Does Committee X have more members than Committee Y

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mehravikas: Legendary Member; Posts: 1161; Joined: Mon May 12, 2008 2:52 amFollowed by: 1 member; Location: Sydney; Thanked: 23 times

by mehravikas » Sat Jul 12, 2008 4:46 pm

DS16-19 Committee X and Committee Y, which have no common members, will combine to form Committee Z. Does Committee X have more members than Committee Y ?

(1) The average (arithmetic mean) age of the members of Committee X is 25.7 years and the average age of the members of Committee Y is 29.3 years.

(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years.

Please help on this one.

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Source: Beat The GMAT — Data Sufficiency | Sat Jul 12, 2008 4:46 pm

tzink: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Thu Jul 10, 2008 9:02 pm; Thanked: 1 times

by tzink » Sat Jul 12, 2008 9:07 pm

I'll take a stab at this one...

(1) clearly does not give enough information. NOT SUFFICIENT

(2) by itself, again, not helpful. NOT SUFFICIENT

(1) and (2) together:

If X and Y have the same number of members, the arithmetic mean age of the new committee Z will be the mean of X and Y.

From (1), Determine the mean age of X and Y by averaging their respective means (to make this easier in your head, the mean of X is .3 lower than 26 and the mean of Y is .3 higher than 29, so you can round each to 26 and 29, respectively).

The average of X and Y is 27.5. This means if X and Y have the same number of members, the average age of Z should be 27.5

From (2), however, the average of the new committee Z is 26.6 years, which means that committee X (which has the lower mean age) must have more members in order to skew down the average age of the new committee Z. SUFFICIENT

Answer is C.

(there's probably a equation method for this as well...)

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ed09: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sat Jul 05, 2008 5:40 pm; Thanked: 2 times

by ed09 » Sat Jul 12, 2008 9:40 pm

Hi guys,

Sorry for my hair-splitting, but, as for me, the answer depends on the meaning of the word combine in the problem statement. In case of the meaning is to unite Committee X and Committee Y the answer should be C. However, if it is something similar to the meaning of to arrange, that is to use all members of both committees X and Y to form Z is not a conditio sine qua non, what is, in my view, also permissible for the word, then the answer must be E. See the weighted average concept.

For the first case, assuming that x and y are the member quantities of committees X and Y, we can draw up and re-arrange an equation
|26.6-25.7|*x-|26.6-29.3|*y=0
0.9x-2.7y=0
0.9x=2.7y
x/y=3/1
That is, in order to satisfy the deviation of means given, in the Committee Z for each member from Committee Y must be three members from Committee X.

See the concept for this solution at
https://www.beatthegmat.com/averages-gma ... 13573.html

Well, what it the answer?

Best!

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tzink: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Thu Jul 10, 2008 9:02 pm; Thanked: 1 times

by tzink » Sun Jul 13, 2008 10:03 pm

ed09 wrote:Hi guys,

Sorry for my hair-splitting, but, as for me, the answer depends on the meaning of the word combine in the problem statement. In case of the meaning is to unite Committee X and Committee Y the answer should be C. However, if it is something similar to the meaning of to arrange, that is to use all members of both committees X and Y to form Z is not a conditio sine qua non, what is, in my view, also permissible for the word, then the answer must be E. See the weighted average concept.

Good point.. If only some of the members from each X and Y joined Z, the answer would definitely be E.
I would think (though I'm not basing this on experience) that if that were the case, the question would be phrased "If some/select/a percentage of members from committees X and Z...etc"

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mehravikas: Legendary Member; Posts: 1161; Joined: Mon May 12, 2008 2:52 amFollowed by: 1 member; Location: Sydney; Thanked: 23 times

by mehravikas » Mon Jul 14, 2008 2:37 am

I think tzinks' explanation is perfect for this question. Answer is 'C'

Thanks guys.

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restero: Newbie | Next Rank: 10 Posts; Posts: 4; Joined: Tue Apr 14, 2009 10:22 am

by restero » Mon Jan 25, 2010 10:31 pm

Always remember, When to sets are combined to give a mean, the distance of each mean from the combined mean will always add up to zero.
Therefore,
(25.7-26.6)X+(29.3-26.6)Y=0
=> -0.9X+2.7Y=0
=>X=3Y. i.e. X>Y!

hence and b together solves the question i.e answer is C.

[b]Extending this, if, sets X and Y have the same number of people, their combined mean will be equal to EACH individual mean!
Also, the combined mean is closer to the mean of the larger set if the individual means are different!
Finally, if the two sets have different number of items, the combined mean will always be in between the individual means of each set!! [/b]

-Addy

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