That would be (E).
It's this week challenging question MGMAT....
http://www.manhattangmat.com/blog/index ... -2nd-2012/

P = {12,16,20,24,28,32,36,40,44,48,52}

(1)
From the given restriction on numbers, there are more than one possible way

Q = {12,16,20,24,32,40,44,48,52}

Q = {12,20,24,28,32,40,44,48,52}


(2)

Not Sufficient

Q = {12,16,20,24,32,40,44,48,52}

Q = {20,24,28,32,36,40,44,48,52}


After combining both the options, there are multiple possible combinations, so answer - E

P = {12,16,20,24,32,40,44,48,52}

P = {12,16,24,28,32,40,44,48,52}
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Start with what is apparent and see where it goes:

P = {12, 16, ..., 52} i.e. 4*3 to 4*13 which means there are 13 - 3 + 1 = 11 numbers here. The mean of these numbers is the middle number i.e. the 8th multiple which is 32

Q = {nine of the above given 11 numbers} Basically, we drop 2 numbers from P and retain the rest in Q. The mean of Q will depend on which 2 numbers we dropped. If we drop 2 numbers which are equidistant from the beginning and the end e.g. 8 and 48, the mean stays 32. Else it changes.

(1) Set Q contains at most 4 consecutive multiples of 4.

You can do this in various ways: 4 _ 4 _ 1 (You pick first 4 numbers, drop the next one, pick another 4, drop the next one and pick the last number)
or 4 _ 3 _ 2 (You pick first 4 numbers, drop the next one, pick another 3, drop the next one and pick the last two)
or 3 _ 4 _ 2 or 3 _ 3 _ 3 etc
Not sufficient.

(2) Set Q contains exactly 1 perfect square.
There are 2 perfect squares in these 11 numbers: 16 and 36. You drop one and take one. You do not know which one you drop and you also do not know the second number that you drop hence you do not know the mean of Q.
Not sufficient.

Using both statements together,
Let's say you drop 16 and retain 36. You get 1 _ 4 _ 4 (You will need to drop the 7th number too which is 36 - doesn't work). Is there any other way in which you can drop 16? 1 _ 5 _ 3 etc will not work so basically, you cannot drop 16 and retain 36.
Let's say you drop 36 and retain 16. You can do it in multiple ways: 3_2_4 or 2_3_4 or 4_1_4. In each of these cases, mean will be different so together, the statements are not sufficient.

Answer (E)
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