Mr. Thomas invested an amount of ₹13,900 divided into two different schemes A and B at simple interest rates of 14% p.a. and 11% p.a. respectively. If the total simple interest earned in 2 years is ₹3,508, what was the amount invested in Scheme B?

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**Solution:**

1. Let the amount invested in Scheme B be *x*.

2. Then, the amount invested in Scheme A is ₹(13,900 - x).

3. Simple interest formula: $$SI = \frac{P 	imes R 	imes T}{100}$$

4. Total interest from both schemes for 2 years is given as ₹3,508:
   $$\frac{(13,900 - x) 	imes 14 	imes 2}{100} + \frac{x 	imes 11 	imes 2}{100} = 3508$$

5. Simplify the equation:
   $$\frac{(13,900 - x) 	imes 28}{100} + \frac{22x}{100} = 3508$$

6. Multiply both sides by 100 to clear denominator:
   $$(13,900 - x) 	imes 28 + 22x = 350,800$$

7. Expand:
   $$389,200 - 28x + 22x = 350,800$$

8. Combine like terms:
   $$389,200 - 6x = 350,800$$

9. Rearrange:
   $$6x = 389,200 - 350,800 = 38,400$$

10. Solve for *x*:
    $$x = \frac{38,400}{6} = 6,400$$

**Answer:** ₹6,400 invested in Scheme B.

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