = (m + 2000) \times 1.44 - (m + 2000)\\ \ \\\text{CI}_Y = 1.44(m + 2000) - (m + 2000)\\ \ \\\text{CI}_Y = (m + 2000)(1.44 - 1) = (m + 2000) \times 0.44\\ \ \\\text{CI}_Y = 0.44(m + 2000)\\ \ \\\frac{\text{SI}_X}{\text{CI}_Y} = \frac{10}{33}\\ \ \\\frac{0.4m}{0.44(m + 2000)} = \frac{10}{33}$$​​

33 × 0.4m = 10 × 0.44 (m+2000)

13.2m = 4.4(m + 2000)

13.2m = 4.4m + 8800

8.8m = 8800

m $$= \frac{8800}{8.8} =$$​ 1000

X's investment is ₹1000.