How many ways to solve this?

This method isn't very rigorous/algebraic, but it's a nice shortcut. First of all, just to clarify, the problem intends to have parentheses around the X+Y, right? So that the whole X+Y is in the denominator? (As opposed to how it's written above, which would have us just simplifying X/X to 1 before we even moved on to the Y.)

Assuming so, this is basically a weighted average problem. The right side of the equation will simplify to (10X + 20Y)/(X+Y) = K ... and that should look a lot like the formula for finding an average. It's as if we've taken X tests and gotten marks of 10 on all of those, and taken Y tests and gotten marks of 20 on all of those, and now we're going to add up all our total points and divide by the number of tests taken (which is X+Y) to come out at our average score K.

We're told that X<Y, so we know that since that means we've gotten 10s on fewer tests, and 20s on more tests, the overall average will be pulled more towards 20, i.e. it will be >15. But it won't land AT 20, because that would only happen if we'd had no tests dragging us down from 20 (not the case, since we've gotten 10s on some tests). So the only remaining possibility among the answer choices is 18.

Added note: the problem hopefully says (or should say, if it doesn't) something like "What is one possible value of k?" We're told that X<Y, which again means that the average must land higher than the midpoint of 10 and 20 (the Y tests on which we got 20 points are pulling the average closer to them, since there are more of them), and we're told that K is a positive integer, but in reality it could be any of the positive integers within the acceptable range, i.e. 16, 17, 18, or 19. (If x and y were 2 and 3 respectively, K would be 16; if x and y were 3 and 7 respectively, K would be 17; if x and y were 1 and 4 respectively, K would be 18; and if x and y were 1 and 9 respectively, K would be 19. Furthermore, if x and y were anything in the *ratio* 2:3, K would be 16, and so on for all the possibilities.)

This is one of the hardest problems I have seen.