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

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

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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
Joined: 17 Jun 2008
Posts: 1374

Kudos [?]: 406 [0], given: 0

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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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Kudos [?]: 406 [0], given: 0

Director
Joined: 20 Sep 2006
Posts: 653

Kudos [?]: 135 [0], given: 7

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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
Joined: 05 Jul 2008
Posts: 1402

Kudos [?]: 437 [0], given: 1

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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.

Kudos [?]: 437 [0], given: 1

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

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

Kudos [?]: 279 [0], given: 0

Re: PS: Average   [#permalink] 22 Sep 2008, 00:56
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