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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]  01 Jun 2009, 17:57
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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]  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.

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Re: Average2 [#permalink]  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]  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 houses is [#permalink]  28 Jan 2010, 15:24
Expert's post
The answer should certainly be B here; it's a weighted average question, and if we know the average of the two groups, to find the overall average we only need the ratio of the sizes of the two groups. Statement 2 tells us the ratio of x to y is 2 to 1, so the combined average will be "twice as close" to the average of the x houses - that is, the combined average will be $206,000. Who is claiming that the answer is C? _________________ GMAT Tutor in Toronto If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com Math Expert Joined: 02 Sep 2009 Posts: 29205 Followers: 4750 Kudos [?]: 50259 [0], given: 7540 Re: If the average (arithmetic mean) of the assessed values of x [#permalink] 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.

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]  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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