BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Average weight of men and women

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sat May 15, 2010 11:24 am

Average weight of men and women

by sitsmegain » Sat May 15, 2010 11:28 am
Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
Top
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 135
Joined: Tue Oct 13, 2009 10:27 am
Thanked: 3 times

by boazkhan » Sat May 15, 2010 11:46 am
IMO B is the answer here. What is the OA?
Top
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Sat May 15, 2010 11:49 am
Yeah Your are very much correct!
We need to apply Weighted Average Concepts:

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.

St 1 gives only number of Men. Insufficient.

St2 : gives a relation between men & women . But not exact figures which we are looking for. Let us equate them as of now. N1=2*N2

Combining both of them, we can fing the Weighted average!

Pick C
Top
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sat May 15, 2010 11:24 am

by sitsmegain » Sat May 15, 2010 11:58 am
gmatmachoman wrote:Yeah Your are very much correct!
We need to apply Weighted Average Concepts:

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.

St 1 gives only number of Men. Insufficient.

St2 : gives a relation between men & women . But not exact figures which we are looking for. Let us equate them as of now. N1=2*N2

Combining both of them, we can fing the Weighted average!

Pick C
Can you please solve this for me in more clarity, i have solved it in simple logical sense. but i want to know how you apply the weighted average concept. Thanks
Top
Master | Next Rank: 500 Posts
Posts: 135
Joined: Tue Oct 13, 2009 10:27 am
Thanked: 3 times

by boazkhan » Sat May 15, 2010 11:58 am
gmatmachoman - As per the statement 2 N1= 2*N2 -- N2 actually cancels out. I don't think we need statement 1. agree?
Top
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sat May 15, 2010 11:24 am

by sitsmegain » Sat May 15, 2010 12:03 pm
boazkhan wrote:gmatmachoman - As per the statement 2 N1= 2*N2 -- N2 actually cancels out. I don't think we need statement 1. agree?
How ? i dint get u? please help me understand. I was expecting this answer since its mentioned that we dont require the statement 1 but im looking out for explanation.
Top
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 am
Thanked: 173 times
Followed by:2 members
GMAT Score:710

by gmatmachoman » Sat May 15, 2010 1:21 pm
sitsmegain wrote:
gmatmachoman wrote:Yeah Your are very much correct!
We need to apply Weighted Average Concepts:

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.

St 1 gives only number of Men. Insufficient.

St2 : gives a relation between men & women . But not exact figures which we are looking for. Let us equate them as of now. N1=2*N2

Combining both of them, we can fing the Weighted average!

Pick C
Can you please solve this for me in more clarity, i have solved it in simple logical sense. but i want to know how you apply the weighted average concept. Thanks
Now that from St 2 we know N1 = 2 *N2

N1 +N2 = 3*N2

Apply this one in the main equation

{(A1 * N1) +(A2 *N2)}/ (N1+N2) eqauls Total weighted average Aw.
{150 * 2N2 + 120 * N2}/3*N2

As Boazkhan said N2 cancels out: (Sorry for my mistake)

(150 * 2 +120) *N2/(3*N2)

=140

Lesson Learnt : Dont be so hasty when it comes to Weighted average concepts..Knowing them is one part but applying them is also as equal as "knowing them".
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Sat May 15, 2010 8:09 pm
sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Tue Sep 28, 2010 11:34 pm
Thanked: 1 times

by bacchewar_prashant » Fri Nov 19, 2010 4:20 am
Nice Question. I got it wrong I thought we need both infomation.
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Nov 19, 2010 5:52 am
sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
When a DS question asks for a specific amount -- in this case, the average weight of the whole group -- a good way to check whether a statement is sufficient:

Plug in TWICE. If the value being sought stays the same, the statement is sufficient. If the value being sought changes, the statement is insufficient.

Looking at statement 2: twice as many men as women
Plug in M=2, W=1
Total weight of M = 2*150 = 300.
Total weight of W = 1*120 = 120.
Average of the whole group = 420/3 = 140.

Plug in M=4, W=2.
Total weight of M = 4*150 = 600.
Total weight of W = 2*120 = 240.
Average of the whole group = 840/6 = 140.

Since in each case the average weight of the whole group is 140, the statement is sufficient.

The correct answer is B.

DS questions become much clearer when we plug in actual values.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sat Apr 03, 2010 8:35 pm
Thanked: 1 times
Followed by:1 members

by thp510 » Sat Nov 20, 2010 10:32 am
sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
I don't know if this is the approved solution on how to solve it but it gave me the right answer.

F=Total of Females
M=Total of Males
TFW = Total Female Weight
TMW = Total Male Weight
Sum/n = Avg

Setup:
(TFW+TMW)/(F+M) = The Avg weight of all competitors (what we're trying to find)

Given: (TFW)/(F) = 120 or TFW = 120F
Given: (TMW)/(M) = 150 or TMW = 150M

They gave us a 2:1 ratio of men to women so I just said 2/1 = M/F, therefore M=2 and F=1

Plug all the numbers into the first equation...
(TFW+TMW)/(F+M) = The Avg weight of all competitors
(120F + 150M)/(F+M) = The Avg weight of all competitors
now plug in 2 & 1 for M & F...
(120*1+150*2)/(3)= The Avg weight of all competitors = 140; since I have an answer I know (B) is sufficient. Answer is B.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 163
Joined: Tue Nov 24, 2009 7:56 am
Location: Philadelphia
Thanked: 13 times
Followed by:4 members
GMAT Score:660

by chendawg » Thu Nov 25, 2010 10:26 am
There is a neat little trick, which I learned in a MGMAT study hall from the great Lunarpower, for this problem. Basically the principle is, if we know the average weight of X and the average weight of Y, and the average weight of X+Y, we know the ratio of X to Y. The ratio of X to Y is the inverse of the distance X is from the average to the distance Y is from the average.

For example, let's say we have a room full of football and basketball players. The average weight of basketball players is 180 pounds and the average weight of football players is 210 pounds. The average of both basketball and football players is 200 pounds. Using the principle listed above, since the weight of basketball players is 20 away from 200, and the weight of basketball players is 10 away from 200, we know that the ratio of basketball players to football players is 10 to 20, or 1:2. We can always figure out the ratio of X to Y, as long as we know the average of X, average of Y, and the average of both.

We can apply that principle to this problem. The problem gives us the average weight of men, 150, and the average weight of women, 120, in the problem.

Statement 1 gives us the total # of men. Just using basic principles of weighted average we know that we need to know the # of women, thus insufficient.

Statement 2 gives us the ratio of men to women, 2 to 1. Since we know the average weight of men as 150, the average weight of women as 120, and we are given a ratio of 2 to 1 for men to women, we know we can calculate the average using the trick we just learned. Instead of figuring out a ratio, we are working "backwards" for an average. Sufficient.


Hopefully you guys can understand my rambling, if you have any questions feel free to ask.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 123
Joined: Tue Nov 23, 2010 7:18 pm
Location: Mumbai, India
Thanked: 5 times
Followed by:4 members

by anirudhbhalotia » Fri Dec 17, 2010 9:29 am
Stuart Kovinsky wrote:
sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.
How did you get 1/3 and 2/3?
Top
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Sun Dec 12, 2010 10:49 pm
Thanked: 1 times

by lukester » Thu Jan 06, 2011 10:26 pm
anirudhbhalotia wrote:
Stuart Kovinsky wrote:
sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.
How did you get 1/3 and 2/3?
1/3 is the ratio of women. 2/3 is for men. since there are twice as many men as women, that is why the ratio is as such.
Top
Master | Next Rank: 500 Posts
Posts: 131
Joined: Fri Jun 18, 2010 10:19 am
Location: New York, NY
Thanked: 10 times

by aleph777 » Fri Jan 07, 2011 8:15 am
Stuart Kovinsky wrote:
sitsmegain wrote:Hello ALL,
As per my solving, we need both (1)and (2) to solve this problem, but i want to be sure that my understanding is right. So what will be the correct answer and how?

****************************************************************
A group of men and women competed in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs, while the average weight of the men was found to be 150 lbs. What was the average weight of all of the competitors?

(1) 100 men competed in the marathon.
(2) There were twice as many men as women competing in the marathon.
*******************************************************************
As you discovered, you don't need the actual number in each group to solve for a weighted average. Here's the rule:

To solve for a weighted average, you need to know the average of each subgroup and the weight of each subgroup.

Weighted Average = (avg group 1)(weight group 1) + (avg group 2)(weight group 2) + .... (avg group n)(weight group n)

When we combine the stem and statement (2), we get:

Weighted Average = (120)(1/3) + (150)(2/3)

which is certainly solvable.

We could also have solved using intuition: if there are twice as many men as women, then if we plot the weights on a number line, the average will be 2/3 of the way between women and men:

120 ---- 2x----avg--x--150

which is also solvable.
Stuart, when you say we need to know the "weight" of each group, you then use the fractions 1/3 and 2/3, which clearly refer to the w:m ratio of 1:2. To clarify, then, by "weight" you mean the ratio of one to the other?

Just want to make sure I'm clear, because it's a really elegant solution.

Thanks!
Top
Post new topic Post Reply