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02 Jun 2016, 11:42
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E
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If d is a positive integer, is \(\sqrt{d}\) greater than 15?
(1) d is divisible by 25.
(2) d is divisible by 40.
(1) d is divisible by 25.
(2) d is divisible by 40.
02 Jun 2016, 11:59
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Bunuel
If d is a positive integer, is \(\sqrt{d}\) greater than 15?
(1) d is divisible by 25.
(2) d is divisible by 40.
(1) d is divisible by 25.
(2) d is divisible by 40.
Given information= d is +ve integer
Question asked= \(\sqrt{d}\) > 15
or d>225
Any statement that gives us value of d will be sufficient to find the solution .
(1) d is divisible by 25.
d has 5*5. It can be 25, 50, and so on. Not sufficient.
(2) d is divisible by 40
d has 5 * 2*2*2 in it. It can be 40 , 80 and so on. not sufficient.
Combining both statements, d has LCM of 5*5*2*2*2= 200. So d can be 200, 400 and so on. Not Sufficient.
IMO, answer is E
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04 Jun 2016, 14:35
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Bunuel
If d is a positive integer, is \(\sqrt{d}\) greater than 15?
(1) d is divisible by 25.
(2) d is divisible by 40.
(1) d is divisible by 25.
(2) d is divisible by 40.
For Statement (1) : d could be 25 or 625 ,\(\sqrt{25}\) or 5<15 but \(\sqrt{625}\) or 25>15,Not Sufficient
For Statement (2) : d could be 40 or 1600,\(\sqrt{40}\) or<7 which is <15 but \(\sqrt{1600}\) or 40>15,Not Sufficient
Combining two Statements together we could find that d could be LCM of 25 and 40 or 200,so \(\sqrt{200}\)<15(since \(\sqrt{225}\)=15) but d could be LCM of 625 and 1600 or 25*25*64 ( also the LCM must be \(\geq{1600}\) because 1600 is the greatest number of the two numbers) too,So \(\sqrt{25*25*64}\) is definitely >15 .So the combination of two statements doesn't work
Correct Answer E
