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by f2001290 » Tue Jun 12, 2007 7:03 am
17. 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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by chatekar » Fri Jul 20, 2007 5:39 am
I think the answer to this question is 25
But to my surprise its not there in the answer choices.

Anyone knows if there is something wrong, either with me or with the question.

Thanks
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by beeparoo » Sun Jul 22, 2007 4:02 pm
chatekar wrote:I think the answer to this question is 25
But to my surprise its not there in the answer choices.
I also came up with 25 as my answer.

Here is my proof:

# of red sticks:
N(red) = W(avg) - 19 = V(avg) + 6

Move terms across the equality sign...

W(avg) - V(avg) = 6 - (-19) = 25

Hey f2001290, did you make a typo? He/she posted this more than a month ago.. Perhaps this person has moved on. Oh well. I support your answer, chatekar.
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