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Is x > y? (1) x > y^3 (2) x/y > 1
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Updated on: 08 Aug 2019, 21:29
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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
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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.



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Re: Is x > y? (1) x > y^3 (2) x/y > 1
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09 Aug 2019, 10:16
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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Re: Is x > y? (1) x > y^3 (2) x/y > 1
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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?



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Re: Is x > y? (1) x > y^3 (2) x/y > 1
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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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Re: Is x > y? (1) x > y^3 (2) x/y > 1
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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 .



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Re: Is x > y? (1) x > y^3 (2) x/y > 1
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09 Aug 2019, 14:06
(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
Answer is (E)
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Re: Is x > y? (1) x > y^3 (2) x/y > 1
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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=> xyy^2>0=>y(xy)> 0, so y> 0 & xy>0 or y<0 & xy <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 case1, that it's insuff. Answer: E




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




