Manhattan GMAT Challenge Problem of the Week – 9 April 2012
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Question
Set P consists of all the multiples of 4 from 12 to 52, inclusive. Set Q consists of 9 different integers drawn from set P. What is the average (arithmetic mean) of the integers in set Q?
(1) Set Q contains at most 4 consecutive multiples of 4.
(2) Set Q contains exactly 1 perfect square.A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.
Answer
Go ahead and build set P by listing the desired multiples of 4.
12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52.
There are 11 numbers in this list. Set Q consists of 9 different integers drawn from set P.
You are asked for the average (arithmetic mean) of set Q, so you need to know which integers are included. That’s the same as knowing which integers have been left out. So you can rephrase the question to ask which integers in set P are NOT in set Q.
Statement 1: INSUFFICIENT. Set Q may contain at most 4 consecutive multiples of 4, but there are many ways to construct such a set. For instance, leave out 24 and 40, or leave out 28 and 40. These two versions will produce sets with different arithmetic means.
Statement 2: INSUFFICIENT. Set Q may contain exactly 1 perfect square, but again, there are many ways to construct such a set. You have to include either 16 or 36 (not both), but the other integer that you leave out is up to you.
Statements 1 and 2 TOGETHER: INSUFFICIENT. It turns out that you have to include 16 in set Q, because if you leave 16 out, you must include 36, and then no matter which other integer you leave out, you will have a string of 5 or more consecutive multiples of 4. However, once you have omitted 36, you have a few choices for the other integer to omit (20, 24, or 28). These different choices will produce sets with different averages.
The correct answer is E.
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