GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2018, 11:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 01 Nov 2007
Posts: 128
Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

29 Jan 2008, 15:34
9
00:00

Difficulty:

45% (medium)

Question Stats:

67% (02:28) correct 33% (02:34) wrong based on 272 sessions

### HideShow timer Statistics

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
Director
Joined: 01 Jan 2008
Posts: 597
Re: Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

29 Jan 2008, 15:49
2
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
VP
Joined: 28 Dec 2005
Posts: 1462
Re: Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

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
Manager
Joined: 17 Aug 2009
Posts: 184
Re: Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

10 Dec 2009, 12:30
2
1
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
Intern
Joined: 17 Oct 2011
Posts: 11
Location: Taiwan
GMAT 1: 590 Q39 V34
GMAT 2: 680 Q47 V35
Re: Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

19 Feb 2012, 06:51
1
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"
Math Expert
Joined: 02 Sep 2009
Posts: 50042
Re: Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

19 Feb 2012, 12:00
3
1
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.
_________________
Manager
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 207
Location: India
Re: Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

14 Nov 2017, 08:05
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
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 267
Box W and Box V each contain several blue sticks and red  [#permalink]

### Show Tags

03 Dec 2017, 16:54
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?

(A) 3
(B) 6
(C) 12
(D) 18
(E) 24

with respect to red

W
+18

and

V
-6

= 18 - (-6)
=24

thanks
Box W and Box V each contain several blue sticks and red &nbs [#permalink] 03 Dec 2017, 16:54
Display posts from previous: Sort by