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Probability - black balls and white balls

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Probability - black balls and white balls

by Brent@GMATPrepNow » Sun Dec 21, 2008 10:54 am
An urn is filled with black balls and white balls only. If the probability of randomly drawing a white ball is 4/5, how many white balls must be added to the urn so that the probability of randomly drawing a white ball is 7/8?
(1) The ratio of white balls to black balls is 4:1
(2) There are 27 more white balls than black balls
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by cramya » Sun Dec 21, 2008 11:06 am
Prob is not my forte but anyways let me get better by giving this a shot.

Is it B)?

white balls = 36 black - 9 So 27 more white balls have to be added to make the ratio 7/8


Stmt I adds no new info. (4:1 part by part ratio can be obtained from the part to whole ratio 4/5 given in the question stem)
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by Brent@GMATPrepNow » Sun Dec 21, 2008 11:13 am
Nice work, cramya! The answer is B.

I placed problem-solving version of this question on the PS side of the forum. [/quote]
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by cramya » Sun Dec 21, 2008 11:45 am
Brent ,thanks for continuing to post good questions!!!

Appreciate your contribution to this forum
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by Brent@GMATPrepNow » Sun Dec 21, 2008 11:50 am
Thanks cramya
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by ronniecoleman » Mon Dec 22, 2008 12:59 am
IMO B

X/X+Y = 4/5

X = 4Y

X = Y+ 27

X = 36
Y = 9

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by vittalgmat » Mon Dec 22, 2008 2:27 am
B for me as well

We have W/(W+B) = 4/5
solving, W = 4B ------------- 1

Stmt 1:
Same as the q stem. Does not add anything more.
so N/S


Stmt 2:

W = B +27
substituting 1 above
4B = B +27

Solving B = 9.
So W = 36.

Now with this any new ratio can be created.

Hence B
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