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by GMATGuruNY » Sat Jun 11, 2016 1:21 pm
sgraves wrote:Is it possible to answer this question by translating the first statement without having Percy, Randy and Quincy all have the same variable?
Image
Statement 1:
Test increasing options for Q until the total number of slices = 15.
If Q=4, then P=2 and R=1, for a total of 7 slices.
If Q=6, then P=3 and R=2, for a total of 11 slices.
If Q=8, then P=4 and R=3, for a total of 15 slices.
Only the case in red yields a total of 15 slices.
Thus, P=4.
SUFFICIENT.

Statement 2:
Thus, P+R=7 and Q=8, for a total of 15 slices.
It's possible that P=6 and R=1.
It's possible that P=5 and R=2.
Since P can be different values, INSUFFICIENT.

The correct answer is A.
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by ceilidh.erickson » Tue Jun 14, 2016 1:14 pm
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.
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by Matt@VeritasPrep » Fri Jul 22, 2016 2:42 am
sgraves wrote:Is it possible to answer this question by translating the first statement without having Percy, Randy and Quincy all have the same variable?
Great question: when there are certain restrictions on the solutions (such as "each variable must be a positive integer"), we CAN sometimes solve an equation with multiple variables despite only having that single equation.
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