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If x < y, is x^2 > y^3 ? (1) |x|y < 0 (2) x^3 < y^3

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If x < y, is x^2 > y^3 ? (1) |x|y < 0 (2) x^3 < y^3  [#permalink]

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New post Updated on: 19 Oct 2018, 04:39
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Question Stats:

65% (01:52) correct 35% (02:14) wrong based on 107 sessions

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If x < y, is x^2 > y^3 ?


(1) |x|y < 0

(2) x^3 < y^3

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Originally posted by push12345 on 19 Oct 2018, 04:36.
Last edited by Bunuel on 19 Oct 2018, 04:39, edited 1 time in total.
Renamed the topic.
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Re: If x < y, is x^2 > y^3 ? (1) |x|y < 0 (2) x^3 < y^3  [#permalink]

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New post 19 Oct 2018, 04:45
push12345 wrote:
If x < y, is x^2 > y^3 ?


(1) |x|y < 0

(2) x^3 < y^3


We know x < y

Statement 1) tells us that y is negative. x cannot be positive because it will violate the question stem condition.

From question stem let y = -1 and x = - 2

The answer is yes. Sufficient.

Statement 2) tells us x^3 < y^3

If y = -1 and x = -2 then the answer is yes

If y = 3 and x = 2 then the answe is no.

Insufficient.

Answer choice A

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Re: If x < y, is x^2 > y^3 ? (1) |x|y < 0 (2) x^3 < y^3  [#permalink]

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New post 19 Oct 2018, 05:44
push12345 wrote:
If x < y, is x^2 > y^3 ?


(1) |x|y < 0

(2) x^3 < y^3


Question: If x < y, is x^2 > y^3 ?

Since we know that x < y, so for \(x^2\) to be greater than \(y^3\) the only possibilities are
1) When both x and y are Negative because \(x^2\) will always be positive in that case while \(y^3\) will be be negative
2) when x and y are both in the range 0 to 1 (with constraints \(x^2 > y^3\)

Statement 1: |x|y < 0

i.e. y < 0 hence
SUFFICIENT

Statement 2: x^3 < y^3
Case 1: x = -2 and y = -1 and answer to the question is YES i.e. \(x^2 > y^3\)
Case 2: x = 1/3 and y = 1/2 and answer to the question is NO i.e. \(x^2 < y^3\)
NOT SUFFICIENT

Answer: Option A
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Re: If x < y, is x^2 > y^3 ? (1) |x|y < 0 (2) x^3 < y^3  [#permalink]

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New post 05 Mar 2019, 08:12
push12345 wrote:
If x < y, is x^2 > y^3 ?


(1) |x|y < 0

(2) x^3 < y^3


#1
lxly<0
only possible when y is -ve
sufficient to say that x^2>y^3
since value of x being + or - ve wont make any difference
#2
x^3<y^3
if both x & y are -ve then x^2>y^3 is sufficient
else if both are +ve then not sufficient
IMO A
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Re: If x < y, is x^2 > y^3 ? (1) |x|y < 0 (2) x^3 < y^3   [#permalink] 05 Mar 2019, 08:12
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