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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative
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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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Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative
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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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If xy < zy < 0, is y positive? 1. x < z 2. x is negative
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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
Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative
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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
Re: If xy < zy < 0, is y positive? 1. x < z 2. x is negative
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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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58% (01:29) correct 
