Mark has a bag that contains 3 black marbles, 6 gold marbles, 2 purple marbles, and 6 red marbles. Mark adds a number of white marbles to the bag and tells Susan that if she now draws a marble at random from the bag, the probability of it being black or gold is $\frac{3}{7}$. The number of white marbles that Mark adds to the bag is
(A) 5
(B) 2
(C) 6
(D) 4
(E) 3
Mark has a bag that contains 3 black marbles, 6 gold marbles, 2 purple marbles, and 6 red marbles. Mark adds a number of white marbles to the bag and tells Susan if she now draws a marble at random from the bag, the probability of it being black or gold is $\frac{3}{7}$. The number of white marbles that Mark adds to the bag is
(A) 5
(B) 2
(C) 6
(D) 4
(E) 3

## Solution

Since the probability of selecting a black or gold marble is $\frac{3}{7}$, this implies that the total number of marbles in the bag is a multiple of 7 . That is to say, there are possibly 7, 14, 21, 28, etc. marbles in the bag. The only acceptable number of marbles in the bag is 21 , since there are 9 marbles in total which are black or gold, and $\frac{9}{21}=\frac{3}{7}$. If there are 21 marbles in the bag, this means that 4 marbles must have been added, since there are 17 already accounted for.

Alternatively, we could say that the number of white marbles in the bag was $w$ (an unknown number), and form the equation

$$
\begin{aligned}
& \frac{6+3}{17+w}=\frac{3}{7} \\
& \frac{9}{17+w}=\frac{3}{7} \\
& \frac{9}{17+w}=\frac{9}{21}, \text { changing to a numerator of } 9 .
\end{aligned}
$$

Thus, $17+w=21$ or $w=4$, and so the number of white marbles is 4 .

ANSwER: (D)