A group of men and women gathered to compete in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs. What percentage of the competitors were women?

(1) The average weight of the men was 150 lb.
(2) The average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women.

Statement 1: The average weight of the men was 150 lb.
It's possible that 50% of the competitors were women
It's possible that 75% of the competitors were women.
INSUFFICIENT.

Statement 2: The average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women.
Use ALLIGATION -- a great way to handle mixture problems.

Let M = men, W = women, G = the entire group.

Step 1: Draw a number line, with the averages for the Men and Women on the ends and the average for the entire group in the middle:
M---------------G---------------W

Step 2: Plot the distances between the averages.
Let x = the distance between M and G.
Since the average for the women is twice as far from G as is the average for the men, the distance between W and G = 2x.
Plotting these distances on the number line, we get:
M-------x-------G------2x-------W

Step 3: Determine the ratio in the mixture.
The ratio of M to W is equal to the RECIPROCAL of the distances in red.
M:W = 2x/x = 2:1.

Implication:
Of every 3 competitors, 2 are men and 1 is a woman, implying that women/total = 1/3 = 33.33%.
SUFFICIENT.

The correct answer is B.

For two other problems solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html