A tank is initially full. Outlets X and Y are opened together and they empty the tank for 8 hours 30 minutes. After that, both outlets are closed and inlet A is opened, which fills the tank completely in 3 hours 30 minutes. If outlet X alone can empty the full tank in 34 hours, in how much time will outlet Y alone empty the full tank?

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**Solution:**

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

2. Given inlet A fills the tank in 4 hours, so rate of A = $\frac{1}{4}$ tank/hour.

3. Inlet A fills the tank in 3 hours 30 minutes (i.e., $3 \frac{1}{2} = \frac{7}{2}$ hours) after outlets X and Y were closed. The part of tank filled by A in this time:
   $$	ext{Part filled} = \frac{7}{2} 	imes \frac{1}{4} = \frac{7}{8}$$

4. Therefore, the part of the tank emptied by outlets X and Y together in 8 hours 30 minutes (i.e., $8.5 = \frac{17}{2}$ hours) is:
   $$1 - \frac{7}{8} = \frac{1}{8}$$

5. Let the time taken by outlet Y alone to empty the tank be $x$ hours.

6. Rate of outlet X = $\frac{1}{34}$ tank/hour.

7. Rate of outlet Y = $\frac{1}{x}$ tank/hour.

8. Combined rate of outlets X and Y:
   $$\frac{1}{34} + \frac{1}{x}$$

9. They empty $\frac{1}{8}$ of the tank in $\frac{17}{2}$ hours, so:
   $$\left( \frac{1}{34} + \frac{1}{x} ight) 	imes \frac{17}{2} = \frac{1}{8}$$

10. Simplify:
    $$\frac{17}{2} 	imes \left( \frac{1}{34} + \frac{1}{x} ight) = \frac{1}{8}$$
    $$\Rightarrow \frac{17}{2} 	imes \frac{1}{34} + \frac{17}{2} 	imes \frac{1}{x} = \frac{1}{8}$$
    $$\Rightarrow \frac{17}{68} + \frac{17}{2x} = \frac{1}{8}$$
    $$\Rightarrow \frac{1}{4} + \frac{17}{2x} = \frac{1}{8}$$

11. Subtract $\frac{1}{4}$ from both sides:
    $$\frac{17}{2x} = \frac{1}{8} - \frac{1}{4} = -\frac{1}{8}$$

12. Since rate cannot be negative, re-check step 10:
    Actually, $\frac{17}{68} = \frac{1}{4}$ is correct, but the sum should equal $\frac{1}{8}$, which is less than $\frac{1}{4}$, so this indicates an error in the assumption about the part emptied.

13. Reconsider step 4:
    The tank was full initially. Outlets X and Y opened for $\frac{17}{2}$ hours emptied some part. Then inlet A filled the tank in $\frac{7}{2}$ hours.

So, the part emptied by outlets X and Y in $\frac{17}{2}$ hours = part filled by A in $\frac{7}{2}$ hours = $\frac{7}{8}$ (from step 3).

14. So, the combined emptying rate of X and Y:
    $$\left( \frac{1}{34} + \frac{1}{x} ight) 	imes \frac{17}{2} = \frac{7}{8}$$

15. Simplify:
    $$\frac{17}{2} 	imes \left( \frac{1}{34} + \frac{1}{x} ight) = \frac{7}{8}$$
    $$\Rightarrow \frac{17}{68} + \frac{17}{2x} = \frac{7}{8}$$
    $$\Rightarrow \frac{1}{4} + \frac{17}{2x} = \frac{7}{8}$$

16. Subtract $\frac{1}{4}$ from both sides:
    $$\frac{17}{2x} = \frac{7}{8} - \frac{1}{4} = \frac{7}{8} - \frac{2}{8} = \frac{5}{8}$$

17. Solve for $x$:
    $$\frac{17}{2x} = \frac{5}{8}$$
    $$\Rightarrow 17 	imes 8 = 5 	imes 2x$$
    $$136 = 10x$$
    $$x = \frac{136}{10} = 13.6 	ext{ hours}$$

18. Convert $13.6$ hours to mixed fraction:
    $$13.6 = 13 \frac{3}{5}$$

**Answer:** $3 \frac{13}{5}$ hours (which is $13 \frac{3}{5}$ hours) for outlet Y to empty the tank alone.

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