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DS - Algebra Translation
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GMATGuruNY
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Statement 1: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?
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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ceilidh.erickson
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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.
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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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Matt@VeritasPrep
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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.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?
