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

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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 21 Apr 2015, 05:59
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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.


Kudos for a correct solution.

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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 21 Apr 2015, 08:01
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Bunuel 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.


Kudos for a correct solution.


1) probability of first blacck marble equal \(\frac{2}{3}\). if we multiply this probability on probability second black marble, we receive common probability \(\frac{5}{12}\)
Let's x be second probability
\(\frac{2}{3} * x = \frac{5}{12}\)
\(x = \frac{5}{8}\) and this probability after we remove one marble, so initially we have 9 marbles in total and \(\frac{1}{3}\) of them white: 3 white marbles
Sufficient

2) let's x be total marbles
\(\frac{1}{3}x-1 = \frac{1}{4}(x-1)\)
\(x = 9\) marbles in total and \(\frac{1}{3}\) of them white: 3 white marbles
Sufficient.

Answer is D
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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 10 Mar 2016, 19:31
1
suppose we have X white, and 2X black
total 3X

now..
1 - probability that 2 black is:
2/3 * Y(variable) = 5/12
y=5/12*3/2
y=5/8
since y must be: (x-1)/(3x-1)
it must be true that there are 9 marbles, out of which 6 black and 2 white. sufficient

2. to draw a second white is 1/4
it must be true that x-1/3x-1 = 1/4
4x-1 = 3x-1
x=3.

sufficient.
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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 03 Jul 2018, 10:06
1
Bunuel 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.


Kudos for a correct solution.


Let \(X\) be the total # of marbles. hence we have \(\frac{2X}{3}\) black marbles & \(\frac{X}{3}\) white marbles.

The question prompt now translates to finding X.

Statement 1: When two marbles are drawn, the probability of both to be black = \(\frac{5}{12}\)

Hence, \({(2X/3)}/X\) * \(((2X/3) - 1)/(X-1)\) = \(\frac{5}{12}\)

We can solve this for X. Hence Statement 1 alone is sufficient.

Statement 2: A white ball is drawn & then a white ball is drawn, the probability of drawing a second white ball is \(\frac{1}{4}\)

Hence we, \(((X/3) - 1)/(X - 1) = 1/4\)

We can solve this for X. Hence statement 2 alone is sufficient.



Answer D.



Thanks,
GyM
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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 10 Jul 2018, 08:47
1) We can say that we have one part of white balls per two parts of black balls. We want to find how much is one part.

The probability of having 2 black balls is:
\(\frac{5}{12}\)=\(\frac{2p}{3p}\).\(\frac{(2p-1)}{(3p-1)}\)

We can cancel out the p in the first fraction \(\frac{2p}{3p}\) so we can find p with the resulting equation. (p=3)

Sufficient.

2) We know that initially the probability of having a white ball is 1:3. To find the probability of having the second white ball, we need to subtract one to each. Let's write down our hypothesis of the initial amount of white balls vs total that meet the 1:3 probability. Then let's write down how it would look like after we removed one white ball and see if the resulting probability is 1:4:

1:3 => this would make it impossible to have a second white ball
2:6 => 1:5 ; not what we want
3:9 => 2:8 which is equivalent to 1:4; what we want

We needed to have 3 out of 9 white balls initially to meet this condition.

Sufficient.
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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 26 Jan 2019, 02:32
GyMrAT wrote:
Bunuel 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.


Kudos for a correct solution.


Let \(X\) be the total # of marbles. hence we have \(\frac{2X}{3}\) black marbles & \(\frac{X}{3}\) white marbles.

The question prompt now translates to finding X.

Statement 1: When two marbles are drawn, the probability of both to be black = \(\frac{5}{12}\)

Hence, \({(2X/3)}/X\) * \(((2X/3) - 1)/(X-1)\) = \(\frac{5}{12}\)

We can solve this for X. Hence Statement 1 alone is sufficient.

Statement 2: A white ball is drawn & then a white ball is drawn, the probability of drawing a second white ball is \(\frac{1}{4}\)

Hence we, \(((X/3) - 1)/(X - 1) = 1/4\)

We can solve this for X. Hence statement 2 alone is sufficient.



Answer D.



Thanks,
GyM


Hi,

You did not solve statement 1. What if x comes out to be in fraction? Then we have to reject this statement? right?
So my ques is : Should be every time not solve 1 degree variable equation in DS questions?
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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 26 Jan 2019, 02:57
Hi nkhl.goyal, Welcome to GMAT club.

It would be a good practice to solve further to ensure that X is not a fraction to be sure that it's value ( which is assured given the liner equation) can be a valid one.
This would be a good thing to spend a couple of seconds to check for.

nkhl.goyal wrote:

Hi,

You did not solve statement 1. What if x comes out to be in fraction? Then we have to reject this statement? right?
So my ques is : Should be every time not solve 1 degree variable equation in DS questions?

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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 26 Jan 2019, 15:10
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nkhl.goyal wrote:
GyMrAT wrote:
Hence, \({(2X/3)}/X\) * \(((2X/3) - 1)/(X-1)\) = \(\frac{5}{12}\)

We can solve this for X. Hence Statement 1 alone is sufficient.


Hi,

You did not solve statement 1. What if x comes out to be in fraction? Then we have to reject this statement? right?
So my ques is : Should be every time not solve 1 degree variable equation in DS questions?


This is a really interesting point. You actually don't have to worry about whether it's a fraction, surprisingly enough. Why? Because if x came out to be a fraction, that would mean there was no valid answer to the question at all! And that's something that never happens in a DS problem on the GMAT (although it might happen in unofficial practice problems.)

In other words:
Sufficient = exactly one valid answer to the question
Insufficient = multiple valid answers to the question
something that never happens on the test = no possible valid answers to the question

So, knowing that there's only going to be one answer is enough. There's never a situation on the GMAT where there's exactly one answer, and it's invalid. So you can stop there without checking - don't spend the extra time! :)
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Re: A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 27 Jan 2019, 08:29
GyMrAT wrote:
Bunuel 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.


Kudos for a correct solution.


Let \(X\) be the total # of marbles. hence we have \(\frac{2X}{3}\) black marbles & \(\frac{X}{3}\) white marbles.

The question prompt now translates to finding X.

Statement 1: When two marbles are drawn, the probability of both to be black = \(\frac{5}{12}\)

Hence, \({(2X/3)}/X\) * \(((2X/3) - 1)/(X-1)\) = \(\frac{5}{12}\)

We can solve this for X. Hence Statement 1 alone is sufficient.

Statement 2: A white ball is drawn & then a white ball is drawn, the probability of drawing a second white ball is \(\frac{1}{4}\)

Hence we, \(((X/3) - 1)/(X - 1) = 1/4\)

We can solve this for X. Hence statement 2 alone is sufficient.



Answer D.



Thanks,
GyM


This was super helpful. Thanks for breaking down each probability into digestible algebra - I found the other responses to be too confusing. Kudos to you.
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A jar contains only black marbles and white marbles. If two thirds of  [#permalink]

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New post 30 Jan 2019, 06:35
Bunuel 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.

\({\rm{jar}}\,\,\,\left\{ \matrix{
\,B = 2x\,\,{\rm{marbles}} \hfill \cr
\,W = x\,\,{\rm{marbles}} \hfill \cr} \right.\,\,\,\,\,\,\,\left( {x \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\,\,\,\left[ {{\mathop{\rm int}} = {\mathop{\rm int}} - {\mathop{\rm int}} = B - W = x} \right]\)

\(? = x\)


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

\(\left( 2 \right)\,\,\,{1 \over 4} = {{x - 1} \over {3x - 1}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,3x - 1 = 4\left( {x - 1} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x\,\,{\rm{unique}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}{\rm{.}}\)


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

Regards,
Fabio.
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A jar contains only black marbles and white marbles. If two thirds of   [#permalink] 30 Jan 2019, 06:35
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