A man invested Rs $X$ in scheme A and Rs $(X + 400)$ in scheme B. The rate of interest for scheme A is 10\% per annum and for scheme B is 12\% per annum. If the total interest earned from scheme A after 4 years and scheme B after 2 years is Rs 640, find the value of $X$.

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Accepted Answer

**Solution:**

1. Let the amount invested in scheme A be Rs $X$.

2. Then, the amount invested in scheme B is Rs $(X + 400)$.

3. The interest from scheme A after 4 years at 10\% p.a. is:
   $$	ext{Interest}_A = \frac{X 	imes 10 	imes 4}{100} = \frac{40X}{100} = 0.4X$$

4. The interest from scheme B after 2 years at 12\% p.a. is:
   $$	ext{Interest}_B = \frac{(X + 400) 	imes 12 	imes 2}{100} = \frac{24(X + 400)}{100} = 0.24(X + 400)$$

5. According to the problem, the sum of these interests is Rs 640:
   $$0.4X + 0.24(X + 400) = 640$$

6. Simplify the equation:
   $$0.4X + 0.24X + 0.24 	imes 400 = 640$$
   $$0.64X + 96 = 640$$

7. Subtract 96 from both sides:
   $$0.64X = 640 - 96 = 544$$

8. Divide both sides by 0.64:
   $$X = \frac{544}{0.64} = 850$$

**Answer:** Rs $850$

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