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

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Math Expert
Joined: 02 Sep 2009
Posts: 51307

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25 Sep 2018, 03:54
00:00

Difficulty:

15% (low)

Question Stats:

100% (02:04) correct 0% (00:00) wrong based on 13 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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27 Sep 2018, 05:49
$$BB$$ = 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, 05:49
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