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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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Box W and Box V each contain several blue sticks and red [#permalink]

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New post 30 Jul 2007, 07:49
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

2. 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 18 inches less than 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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Re: average [#permalink]

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New post 30 Jul 2007, 08:05
jet1445 wrote:
2. 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 18 inches less than 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


Getting E.

Avg. length in box W = w
Avg. length in box V = v

Therefore, w-18=v+6
So, w-v = 24.
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Re: average [#permalink]

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New post 30 Jul 2007, 10:32
jet1445 wrote:
2. 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 18 inches less than 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


I get E as well. Box W avg - 18 = Box V avg + 6
Re: average   [#permalink] 30 Jul 2007, 10:32
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Box W and Box V each contain several blue sticks and red

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