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# A group of men and women gathered to compete in a marathon.

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A group of men and women gathered to compete in a marathon.  [#permalink]

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17 Nov 2011, 16:42
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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.

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A group of men and women gathered to compete in a marathon.  [#permalink]

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Updated on: 25 Oct 2015, 04:27
9
10
Responding to a pm:

Question:
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.

Solution:

Avg weight of women = 120 lbs

Stmnt 1: Avg weight of men = 150 lbs
This statement doesn't tell us the average weight of the entire group. The avg weight of men is not helpful till we know the avg weight of the entire group. Not sufficient

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women So women must be 1/3 of the total group. Think of the scale method we use. Ratio of the distances is 1:2 (avg is closer to men's avg than women's avg) so number of men: number of women = 2:1.
Sufficient.
Check out this link for scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/

Answer (B)
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Originally posted by VeritasKarishma on 31 Jul 2012, 09:09.
Last edited by Bunuel on 25 Oct 2015, 04:27, edited 3 times in total.
Renamed the topic and added OA.
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Re: Weight of people  [#permalink]

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23 Nov 2011, 20:48
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This question deals with weighted averages.
The two subgroups are the men and women. Each subgroup has an average weight.
We need the averages of each subgroup along with the ratio of men to women to calculate the overall average weight of the group. The ratio of men to women would determine the weight to give to each subgroup's average.
This question 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; hence we don't know the ratio of the women to men.

(2) SUFFICIENT
If the average weight of the entire group were twice as close to the average weight of the men as it were 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.

B
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Re: Weight of people  [#permalink]

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18 Nov 2011, 05:40
2
1
Hi,

theres a derieved formula* that goes like this:
RATIO OF{( avg of group1-average of entire group) to (average of entire group- avg of group1)}=RATIO of {(number of units in group 2) to (number of units in group1).
(120-x)/(x-men(avg))=men/women

using the above, we can now analyze the 2 statements as follows:
1. men(avg)=150. replacing in formula: (120-x)/(x-150)=men/women. x is still unknown....cannot find the ratio to men and women. INSUFFECIENT
2. 'closeness' means difference......according to this statement (x-men(avg))=2*(120-x)
therefore, (120-x)/(x-men(avg))=men/women=1:2...percentage of women is 2/3*100=66.66%. SUFFECIENT
hence answer is B

* you can easily derive the formula from the basic formulae of averages. if you need help with that let me know
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Re: Weight of people  [#permalink]

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24 Nov 2011, 21:02
3
1
upasanadatta wrote:
Hi,

theres a derieved formula* that goes like this:
RATIO OF{( avg of group1-average of entire group) to (average of entire group- avg of group1)}=RATIO of {(number of units in group 2) to (number of units in group1).
(120-x)/(x-men(avg))=men/women

using the above, we can now analyze the 2 statements as follows:
1. men(avg)=150. replacing in formula: (120-x)/(x-150)=men/women. x is still unknown....cannot find the ratio to men and women. INSUFFECIENT
2. 'closeness' means difference......according to this statement (x-men(avg))=2*(120-x)
therefore, (120-x)/(x-men(avg))=men/women=1:2...percentage of women is 2/3*100=66.66%. SUFFECIENT
hence answer is B

* you can easily derive the formula from the basic formulae of averages. if you need help with that let me know

Responding to a pm:
Check out this post: http://www.veritasprep.com/blog/2011/03 ... -averages/

It discusses the scale method i.e. using a number line. We see that the group that has more number of people pulls the weighted average toward it.

The number of women:number of men = Distance of men's average from weighted average:Distance of weighted average from women's average.
If the distance of men's average from weighted average is half of the distance of weighted average from women's average (because men's average is twice as close), number of women:number of men = 1:2
So women must be (1/3)rd = 33.33% of total competitors.

You don't need to know their exact averages. Just knowing where the weighted average lies relative to their averages is enough to know the ratio of the number of people in each group.
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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28 Oct 2012, 02:14
VeritasPrepKarishma wrote:
Responding to a pm:

Question:
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.

Solution:

Avg weight of women = 120 lbs

Stmnt 1: Avg weight of men = 150 lbs
This statement doesn't tell us the average weight of the entire group. The avg weight of men is not helpful till we know the avg weight of the entire group. Not sufficient

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women So women must be 1/3 of the total group. Think of the scale method we use. Ratio of the distances is 1:2 (avg is closer to men's avg than women's avg) so number of men: number of women = 2:1.
Sufficient.
Check out this link for scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/

Answer (B)

how can it be B ?

To use the average weight of men we need to consider statement 1 .

So B alone is insufficient, any thoughts ?

Thanks,
Ankit
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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28 Oct 2012, 05:28
ankit0411 wrote:
VeritasPrepKarishma wrote:
Responding to a pm:

Question:
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.

Solution:

Avg weight of women = 120 lbs

Stmnt 1: Avg weight of men = 150 lbs
This statement doesn't tell us the average weight of the entire group. The avg weight of men is not helpful till we know the avg weight of the entire group. Not sufficient

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women So women must be 1/3 of the total group. Think of the scale method we use. Ratio of the distances is 1:2 (avg is closer to men's avg than women's avg) so number of men: number of women = 2:1.
Sufficient.
Check out this link for scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/

Answer (B)

how can it be B ?

To use the average weight of men we need to consider statement 1 .

So B alone is insufficient, any thoughts ?

Thanks,
Ankit

The question asks for the % of women. If you have the ratio of men to women, you can answer this question. Statement II gives you the relation between weighted avg of the group and the weights of men and women. This information is enough to find the ratio of men and women. Check out the link I have given in the first post.
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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12 Sep 2013, 10:34
1
VeritasPrepKarishma wrote:
Responding to a pm:

Question:
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.

Solution:

Avg weight of women = 120 lbs

Stmnt 1: Avg weight of men = 150 lbs
This statement doesn't tell us the average weight of the entire group. The avg weight of men is not helpful till we know the avg weight of the entire group. Not sufficient

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women So women must be 1/3 of the total group. Think of the scale method we use. Ratio of the distances is 1:2 (avg is closer to men's avg than women's avg) so number of men: number of women = 2:1.
Sufficient.
Check out this link for scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/

Answer (B)

Hi, thanks in advance.

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
I couldn't really understand the second statement & how you arrived at this conclusion from the second statement:
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women

What I could gather from this statement is that the difference between the avg wt of the group & avg wt of men is twice the difference between avg wt of group and avg wt. of women
i.e. |avg wt. of group - avg wt. of men| = 2 x |avg wt. of group - avg wt. of women| where || represent mod

Could you please help me understand the language of the statement?
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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12 Sep 2013, 19:13
2
1
divineacclivity wrote:

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
I couldn't really understand the second statement & how you arrived at this conclusion from the second statement:
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women

What I could gather from this statement is that the difference between the avg wt of the group & avg wt of men is twice the difference between avg wt of group and avg wt. of women
i.e. |avg wt. of group - avg wt. of men| = 2 x |avg wt. of group - avg wt. of women| where || represent mod

Could you please help me understand the language of the statement?

Statement 2 is a direct reference to the number line method. Or you can say that it refers to the core concept of average weight.
It says that the group avg is twice as close to the men's average as the women's average. We know that group average lies between the two averages. If men are more, the group average will be closer to men's average. If women are more, the group average will be closer to women's average

....................|.......................................|..........................|
.........Women's avg.....................Group Avg...................Men's Avg

The statement says that group average is twice as close to men's avg so it means that number of men is twice of number of women. If you look at the scale method discussed in the link, you will understand it even better. We are given that the distance between group avg and men's avg is half the distance between group avg and women's avg. So men must be twice of women.
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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29 Dec 2013, 08:34
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Here you go hope it helps.
Attachments

Photo 29-12-13 16 19 21.jpg [ 1.38 MiB | Viewed 13452 times ]

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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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23 Jul 2014, 16:07
I was confused about the explanation, but think about this simple example:

Women's average = 2
Group average = 4
Men's average = 5

Notice that the group average is twice as far from women's average than it is from men's.

To get group average of 4, we must have ONE Woman and TWO Men --> (1*2+2*5) / 3 = 4.
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A group of men and women gathered to compete in a marathon.  [#permalink]

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10 Sep 2015, 22:01
1
For someone like me, whose not aware of the weighted avg rules, (i.e. if avg wt of the group is twice as close to the avg wt of men that women, then the number of men is twice that of women) u can use the following method:

lets say the difference between the avg wt of the group and the avg wt of men be x,
then the difference between the avg wt of the group and the avg wt of women will be 2x,
now we can write avg wt of group = 120 + 2x
and avg wt of men = 120 + 3x

now we can put these values in the weighted avg equation:
{120f + (120 + 3x)m}/(f +m) = 120 + 2x (where f and m are number of women and men respectively)
=> 120f + 120m +3xm = (120 + 2x)(f + m)
=> 120f + 120m +3xm = 120f + 120m + 2xf + 2xm
=> xm = 2xf
=> m = 2f

From this we can get the required % = [f/(m + f)] *100 = [f/(f +2f)] *100 = (f/3f)*100 = 33.33%

I know this is a bit long, but if u r not aware of the rule, like i was, this can be helpful.
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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04 Oct 2017, 18:08
VeritasPrepKarishma wrote:
Responding to a pm:

Question:
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.

Solution:

Avg weight of women = 120 lbs

Stmnt 1: Avg weight of men = 150 lbs
This statement doesn't tell us the average weight of the entire group. The avg weight of men is not helpful till we know the avg weight of the entire group. Not sufficient

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women So women must be 1/3 of the total group. Think of the scale method we use. Ratio of the distances is 1:2 (avg is closer to men's avg than women's avg) so number of men: number of women = 2:1.
Sufficient.
Check out this link for scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/

Answer (B)

if the average of men and women were equal, we could say the ratio of the distance between total average with men and women indicate the proportion
but it could be so different! for example, the average weight of men is 500, so the total average will be the great number and close to men.

am I wrong?
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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04 Oct 2017, 22:41
1
soodia wrote:
VeritasPrepKarishma wrote:
Responding to a pm:

Question:
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.

Solution:

Avg weight of women = 120 lbs

Stmnt 1: Avg weight of men = 150 lbs
This statement doesn't tell us the average weight of the entire group. The avg weight of men is not helpful till we know the avg weight of the entire group. Not sufficient

Stmnt 2: Avg weight of group was twice as close to the average weight of the men as it was to the average weight of the women.
Since avg weight is twice as close to avg weight of men, the number of men must be twice of the number of women So women must be 1/3 of the total group. Think of the scale method we use. Ratio of the distances is 1:2 (avg is closer to men's avg than women's avg) so number of men: number of women = 2:1.
Sufficient.
Check out this link for scale method: http://www.veritasprep.com/blog/2011/03 ... -averages/

Answer (B)

if the average of men and women were equal, we could say the ratio of the distance between total average with men and women indicate the proportion
but it could be so different! for example, the average weight of men is 500, so the total average will be the great number and close to men.

am I wrong?

"if the average of men and women were equal,"

If the avg weight of men was 120 lbs too, the average of the entire group would be 120 too. It wouldn't matter what the proportion of women is in the group. The average of the two would always be 120 lbs.
I am not sure what the rest of your comment means.
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Re: A group of men and women gathered to compete in a marathon.  [#permalink]

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23 Jun 2018, 23:21
How can you say that the number of men is more just because the group's avg is closer to that of men? Shouldn't it depend on both the number and the actual weight itself?

And how can u assume the women weigh less? Seriously!

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Re: A group of men and women gathered to compete in a marathon. &nbs [#permalink] 23 Jun 2018, 23:21
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# A group of men and women gathered to compete in a marathon.

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