When all 3 pipes were opened simultaneously, volume of tank S filled in 1 hour is 3575 cubic meters. Pipe B alone can fill 25\% of tank S in 3.5 hours. In how much time can pipe A fill 40\% of tank S alone?

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**Solution:**

1. Given that pipe B fills 25\% of tank S in 3.5 hours.
   - So, pipe B fills the whole tank in:
     $$\text{Time}_B = \frac{3.5}{0.25} = 14 \text{ hours}$$

2. Pipe C empties the tank in 35 hours (given by the relation in the solution).

3. Let pipe A alone fill the tank in $T$ hours.

4. The combined rate of filling the tank by pipes A and B and emptying by pipe C is given by:
   $$\frac{1}{T} + \frac{1}{14} - \frac{1}{35} = \frac{3575}{19250}$$

5. Simplify the right side:
   $$\frac{3575}{19250} = \frac{3575 \div 3575}{19250 \div 3575} = \frac{1}{5.38} \approx 0.186$$

6. Calculate $\frac{1}{14} - \frac{1}{35} = \frac{5 - 2}{70} = \frac{3}{70} = 0.042857$

7. So,
   $$\frac{1}{T} = 0.186 - 0.042857 = 0.143143$$

8. Therefore,
   $$T = \frac{1}{0.143143} \approx 7 \text{ hours}$$

9. Time taken by pipe A to fill 40\% of tank S alone:
   $$7 \times 0.4 = 2.8 \text{ hours}$$

**Answer:** 2.8 hours

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