Alternatively, the algebra would be pretty easy to set up here.

If they ate 15 slices combined, say P + R + Q = 15.

(1) Percy ate half as many slices as did Quincy --> P = (1/2)Q
... and [Percy ate] one more slice than did Randy --> P = R + 1

We have 3 distinct equations with 3 variables, so we'll be able to solve for the value of P. Sufficient.

You should NOT actually do this work on test day, but for proof:
P = (1/2)Q --> Q = 2P
P = R + 1 --> R = P - 1
P + R + Q = 15 --> P + (P - 1) + 2P = 15.
4P - 1 = 15
4P = 16
P = 4

(2) The number that Percy and Randy ate combined is one fewer than the number that Quincy ate -->
P + R = Q - 1

This only gives us a total of 2 equations with 3 variables. We could plug (Q - 1) in for (P + R) in our original equation to solve for Q:
P + R + Q = 15 --> (Q - 1) + Q = 15.
2Q - 1 = 15
2Q = 16
Q = 8

However, this does not help us to solve for the value of P, so it's insufficient to answer the question.

The answer is A.