A tank when full can be emptied by an outlet pipe A in $5.6$ hours, while an inlet pipe B can fill the same empty tank in $7$ hours. If pipes A and B are turned on alternatively for $1$ hour each starting with pipe A when the tank is full, how long will it take to empty the tank?

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**Solution:**

1. Let the capacity of the tank be the LCM of $5.6$ and $7$ to simplify calculations:
   $$	ext{LCM}(5.6, 7) = 28 	ext{ units}$$

2. Calculate the efficiency (work done per hour) of each pipe:
   - Pipe A empties the tank in $5.6$ hours, so efficiency of A:
     $$\frac{28}{5.6} = 5 	ext{ units/hour (emptying)}$$
   - Pipe B fills the tank in $7$ hours, so efficiency of B:
     $$\frac{28}{7} = 4 	ext{ units/hour (filling)}$$

3. Since pipe A empties and pipe B fills, consider emptying as negative work and filling as positive work:
   - Efficiency of A = $-5$ units/hour
   - Efficiency of B = $+4$ units/hour

4. Pipes work alternately for 1 hour each starting with pipe A:
   - In 2 hours (1 hour A + 1 hour B), net work done:
     $$-5 + 4 = -1 	ext{ unit}$$

5. After $n$ such 2-hour cycles, total work done is $-1 	imes n$ units.

6. When the tank is full (28 units), after $n$ cycles, the remaining work is:
   $$28 - n 	imes 1 = 28 - n$$

7. The last hour is worked by pipe A alone (emptying 5 units):
   - So, after $n$ cycles (each 2 hours), pipe A empties 5 units in the last hour.

8. Remaining work after last hour by pipe A:
   $$28 - 5 = 23 	ext{ units}$$

9. Since each 2-hour cycle empties 1 unit, to empty 23 units:
   $$23 	imes 2 = 46 	ext{ hours}$$

10. Total time taken:
    $$46 + 1 = 47 	ext{ hours}$$

**Answer:** $47$ hours

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