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

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Director
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A group of men and women competed in a marathon. Before the [#permalink]

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22 Jan 2008, 16:06
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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.

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Intern
Joined: 02 Mar 2007
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22 Jan 2008, 16:55
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

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Director
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22 Jan 2008, 17:00

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

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Director
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22 Jan 2008, 19:07
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?

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Director
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22 Jan 2008, 19:19
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?

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Director
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22 Jan 2008, 20:36
This is correct:

$formdata=\frac{120x+150y}{x+y}$

Since there are twice as many men as women competing: $formdata=y=2x$

Substitute 2x for y in the equation: $formdata=\frac{120x+150(2x)}{x+2x}+=+\frac{120x+300x}{3x}+=+\frac{420x}{3x}+=+140$

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.

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Director
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22 Jan 2008, 21:11
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.
_________________

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Director
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22 Jan 2008, 21:18
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

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VP
Joined: 22 Nov 2007
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22 Jan 2008, 21:47
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

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Intern
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24 Jan 2008, 01:33
you guys are right! i did the problem in my head without writing it out. tricky tricky.

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Re: DS: Ratios!   [#permalink] 24 Jan 2008, 01:33
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