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

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Senior Manager
Joined: 08 Jun 2004
Posts: 499
Location: Europe
Followers: 1

Kudos [?]: 20 [0], given: 0

Box W and Box V each contain several blue sticks and red [#permalink]  24 Apr 2006, 11:51
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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: 20 Nov 2004
Posts: 108
Followers: 0

Kudos [?]: 2 [0], given: 0

Don't be confused by the stem.

I'd go for 24 and E.
Manager
Joined: 13 Dec 2005
Posts: 224
Location: Milwaukee,WI
Followers: 1

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

E 24

let R be lenght of any red stick

W be average length of blue and red stick in box w
V be average length of blue and red stick in box v

according to question R = W -18 ----- eqn 1
also R = V +6 ------ eqn 2

equating the both equation we get W-18 =V +6
===> W -V = 24

hence 24 .
VP
Joined: 29 Apr 2003
Posts: 1405
Followers: 2

Kudos [?]: 17 [0], given: 0

r = Aw - 18
r = Av + 6

where r is the length of red sticks and Aw and Av are the avg for box W and V respectively.

Thus subtracting the first equation from the seconds, we get Aw - Av = 24!

The Problem looks complicated... When I first started working I was sweating... but, its a paper tiger!
Senior Manager
Joined: 08 Jun 2004
Posts: 499
Location: Europe
Followers: 1

Kudos [?]: 20 [0], given: 0

OA is 'E'.
Well done guys.
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