There are a certain number of red balls, green balls and blue balls in a bag. Of the balls in the bag, $\frac{1}{3}$ are red and $\frac{2}{7}$ are blue. The number of green balls in the bag is 8 less than twice the number of blue balls. The number of green balls in the bag is
(A) 12
(B) 16
(C) 20
(D) 24
(E) 28
Solution 1

Suppose there are $x$ balls in total in the bag.

Then there are $\frac{1}{3} x$ red balls and $\frac{2}{7} x$ blue balls.

This tells us that the number of green balls is $x-\frac{1}{3} x-\frac{2}{7} x=\frac{21}{21} x-\frac{7}{21} x-\frac{6}{21} x=\frac{8}{21} x$.

But we know that the number of green balls is $2 \times \frac{2}{7} x-8$.

Thus, $\frac{8}{21} x=2 \times\left(\frac{2}{7} x\right)-8$ or $\frac{8}{21} x=\frac{12}{21} x-8$ or $\frac{4}{21} x=8$ or $x=42$.

Since $x=42$, the number of green balls is $\frac{8}{21} x=\frac{8}{21}(42)=16$.

## Solution 2

Suppose that there were 21 balls in the bag. (We choose 21 since there are fractions with denominator 3 and fractions with denominator 7 in the problem.)

Since $\frac{1}{3}$ of the balls are red, then 7 balls are red.

Since $\frac{2}{7}$ of the balls are blue, then 6 balls are red.

Thus, there are $21-7-6=8$ green balls in the bag. However, this is only 4 less than twice the number of blue balls, so there cannot be 21 balls in the bag.

To get from " 4 less" to " 8 less", we try doubling the number of balls in the bag.

If there are 42 balls in the bag, then 14 are red and 12 are blue, so 16 are green, which is 8 less than twice the number of blue balls.

Therefore, the number of green balls is 16 .

ANSwER: (B)