Bunuel wrote:
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.
Can't they have same number of members?
For example, say X has two members of ages: 20 and 66. (Average= 43)
Similarly, say Y also has two members of ages: 20 and 20. (Average= 20)
Committee Z: 20+66+20+20/4= 32 (Average of Z)
(All of these twenty year olds are three different individuals, as the two set don't share any member.)
Now, as Bunuel did: 43-32=11; 32-20=12. So the two groups, X and Y, would be in the ratio 12:11.
As we can see, if we go by the logic above, then X should have more members than Y as the ratio suggests that. But in reality, X and Y have two members each.
What am I missing?