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Is xy > 0 (1) x + y > 0 (2) x + 3y < 0

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Rakesh1987
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Is xy > 0 (1) x + y > 0 (2) x + 3y < 0 [#permalink] New post   21 May 2019, 03:14
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Is xy > 0

(1) x + y > 0
(2) x + 3y < 0

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C
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Bunuel
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Re: Is xy > 0 (1) x + y > 0 (2) x + 3y < 0 [#permalink] New post   21 May 2019, 03:34
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Is xy > 0

The question asks whether x and y have the same sign.

(1) x + y > 0. The sum of two numbers is greater than 0. From this it's impossible to say whether they have the same sign (for example: 1 + 1 > 0 and 2 + (-1) > 0). Not sufficient.

(2) x + 3y < 0. The same here. Not sufficient.


(1)+(2) Subtract (1) from (2) (we can safely do that since the signs are in opposite direction): x + 3y - (x + y) < 0. This gives: 2y < 0. So, y is negative. Since y is negative, then from (1) x must be positive (if x were negative too then the sum of two negative numbers would be negative not positive as stated in (1)). x > 0 and y < 0 means that xy < 0. Sufficient.

Answer: C.

Hope it helps.
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Re: Is xy > 0 (1) x + y > 0 (2) x + 3y < 0 [#permalink] New post   21 May 2019, 03:28
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Rakesh1987 wrote:
Is xy > 0

(1) x + y > 0
(2) x + 3y < 0

Source: Expert's Global


Very good question.

Statement 1.
X+Y>0
X Y [Taking random Values]
2 3
3 -2. Hence insufficient.


Statement 2.
x + 3y<0
X Y [Taking random Values]
-2 -3
2 -3. Hence insufficient.

Combining Both 1+2

x+3y < 0
(x+y) + (2y) <0; The RED portion is +ve
then 2y must be -ve, so y is -ve. Confirmed.

Now

x+y >0 and y is -ve; then x has to be a positive value more than y. So product of x and y is -ve. C
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Re: Is xy > 0 (1) x + y > 0 (2) x + 3y < 0 [#permalink] New post   08 Mar 2021, 07:02
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rudywip wrote:
Bunuel wrote:
Is xy > 0

The question asks whether x and y have the same sign.

(1) x + y > 0. The sum of two numbers is greater than 0. From this it's impossible to say whether they have the same sign (for example: 1 + 1 > 0 and 2 + (-1) > 0). Not sufficient.

(2) x + 3y < 0. The same here. Not sufficient.


(1)+(2) Subtract (1) from (2) (we can safely do that since the signs are in opposite direction): x + 3y - (x + y) < 0. This gives: 2y < 0. So, y is negative. Since y is negative, then from (1) x must be positive (if x were negative too then the sum of two negative numbers would be negative not positive as stated in (1)). x > 0 and y < 0 means that xy < 0. Sufficient.

Answer: C.

Hope it helps.


Hey Bunuel

I am facing a minor logical issue here. What if we subtract (2) from (1)? IMO the result will be x + y - ( x + 3y ) > 0 --> -2y > 0 --> y > 0, but it should be the other way around according to your calculations. Did I apply the rule of subtracting inequalities in a wrong way?

Please, help out.

Cheers
Rudolf


-2y > 0 means that y < 0. Divide by -2 and flip the sign or add 2y to both sides to get 0 > 2y --> y < 0.
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Re: Is xy > 0 (1) x + y > 0 (2) x + 3y < 0 [#permalink] New post   08 Mar 2021, 06:57
Bunuel wrote:
Is xy > 0

The question asks whether x and y have the same sign.

(1) x + y > 0. The sum of two numbers is greater than 0. From this it's impossible to say whether they have the same sign (for example: 1 + 1 > 0 and 2 + (-1) > 0). Not sufficient.

(2) x + 3y < 0. The same here. Not sufficient.


(1)+(2) Subtract (1) from (2) (we can safely do that since the signs are in opposite direction): x + 3y - (x + y) < 0. This gives: 2y < 0. So, y is negative. Since y is negative, then from (1) x must be positive (if x were negative too then the sum of two negative numbers would be negative not positive as stated in (1)). x > 0 and y < 0 means that xy < 0. Sufficient.

Answer: C.

Hope it helps.


Hey Bunuel

I am facing a minor logical issue here. What if we subtract (2) from (1)? IMO the result will be x + y - ( x + 3y ) > 0 --> -2y > 0 --> y > 0, but it should be the other way around according to your calculations. Did I apply the rule of subtracting inequalities in a wrong way?

Please, help out.

Cheers
Rudolf
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Re: Is xy > 0 (1) x + y > 0 (2) x + 3y < 0 [#permalink]
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