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Committee X and Committee Y, which have no....

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Committee X and Committee Y, which have no....

by factor26 » Sat Dec 17, 2011 1:55 pm
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

i'm a little stuck on this one ... i'm assuming this here is a weighted average question ... but can someone explain how we can find out whehter x or y has a greater number of members?

OG answer is c any help would be much appreciated. Thanks all!
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by adichill » Sat Dec 17, 2011 2:20 pm
factor26 wrote: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

i'm a little stuck on this one ... i'm assuming this here is a weighted average question ... but can someone explain how we can find out whehter x or y has a greater number of members?

OG answer is c any help would be much appreciated. Thanks all!
Avg Funda : Assume a group has 10 members and their average age is 20 years ...now if you move any members with age greater than 20 out of the group..the avg of the group will fall down. On the contrary if you remove any members of age below 20 years .. the avg will rise.
statement 1 ) ---> not sufficent as it does not give us the avg age of Z..
statement 2 ) ---> not sufficent as the avg age of X and Y are not given

combining both statements : iF you see the avg of Z..you will find that the avg is tilted more towards the avg of X...what this means is that the average is contributed by more members from X than Y and hence there are more members from X in the group Z...hence both statements are sufficent to answer the question..

Please correct me if i am wrong...i am also a novice in GMAT..:)
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by neelgandham » Sat Dec 17, 2011 2:23 pm
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.

Insufficient as it doesn't speak about the average (arithmetic mean) age of the members of Committee Z
(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years
Insufficient as it doesn't speak about the average (arithmetic mean) ages of the members of Committee X and Y

Alright let me solve it using variables first.
Let X be the number of members in committee X and Y, the number of members in committee Y
Sum of ages of members in committee X = Average age of the members of Committee X * number of members in committee X = 25.7*X
Sum of ages of members in committee Y = Average age of the members of Committee Y * number of members in committee Y = 29.3*Y
Sum of ages of members in committee Z = Average age of the members of Committee Z * number of members in committee Z = 26.6*(X+Y)
Sum of ages of members in committee Z = Sum of ages of members in committee X + Sum of ages of members in committee Y
Implies 26.6X + 26.6Y = 25.7*X + 29.3*Y
Implies 0.9X = 2.7Y, Implies X>Y(X=3Y). Hence Sufficient!

You can also use mixture and allegations concepts as shown in the attachment.

GMAT way: The value of average age of the members of Z is closer to the value of the average age of the members of X, which implies that the members of x are greater than y in number. Hence C
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by factor26 » Sun Dec 18, 2011 10:55 am
@ neel,

Thanks for your help ... i also solved using the GMAT way but was just a little unsure about the relationship between age and the # of individuals in each group ... well explained...Cheers!
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