Find all School-related info fast with the new School-Specific MBA Forum

It is currently 25 Oct 2014, 05:55

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A group of men and women competed in a marathon. Before the

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Director
Director
avatar
Joined: 01 May 2007
Posts: 795
Followers: 1

Kudos [?]: 79 [0], given: 0

A group of men and women competed in a marathon. Before the [#permalink] New post 22 Jan 2008, 15:06
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (01:00) correct 0% (00:00) wrong based on 0 sessions
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.
Intern
Intern
avatar
Joined: 02 Mar 2007
Posts: 27
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 15:55
C

Average = Sum of weights / Number of people

1. 100 men

this tells you the total weight of men in marathon.

insufficent

2. insufficent because you don't know the total number of people. nothing to plug into the average equation.

together...sufficent
Director
Director
User avatar
Joined: 12 Jul 2007
Posts: 867
Followers: 12

Kudos [?]: 202 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 16:00
Answer B

We don't need exact figures, just the ratio. Check and see for yourself:

(2*150)+(1*120)=420/3 = 140 average
(6*150)+(3+120)=1260/9 = 140 average
(100*150)+(50*120) = 21,000/150 = 140 average
Director
Director
avatar
Joined: 01 May 2007
Posts: 795
Followers: 1

Kudos [?]: 79 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 18:07
eschn3am, Correct answer is B. Can you give a detailed explanation why this ratio approach works? Does it have something to do with weighted averages?
Director
Director
avatar
Joined: 01 May 2007
Posts: 795
Followers: 1

Kudos [?]: 79 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 18:19
Oh I see...Is this accurate....

150 = Sum of males weight / y of males

so...150y = Sum of males

120 = sum of females weight / x of females

so...120x = Sum of females

So 120x + 150y/(x+y)

We know m/f = 2/1 so assume x = 2 and y = 1?

Is this the correct explanation?
Director
Director
User avatar
Joined: 12 Jul 2007
Posts: 867
Followers: 12

Kudos [?]: 202 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 19:36
This is correct:

Image

Since there are twice as many men as women competing: Image

Substitute 2x for y in the equation: Image


As you can see the ratio is enough to get your answer. As long as these ratio is constant the average will always be 140, regardless of how many people are running.
Director
Director
avatar
Joined: 05 Jan 2008
Posts: 709
Followers: 2

Kudos [?]: 110 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 20:11
You don't need absolute numbers in this question, as it is ratio, it is enough if you know the ratio

Considering the number of men as 2x => # of women would be x

total weight/total number => (240X+150x)/(2x+x) =>130

Thus B alone is sufficient.

----------------------------------------------------------------------------------------

In these kind of questions, it is sufficient to know one of the values, i.e in this specific case weight and number are the variables and weight is already given, and the ratio of numbers is sufficient.

Think of a scenario where weight and number both are given as ratios, then you will have two unknowns, i.e let us say the problem says the weight of men is twice the weight of women, the rest of the question being same, we will have

2y is the weight of men => y would be the weight of women

2x men and x women, thus (4xy+xy)/3x =>5y/3 thus it we need to know one of the values of the two unknowns in the questions of ratios to find out the absolute ratio.
_________________

Persistence+Patience+Persistence+Patience=G...O...A...L

Director
Director
User avatar
Joined: 03 Sep 2006
Posts: 889
Followers: 6

Kudos [?]: 171 [0], given: 33

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 20:18
madcowudub wrote:
C

Average = Sum of weights / Number of people

1. 100 men

this tells you the total weight of men in marathon.

insufficent

2. insufficent because you don't know the total number of people. nothing to plug into the average equation.

together...sufficent



(150*m + 120*f)/(m+f)= average

From (ii) m=2f

Thus sufficient

Answer is "B"
VP
VP
avatar
Joined: 22 Nov 2007
Posts: 1102
Followers: 6

Kudos [?]: 136 [0], given: 0

Re: DS: Ratios! [#permalink] New post 22 Jan 2008, 20:47
jimmyjamesdonkey wrote:
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.



B it is. it only needs to substitute men=2 females

so, 120f+300f/3f= average
Intern
Intern
avatar
Joined: 02 Mar 2007
Posts: 27
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: DS: Ratios! [#permalink] New post 24 Jan 2008, 00:33
you guys are right! i did the problem in my head without writing it out. tricky tricky.
Re: DS: Ratios!   [#permalink] 24 Jan 2008, 00:33
    Similar topics Author Replies Last post
Similar
Topics:
8 Experts publish their posts in the topic A group of men and women gathered to compete in a marathon. VeritasPrepKarishma 8 31 Jul 2012, 08:09
2 Experts publish their posts in the topic A group of men and women gathered to compete in a marathon. enigma123 4 17 Nov 2011, 15:42
A group of men and women competed in a marathon. Before the bmwhype2 2 28 Dec 2007, 13:43
A group of men and women competed in a marathon. Before the terp26 3 03 Apr 2007, 09:40
The ratio of the women to the men in a group of people is 2 getzgetzu 8 05 May 2006, 23:40
Display posts from previous: Sort by

A group of men and women competed in a marathon. Before the

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.