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15 May 2018, 03:50
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
86% (00:52) correct
14%
(01:23)
wrong
based on 290
sessions
History
Date
Time
Result
Not Attempted Yet
If m is a positive integer, is \(\sqrt{m}\) an integer?
(1) \(\sqrt{49m}\) is an integer.
(2) \(\sqrt{7m}\) is not an integer.
(1) \(\sqrt{49m}\) is an integer.
(2) \(\sqrt{7m}\) is not an integer.
15 May 2018, 04:26
Kudos
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Statement 1: -
\(\sqrt{49m}\) is an integer
Therefore, \(m = integer^2\)
hence, \(\sqrt{m} = \sqrt{integer^2}\)
\(= integer\)
SUFFICIENT
Statement 2: -
\(\sqrt{7m}\) is not an integer
if \(m = 2\) , then \(\sqrt{7m}\) is not an integer and \(\sqrt{2}\) is also not an integer
if \(m = 16\), then \(\sqrt{7m}\) is not an integer but \(\sqrt{16}\) is an integer integer
INSUFFICIENT
Hence answer is A
\(\sqrt{49m}\) is an integer
Therefore, \(m = integer^2\)
hence, \(\sqrt{m} = \sqrt{integer^2}\)
\(= integer\)
SUFFICIENT
Statement 2: -
\(\sqrt{7m}\) is not an integer
if \(m = 2\) , then \(\sqrt{7m}\) is not an integer and \(\sqrt{2}\) is also not an integer
if \(m = 16\), then \(\sqrt{7m}\) is not an integer but \(\sqrt{16}\) is an integer integer
INSUFFICIENT
Hence answer is A
15 May 2018, 04:26
Kudos
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Bunuel
m is a positive integer & if \(\sqrt{m}\) also has to be an integer , then m is a perfect square number.
Lets assume \(m = a^x\), Thus x has to be even.
(1) \(\sqrt{49m}\) is an integer
now, \(\sqrt{49m}\)=\(\sqrt{7^2*a^x}\)
Thus \(\sqrt{7^2*a^x}\) is an integer .
Now as \(\sqrt{7^2}\) is an integer (+7 or -7), \(\sqrt{a^x}\) has to be an integer ( int*non-int= non-int)
Thus m is an integer..........................................................Sufficient.
(2) \(\sqrt{7m}\) is not an integer.
The expression to be integer m= 7^y*a^x when y is +ve odd & x is +ve odd& 0.
All other cases it would be non integer.................................................................Not sufficient.
Hence I would go for option A.
