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# If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0

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Intern
Joined: 22 Mar 2018
Posts: 4
If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0  [#permalink]

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27 Mar 2018, 02:30
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15% (low)

Question Stats:

73% (00:47) correct 28% (00:58) wrong based on 119 sessions

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If xy ≠ 0, is x = y?

(1) |x| = |y|
(2) xy > 0

Statement (1) is not sufficient as x could be either >0 or <0 and same for y

Statement (2) is also insufficient as if x>0 then y must be >0 to have xy>0 and if x<0 then y must be <0 to have xy>0
but x could differ from y

However, I do not understand why taking together statement (1) and statement (2) is sufficient as we have two possibilities:
x>0 y>0 , lxl=lyl , then x=y
x<0 y<0, lxl=lyl, then x=y
Math Expert
Joined: 02 Sep 2009
Posts: 58453
Re: If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0  [#permalink]

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27 Mar 2018, 06:06
HIUY wrote:
If xy ≠ 0, is x = y?

(1) |x| = |y|
(2) xy > 0

Statement (1) is not sufficient as x could be either >0 or <0 and same for y

Statement (2) is also insufficient as if x>0 then y must be >0 to have xy>0 and if x<0 then y must be <0 to have xy>0
but x could differ from y

However, I do not understand why taking together statement (1) and statement (2) is sufficient as we have two possibilities:
x>0 y>0 , lxl=lyl , then x=y
x<0 y<0, lxl=lyl, then x=y

If xy ≠ 0, is x = y?

(1) |x| = |y|. This implies that x and y are same distance from 0: either x = y or x = -y. Not sufficient.

(2) xy > 0. This implies that x and y have the same sign. Not sufficient.

(1)+(2) Since from (2) we know that x and y have the same sign and from (1) we know that x and y are same distance from 0, then it must be true that x = y. Sufficient.

HIUY wrote:
If xy ≠ 0, is x = y?

(1) |x| = |y|
(2) xy > 0

Statement (1) is not sufficient as x could be either >0 or <0 and same for y

Statement (2) is also insufficient as if x>0 then y must be >0 to have xy>0 and if x<0 then y must be <0 to have xy>0
but x could differ from y

However, I do not understand why taking together statement (1) and statement (2) is sufficient as we have two possibilities:
x>0 y>0 , lxl=lyl , then x=y
x<0 y<0, lxl=lyl, then x=y

Are both cases you have there the same? The question asks whether x = y and both cases give an YES answer to the question.
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Re: If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0  [#permalink]

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27 Mar 2018, 13:43
HIUY wrote:
If xy ≠ 0, is x = y?

(1) |x| = |y|
(2) xy > 0

Statement (1) is not sufficient as x could be either >0 or <0 and same for y

Statement (2) is also insufficient as if x>0 then y must be >0 to have xy>0 and if x<0 then y must be <0 to have xy>0
but x could differ from y

However, I do not understand why taking together statement (1) and statement (2) is sufficient as we have two possibilities:
x>0 y>0 , lxl=lyl , then x=y
x<0 y<0, lxl=lyl, then x=y

Straight C.

1- We do not know anything about their sign. X can be +5 and Y can be -5. Not sufficient.
2- X and Y can be any number other than 0. This statement does tell us that both x and y have the same sign.

Statement 1 and 2 together are sufficient. Hence C.
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Math Expert
Joined: 02 Sep 2009
Posts: 58453
Re: If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0  [#permalink]

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30 Mar 2018, 08:44
HIUY wrote:
If xy ≠ 0, is x = y?

(1) |x| = |y|
(2) xy > 0

Statement (1) is not sufficient as x could be either >0 or <0 and same for y

Statement (2) is also insufficient as if x>0 then y must be >0 to have xy>0 and if x<0 then y must be <0 to have xy>0
but x could differ from y

However, I do not understand why taking together statement (1) and statement (2) is sufficient as we have two possibilities:
x>0 y>0 , lxl=lyl , then x=y
x<0 y<0, lxl=lyl, then x=y

10. Absolute Value

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Re: If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0  [#permalink]

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30 Mar 2018, 09:02
HIUY wrote:
If xy ≠ 0, is x = y?

(1) |x| = |y|
(2) xy > 0

Statement 1: It tells you only that distance of x and y from 0 on the number line equal. Insufficient

Statement 2: It tells you only that sign of x and y are same. Insufficient

Combining we know that x = y

(C)
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Re: If xy ≠ 0, is x = y? (1) |x| = |y| (2) xy > 0   [#permalink] 30 Mar 2018, 09:02
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