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Weighted average (lost in 2 second option)

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Weighted average (lost in 2 second option)

by El Cucu » Sat Mar 28, 2009 7:48 am
The average weight of the women in a room is 120 lbs, and the average weight of the men in the room is 150 lbs. What is the average weight of the people in the room?

(1) There are twice as many men as women in the room.

(2) There are a total of 120 people in the room.


(2) I can't figure out why we can not make an equation Boys+Girls=120 So Men=120-W and substitute= w*120+(120-W)*150/(120-W)+W=
Tks for you help!!
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by DanaJ » Sat Mar 28, 2009 8:54 am
Think of it this way:
w = number of women in the room
m = number of men in the room.

1 tells you that m = 2w.
Now, the average weight will be (120w + 150m)/(m + w). Substitute m = 2w to get that (120w + 300w)/3w = 420w/3w = 140 is the average weight.

2 is insufficient. While you can substitute m for 120 - w, you can go no further with your calculations: notice that even in you equation you can't get rid of that w.
Think of it this way: there could be 60 men and 60 women, but there could also be 119 men and one woman. There is no way of distinguishing between the cases.
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by mjjking » Sat Mar 28, 2009 9:01 am
Let's say women number = x
Let's also say that men number = x

The problem tells us that (120x+150y)/(x+y)=average weight

From stm. 1 we get that that y=2x
By substituting we get:

[120(1/2y)+150y]/3/2y= average
solving we get 210y/(3/2)y
let's divide num and den by y and we have 70(2/3) = 140

voilà, we have found our average

from b, we know a+b, but we cannot determine weather there are 40 males or 70 males. Even assigning variables like you says we don't get to a definite conclusion, hence IMO A.
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