The monthly income of A is 60\% more than that of B. Both A and B pay the same amount for EMI. B's EMI payment is 20\% of his monthly income. A spends $n$ times the EMI amount on house rent. The amount A spends on house rent is Rs. 5000 more than B's house rent. Also, A's house rent is Rs. 3000 more than his EMI payment. If B's house rent is 1/3 of A's EMI payment, what is the possible monthly income of A?

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Accepted Answer

**Solution:**

1. Let B's monthly income be $x$. Then A's monthly income is $1.6x$ (since A's income is 60\% more than B's).

2. Both pay the same EMI amount. Given B pays 20\% of his income as EMI, so EMI = $0.2x$.

3. Let A's EMI amount = $0.2x$ (same as B).

4. A spends $n$ times the EMI amount on house rent, so A's house rent = $n \times 0.2x = 0.2nx$.

5. B's house rent is $\frac{1}{3}$ of A's EMI, so B's house rent = $\frac{1}{3} \times 0.2x = \frac{0.2x}{3} = \frac{0.2x}{3}$.

6. The difference between A's and B's house rent is Rs. 5000:
   $$
   0.2nx - \frac{0.2x}{3} = 5000
   $$

7. Also, A's house rent is Rs. 3000 more than his EMI:
   $$
   0.2nx - 0.2x = 3000
   $$

8. From equation (7), factor out $0.2x$:
   $$
   0.2x(n - 1) = 3000 \implies n - 1 = \frac{3000}{0.2x} = \frac{3000}{0.2x} = \frac{3000}{0.2x}
   $$

9. From equation (6):
   $$
   0.2nx - \frac{0.2x}{3} = 5000 \implies 0.2x n = 5000 + \frac{0.2x}{3}
   $$

10. Substitute $0.2x n$ from (9) into (7):
    $$
    0.2x n - 0.2x = 3000 \implies (5000 + \frac{0.2x}{3}) - 0.2x = 3000
    $$

11. Simplify:
    $$
    5000 + \frac{0.2x}{3} - 0.2x = 3000 \implies 5000 - 3000 = 0.2x - \frac{0.2x}{3}
    $$

$$
    2000 = 0.2x \left(1 - \frac{1}{3}\right) = 0.2x \times \frac{2}{3} = \frac{0.4x}{3}
    $$

12. Solve for $x$:
    $$
    2000 = \frac{0.4x}{3} \implies 2000 \times 3 = 0.4x \implies 6000 = 0.4x \implies x = \frac{6000}{0.4} = 15000
    $$

13. Recall A's income is $1.6x$:
    $$
    1.6 \times 15000 = 24000
    $$

**Answer:** Rs. 24000

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