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Intern
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Joined: 16 Mar 2017
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Kudos [?]: 9 [0], given: 15

x/y > 0? [#permalink]

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New post 20 Sep 2017, 05:28
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Difficulty:

  55% (hard)

Question Stats:

61% (00:56) correct 39% (01:14) wrong based on 31 sessions

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x/y > 0?

1) x/(x+y) > 0

2) y/(x+y) > 0
[Reveal] Spoiler: OA

Kudos [?]: 9 [0], given: 15

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Senior Manager
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Joined: 02 Jul 2017
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Re: x/y > 0? [#permalink]

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New post 20 Sep 2017, 05:46
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is \(\frac{x}{y}\)> 0?

=> Above equation to be true both x and y should hv same sign => either x and y both +ve or both -ve

1) \(\frac{x}{x+y} > 0\)
=> x and (x+y) both have same sign . both +ve or -ve.
But here we cannot find relation between x and y
eg for below values \(\frac{x}{x+y} > 0\) is True
x= 10 and y =-5 => \(\frac{x}{y} > 0\) False

x= 5 and y =5 => \(\frac{x}{y} > 0\) True
Insufficient


2) \(\frac{y}{x+y} > 0\)
=> Like above equation :
=> y and (x+y) both have same sign . both +ve or -ve.
But here we cannot find relation between x and y
Insufficient

(1)+(2)
=> By combining both equations we get : x , y and (x+y) all three have same sign, +ve or -ve.
=> x and y have same sign
=> \(\frac{x}{y}\)> 0?
Sufficient

Answer: C

Kudos [?]: 59 [1], given: 60

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Senior Manager
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Joined: 25 Feb 2013
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x/y > 0? [#permalink]

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New post 20 Sep 2017, 05:52
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petrified17 wrote:
x/y > 0?

1) x/(x+y) > 0

2) y/(x+y) > 0


Statement 1: the inequality is positive but nothing can be inferred about \(x\) & \(y\) individually, they may both be positive or they may both be negative or one may be positive and the other may be negative. Hence Insufficient

Statement 2: Same scenario as Statement 1. Hence Insufficient

Combining 1 & 2: we know LHS of both the inequalities are positive and RHS is \(0\), so we can divide the inequalities with each other to get

\(\frac{x}{(x+y)}*\frac{(x+y)}{y}>0\) or \(\frac{x}{y}>0\). Sufficient

Option C

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Re: x/y > 0? [#permalink]

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New post 20 Sep 2017, 06:58
We need to know whether x is not 0 and whether x and y have same sign .. both conditions are needed.

From one and two each alone.
We can't deduct that both conditions are a must.

Both together are sufficient



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Intern
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Joined: 19 Sep 2016
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Re: x/y > 0? [#permalink]

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New post 20 Sep 2017, 07:28
X/y>0, is possible when both x and y are either positive or both are negative.
Statement 1 - x/(x+y)>0, simplifying we get x*(x+y)>0, which means that x>0 and x+y>0 or x<0 and x+y<0 which means that x/y can be both > and < 0 , hence insufficient.
Statement 2 - y/(x+y)>0, or y*(x+y)>0, which again means that either y and x+y>0 or y and x+y <0 so again insufficient.
Statement 1+2 - clearly states that either x & y both together >0 or both together <0, either ways x/y is >0

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Re: x/y > 0?   [#permalink] 20 Sep 2017, 07:28
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