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

Question

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

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

Box W and box Z 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 than the average length of the sticks in Box W and 6 inches greater than the average length of the sticks in Box Z. 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 Z?

zaarathelab · Accepted Answer

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

Bunuel · Answer

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.

maratikus · Answer

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

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

pmenon · Answer

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

wizard · Answer

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"

longhaul123 · Answer

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

testcracker · Answer

with respect to red W +18 and V -6 = 18 - (-6) =24 thanks

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

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

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