Quant Review Guide Data Sufficiency Problem #31

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Problem explained here: https://www.youtube.com/watch?v=GcsEQsLt ... C&index=31

Would be pretty simple except couldn't x = 1 thus throwing off the answer that 1. is sufficient?

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by ceilidh.erickson » Wed Feb 06, 2013 8:22 am
If the question is asking "is sqrt(x) an integer?", what it's really asking is - is x a perfect square? If it is a square, we'll get an integer for a root. If not, we won't.

Statement (1) tells us that sqrt(4x) is an integer. In other words, 4x is a perfect square. Since 4 is already a perfect square, that means that x would also have to be a perfect square to get a product that is also a perfect square. For example, (4)(9) = 36. Square * square = square. If x is any non-square integer, then sqrt(4x) would not be an integer. Sufficient.

If, as you suggest, x = 1, that would satisfy the statement. 1 is a perfect square, as is (4)(1). If x = 1, then sqrt(1) is definitely an integer, so this is sufficient.

Statement (2) tells us that sqrt(3x) is not an integer. This means that 3x is not a perfect square. But does that necessarily mean that x is not a perfect square? x could be a non-square, such as 5: sqrt(3*5) = sqrt(15), which is not an integer. This satisfies the statement, and gives us a "no" answer - sqrt(5) is not an integer.

But could x be a square? Try x = 1. sqrt(3*1) = sqrt(3), which is not an integer. This satisfies the statement, but gives us a "yes" answer to the question - sqrt(1) IS an integer. Insufficient.

Choosing x=1 does not change the fact that statement (1) is sufficient and (2) is not. The answer is A.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education