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20 Jun 2019, 03:53
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:37) correct
30%
(01:38)
wrong
based on 291
sessions
History
Date
Time
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Not Attempted Yet
20 Jun 2019, 10:10
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Pls explain answer...how both r sufficient together? And what is concept for this question to answer
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Posted from my mobile device
20 Jun 2019, 10:23
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prashanths
Interesting problem! Here's how I approached it:
First, I noticed that both statements together would definitely allow me to answer the question. That's because both statements together will give you the values of m and n, and that would let you calculate the exact answer, if you really wanted to. Here's what that made me think: uh-oh, this might be a C trap. I'm going to try very carefully to prove that each statement alone is insufficient before I assume the answer is C.
Then, I rewrote the question to look a little clearer:
Is \(m^\frac{1}{2n}\) an integer?
Statement 1: Let's start writing out cases and see if we can find some integers and non-integers.
First case: n = 1, m = 15. Is \(15^\frac{1}{2}\) an integer? NO.
Second case: n = 2, m = 16. Is \(16^\frac{1}{4}\) an integer? YES, and its value is 2.
The answer can be either YES or NO, so this statement is insufficient. Eliminate A and D.
Statement 2: Let's write out some more cases.
First case: If n = 1, m = 5(1) + 6 = 11. Is Is \(11^\frac{1}{2}\) an integer? NO.
Second case: n = 2, m = 16. Is \(16^\frac{1}{4}\) an integer? YES, and its value is 2.
The answer can be either YES or NO, so this statement is insufficient. Eliminate B.
Statements together: We can solve and find that m = 16, n = 2, and the answer is definitely YES. That's why the answer is C.
