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X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y

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X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post 13 Jan 2019, 22:13
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

53% (00:58) correct 48% (01:02) wrong based on 40 sessions

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X and Y are real numbers. Is X*Y negative?

(1) (X+Y)^2 > 0

(2) X = Y
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X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post Updated on: 14 Jan 2019, 10:07
amanvermagmat wrote:
X and Y are real numbers. Is X*Y negative?

(1) (X+Y)^2 > 0

(2) X = Y


IMO B

(1) (X+Y)^2 > 0
X & Y can take any value, X*Y can be -ive or not

(2)X=Y
This will always be > 0, for +ive and -ive values, and = 0 for x=y=0

Edit -> *Correcting after Bunnel's input*

If x and y are equal, x^2 will never be less than 0

Answer B
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Originally posted by KanishkM on 13 Jan 2019, 22:27.
Last edited by KanishkM on 14 Jan 2019, 10:07, edited 2 times in total.
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Re: X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post 14 Jan 2019, 01:39
KanishkM wrote:
amanvermagmat wrote:
X and Y are real numbers. Is X*Y negative?

(1) (X+Y)^2 > 0

(2) X = Y


IMO B

(1) (X+Y)^2 > 0
X & Y can take any value, X*Y can be -ive or not

(2)X=Y
This will always be > 0, for +ive and -ive values.

Answer B



x and y both can be 0. it is stated that x and y are real numbers. real number includes 0 too. in this case, xy yields 0, which is nither positive not negative.
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Re: X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post 14 Jan 2019, 04:08
selim wrote:
KanishkM wrote:
amanvermagmat wrote:
X and Y are real numbers. Is X*Y negative?

(1) (X+Y)^2 > 0

(2) X = Y


IMO B

(1) (X+Y)^2 > 0
X & Y can take any value, X*Y can be -ive or not

(2)X=Y
This will always be > 0, for +ive and -ive values.

Answer B



x and y both can be 0. it is stated that x and y are real numbers. real number includes 0 too. in this case, xy yields 0, which is nither positive not negative.


You are right, i missed out that.

Then the answer becomes E.
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Re: X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post 14 Jan 2019, 04:25
2
KanishkM wrote:
selim wrote:
KanishkM wrote:
X and Y are real numbers. Is X*Y negative?

(1) (X+Y)^2 > 0

(2) X = Y

IMO B

(1) (X+Y)^2 > 0
X & Y can take any value, X*Y can be -ive or not

(2)X=Y
This will always be > 0, for +ive and -ive values.

Answer B



x and y both can be 0. it is stated that x and y are real numbers. real number includes 0 too. in this case, xy yields 0, which is nither positive not negative.


You are right, i missed out that.

Then the answer becomes E.


No, the answer is still B. The question asks whether xy < 0. From (2) the question becomes: is x^2 < 0? The answer to this question is always NO, because the square of a number cannot be negative.
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Re: X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post 14 Jan 2019, 05:10
X and Y are real numbers. Is X*Y negative?

(1) \((X+Y)^2\) > 0

Let X = -1 and Y = 2, then asnwer is YES

Let X = -2 and Y = -3, then answer is NO

INSUFFICIENT

(2) X = Y

If X is positive then Y is positive, If X is negative then Y is also negative.

In either case, X*Y is positive and answer to the question is NO.

SUFFICIENT

OPTION: B
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Re: X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y  [#permalink]

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New post 04 Jun 2019, 03:44
amanvermagmat wrote:
X and Y are real numbers. Is X*Y negative?

(1) (X+Y)^2 > 0

(2) X = Y


again a wrong answer ,the OA MUST BE E
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Re: X and Y are real numbers. Is X*Y negative? (1) (X+Y)^2 > 0 (2) X = Y   [#permalink] 04 Jun 2019, 03:44
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