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

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29 Jan 2008, 14:34

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71% (01:37) correct 29% (01:56) wrong based on 241 sessions

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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 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?

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

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29 Jan 2008, 14:49

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JCLEONES wrote:

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

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

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10 Dec 2009, 11:30

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Why do test Makers unnecessarily complicate matters????

Simply put, let the length of each red stick be R

Now R = Average length of sticks in Box W – 18------------- (1) Also, R = Average length of sticks in Box V +6---------------(2)

Avg. length of sticks in Box W - 18 = Avg. length of sticks in Box V + 6 Avg. length of sticks in Box W - Avg. length of sticks in Box V = 24-----which is what we had to calculate

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

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19 Feb 2012, 05:51

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The right number is not 19 but 18. "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"

The right number is not 19 but 18. "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"

For the question with 19 and 6 inches correct answer is 25: {Average lengths of sticks in W} - 19 = Length of red; {Average lengths of sticks in V} + 6 = Length of red;

For the question with 18 and 6 inches correct answer is 24: {Average lengths of sticks in W} - 1 = Length of red; {Average lengths of sticks in V} + 6 = Length of red;

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

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14 Nov 2017, 07:05

this is how i solved this question . So let the number of blue sticks in Box W be BW and red sticks be RW and let the number of blue sticks in Box V be BV and red sticks be RV. Also let the length of each red stick be x and the length of each blue stick be y. Therefore as per the statements given , x= RW(x)+ BW(y) / RW+BW - 18 and x = 6+ RV(x)+BV(y) / RV+BV Therefore the question is asking to find the difference between the two.So RW(x)+BW(y)/BW+RW - RV(x)+BV(y)/ BV+RV (that is the difference between the averages of the two boxes), So (x+18)-(x-6)= x+18-x+6=24 . The answer is E

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

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03 Dec 2017, 15:54

JCLEONES wrote:

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 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?