Data Sufficiency

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.

Answer: D

Source: Veritas Prep

Let total marbles = x
Target question: how many white marbles are in the jar?
Therefore, black marbles = 2x/3 and white marbles = 1 - (2x/3) = x/3

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.
Probability of ball drawn both to be black = 5/12
$$\left(\frac{\frac{2x}{3}}{x}\right)\cdot\left(\frac{\left(\frac{2x}{3}-1\right)}{x-1}\right)=\frac{5}{12}$$
$$\left(\frac{2x}{3}\cdot\frac{1}{x}\right)\cdot\left(\frac{2x-3}{3}\right)\cdot\left(\frac{1}{x-1}\right)=\frac{5}{12}$$
$$\left(\frac{2}{3}\right)\cdot\left(\frac{2x-3}{3x-3}\right)=\frac{5}{12}$$
$$\frac{4x-6}{9x-9}=\frac{5}{12}$$
$$12\left(4x-6\right)=5\left(9x-9\right)$$
$$48x-72=45x-45$$
$$48x-45x=-45+72$$
$$3x=27$$
$$x=\frac{27}{3}=9$$
From the question stem,
$$White=\frac{x}{3}=\frac{9}{3}=3$$
White = 3 marbles. Therefore, statement 1 is sufficient.

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.
If (white marbles - 1) probability of next ball being white = 1/4
$$\frac{\left(\frac{x}{3}-1\right)}{x-1}=\frac{1}{4}$$
$$\frac{\left(x-3\right)}{3}\cdot\frac{1}{x-1}=\frac{1}{4}$$
$$\frac{x-3}{3x-3}=\frac{1}{4}$$
$$4\left(x-3\right)=3x-3$$
$$4x-12=3x-3$$
$$4x-3x=-3+12$$
$$x=9$$
From the question stem,
$$White=\frac{x}{3}=\frac{9}{3}=3$$
White = 3 marbles. Therefore, statement 2 is sufficient.

Conclusively, since each statement alone is SUFFICIENT, answer = option D.