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Is xy < 1 ? (1) x > 1 and y < 1 (2) 2x – y > 1

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Is xy < 1 ? (1) x > 1 and y < 1 (2) 2x – y > 1 [#permalink]

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New post 10 Mar 2017, 06:00
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Re: Is xy < 1 ? (1) x > 1 and y < 1 (2) 2x – y > 1 [#permalink]

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Bunuel wrote:
Is xy < 1 ?

(1) x > 1 and y < 1
(2) 2x – y > 1



(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
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Is xy < 1 ? (1) x > 1 and y < 1 (2) 2x – y > 1 [#permalink]

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New post 10 Mar 2017, 07:32
rohit8865 wrote:
Bunuel wrote:
Is xy < 1 ?

(1) x > 1 and y < 1
(2) 2x – y > 1



(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



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.
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Is xy < 1 ? (1) x > 1 and y < 1 (2) 2x – y > 1   [#permalink] 10 Mar 2017, 07:32
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