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

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VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
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Box W and Box V each contain several blue sticks and red [#permalink]

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18 Oct 2004, 19:10
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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 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
Manager
Joined: 18 Sep 2004
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Location: Dallas, TX
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18 Oct 2004, 20:11
18 + 6 = 24 The difference between avg length of Box w and v
Intern
Joined: 07 Oct 2004
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18 Oct 2004, 22:16
According to ur question
Length(red)=w(avg) - 18 = v(avg.) + 6

=> w(avg.) - v(avg.) = 18+6 =24
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VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 36 [0], given: 0

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19 Oct 2004, 01:40
OA is E but I really had difficulties to translate this problem into a correct equation. Thanks.
[#permalink] 19 Oct 2004, 01:40
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# Box W and Box V each contain several blue sticks and red

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