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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]

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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, 23:52, edited 2 times in total.
Corrected the OA
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Re: Average2 [#permalink]

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New post 01 Jun 2009, 19:18
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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]

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New post 01 Jun 2009, 23:33
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]

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New post 01 Jun 2009, 23:48
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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]

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New post 28 Jan 2010, 16:24
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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?
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Re: If the average (arithmetic mean) of the assessed values of x [#permalink]

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New post 26 Jan 2012, 23:55
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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]

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Re: If the average (arithmetic mean) of the assessed values of x [#permalink]

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New post 03 Dec 2015, 08:25
Expert's post
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

This is a '2by2'question, one of the most common type of questions in GMAT math
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We have 2 variables (x,y) and 2 equations are given by the 2 conditions, which gives (C) high chance of being the answer
Looking at the conditions together, x=24, y=12 so this seems sufficient, but this is a commonly made mistake; if we look at the conditions separately,
condition 2 contains ratios and when one condition contains numbers and another contains ratios, there is high chance the one with ratio is the answer, so if condition 2 is examined,

average assessed value= (212,000x+194,000y)/(x+y)=(212,000*2y+194,000y)/(2y+y).
y can be canceled out, so the condition is sufficient and the answer becomes (B).

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If the average (arithmetic mean) of the assessed values of x   [#permalink] 03 Dec 2015, 08:25
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