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

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

29 Jan 2008, 15:34

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Question Stats:

25% (02:36) correct

75% (01:29) wrong

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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 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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Re: PS- Blue and red sticks [#permalink]

29 Jan 2008, 15:49

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

length(red) = average(W) - 19
length(red) = average(V) + 6

average(W) - average(V) = 6+19=25 -> closest to E

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Re: PS- Blue and red sticks [#permalink]

29 Jan 2008, 20:26

concur with E ... closest to 25.

W has x red and y blue. V has a red and b blue

length(x)=length(a) = length(x)*x + length(y)*y / x+y + 19 = length(a)*a+length(b)*b/a+b + 6

shifting terms, we get the first term minus second term is 6+19=25

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Re: PS- Blue and red sticks [#permalink]

10 Dec 2009, 12:30

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

Hence E

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

19 Feb 2012, 06:51

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"

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

19 Feb 2012, 12:00

wizard wrote:

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;

{Average W}-19={Average V}+6 --> {Average W}-{Average V}=25.

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;

{Average W}-18={Average V}+6 --> {Average W}-{Average V}=24.

Hope it's clear.
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Re: Box W and Box V each contain several blue sticks and red [#permalink] 19 Feb 2012, 12:00

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

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

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

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