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Is x < y ? (1) 1/x < 1/y (2) x/y < 0

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vishalrastogi
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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   Updated on: 05 Nov 2020, 06:39
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66% (01:46) correct 34% (01:36) wrong based on 323 sessions
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Is x < y ?

(1) 1/x < 1/y

(2) x/y < 0
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C

Originally posted by vishalrastogi on 18 Nov 2013, 03:45.
Last edited by Bunuel on 05 Nov 2020, 06:39, edited 3 times in total.
Edited the question.
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   18 Nov 2013, 04:00
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Is x < y ?

(1) 1/x < 1/y. If \(x=2\) and \(y=1\), then the answer is NO but if \(x=-2\) and \(y=1\), then the answer is YES. Not sufficient.
Note that here we cannot cross-multiply and write: y<x, since we don't know the signs of x and y, thus don't know whether we should flip the sign of the inequality.

(2) x/y < 0. This implies that x and y have the opposite signs. Not sufficient.

(1)+(2) From (1): \(\frac{1}{y} - \frac{1}{x}>0\) --> \(\frac{x-y}{xy}>0\) --> \(x-y\) and \(xy\) have the same sign. From (2) we can get that \(xy<0\), thus \(x-y<0\) --> \(x<y\). Sufficient.

Answer: C.

Hope it's clear.
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   21 Jan 2014, 05:01
Bunuel wrote:
Is x < y ?

(1) 1/x < 1/y. If \(x=2\) and \(y=1\), then the answer is NO but if \(x=-2\) and \(y=1\), then the answer is YES. Not sufficient.
Note that here we cannot cross-multiply and write: y<x, since we don't know the signs of x and y, thus don't know whether we should flip the sign of the inequality.

(2) x/y < 0. This implies that x and y have the opposite signs. Not sufficient.

(1)+(2) From (1): \(\frac{1}{y} - \frac{1}{x}>0\) --> \(\frac{x-y}{xy}>0\) --> \(x-y\) and \(xy\) have the same sign. From (2) we can get that \(xy<0\), thus \(x-y<0\) --> \(x<y\). Sufficient.

Answer: C.

Hope it's clear.



Hey Bunuel,

I did just as you, but chose E because I thought both together wouldn't be sufficient. I also don't understand your approach/explanation of why C is correct. Could you write it down more detailed? I would appreciate it!

Thanks!!
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   21 Jan 2014, 05:10
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unceldolan wrote:
Bunuel wrote:
Is x < y ?

(1) 1/x < 1/y. If \(x=2\) and \(y=1\), then the answer is NO but if \(x=-2\) and \(y=1\), then the answer is YES. Not sufficient.
Note that here we cannot cross-multiply and write: y<x, since we don't know the signs of x and y, thus don't know whether we should flip the sign of the inequality.

(2) x/y < 0. This implies that x and y have the opposite signs. Not sufficient.

(1)+(2) From (1): \(\frac{1}{y} - \frac{1}{x}>0\) --> \(\frac{x-y}{xy}>0\) --> \(x-y\) and \(xy\) have the same sign. From (2) we can get that \(xy<0\), thus \(x-y<0\) --> \(x<y\). Sufficient.

Answer: C.

Hope it's clear.



Hey Bunuel,

I did just as you, but chose E because I thought both together wouldn't be sufficient. I also don't understand your approach/explanation of why C is correct. Could you write it down more detailed? I would appreciate it!

Thanks!!


From (1): \(\frac{1}{y} - \frac{1}{x}>0\) --> \(\frac{x-y}{xy}>0\). Now, in order this to be true the numerator and and the denominator must have the same sign: (positive)/(positive)>0 and (negative)/(negative)>0.

From (2): \(\frac{x}{y} < 0\) implies that x and y have the opposite signs, thus \(xy<0\).

Finally, since \(xy<0\) (the denominator of \(\frac{x-y}{xy}\) is negative), then the numerator must also be negative: \(x-y<0\) --> \(x<y\).

Hope it's clear.
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   21 Jan 2014, 05:19
Bunuel wrote:
unceldolan wrote:
Bunuel wrote:
Is x < y ?

(1) 1/x < 1/y. If \(x=2\) and \(y=1\), then the answer is NO but if \(x=-2\) and \(y=1\), then the answer is YES. Not sufficient.
Note that here we cannot cross-multiply and write: y<x, since we don't know the signs of x and y, thus don't know whether we should flip the sign of the inequality.

(2) x/y < 0. This implies that x and y have the opposite signs. Not sufficient.

(1)+(2) From (1): \(\frac{1}{y} - \frac{1}{x}>0\) --> \(\frac{x-y}{xy}>0\) --> \(x-y\) and \(xy\) have the same sign. From (2) we can get that \(xy<0\), thus \(x-y<0\) --> \(x<y\). Sufficient.

Answer: C.

Hope it's clear.



Hey Bunuel,

I did just as you, but chose E because I thought both together wouldn't be sufficient. I also don't understand your approach/explanation of why C is correct. Could you write it down more detailed? I would appreciate it!

Thanks!!


From (1): \(\frac{1}{y} - \frac{1}{x}>0\) --> \(\frac{x-y}{xy}>0\). Now, in order this to be true the numerator and and the denominator must have the same sign: (positive)/(positive)>0 and (negative)/(negative)>0.

From (2): \(\frac{x}{y} < 0\) implies that x and y have the opposite signs, thus \(xy<0\).

Finally, since \(xy<0\) (the denominator of \(\frac{x-y}{xy}\) is negative), then the numerator must also be negative: \(x-y<0\) --> \(x<y\).

Hope it's clear.


Ok now I get it! I solved statement 1 without alternating the inequality. This makes perfect sense now. Hope I remember it the next time! Thanks :)
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   14 May 2017, 02:20
Is x < y?

(1) 1/x < 1/y
(2) x/y < 0
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   14 May 2017, 02:26
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ziyuen wrote:
Is x < y?

(1) 1/x < 1/y
(2) x/y < 0


Merging topics. Please refer to the discussion above.
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   24 Jun 2021, 13:30
Here is my 30 sec approach! Please correct if I lag in reasoning!

Statement 1 : Clearly insufficient.. if x=-1, y =2, answer is yes, if x=5 , y=4 answer is no

Statement 2 : Again insufficient .. X can be < or > y depending on signs

Statement 1 + 2 : We can infer that x and y have negative signs. Also from st. 1 1/x<1/y which is only possible when x<0 and Y>0.. this implies that X<Y .. Sufficient

Ans : C
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink] New post   04 May 2023, 05:02
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Re: Is x < y ? (1) 1/x < 1/y (2) x/y < 0 [#permalink]
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