A and B invest in a business in the ratio of $10:7$. After $T$ months, B increases his investment by $40\%$ while A withdraws $50\%$ of his investment. If at the end of the year, the ratio of their profit shares is $95:98$, find the value of $T$.

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**Solution:**

1. Let the initial investments of A and B be $10x$ and $7x$ respectively.

2. For the first $T$ months, both invest the original amounts:
   - A's investment for $T$ months = $10x 	imes T$
   - B's investment for $T$ months = $7x 	imes T$

3. After $T$ months, A withdraws $50\%$ of his investment, so A's new investment = $10x 	imes 0.5 = 5x$.

4. B increases his investment by $40\%$, so B's new investment = $7x 	imes 1.4 = 9.8x$.

5. For the remaining $(12 - T)$ months, the investments are:
   - A: $5x 	imes (12 - T)$
   - B: $9.8x 	imes (12 - T)$

6. Total investment (investment × time) for A and B over the year:
   - A's total = $10xT + 5x(12 - T) = 10xT + 60x - 5xT = (5T + 60)x$
   - B's total = $7xT + 9.8x(12 - T) = 7xT + 117.6x - 9.8xT = (117.6 - 2.8T)x$

7. The ratio of their profits equals the ratio of their total investments:
   $$\frac{5T + 60}{117.6 - 2.8T} = \frac{95}{98}$$

8. Cross-multiply and solve for $T$:
   $$98(5T + 60) = 95(117.6 - 2.8T)$$
   $$490T + 5880 = 11172 - 266T$$
   $$490T + 266T = 11172 - 5880$$
   $$756T = 5292$$
   $$T = \frac{5292}{756} = 7$$

**Answer:** $7$

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