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

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Manager
Joined: 01 Nov 2007
Posts: 147

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

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29 Jan 2008, 15:34
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35% (medium)

Question Stats:

72% (01:38) correct 28% (01:51) wrong based on 212 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?

(A) 3
(B) 6
(C) 12
(D) 18
(E) 24
[Reveal] Spoiler: OA

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Director
Joined: 01 Jan 2008
Posts: 618

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

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29 Jan 2008, 15:49
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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

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

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

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SVP
Joined: 28 Dec 2005
Posts: 1545

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

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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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Manager
Joined: 17 Aug 2009
Posts: 228

Kudos [?]: 293 [1], given: 25

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

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10 Dec 2009, 12:30
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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

Hence E

Kudos [?]: 293 [1], given: 25

Intern
Joined: 17 Oct 2011
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Location: Taiwan
GMAT 1: 590 Q39 V34
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Re: Box W and Box V each contain several blue sticks and red [#permalink]

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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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Math Expert
Joined: 02 Sep 2009
Posts: 41873

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

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19 Feb 2012, 12:00
1
KUDOS
Expert's post
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]

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04 Jul 2017, 11:12
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Re: Box W and Box V each contain several blue sticks and red   [#permalink] 04 Jul 2017, 11:12
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