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Master | Next Rank: 500 Posts
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by HPengineer » Sat Oct 02, 2010 6:14 pm
In a certain group, the average (arithmetic mean) age of the males is 28, and the average age of the females is 30. If there are 100 people in the group, how many of them are males?

(1) The average age of all 100 people is 28.9

(2) There are 10 more males than there are females.



How to use weighted average to solve?
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Source: — Data Sufficiency |
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by mj78ind » Sat Oct 02, 2010 6:21 pm
HPengineer wrote:In a certain group, the average (arithmetic mean) age of the males is 28, and the average age of the females is 30. If there are 100 people in the group, how many of them are males?

(1) The average age of all 100 people is 28.9

(2) There are 10 more males than there are females.



How to use weighted average to solve?
One cool trick is we don't have to solve it but ensure that it can be solved and only one value will result.

1 - 28.9 = {28*x + 30*(100-x)}/100 will give us a unique x, which is the number of men

2 - x + x+ 10 = 100. x+10 is the number of males can solve

Hence D
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by HPengineer » Sat Oct 02, 2010 6:31 pm
cool i was on the right track!! good to know thanks!
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