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by ketkoag » Sat May 16, 2009 1:19 am
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

please lemme know if the answer to this question is 25 as when i solved it, i got 25 but ther is no option here..
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by DanaJ » Sat May 16, 2009 2:22 am
Start off by making some convenient notations. This is important in the real test as well, since it saves time and space on that annoying erasable pad...
r = length of red sticks
w = average length for sticks in box W
v = average length for sticks in box V

You are looking for w - v.

Now, the problem tells you that:

r + 19 = w --- length of each red stick is 19 inches less that the average length of the sticks in Box W

r - 6 = v --- 6 inches greater than the average length of the sticks in Box V

Now, replace those in w - v to get:

r + 19 - r + 6 = 25.

I'm getting 25 too... What's the source of this problem?
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by dadagreat » Sat May 16, 2009 3:14 am
I also got 25 and it was pretty much the same method which DanaJ has posted. Cant think of any other method!!

is the source credible?
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by dumb.doofus » Sat May 16, 2009 7:57 am
I searched a bit on the net.. the correct question is where the number given is 18 and not 19..

with that the answer is 24.
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by ketkoag » Sat May 16, 2009 10:24 am
thanks for ur reply guys.....
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