The time taken to fill an empty tank completely when inlet A and outlet X are opened together is 6 hours 40 minutes less than the time taken by inlet B and outlet W together. Given that inlet A alone fills the tank in 4 hours, inlet B alone fills it in 8 hours, and outlet X alone empties the tank in 34 hours, how much time will outlet W alone take to empty the full tank?

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**Solution:**

1. Let the time taken by outlet W alone to empty the full tank be *x* hours.

2. Calculate the combined rate of inlet A and outlet X:
   - Rate of inlet A = $ \frac{1}{4} $ tank/hour
   - Rate of outlet X = $ \frac{1}{34} $ tank/hour (emptying)
   - Combined rate = $ \frac{1}{4} - \frac{1}{34} = \frac{15}{68} $ tank/hour

3. Therefore, time taken by inlet A and outlet X together to fill the tank:
   $$
   \text{Time} = \frac{1}{\text{Rate}} = \frac{68}{15} \text{ hours}
   $$

4. Calculate the combined rate of inlet B and outlet W:
   - Rate of inlet B = $ \frac{1}{8} $ tank/hour
   - Rate of outlet W = $ \frac{1}{x} $ tank/hour (emptying)
   - Combined rate = $ \frac{1}{8} - \frac{1}{x} = \frac{x - 8}{8x} $ tank/hour

5. Time taken by inlet B and outlet W together to fill the tank:
   $$
   \frac{1}{\text{Combined rate}} = \frac{8x}{x - 8} \text{ hours}
   $$

6. According to the problem, the time taken by inlet A and outlet X together is 6 hours 40 minutes (which is $ \frac{20}{3} $ hours) less than the time taken by inlet B and outlet W together:
   $$
   \frac{8x}{x - 8} - \frac{68}{15} = \frac{20}{3}
   $$

7. Multiply both sides by $ 15(x - 8) $ to clear denominators:
   $$
   15 \times 8x - 68(x - 8) = 15(x - 8) \times \frac{20}{3}
   $$

Simplify:
   $$
   120x - 68x + 544 = 100(x - 8)
   $$

$$
   52x + 544 = 100x - 800
   $$

8. Rearranging terms:
   $$
   544 + 800 = 100x - 52x
   $$

$$
   1344 = 48x
   $$

$$
   x = \frac{1344}{48} = 28 \text{ hours}
   $$

**Answer:** Outlet W alone will take $ \boxed{28} $ hours to empty the full tank.

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