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# The average (arithmetic mean) of x and z is greater than y,

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The average (arithmetic mean) of x and z is greater than y, [#permalink]  15 Nov 2019, 16:39
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Question Stats:

75% (00:22) correct 25% (00:59) wrong based on 20 sessions
The average (arithmetic mean) of x and z is greater than y, and $$x < y < z$$.

 Quantity A Quantity B The average of x, y, and z The median of x, y, and z

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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[Reveal] Spoiler: OA

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Re: The average (arithmetic mean) of x and z is greater than y, [#permalink]  16 Nov 2019, 07:11
Expert's post
Carcass wrote:
The average (arithmetic mean) of x and z is greater than y, and $$x < y < z$$.

 Quantity A Quantity B The average of x, y, and z The median of x, y, and z

Since $$x < y < z$$, we can see that the median of the set must be y. So Quantity B = y

We have:
Quantity A: $$\frac{x + y + z}{3}$$

Quantity B: $$y$$

Multiply both quantities by 3 to get:
Quantity A: $$x + y + z$$
Quantity B: $$3y$$

Subtract $$y$$ from both quantities to get
Quantity A: $$x + z$$
Quantity B: $$2y$$

GIVEN: The average (arithmetic mean) of x and z is greater than y

We can write: $$\frac{x + y}{2} > y$$

Multiply both sides of the inequality by 2 to get: $$x + y > 2y$$

Perfect! This last bit of information tells us that Quantity A must be greater

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Re: The average (arithmetic mean) of x and z is greater than y,   [#permalink] 16 Nov 2019, 07:11
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