Averages

[Striver]

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 than 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

[GMATGuruNY]

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 than 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

Let the length of each red stick = 10.
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.
Thus:
W = 10+18 = 28.
V = 10-6 = 4.
W-V = 28-4 = 24.

The correct answer is E.

[Anurag@Gurome]

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 than 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

Say, the length of each red stick = L inches

Hence, the average length of the sticks in Box W = MW = (L + 18)
And, the average length of the sticks in Box V = MV = (L - 6)

Hence, MW - MV = (L + 18) - (L - 6) = (28 + 6) = 24

The correct answer is E.