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Cask A and Cask B each contain many white balls and black balls. All

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Cask A and Cask B each contain many white balls and black balls. All  [#permalink]

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New post 25 Sep 2018, 04:54
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A
B
C
D
E

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  15% (low)

Question Stats:

100% (01:53) correct 0% (00:00) wrong based on 10 sessions

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Cask A and Cask B each contain many white balls and black balls. All of the black balls have the same diameter. The diameter of each black ball is 4 inches less than the average diameter of the balls in Cask A and 2 inches greater than the average diameter of the balls in Cask B. What is the difference between average (arithmetic mean) diameter, in inches, of the balls in Cask A and of the balls in Cask B?

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

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Cask A and Cask B each contain many white balls and black balls. All  [#permalink]

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New post 27 Sep 2018, 06:49
\(BB\) = Black balls
\(BBd\) = Black Balls diameter
\(Ad\) = Average diameter in Cask A
\(Bd\) = Average diameter in Cask B

We know that every Black Ball has the same diameter.

Therefore, from the question stem we get:
\(BBd-4=Ad\) --> "The diameter of each black ball is 4 inches less than the average diameter of the balls in Cask A"
and
\(BBd+2=Bd\) --> "The diameter of each black ball is [...] and 2 inches greater than the average diameter of the balls in Cask B."

The questions asks for difference between \(Ad\) and \(Bd\).
--> \((BBd+2)-(BBd+4)=-6\)
The difference is 6. Hence, B.
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Cask A and Cask B each contain many white balls and black balls. All &nbs [#permalink] 27 Sep 2018, 06:49
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