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Que: Box P and Box Q each contain many red balls and white balls.....

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by Max@Math Revolution » Sat Jun 12, 2021 8:40 pm

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Que: Box P and Box Q each contain many red balls and white balls. All of the white balls have the same radius. The radius of each white ball is 6 inches less than the average radius of the balls in Box P and 3 inches greater than the average radius of the balls in Box Q. What is the difference between the average (arithmetic mean) radius, in inches, of the balls in Box P and of the balls in Box Q?

(A) 3
(B) 4
(C) 6
(D) 8
(E) 9
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Re: Que: Box P and Box Q each contain many red balls and white balls.....

by Max@Math Revolution » Sun Jun 13, 2021 9:03 pm
Solution: Let the radius of each white ball = x inches.

Each green ball is 6 inches less than the average radius of the balls in Box P.

Thus, the average radius of balls in Box P = (x + 6) inches.

Also, each white ball is 3 inches greater than the average radius of the balls in Box Q.

Thus, the average radius of balls in Box Q = (x − 3) inches.

Thus, the required difference = ((x + 6) − (x − 3)) = 9 inches.

Therefore, E is the correct answer.

Answer E
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