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If the average (arithmetic mean) of the assessed values of x

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If the average (arithmetic mean) of the assessed values of x [#permalink] New post 01 Jun 2009, 17:57
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64% (01:45) correct 36% (01:11) wrong based on 68 sessions
If the average (arithmetic mean) of the assessed values of x houses is $212,000 and the average of the assessed values of y other houses is $194,000, what is the average of the assessed values of the x+y houses?

(1) x+y=36
(2) x=2y
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Jan 2012, 22:52, edited 2 times in total.
Corrected the OA
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Re: Average2 [#permalink] New post 01 Jun 2009, 18:18
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Value of X houses = 212000*X
Value of Y houses = 194000*Y


Avg of X+Y houses = \frac{(212000*X + 194000*Y)}{X+Y}


(1) Avg of X+Y houses = \frac{(212000*X + 194000*Y)}{36} \Rightarrow INSUFFICIENT.


(2) Avg of X+Y houses = \frac{(212000*2Y + 194000*Y)}{2Y+Y} \Rightarrow Y's cancel out and it is SUFFICIENT.


Answer B.
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Re: Average2 [#permalink] New post 01 Jun 2009, 22:33
sondenso wrote:
If the average (arithmetic mean) of the assessed values
of x houses is $212,000 and the average of the assessed
values of y other houses is $194,000, what is the average
of the assessed values of the x+y houses?
(1) x+y=36
(2) x=2y

OA is wrong!


SumX/X = 212 ==> SumX = 212X
SumY/Y = 194 ==> SumY = 194Y

Question:(\frac{SumX + SumY}{X+Y})?
Question:(\frac{212X + 194Y}{X+Y})?

(1) X+Y=36

Substitute into question:
Question:(\frac{212X + 194Y}{36})?

Insuficient

(2) X=2Y

Substitute into question
Question:(\frac{212(2Y) + 194Y}{2Y+Y)})?
Question:(\frac{618Y}{3Y})?
Question:(\frac{618}{3})?
Question:(206)?

Wait a second-- do you know what 206 is? Of course we do... Sufficient!

Final answer, B.
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Re: Average2 [#permalink] New post 01 Jun 2009, 22:48
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This is a very frequently tested type of question on the GMAT.

Anytime you're given two averages and you know the ratio/proportion between the two groups/sets, sufficient (it's usually A/B).
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Re: If the average (arithmetic mean) of the assessed values of x [#permalink] New post 26 Jan 2012, 22:44
I calculated B too.
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Re: If the average (arithmetic mean) of the assessed values of x [#permalink] New post 26 Jan 2012, 22:55
Expert's post
Corrected the OA it should be B, not C.

If the average (arithmetic mean) of the assessed values of x houses is $212,000 and the average of the assessed values of y other houses is $194,000, what is the average of the assessed values of the x+y houses?

This is a simple weighted average question, and if we knew the ratio of x to y it would be sufficient to answer the question

(1) x+y=36. Not sufficient.
(2) x=2y. Sufficient.

Answer: B.

P.S. General formula for weighted average: weighted \ average=\frac{total \ weight}{sum \ of \ values} --> weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}.

So, as you can see the ratio of values (the ratio of x to y) would give us the answer for the given question.
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Re: If the average (arithmetic mean) of the assessed values of x [#permalink] New post 08 Oct 2013, 05:48
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Re: If the average (arithmetic mean) of the assessed values of x   [#permalink] 08 Oct 2013, 05:48
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