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in the Box B?

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in the Box B?

by sanju09 » Fri Mar 26, 2010 1:58 am
The ratio in the number of black balls to the white balls in Box A is 0.8 and that in Box B is 0.7. If the number of black balls in Box B is 120 percent more than the number of white balls in Box A, and that the Box A contains 56 black balls; then how many white balls are there in the Box B?
(A) 220
(B) 154
(C) 120
(D) 70
(E) 22
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by harshavardhanc » Fri Mar 26, 2010 2:13 am
sanju09 wrote:The ratio in the number of black balls to the white balls in Box A is 0.8 and that in Box B is 0.7. If the number of black balls in Box B is 120 percent more than the number of white balls in Box A, and that the Box A contains 56 black balls; then how many white balls are there in the Box B?
(A) 220
(B) 154
(C) 120
(D) 70
(E) 22
A contains 56 black balls. Therefore, A contains 56/.8 = 70 White balls.

Now, Black balls in B is 120 % more than 70. => (100% + 20 % ) more => 70 + 14 = 84 more => 154 balls.

Given . Black/White in box B = 7/10 => white balls in B = (154/7) * 10 = 220 . IMO A.
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by outreach » Fri Mar 26, 2010 2:18 am
A->bl/wh=0.8
B->bl/wh=0.7

bl(B)=2.2wh(A)
bl(A)=56

=> wh(A)=bl(A)/0.8
=70

bl(B)=2.2*wh(A)
=154

wh(B)=bl(B)/0.7=220


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