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

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Intern
Joined: 08 Nov 2008
Posts: 29
If the average (arithmetic mean) of the assessed values of x [#permalink]

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19 Nov 2008, 18:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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Joined: 30 Apr 2008
Posts: 1855
Location: Oklahoma City
Schools: Hard Knocks

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19 Nov 2008, 18:55
B

(1) Is insufficient. First we need to figure out what information we need to be able to answer the question. We need to know how many of each group (x group and y group) there are, or at least the relationship to each other. If you have 2 houses of \$90,000 and \$110,000, their average is \$100,000. But if the other goup has 1,000 houses averaging \$150,000, those 1,000 houses outweigh the 2 houses in the first group. Because we only know the total number of houses is 36, we don't know if x is 1 and y = 35, or if x = 18 and y = 18. Because we don't know this, (1) is insufficient.

(2) is sufficient because we're told the relationship between x and y. It might be easiest to use number picking here to illustrate how the relationship means we can determine the assessed values of x+y houses.

Lets use K.I.S.S. -> Keep It Simple Stupid.
x = 4; therefore, y=2.

First, how do we figure up the average of a group? Total all the values and divide by the number of houses in the group. If we're finding the average of all houses x+y, then we need to take the sum of all houses in x and y and divide by the number of houses (6 for the numbers we picked). We know that the average is \$212,000 and if we've picked 4 for the number of houses in this group, then regardless of the individual house values, we know that the sum of these values is \$212,000 * 4 = \$848,000. Because x = 2y, we know that y must = 2 if x =4. So we need to fnd the sum of the values of this group of houses. Like x, we know that this will be y * the average, or 2 * \$194,000 = \$388,000. So we have \$388,000 + \$848,000 = \$1,236,000 and that value needs to be divided by the number of houses (6), so this = \$206,000. Now we need to figure out if this will be the same average for all possible values of x and y. We can do a few more number picking examples to make sure the info in Statement 2 is sufficient.

So x = 10, y = 5.

10 * 212,000 = 2,120,000 and 5 * 194,000 = \$970,000....2,120,000 + 970,000 = 3,090,000. Now divide \$3,090,000 by 15. = \$206k. We know that we're right that Statement (2) is sufficient since we know the relationship between x and y.

We need to take into account the different number of houses in each group. First

petercao 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

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a\$\$.

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Re: house--25   [#permalink] 19 Nov 2008, 18:55
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# If the average (arithmetic mean) of the assessed values of x

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