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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.
(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.
16 Mar 2012, 01:23
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There is actually no need for variables and formulas here.
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 on its own.
(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years. Insufficient on its own.
(1)+(2) The difference between the averages of Z and X is 26.6-25.7=0.9 and the difference between the averages of Z and Y is 29.3-26.6=2.7. Since the average of Z is closer to the average of X then X must have more members than Y (to draw the overall average closer to its own average). Sufficient.
Notice that we cannot find # of members in X or Y, though based on the difference in averages (0.9 and 2.7) we can deduce that the ratio of members in X and Y is 3:1.
Answer: C.
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 on its own.
(2) The average (arithmetic mean) age of the members of Committee Z will be 26.6 years. Insufficient on its own.
(1)+(2) The difference between the averages of Z and X is 26.6-25.7=0.9 and the difference between the averages of Z and Y is 29.3-26.6=2.7. Since the average of Z is closer to the average of X then X must have more members than Y (to draw the overall average closer to its own average). Sufficient.
Notice that we cannot find # of members in X or Y, though based on the difference in averages (0.9 and 2.7) we can deduce that the ratio of members in X and Y is 3:1.
Answer: C.
01 Jul 2010, 09:39
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OK, i replied to this topic 3 times but failed all, this is the last time! Kudo me if I'm right
Let X be sums of ages of mem of Com X, n be the number of mems of Com X
Let Y be sums of ages of mem of com Y, m be the number of mems of Com Y
X/n=25.7 so X=25.7*n
Y/m=29.3 so Y=29.3*m
(X+Y)/(m+n)= (25.7*n+29.3*m)/(m+n)=26.6
So 25.7*n+29.3*m=26.6*m+26.6*n
=> 2.7m=0.9n
=>3m=n
=>n>m
C suff
Let X be sums of ages of mem of Com X, n be the number of mems of Com X
Let Y be sums of ages of mem of com Y, m be the number of mems of Com Y
X/n=25.7 so X=25.7*n
Y/m=29.3 so Y=29.3*m
(X+Y)/(m+n)= (25.7*n+29.3*m)/(m+n)=26.6
So 25.7*n+29.3*m=26.6*m+26.6*n
=> 2.7m=0.9n
=>3m=n
=>n>m
C suff
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