GMAT paper test DS

[mlane25269]

Having some trouble understading the logic behind this question. OA is C but I picked E.

[vittalgmat]

Not sure of the answer but here is my attempt.

Stmt 1 tells us the avg age of x and y committees
ie. 25.7 and 29.3. cant find the size of x, y or z committees.

So not sufficient.

Stmt 2 tells us the avg age of committee z: 26.6
insufficient.

Lets combine.
Lets assume that both x and y committees have equal number of ppl.
So if we combine such a committe, the avg of new committee should have the avg age = avg of x and y
ie avg age of z should have been (25.7 + 29.3)/2 = 28

But the avg age of z is given to be 26.6. This can only happen if number of ppl in committee x is > number of ppl in y.

Picture this problem as a see-saw with one end having x ppl each aged 25.7 yrs; other end has y ppl, each age is 29.3.
If y > x then see-saw tilts towards y. else it tilts towards x.

Ht Helps

[cramya]

Weighted average.

25.7------ 0.9 ----- 26.6 ------2.7------29.3

We can find the ratio from distance bwteen the combined mean and the individual means

.9 and 2.7

.9/2.7 = 1/3

There are 3 times as many x members as y members which means more x than y

Choose C

[ketkoag]

cramya, could u please elaborate some more on this solution that how .9/2.7 proves that x>y. is the sum of ages of both the committees same?
please explain..
thanks