Set P contains three consecutive multiples of 9, and set Q contains four consecutive even numbers. The largest number in set P is equal to the smallest number in set Q. The average of the smallest numbers in sets P and Q is 27. Find the difference between the second smallest number in set P and the second largest number in set Q.

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**Solution:**

1. Let the three consecutive multiples of 9 in set P be:
   $$9n, 9(n+1), 9(n+2)$$

2. Let the four consecutive even numbers in set Q be:
   $$x, x+2, x+4, x+6$$

3. Given that the largest number in set P equals the smallest number in set Q:
   $$9(n+2) = x$$

4. The average of the smallest numbers in sets P and Q is 27:
   $$\frac{9n + x}{2} = 27$$

5. Substitute $x$ from step 3 into step 4:
   $$\frac{9n + 9(n+2)}{2} = 27$$

6. Simplify:
   $$\frac{9n + 9n + 18}{2} = 27$$
   $$\frac{18n + 18}{2} = 27$$
   $$9n + 9 = 27$$

7. Solve for $n$:
   $$9n = 18$$
   $$n = 2$$

8. Find $x$:
   $$x = 9(n+2) = 9(2+2) = 36$$

9. Now, set P is:
   $$9 	imes 2 = 18, 9 	imes 3 = 27, 9 	imes 4 = 36$$

10. Set Q is:
    $$36, 38, 40, 42$$

11. The second smallest number in set P is 27.

12. The second largest number in set Q is 40.

13. Calculate the difference:
    $$40 - 27 = 13$$

**Answer:** 13

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