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If x and y are integers, is x less than y ?

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If x and y are integers, is x less than y ? [#permalink]

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New post 06 Feb 2018, 05:24
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A
B
C
D
E

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76% (00:36) correct 24% (00:59) wrong based on 16 sessions

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Re: If x and y are integers, is x less than y ? [#permalink]

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New post 06 Feb 2018, 06:11
IMO the answer should be A
from statement 1 :
x^3 < y^3
basically it says x<y (sign doesn't have the effect) hence sufficient
From statement 2
x^2 < y^2
consider 4<9
hence 2<3 ( yes)
but -2>-3(no)... insufficient
answer : A
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Re: If x and y are integers, is x less than y ? [#permalink]

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New post 06 Feb 2018, 06:21
If x and y are integers, is x less than y ?

(a) The cube of x is less than the cube of y.
- This option works well with both positive and negative values. If X^3 < Y^3 , then X < Y. - Sufficient.

(b) The square of x is less than the square y.
If the x and y are negative values, then square of X is not always less than square of Y. - Not Sufficient.

Ans - Option A
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Re: If x and y are integers, is x less than y ? [#permalink]

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New post 06 Feb 2018, 06:33
Bunuel wrote:
If x and y are integers, is x less than y ?

(1) The cube of x is less than the cube of y.

(2) The square of x is less than the square y.


IMO A

Stmt 1: \(x^3\)<\(y^3\)
=> \(x^3\)-\(y^3\)<0
=> (x-y)(\(x^2\)+y^2\(\)+xy)<0; Now, (\(x^2\)+\(y^2\)+xy) will be >0
Thus, x-y<0 => x<y; Sufficient

Stmt 2: \(x^2\)<\(y^2\)
=> (x+y)(x-y)<0; Now either could be <0; Insufficient
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If x and y are integers, is x less than y ? [#permalink]

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New post 06 Feb 2018, 06:59
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If x and y are integers, is x less than y ?   [#permalink] 06 Feb 2018, 06:59
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