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28 Dec 2007, 14:43
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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 if there were twice as many men as women competing in the marathon?
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28 Dec 2007, 14:55
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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
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
28 Dec 2007, 15:02
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very nice.
i wanted to see if anyone else caught that the # of elements was irrelevant. :-D
i approached it like a weighed avg question
%*Amount A + %AmountB = proportion combined amt
120(1/3) + 150 (2/3) = 140
i wanted to see if anyone else caught that the # of elements was irrelevant. :-D
i approached it like a weighed avg question
%*Amount A + %AmountB = proportion combined amt
120(1/3) + 150 (2/3) = 140
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
