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vishalrastogi
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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 Updated on: 05 Nov 2020, 06:39
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.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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55% (hard)

Question Stats:

64% (01:47) correct 36% (01:34) wrong based on 390 sessions

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Is x < y ?

(1) 1/x < 1/y

(2) x/y < 0
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C
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Bunuel
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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 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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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 21 Jan 2014, 05:01
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Bunuel
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.
Show more


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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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 21 Jan 2014, 05:10
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unceldolan
Bunuel
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!!
Show more

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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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 21 Jan 2014, 05:19
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Bunuel
unceldolan
Bunuel
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.
Show more

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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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 14 May 2017, 02:20
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Is x < y?

(1) 1/x < 1/y
(2) x/y < 0
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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 14 May 2017, 02:26
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ziyuen
Is x < y?

(1) 1/x < 1/y
(2) x/y < 0
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Merging topics. Please refer to the discussion above.
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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 24 Jun 2021, 13:30
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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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Is x < y ? (1) 1/x < 1/y (2) x/y < 0 22 May 2025, 03:56
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we don't know the whether x & y are -ve or +ve
so from 1
we just can derive
y-x/xy < 0
doesn't mean anything ( maybe x -ve and y = +ve or both are positive and y<x)

2nd :
x/y < 0
either x or Y are negative

C :
Y-X / XY < 0
XY = Negative so y-x has to be positive
hence y-x > 0
y>x
vishalrastogi
Is x < y ?

(1) 1/x < 1/y

(2) x/y < 0
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