Que: Box X and Box Y each contain many yellow balls and green balls......

[Max@Math Revolution]

Que: Box X and Box Y each contain many yellow balls and green balls. All of the green balls have the same radius. The radius of each green ball is 4 inches less than the average radius of the balls in Box X and 2 inches greater than the average radius of the balls in Box Y. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box X and of the balls in Box Y?

(A) 4
(B) 10
(C) 7
(D) 8
(E) 6

[Max@Math Revolution]

Solution: Let the radius of each green ball = x inches.

Each green ball is 4 inches less than the average radius of the balls in Box X.

Thus, the average radius of balls in Box X = (x + 4) inches.

Also, each green ball is 2 inches greater than the average radius of the balls in Box Y.

Thus, the average radius of balls in Box Y = (x − 2) inches.

Thus, the required difference = ((x + 4) − (x − 2)) = 6 inches.

Therefore, E is the correct answer.

Answer E

[Scott@TargetTestPrep]

Que: Box X and Box Y each contain many yellow balls and green balls. All of the green balls have the same radius. The radius of each green ball is 4 inches less than the average radius of the balls in Box X and 2 inches greater than the average radius of the balls in Box Y. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box X and of the balls in Box Y?

(A) 4
(B) 10
(C) 7
(D) 8
(E) 6

Solution:

If we let the radius of the green balls be 10 inches, then the average radius of the balls in Box X is 14 inches, and the average radius of the balls in Box Y is 8 inches. Therefore, the difference between the average radius of the balls in Box X and of the balls in Box Y is 14 - 8 = 6 inches.

Answer: E