Pipe A and B together can fill a tank in 24 hours. Pipe A and C together can fill the same tank in 16 hours. Pipe A alone can empty the tank in 20 hours. What is the time taken by pipes B and C together to fill the tank?

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**Solution:**

1. Let the capacity of the tank be 1 unit.

2. Define the rates of work:
   - Rate of A filling the tank = $a$ units/hour
   - Rate of B filling the tank = $b$ units/hour
   - Rate of C filling the tank = $c$ units/hour

3. Given:
   - A and B together fill the tank in 24 hours, so:
     $$a + b = \frac{1}{24}$$

- A and C together fill the tank in 16 hours, so:
     $$a + c = \frac{1}{16}$$

- A alone empties the tank in 20 hours, so A's emptying rate is:
     $$-a = -\frac{1}{20} \implies a = -\frac{1}{20}$$

(Since A empties the tank, its rate is negative when considering filling.)

4. Substitute $a = -\frac{1}{20}$ into the first two equations:
   - From $a + b = \frac{1}{24}$:
     $$-\frac{1}{20} + b = \frac{1}{24} \implies b = \frac{1}{24} + \frac{1}{20} = \frac{5 + 6}{120} = \frac{11}{120}$$

- From $a + c = \frac{1}{16}$:
     $$-\frac{1}{20} + c = \frac{1}{16} \implies c = \frac{1}{16} + \frac{1}{20} = \frac{5 + 4}{80} = \frac{9}{80}$$

5. The combined rate of B and C filling the tank is:
   $$b + c = \frac{11}{120} + \frac{9}{80} = \frac{11 	imes 2}{240} + \frac{9 	imes 3}{240} = \frac{22 + 27}{240} = \frac{49}{240}$$

6. Therefore, the time taken by B and C together to fill the tank is the reciprocal of their combined rate:
   $$	ext{Time} = \frac{1}{b + c} = \frac{240}{49} 	ext{ hours}$$

**Answer:** $\frac{240}{49}$ hours

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