There are two inlets A and B connected to a tank. A and B can fill the tank in 32 hours and 28 hours, respectively. If both the pipes are opened alternately for 1 hour each, starting with A, then in how much time (in hours, to the nearest integer) will the tank be filled?

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**Solution:**

1. Pipe A fills the tank in 32 hours, so in 1 hour it fills $ \frac{1}{32} $ of the tank.

2. Pipe B fills the tank in 28 hours, so in 1 hour it fills $ \frac{1}{28} $ of the tank.

3. Both pipes are opened alternately for 1 hour each, starting with A. So, in 2 hours, the amount filled is:
   
   $$\frac{1}{32} + \frac{1}{28} = \frac{7}{224} + \frac{8}{224} = \frac{15}{224}$$

4. Let the total time to fill the tank be $ T $ hours. Since the pipes work in 2-hour cycles, consider $ n $ such cycles:

- Amount filled in $ n $ cycles (i.e., $ 2n $ hours) = $ n \times \frac{15}{224} $

5. We want to find $ n $ such that the tank is almost full:

$$ n \times \frac{15}{224} \approx 1 $$

Solving for $ n $:

$$ n = \frac{224}{15} \approx 14.93 $$

6. Since $ n $ must be an integer number of cycles, after 14 full cycles (28 hours), the amount filled is:

$$ 14 \times \frac{15}{224} = \frac{210}{224} = \frac{15}{16} $$

So, after 28 hours, $ \frac{15}{16} $ of the tank is filled.

7. Remaining part to fill:

$$ 1 - \frac{15}{16} = \frac{1}{16} $$

8. The next pipe to open is A (since we started with A and alternate every hour). Pipe A fills $ \frac{1}{32} $ of the tank per hour.

9. Time taken by pipe A to fill the remaining $ \frac{1}{16} $ of the tank:

$$ \text{Time} = \frac{\text{Remaining part}}{\text{Rate of A}} = \frac{\frac{1}{16}}{\frac{1}{32}} = 2 \text{ hours} $$

10. Total time to fill the tank:

$$ 28 + 2 = 30 \text{ hours} $$

**Answer:** 30 hours

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