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I must be reading this wrong... averaging question

Expert replies
by shiftcourse » Wed Apr 19, 2017 6:16 pm
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 x + y houses?

1) x + y = 36

2) x = 2y

Answer: B

The way I read this question it seems like the average of x + y houses would just be ($212,000 + $194,000)/2.
Why is that not true?
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Source: — Data Sufficiency |

by [email protected] » Wed Apr 19, 2017 8:35 pm
Hi shiftcourse,

IF... the values of X and Y were EQUAL, then the average of the X houses (at $212,000 each) and the other Y houses (at $194,000 each) would simply be ($212,000 + $194,000)/2. However, if X and Y are NOT equal, then you have a weighted average, and that calculation would be...

[X(212,000) + Y(194,000)] / (X+Y)

Knowing that, how would you proceed with this question?

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by Jay@ManhattanReview » Wed Apr 19, 2017 9:30 pm
shiftcourse 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 x + y houses?

1) x + y = 36

2) x = 2y

Answer: B

The way I read this question it seems like the average of x + y houses would just be ($212,000 + $194,000)/2.
Why is that not true?
Hi shiftcourse,

Computing the average as ($212,000 + $194,000)/2 underlies an assumption that x = y, ie, the number of type 1 houses (x in number) is equal to the number of type 2 houses (y in number); however, it is not stated in the question.

Had this been the case, you do not need any statement at all. On a funny note, the question is self-sufficient to be solved! :)

This question tests the concept of Weighted Average. When the number of quantities is not equal, one must take into account their relative importance, ie, their count!

Weighted average = Total value / Total numbers

Total value of (x+y) number of houses = 212000*x + 194000*y

We have, Total number of houses = (x+y)

Thus, Weighted average = (212000*x + 194000*y) / (x+y)

Statement 1: x+y = 36

In Weighted average = (212000*x + 194000*y) / (x+y), we only know the value of the denominator (x+y), thus we cannot calculate the value of Weighted average. Insufficient!

Statement 2: x = 2y

We can put x = 2y in Weighted average = (212000*x + 194000*y) / (x+y)

=> Weighted average = (212000*x + 194000*y) / (x+y) = (212000*2y + 194000*y) / (2y+y)

=> Weighted average = (414000y + 194000y) / 3y

=> Weighted average = (414000 + 194000) / 3; y cancels

Weighted average = a unique value. There is no need to calculate if you are satisfied that you get a unique value. Sufficient.

The correct answer: B

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by shiftcourse » Thu Apr 20, 2017 4:05 am
Ohhhh, right! Thank you for the responses. Much better explanations than the GMAT prep software, which was the source for the question.
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by Jay@ManhattanReview » Thu Apr 20, 2017 9:07 pm
shiftcourse wrote:Ohhhh, right! Thank you for the responses. Much better explanations than the GMAT prep software, which was the source for the question.
Thank you! You are welcome.

-Jay
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