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

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Manager
Joined: 20 Mar 2005
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If the average (arithmetic mean) of the assessed values of x [#permalink]

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01 Jul 2007, 07:46
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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

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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Senior Manager
Joined: 04 Jun 2007
Posts: 345

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01 Jul 2007, 07:58
Balvinder 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

Average of (x+y) houses = (212,000*x + 194,000*y)/(x+y)

Stmt 1: Insufficient since x and y can have many possible values.
Stmt 2: Sufficient. Plugging in 2y for x we can find the average. I think it is safe to assume y cannot be 0, since the average of y houses is defined.

Hence, B for me.

Kudos [?]: 33 [0], given: 0

Re: DS - mean   [#permalink] 01 Jul 2007, 07:58
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