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A group of men

by atulmangal » Sun May 01, 2011 7:42 am
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.

I'm unable to understand the meaning of the second statement...please explain exactly what does Op B convey???
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by Chaitanya_1986 » Sun May 01, 2011 8:06 am
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.

Ans)
Let,
Average weight of men = AVM , Average weight of Women = ANW, No of male=m, No of female= w

total no of competitors= (m+w)
Given in question stem
AVM= 120
-->> total weight of women = 120 * w

we need to fine % of women in (m+w)

A) AVM= 150
-->>> total weight of men= 150 * m

120w+150m/(w+m)= x?? , x= avg weight of all the competitors


we cannot get any solution by this alone because we dunno the total wight of people or total average weight

b)
total weight of men+total weight of women/ (m+w)= 2 * AVM (or) 2*AVW

by this statement alone we cannot get the solution because we dont know the average weight of men in group.

BY combining both we get

(120*m+150*n)/(m+n)=2*m
120m+150n=2 m*m+ 2 mn----1

(120*m+150*n)/(m+n)=2*n
120m+150n=2 n*n+ 2 mn---2

from 1 ,2

we get m*m=n*n

m=n

by substituting this in 1,2 we get the answer

So answer is C

OA????
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by sharmasumitn1 » Sun May 01, 2011 11:25 am
atulmangal wrote: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.

I'm unable to understand the meaning of the second statement...please explain exactly what does Op B convey???

(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.

This implies that
|the avg of entire group - avg of men| = 2 |the avg of entire group-the avg of women|
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by atulmangal » Sun May 01, 2011 12:13 pm
sharmasumitn1 wrote:
atulmangal wrote: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.

I'm unable to understand the meaning of the second statement...please explain exactly what does Op B convey???

(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.

This implies that
|the avg of entire group - avg of men| = 2 |the avg of entire group-the avg of women|
R u sure about this: can u please solve the question then...whats your answer???
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by GMATGuruNY » Sun May 01, 2011 2:03 pm
atulmangal wrote: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.

I'm unable to understand the meaning of the second statement...please explain exactly what does Op B convey???
Statement 1: Average weight of the men = 150.
No way to determine the ratio of men to 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.
Statement 2 describes the very concept behind a weighted average.
For the average weight of the whole group to be "twice as close" to the average weight of the men, there must be twice as many men as women.
Thus, men:women = 2:1.
Since 2+1 = 3, the women are 1/3 = 33.34% of the whole group.
Sufficient.

The correct answer is B.

Examples that illustrate statement 2:
Let average weight of the men = 150 pounds.
Let men = 2, women = 1.
Total weight of 2 men = 2*150 = 300.
Total weight of 1 woman = 1*120 = 120.
Total weight of the whole group = 300+120 = 420.
Average weight of the whole group = 420/3 = 140.
Difference between the average weight of the men and the average weight of the whole group = 150-140 = 10.
Difference between the average weight of the whole group and the average weight of the women = 140-120 = 20.

Let average weight of the men = 180 pounds.
Let men = 2, women = 1.
Total weight of 2 men = 2*180 = 360.
Total weight of 1 woman = 1*120 = 120.
Total weight of the whole group = 360+120 = 480.
Average weight of the whole group = 480/3 = 160.
Difference between the average weight of the men and the average weight of the whole group = 180-160 = 20.
Difference between the average weight of the whole group and the average weight of the women = 160-120 = 40.

In each case -- regardless of the average weight of the men -- the weighted average is "twice as close" to the average weight of the men.
Last edited by GMATGuruNY on Sun May 01, 2011 2:31 pm, edited 2 times in total.
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by parkman » Sun May 01, 2011 2:09 pm
G=Group Average
M=Men Average
W=Women Average

According to (2)

(Group average - Men Average) + 2(Group average - Women Average) = 0

The two distances from the group average will cancel each other out, since the men's average being 2x closer to the group average is accounted for by multiplying the distance of the women's average from the group average by 2.

Substitute values, using the question and information from (1)

(G-150) + 2(G-120)=0

G - 150 + 2G -240=0
3G - 390 = 0
3G = 390
G = 130

So group Average = 130 Pounds

Let X = #men
Let y = #women

Average Weight = (#men(average male weight) + #women(average women weight))/(#men+#women)
= (x(150)+y(120))/(x+y)
130 = (150x+120y)/(x+y)
130x +130y = 150x +120y
10y = 20x
y = 2x

# women = 2 times # men
Women = 66% of the participants

Answer C

NVM this is wrong lol i suck, statement carryover
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by atulmangal » Sun May 01, 2011 2:23 pm
Thanks guys

OA is B
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by atulmangal » Sun May 01, 2011 2:24 pm
GMATGuruNY wrote:
atulmangal wrote: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.

I'm unable to understand the meaning of the second statement...please explain exactly what does Op B convey???
Statement 1: Average weight of the men = 150.
No way to determine the ratio of men to 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.
Statement 2 describes the very concept behind a weighted average.
For the average weight of the whole group to be "twice as close" to the average weight of the men, there must be twice as many men as women.
Thus, men:women = 2:1.
Since 2+1 = 3, the women are 1/3 = 33.34% of the whole group.
Sufficient.

The correct answer is B.

Examples that illustrate statement 2:
Let average weight of the men = 150 pounds.
Let men = 2, women = 1.
Total weight of 2 men = 2*150 = 300.
Total weight of 1 woman = 1*120 = 120.
Total weight of the whole group = 300+120 = 420.
Average weight of the whole group = 420/3 = 140.
Average weight of the men - average weight of the whole group = 150-140 = 10.
Average weight of the whole group - average weight of the women = 140-120 = 20.

Let average weight of the men = 180 pounds.
Let men = 2, women = 1.
Total weight of 2 men = 2*180 = 360.
Total weight of 1 woman = 1*120 = 120.
Total weight of the whole group = 360+120 = 480.
Average weight of the whole group = 480/3 = 160.
Average weight of the men - average weight of the whole group = 180-160 = 20.
Average weight of the whole group - average weight of the women = 160-120 = 40.

In each case -- regardless of the average weight of the men -- the weighted average is "twice as close" to the average weight of the men.
Thanks a Lot Mitch for explaining the concept...
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by mk101 » Mon May 02, 2011 4:52 am
What option B says mathematically is -

2(Wm-Wa) = (wa - ww) ( Wa = average wt of the whole group, ww = average wt of women, wm = average wt of men)

(Wm -wa)/(wa-ww) = 2/1

On simplifying the above equation - 2 Wm + Ww = 3 Wa -----(1)

the above equation is similar to the weighted avg equation

Nm*Wm + Nw * Ww = (Nm + Nw)Wa --------(2)

on comparing 1 and 2 we have

Nm : Nw = 2:1 ( Nm = number of men, Nw = number of women)

Thus B is sufficient
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