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

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CEO
Joined: 21 Jan 2007
Posts: 2734

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Location: New York City
A group of men and women competed in a marathon. Before the [#permalink]

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28 Dec 2007, 14:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 if there were twice as many men as women competing in the marathon?

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Director
Joined: 12 Jul 2007
Posts: 857

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28 Dec 2007, 14:55
super simple!

2(150)+1(120)=420/3=140 pounds! the average of a set of integers will remain the same no matter how many integers (runners in this case) are involved as long as the ratios remain the same.

Sometimes I like to solve these problems with a little shortcut I made up (disclaimer: I'm sure this strategy is older than the GMAT, but I've never read it anywhere)

120-----150

there is a difference of 30 pounds and a 2:1 ratio between men:women. Put 2:1 into the 30 pound difference 20:10. 120+20=140 or 150-10=140. You think "there are more men then women, so the average will be much closer to the heavier side." then it's just applying the ratio to the difference to get specific.

sometimes this strategy is easier, sometimes it's harder. take it on a case by case basis

Kudos [?]: 330 [0], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1043 [0], given: 4

Location: New York City

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28 Dec 2007, 15:02
very nice.

i wanted to see if anyone else caught that the # of elements was irrelevant.

i approached it like a weighed avg question

%*Amount A + %AmountB = proportion combined amt
120(1/3) + 150 (2/3) = 140

Kudos [?]: 1043 [0], given: 4

28 Dec 2007, 15:02
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