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

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

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.

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?

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.

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!

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.

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

\(? = x\)


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

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


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

Regards,
Fabio.

How do you get the number 9 on statement 1?

isnt there any chance of getting 2 values of X?
shouldn't we solve the equation completely?

Posted from my mobile device

Alternate approach:

Since 2/3 of the marbles are black -- implying that 1/3 are white -- the total number of marbles must be a MULTIPLE OF 3.

Statement 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.
When the total number of marbles is reduced by 1, the result must be a multiple of 4, since 1/4 of the remaining marbles will be white:
T-1 = 4a
T = 4a+1 = 5, 9, 13, 17, 21, 25, 29, 33...
In the list above, only the values in blue are viable, since the total number of marbles must be a multiple of 3.

Test T=9, which implies B=6 and W=3.
If one white marble is removed -- leaving 2 white marbles out of a new total of 8 -- P(W) = 2/8 = 1/4.

Test T=21, which implies B=14 and W=7.
If one white marble is removed -- leaving 6 white marbles out of a new total of 20 -- P(W) = 6/20 = 3/10.

As the value of T increases from 9 to 21, the value of P(W) increases from 1/4 to 3/10.
Implication:
Only T=9 will yield a value of 1/4 for P(W).
Since T=9, W=3.
SUFFICIENT.

Rule:
There must be AT LEAST ONE CASE that satisfies both statements.

Statement 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.
Let P(BB) = the probability that both marbles are black.

Since T=9 is the only case that satisfies Statement 2, T=9 must also satisfy Statement 1; otherwise, the rule above will be violated.
Since T=9 must satisfy Statement 1, we know the following:
When T=9, P(BB) = 5/12.

Test whether the value of T can change.
If T=6 -- implying B=4 and W=2 -- then P(BB) = 4/6 * 3/5 = 2/5.

As the value of T decreases from 9 to 6, the value of P(BB) decreases from 5/12 to 2/5.
Implication:
Only T=9 will yield a value of 5/12 for P(BB).
Since T=9, W=3.
SUFFICIENT.

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