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If xy < zy < 0, is y positive? 1. x < z 2. x is negative

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If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 08 Jan 2016, 08:41
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If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative

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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 08 Jan 2016, 22:48
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2
shasadou wrote:
If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


xy < zy < 0
Both xy and zy are negative.
Either y is negative, x and z both are positive and x > z.
Or y is positive, x and z both are negative and x < z.

Question: Is y positive?

(1) x < z
If x < z, x and z are both negative and y is positive. Sufficient.

(2) x is negative
If x is negative, z is also negative and then y must be positive. Sufficient.

Answer (D)
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If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 08 Jan 2016, 08:50
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shasadou wrote:
If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


Typical inequalities question where a step by step method is a must in order to reach the correct answer.

You are given xy<zy ---> y(x-z)<0 , 2 cases possible here

1. y<0 and x>z
2. y>0 and x<z

The question is asking us whether y >0

Per statement 1, x<z ---> case 2 comes into the picture, providing a definite answer to the question asked. Sufficient.

Per statement 2, x< 0 ---> use the given information that xy<0 --> y must be >0 for xy<0 with x<0. Thus sufficient as well.

Both statements are sufficient ----> D is the correct answer.

Hope this helps.
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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 10 Jan 2016, 18:13
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


When you modify the original condition and the question, it becomes (x-z)y<0에서 y>0? --> x-z<0?. x<Z에서 1) is yes, which is sufficient.
For 2), if x<0, y>0 derived from xy<0, which is yes and sufficient. Therefore, the answer is D.


 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 12 Jun 2016, 14:54
Can you just divide the original equation by Y?
XY < ZY < 0 --> X< Z < 0?

If not, why not?
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If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 12 Jun 2016, 15:21
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glochou wrote:
Can you just divide the original equation by Y?
XY < ZY < 0 --> X< Z < 0?

If not, why not?


Dividing an inequality by a variable whose sign is UNKNOWN is the biggest mistake you can make in inequality questions.

The sign of the inequality changes if you divide or multiply by a negative number. This is the reason why you CAN NOT divide by y.

If xy<zy ---> y*(x-z)<0 --> this means that either

Case 1: y>0 and x-z<0 OR
Case 2: y<0 and x-z>0

When you divide by y WITHOUT reversing the signs, you are inherently assuming that y>0. This is the very question asked.
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If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 23 Jul 2018, 10:44
KarishmaB wrote:
shasadou wrote:
If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


xy < zy < 0
Both xy and zy are negative.
Either y is negative, x and z both are positive and x > z.
Or y is positive, x and z both are negative and x < z.

Question: Is y positive?

(1) x < z
If x < z, x and z are both negative and y is positive. Sufficient.

(2) x is negative
If x is negative, z is also negative and then y must be positive. Sufficient.

Answer (D)


as per the question, it asks y(x-z)<0;
I agree for option 1 is sufficient on the basis that x<z;
for the second option, since x is -ve
x=-5 and z=-4 then y(-5-(-4))<0; hence y is +ve
x=-5 and z=-10 then y(-5-(-10))<0; here y should be -ve

there are conflicting solutions whether y is +ve/-ve

where am I wrong??
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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 23 Jul 2018, 10:54
harish1986 wrote:
KarishmaB wrote:
shasadou wrote:
If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


xy < zy < 0
Both xy and zy are negative.
Either y is negative, x and z both are positive and x > z.
Or y is positive, x and z both are negative and x < z.

Question: Is y positive?

(1) x < z
If x < z, x and z are both negative and y is positive. Sufficient.

(2) x is negative
If x is negative, z is also negative and then y must be positive. Sufficient.

Answer (D)


as per the question, it asks y(x-z)<0;
I agree for option 1 is sufficient on the basis that x<z;
for the second option, since x is -ve
x=-5 and z=-4 then y(-5-(-4))<0; hence y is +ve
x=-5 and z=-10 then y(-5-(-10))<0; here y should be -ve

there are conflicting solutions whether y is +ve/-ve

where am I wrong??


From Statement 2, we don't get any information for z & we are not bothered about z, all we need to figure out is whether y > 0

We know from question prompt that xy < 0, hence if x < 0, then y > 0

I hope that helps


Thanks,
GyM
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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 23 Jul 2018, 10:56
harish1986 wrote:
KarishmaB wrote:
shasadou wrote:
If xy < zy < 0, is y positive?

(1) x < z
(2) x is negative


xy < zy < 0
Both xy and zy are negative.
Either y is negative, x and z both are positive and x > z.
Or y is positive, x and z both are negative and x < z.

Question: Is y positive?

(1) x < z
If x < z, x and z are both negative and y is positive. Sufficient.

(2) x is negative
If x is negative, z is also negative and then y must be positive. Sufficient.

Answer (D)


as per the question, it asks y(x-z)<0;
I agree for option 1 is sufficient on the basis that x<z;
for the second option, since x is -ve
x=-5 and z=-4 then y(-5-(-4))<0; hence y is +ve
x=-5 and z=-10 then y(-5-(-10))<0; here y should be -ve

there are conflicting solutions whether y is +ve/-ve

where am I wrong??


The #'s plugged in are not correct x < z, hence x=-5 and z=-10 is not valid
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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative  [#permalink]

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New post 03 Sep 2018, 03:20
xy<zy<0
1: x<z, then -x>-z, if we subtract it from above inequality we will get y>0
2: because xy<0 and x<0, therefore y>0
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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative &nbs [#permalink] 03 Sep 2018, 03:20
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