700+ Challenge Problem, July 10 2006

[KevinFitzgerald]

Most Manhattan GMAT students are trying to break the 700 barrier. As a result, we've developed our own math problems written at the 700+ level; these are the types of questions you'll WANT to see, when you are working at that level. Try to solve this 700+ level problem (I'll post the solution next Monday).

Question: Weighted Weights
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 150lb.
(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.

(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.

[ahmadazeem777]

Originally Posted by KevinFitzgerald

Question: Weighted Weights
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?

I think it is B .

[puneet_gmat]

I think its - C

Let number of women be W
Let number of men be M
We need to calculate W/(M+W)

A- Alone tells nothing M or W

Trying B -
Let ave wt of M ge wt of be k
then
J=(Mk + W120)/(M+W) is the ave wt of the group
Given the average wt of the group needs to be in between ave wt of men and women from B we get
2(k-J) = J - 120

this equation has k, M and W and k does not cancel out. hence using A for k=150

we get
2*150 -2j = j -120
J = 420/3 = 140

=> 150M + 120W = 140M + 140W
=> M = 2W
Hence
W/(M+W) = 1/3
%age = 1/3 *100

[skarya]

C - both statements required

(1) alone does say much except the average value for second group
(2) correlates overall average with individual averages

with (1) and (2) together overall average value is possible...and that leads to ratio of two groups...

Xw=120
Xm=150
Xw+m - Xw = 2 * (Xm-Xw+m) or 3Xw+m = 2Xm+Xw = 300+120=420 or Xw+m=140

150M+120W=140 (M+W) or 10M=20W or M=2W or M+W = 3W or W % = W/(W+M)*100 % = 1/3*100 %

[abhi_luthra]

I believe its B....statement 2 is sufficient.

Let the average of men be A1, the avg. of women be A2 and the overall average be A.

Statement B says that A - A2 = 2(A1-A)
-> A = (2A1 + A2)/3

which implies no. of men is twice the no. of women.

[amitnsitian]

my answer is also B

[skarya]

I agree 2 alone is sufficient to answer.

Correct answer = B

[KevinFitzgerald]

Answer
This question deals with weighted averages. A weighted average is used to combine the averages of two or more subgroups and to compute the overall average of a group. The two subgroups in this question are the men and women. Each subgroup has an average weight (the women’s is given in the question; the men’s is given in the first statement). To calculate the overall average weight of the group, we would need the averages of each subgroup along with the ratio of men to women. The ratio of men to women would determine the weight to give to each subgroup’s average. However, this question is not asking for the weighted average, but is simply asking for the ratio of women to men (i.e. what percentage of the competitors were women).

(1) INSUFFICIENT: This statement merely provides us with the average of the other subgroup – the men. We don’t know what weight to give to either subgroup; therefore we don’t know the ratio of the women to men.

(2) SUFFICIENT: If 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, there must be twice as many men as women. With a 2:1 ratio of men to women of, 33 1/3% (i.e. 1/3) of the competitors must have been women. Consider the following rule and its proof.

RULE: The ratio that determines how to weight the averages of two or more subgroups in a weighted average ALSO REFLECTS the ratio of the distances from the weighted average to each subgroup’s average.

Let’s use this question to understand what this rule means. If we start from the solution, we will see why this rule holds true. The average weight of the men here is 150 lbs, and the average weight of the women is 120 lbs. There are twice as many men as women in the group (from the solution) so to calculate the weighted average, we would use the formula [1(120) + 2(150)] / 3. If we do the math, the overall weighted average comes to 140.

Now let’s look at the distance from the weighted average to the average of each subgroup.
Distance from the weighted avg. to the avg. weight of the men is 150 – 140 = 10.
Distance from the weighted avg. to the avg. weight of the women is 140 – 120 = 20.

Notice that the weighted average is twice as close to the men’s average as it is to the women’s average, and notice that this reflects the fact that there were twice as many men as women. In general, the ratio of these distances will always reflect the relative ratio of the subgroups.

The correct answer is (B), Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.