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Is the average (arithmetic mean) of the numbers \(x, y,\) and \(z\) greater than \(z ?\)

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by Gmat_mission » Sun May 31, 2020 1:51 pm

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Is the average (arithmetic mean) of the numbers \(x, y,\) and \(z\) greater than \(z ?\)

(1) \(z − x < y − z\)
(2) \(x < z < y\)

[spoiler]OA=A[/spoiler]

Source: Official Guide
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Source: — Data Sufficiency |

Gmat_mission wrote:
Sun May 31, 2020 1:51 pm
Is the average (arithmetic mean) of the numbers \(x, y,\) and \(z\) greater than \(z ?\)

(1) \(z − x < y − z\)
(2) \(x < z < y\)

[spoiler]OA=A[/spoiler]

Source: Official Guide
Target question: Is the average (arithmetic mean) of the numbers x, y, and z greater than z?
This is a good candidate for rephrasing the target question.
Rewrite the question as "Is (x + y + z)/3 > z?"
Multiply both sides by 3 to get: "Is x + y + z > 3z?"
Subtract z from both sides to get: "Is x + y > 2z?"

REPHRASED target question: Is 2z less than x + y?

Aside: Here’s a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Statement 1: z − x < y − z
Add z to both sides to get: 2z − x < y
Add x to both sides to get: 2z < x + y
PERFECT!
The answer to the REPHRASED target question is YES, 2z IS less than x+y
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: x < z < y
There are several values of x, y and z that satisfy statement 2. Here are two:
Case a: x = 0, y = 3 and z = 1. In this case, 2z = 2(1) = 2 and x + y = 0 + 3 = 3. So, the answer to the REPHRASED target question is YES, 2z IS less than x+y
Case b: x = 0, y = 3 and z = 2. In this case, 2z = 2(2) = 4 and x + y = 0 + 3 = 3. So, the answer to the REPHRASED target question is NO, 2z is NOT less than x+y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A
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