Divyam has Rs. 40000. He invests a part of this amount in scheme A, which offers 30\% per annum simple interest, and the remaining amount in scheme B, which offers 20\% compound interest compounded annually. After 3 years, the interest earned from scheme A exceeds the interest earned from scheme B by Rs. 9952. What is the amount invested in scheme B?

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Accepted Answer

**Solution:**

1. Let the amount invested in scheme A be $ x $ rupees. Then, the amount invested in scheme B will be $ 40000 - x $.

2. Interest from scheme A (Simple Interest) for 3 years at 30\% per annum:
   $$
   $SI_A = \frac{x \times 30 \times 3}{100} = 0.9x$
   $$

3. Interest from scheme B (Compound Interest) for 3 years at 20\% per annum:
   The compound amount after 3 years is:
   $$
   $$A_B = (40000 - x) \times \left(1 + \frac{20}{100}\right)^3 = (40000 - x) \times (1.2)^3 = (40000 - x) \times 1.728$$
   $$

Compound interest from scheme B:
   $$
   $CI_B = A_B - (40000 - x) = (40000 - x)(1.728 - 1) = (40000 - x) \times 0.728$
   $$

4. According to the problem, the interest from scheme A exceeds the interest from scheme B by Rs. 9952:
   $$
   $SI_A - CI_B = 9952$
   $$

Substitute the values:
   $$
  $ 0.9x - 0.728(40000 - x) = 9952$
   $$

5. Simplify the equation:
   $$
  $ 0.9x - 29120 + 0.728x = 9952$
   $$
   $$
   $(0.9 + 0.728)x = 9952 + 29120$
   $$
   $$
  $ 1.628x = 39072$
   $$

6. Solve for $ x $:
   $$
  $ x = \frac{39072}{1.628} = 24000$
   $$

7. Amount invested in scheme B:
   $$
   40000 - 24000 = 16000
   $$

**Answer:** Rs. 16000

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