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The average (arithmetic mean) of 12 and 20 is equal to the a

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The average (arithmetic mean) of 12 and 20 is equal to the a [#permalink] New post 14 May 2018, 12:13
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The average (arithmetic mean) of 12 and 20 is equal to the average (arithmetic mean) of 15 and x.

Quantity A
Quantity B
x
16


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.
[Reveal] Spoiler: OA

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Re: The average (arithmetic mean) of 12 and 20 is equal to the a [#permalink] New post 03 Aug 2018, 21:52
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Carcass wrote:
The average (arithmetic mean) of 12 and 20 is equal to the average (arithmetic mean) of 15 and x.

Quantity A
Quantity B
x
16


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.



From the statement we can write as:

\(\frac{12 + 20}2 = \frac{15 + x}2\)

or \(x = 12 + 5 = 17\)

QTY A > QTY B
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Re: The average (arithmetic mean) of 12 and 20 is equal to the a [#permalink] New post 01 Oct 2019, 09:14
Expert's post
Carcass wrote:
The average (arithmetic mean) of 12 and 20 is equal to the average (arithmetic mean) of 15 and x.

Quantity A
Quantity B
x
16



Given: The average (arithmetic mean) of 12 and 20 is equal to the average (arithmetic mean) of 15 and x

We can write: \(\frac{12+20}{2} = \frac{15+x}{2}\)

Multiply both sides by 2 to get: \(12+20 = 15+x\)

Simplify to get: \(32 = 15+x\)

Solve: \(x = 17\)

we get:
Quantity A: 17
Quantity B: 16

Answer: A

Cheers,
Brent
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Re: The average (arithmetic mean) of 12 and 20 is equal to the a   [#permalink] 01 Oct 2019, 09:14
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