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A jar contains only black marbles and white marbles. If two

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A jar contains only black marbles and white marbles. If two

by BTGmoderatorDC » Wed Jan 30, 2019 4:55 am

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A jar contains only black marbles and white marbles. If two thirds of the marbles are black, how many white marbles are in the jar?

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.

OA D

Source: Veritas Prep
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A jar contains only black marbles and white marbles. If

by fskilnik@GMATH » Wed Jan 30, 2019 5:22 am

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BTGmoderatorDC wrote:A jar contains only black marbles and white marbles. If two thirds of the marbles are black, how many white marbles are in the jar?

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that a randomly drawn marble (taken from the modified jar) would be white.
Source: Veritas Prep
$${\rm{jar}}\,\,\,\left\{ \matrix{
\,B = 2x\,\,{\rm{marbles}} \hfill \cr
\,W = x\,\,{\rm{marbles}} \hfill \cr} \right.\,\,\,\,\,\,\,\left( {x \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\,\,\,\left[ {{\mathop{\rm int}} = {\mathop{\rm int}} - {\mathop{\rm int}} = B - W = x} \right]$$
$$? = x$$

$$\left( 1 \right)\,\,\,{5 \over {12}} = {{C\left( {2x,2} \right)} \over {C\left( {2x + x,2} \right)}}\,\, = \,\,{{\,{{2x\left( {2x - 1} \right)} \over 2}\,} \over {\,{{3x\left( {3x - 1} \right)} \over 2}\,}}\,\, = \,\,{{2\left( {2x - 1} \right)} \over {3\left( {3x - 1} \right)}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,15\left( {3x - 1} \right) = 24\left( {2x - 1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.$$
$$\left( 2 \right)\,\,\,{1 \over 4} = {{x - 1} \over {3x - 1}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,3x - 1 = 4\left( {x - 1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}$$


The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Last edited by fskilnik@GMATH on Thu Jan 31, 2019 11:04 am, edited 1 time in total.
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by Jay@ManhattanReview » Wed Jan 30, 2019 10:41 pm

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BTGmoderatorDC wrote:A jar contains only black marbles and white marbles. If two thirds of the marbles are black, how many white marbles are in the jar?

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.

OA D

Source: Veritas Prep
Say there are a total of 3x numbers of marbles; thus, there are 2x numbers of black marbles and x numbers of white marbles.

We have to get the value of x.

Let's take each statement one by one.

(1) If two marbles were to be drawn, simultaneously and at random, from the jar, there is a 5/12 probability that both would be black.

Required probability = (# of ways of choosing two black marbles) / (# of ways of choosing two marbles)

# of ways of choosing two black marbles = 2xC2 = [(2x)*(2x - 1)]/(1.2)
# of ways of choosing two marbles = 3xC2 = [(3x)*(3x - 1)]/(1.2)

Thus,

Required probability =[ [(2x)*(2x - 1)]/(1.2)] / [[(3x)*(3x - 1)]/(1.2)] = (2x - 1) / 3(3x - 1) = 5/12

Upon solving, we get x = 3. Sufficient.

(2) If one white marble were removed from the jar, there would be a 1/4 probability that the next randomly-drawn marble would be white.

Thus,

# of white marbles = x - 1;
# of black marbles = 2x;

and, total # of marbles = 3x - 1

Required probability = (x - 1) / (3x - 1) = 1/4

Upon solving, we get x = 3. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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