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Committee X and Committee Y, which have no common members,

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Manager
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Committee X and Committee Y, which have no common members, [#permalink]

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28 Apr 2009, 00:56
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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.
Director
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28 Apr 2009, 01:20
Hi,
I believe the ans is C

Stmnt 1

It gives us the averages of both committees but we do not know how many members these committtees have

Stmnt 2 is not suff by itself since we know nothing about the committees that make com Z

From both statements if comtes have equal number of peple then the avg should be (25.7+29.3)/2=27.5.
It is given that the avg of Z is 26.6 which means that the committee with lower average ( 25.7) is bigger.

Regards
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28 Apr 2009, 17:36
Interesting problem. An opportunity to learn, I hope.
IMO E.
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29 Apr 2009, 02:34
milind1979 wrote:
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.

x: number of members of committee X
y: number of members of committee Y

(1) insufficient
(2) insufficient
(1) and (2)
(1) => sum of ages of all members of Com X is 25.7 * x
sum of ages of all members of Com Y is 29.3 * y
(2) average age of members of Com Z = (sum of ages of all X members + sum of ages of all Y members) / (number of members of committee Z)
= (25.7 * x + 29.3 * y) / (x+y)
= 26.6
So 25.7 * x + 29.3 *y = 26.6 * x + 26.6 * y
27 * x = 9 * y
x = 3y

C for me
Manager
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29 Apr 2009, 02:37
BG wrote:
Hi,
I believe the ans is C

Stmnt 1

It gives us the averages of both committees but we do not know how many members these committtees have

Stmnt 2 is not suff by itself since we know nothing about the committees that make com Z

From both statements if comtes have equal number of peple then the avg should be (25.7+29.3)/2=27.5.
It is given that the avg of Z is 26.6 which means that the committee with lower average ( 25.7) is bigger.

Regards

Very good answer in my opinion.
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29 Apr 2009, 05:58
DavidArchuleta wrote:
milind1979 wrote:
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.

x: number of members of committee X
y: number of members of committee Y

(1) insufficient
(2) insufficient
(1) and (2)
(1) => sum of ages of all members of Com X is 25.7 * x
sum of ages of all members of Com Y is 29.3 * y
(2) average age of members of Com Z = (sum of ages of all X members + sum of ages of all Y members) / (number of members of committee Z)
= (25.7 * x + 29.3 * y) / (x+y)
= 26.6
So 25.7 * x + 29.3 *y = 26.6 * x + 26.6 * y
27 * x = 9 * y
x = 3y

C for me

Excellent!
Thank you.
Re: Commitee age..........   [#permalink] 29 Apr 2009, 05:58
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