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# Is x > y? (1) x > y^3 (2) x/y > 1

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Manager
Joined: 01 Nov 2017
Posts: 64
Location: India
Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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Updated on: 08 Aug 2019, 21:29
1
5
00:00

Difficulty:

75% (hard)

Question Stats:

48% (01:53) correct 52% (01:24) wrong based on 91 sessions

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Is x > y?

(1) x > y^3

(2) x/y > 1

Originally posted by raghavrf on 08 Aug 2019, 20:39.
Last edited by Bunuel on 08 Aug 2019, 21:29, edited 2 times in total.
Renamed the topic and edited the question.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 2111
Re: Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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09 Aug 2019, 10:16
1
When x and y are positive, we can easily get a 'yes' answer to the question (if x = 4 and y = 1, say). But x and y can be negative, and then we can get a 'no' answer to the question. If we have x = -4, and y = -2, then x/y > 1, and x > y^3, but x is not greater than y. So the answer is E.
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Intern
Joined: 01 Dec 2018
Posts: 42
Re: Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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09 Aug 2019, 10:25
IanStewart wrote:
When x and y are positive, we can easily get a 'yes' answer to the question (if x = 4 and y = 1, say). But x and y can be negative, and then we can get a 'no' answer to the question. If we have x = -4, and y = -2, then x/y > 1, and x > y^3, but x is not greater than y. So the answer is E.

Hi! Why is B wrong? If X/Y > 1 it should mean that x > y ?If x/y > 0 then we could have said that there can be values between 0 and 1?
GMAT Tutor
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Re: Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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09 Aug 2019, 10:52
kunalbean wrote:
Hi! Why is B wrong? If X/Y > 1 it should mean that x > y ?If x/y > 0 then we could have said that there can be values between 0 and 1?

x/y > 1 does not mean x > y. I gather you're multiplying by y on both sides of the inequality. But you can't do that unless you know whether y is positive or negative. If y is positive, then we can multiply by y and we do not need to flip the inequality. Then it is true that x > y. But if y is negative, then when you multiply by y on both sides, you need to flip the inequality. So if x/y > 1 is true, and y < 0 is true, then x < y is true. You can confirm that by testing any negative numbers at all -- let x = -3 and y = -1, for example. Then x/y > 1 is true, and x < y is true.
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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Intern
Joined: 01 Dec 2018
Posts: 42
Re: Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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09 Aug 2019, 12:14
IanStewart wrote:
kunalbean wrote:
Hi! Why is B wrong? If X/Y > 1 it should mean that x > y ?If x/y > 0 then we could have said that there can be values between 0 and 1?

x/y > 1 does not mean x > y. I gather you're multiplying by y on both sides of the inequality. But you can't do that unless you know whether y is positive or negative. If y is positive, then we can multiply by y and we do not need to flip the inequality. Then it is true that x > y. But if y is negative, then when you multiply by y on both sides, you need to flip the inequality. So if x/y > 1 is true, and y < 0 is true, then x < y is true. You can confirm that by testing any negative numbers at all -- let x = -3 and y = -1, for example. Then x/y > 1 is true, and x < y is true.

Yes. didnt think of 1 as a possibility .
Director
Joined: 16 Jan 2019
Posts: 612
Location: India
Concentration: General Management
WE: Sales (Other)
Re: Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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09 Aug 2019, 14:06
1
(1) x > y^3

If x=-1, y=-2

x>y and x>y^3

If x=-3, y=-2

x<y and x>y^3

1 is insufficient

(2) x/y > 1

If x=-2 and y=-1

x<y and x/y>1

If x=2 and y=1

x>y and x/y>1

2 is insufficient

(1)+(2)

If x=-3, y=-2

x<y and x>y^3 and x/y>1

If x=2, y=1

x>y and x>y^3 and x/y>1

(1)+(2) is insufficient

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Director
Joined: 12 Dec 2015
Posts: 500
Re: Is x > y? (1) x > y^3 (2) x/y > 1  [#permalink]

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06 Mar 2020, 07:06
Is x > y?

(1) x > y^3 --> insuff: if x = 10 & y =2, then x >y^3 & x> y: yes. But if x = 0.1 & y =0.2, then x >y^3 & x< y: no

(2) x/y > 1--> insuff: x/y > 1 => xy/y^2>1 => xy > y^2=> xy-y^2>0=>y(x-y)> 0, so y> 0 & x-y>0 or y<0 & x-y <0 => 0<y<x i.e if y> 0: yes, but x<y<0 i.e if y< 0: no

combining (1) & (2) also, we can say if y> 0 also, we can say from case-1, that it's insuff.

Re: Is x > y? (1) x > y^3 (2) x/y > 1   [#permalink] 06 Mar 2020, 07:06

# Is x > y? (1) x > y^3 (2) x/y > 1

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