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Box W and Box V each contain several blue sticks and red

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Manager
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Joined: 26 Mar 2008
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Box W and Box V each contain several blue sticks and red [#permalink]

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New post 21 Sep 2008, 14:35
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Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24

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VP
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Joined: 17 Jun 2008
Posts: 1374

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Re: PS: Average [#permalink]

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New post 21 Sep 2008, 18:39
arorag wrote:
Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24


BW=ave length of sticks in Box W
BV= ave length of sticks in BOX V

Let the length of red stick be R
BW=R+19
BV=R-6
HENCE BW-BV=19+6=25
BUT somehow i found no such option here am i overlooiking something !!help me out
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Director
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Re: PS: Average [#permalink]

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New post 21 Sep 2008, 19:09
Length of Red = AVG_W - 19 = AVG_V + 6

AVG_W - AVG_V = 25

Humm Interesting ... 25 is not even an ans choice

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VP
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Joined: 05 Jul 2008
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Re: PS: Average [#permalink]

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New post 21 Sep 2008, 22:07
spriya wrote:
arorag wrote:
Box W and Box V each contain several blue sticks and red sticks, and all of the red sticks have the same length. The length of each red stick is 19 inches less that the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box V. What is the average (arithmetic mean) length, in inches, of the sticks in Box W minus the average length, in inches, of the sticks in Box V?
(A) 3
(B) 6
(C) 12
(D) 18
(E) 24


BW=ave length of sticks in Box W
BV= ave length of sticks in BOX V

Let the length of red stick be R
BW=R+19
BV=R-6
HENCE BW-BV=19+6=25
BUT somehow i found no such option here am i overlooiking something !!help me out


Such an annoying word problem. I got 25 as well and there is no such answer choice.

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SVP
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Joined: 17 Jun 2008
Posts: 1534

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Re: PS: Average [#permalink]

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New post 22 Sep 2008, 00:56
Is there a typo in the question? I also get 25 as the answer.

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Re: PS: Average   [#permalink] 22 Sep 2008, 00:56
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Box W and Box V each contain several blue sticks and red

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