A set of nonnegative integers consists of {x, x + 7, 2x, y, y + 5}.

A set of nonnegative integers consists of {x, x + 7, 2x, y, y + 5}. The numbers of this set have four distinct values. What is its average (arithmetic mean)?

Notice couple of things:
1. We are told that all numbers in the set are integers;
2. We are told that all those integers are non-negative;
3. We are told that the set contains four distinct values out of five (so two integers out of 5 are alike and other three are distinct).

(1) x ≠ 5. Clearly insufficient.

(2) 4y + 12 = 6(y + 2) --> 4y+12=6y+12 --> 2y=0 --> y=0. So, we have that our set is {x, x + 7, 2x, 0, 5}. Since we know that two integers out of 5 are alike then:

1. x, x+7 or 2x is 0.
If x=0, then 2x=0 too, so in this case we'll have three alike integers x, 2x, and y. Thus this case is out.
If x+7=0, then x=-7 and we know that all integers must be non-negative so this case is out too.

2. x, x+7 or 2x is 5
If x=5, then x+7=12 and 2x=10. This scenario is OK. The set in this case would be: {5, 12, 10, 0, 5}
If x+7=5, then x=-2 and we know that the integers in the set must be non-negative. Thus this case is out.
If 2x=5, then x=5/2 and we know that the numbers in the set must be integers. Thus this case is out.

3. Two out of x, x+7 and 2x are alike.
x=2x is not possible, since in this case x=2x=0 and in this case we'll have three alike integers x, 2x, and y.
x=x+7 has no solution.
2x=x+7 --> x=7. This scenario is OK. The set in this case would be: {7, 14, 14, 0, 5}.

Two cases are possible. Not sufficient.

(1)+(2) Since from (1) x ≠ 5, then from (2) x=7. So, the set is {7, 14, 14, 0, 5}. Sufficient.

Answer: C.

Hope it's clear.