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10 Mar 2017, 07:00
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (01:58) correct
37%
(01:57)
wrong
based on 193
sessions
History
Date
Time
Result
Not Attempted Yet
10 Mar 2017, 08:00
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Bunuel
(1) clearly insuff
if x=15 , y=0.5 then NO
if x =2 , y = -1 then YES....insuff
(2) 2x-(y+1)>0
if x = 2 , y = 0 then YES
if x= 10 , y = 0.9 then NO...insuff
combining again take above examples as in (2)
insuff
Ans E
10 Mar 2017, 08:32
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rohit8865
I also reached the same answer, but not sure if we're both right. I tried it slightly differently for 2.
1- \(x > 1, y < 1\)
If \(x = 2, y = -1 , xy < 1\)
If \(x = 2, y = 0.99, xy > 1\) -- Assuming decimal value since we don't see defined anywhere x and y have to be integers
-- clearly not sufficient.
2- \(2x - y > 1 --> 2x > y + 1 --> x > y/2 + 1/2\)
Taken alone, for \(y = 2, x > 2/2 + 1/2 --> x > 3/2\) and thus \(xy > 1\).
However, for \(y = -8, x > -8/2 + 1/2 --> x > -7/2\) which could make \(xy < 1\) in case x is a positive number, or \(xy > 1\) for certain negative x values.
-- clearly not sufficient.
Even combined, \(x > y/2 + 1/2\) assuming close to max value of \(y = 0.98, --> x > 0.98/2 + 1/2 --> x > 0.49 + 0.50\) --> \(x > 0.99\) which does not restrict \(x\)'s range from what we already know. \(y\) could be a negative value and it would again expand the range of \(x\) to either negative of positive thus leading to no real confirmation of the question.
