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# If x and y are non-zero integers. Is

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Retired Moderator
Joined: 27 Oct 2017
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If x and y are non-zero integers. Is  [#permalink]

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14 Mar 2020, 18:46
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GMATBusters’ Quant Quiz Question -10

If x and y are non-zero integers. Is $$\frac{(x+y)}{(x-y)}$$>0?
1) x > y
2) |x| > |y|

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Joined: 22 Feb 2018
Posts: 764
Re: If x and y are non-zero integers. Is  [#permalink]

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14 Mar 2020, 18:57
1
1
If x and y are non-zero integers. Is (x+y)/(x−y)>0?

The question basically asks are both x+y and x-y positive or negative?

(1) x > y
x-y> 0 => x-y is positive
says nothing about x+y which can either be positive or negative
Insufficient

(2) |x| > |y|
√(x^2 ) > . √(y^2 )
x^2-y^2 > 0
(x+y)(x-y) > 0
so both x+y and x-y can be positive or both x+y and x-y can be negative
This means that x+y /x-y will always be positive and greater than 0
Sufficient

Imo. B
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Re: If x and y are non-zero integers. Is  [#permalink]

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14 Mar 2020, 20:46
Given: X and Y are Non-Zero Integers

We know that X and Y are Integers. X and Y is not equal to 0.

We do not know X and Y is either Positive / Negative ir Even / Odd.

1) X > Y

Both x and y positives - Lets consider x=3, y= 2 , (x+y)/(x-y) > 0
Both x and y negatives - lets consider x=-2, y=-3 , (x+y)/(x-y) < 0

We get 2 different answers, so option 1 is not sufficient.

2) |x| > |y|

Both x and y positives , lets consider x=2,y=1 , (x+y)/(x-y) > 0
Both x and y negatives, lets consider x=-3 , y=-2, (x+y)/(x-y) > 0

In both the choices we get (x+y) / (x-y) > 0

Option B alone is sufficient.

Ans: B
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Re: If x and y are non-zero integers. Is  [#permalink]

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14 Mar 2020, 21:42
OA is A.
Statement one is alone sufficient.
Statement 1: (x+y)/(x-y)>0
X>Y (given)
Let X = 2 and Y=1
2+1/2-1=3
Which is greater than zero. (Sufficient)
Statement 2: |x|>|y|
Case 1: x>-y
Case 2: x>y
Case 3: -x>-y
Case 4: -x>y (not possible)
If we put X=2 and Y=1 in above cases in the given solution. It's not confirmed that the answer will always be greater than zero.(insufficient)
So Statement A is sufficient alone.
OA is A

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If x and y are non-zero integers. Is  [#permalink]

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Updated on: 15 Mar 2020, 21:10
If x and y are non-zero integers. $$Is \frac{(x+y)}{(x−y)}>0?$$

1) x > y
x-y>0
If y>0; $$\frac{(x+y)}{(x−y)}>0$$
But if y<0; $$\frac{(x+y)}{(x−y)} may not be>0$$
NOT SUFFICIENT

2) |x| > |y|
If x = 3; y = -2 ; $$\frac{(x+y)}{(x−y)}= 1/5>0$$
But If x = -3; y = -2 ; $$\frac{(x+y)}{(x−y)}= (-5/-1)>0$$
We see that if |x|>|y| then signs of (x+y) & (x-y) are determined by signs of x and are either both +ve or both -ve
SUFFICIENT

IMO B
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com

Originally posted by Kinshook on 15 Mar 2020, 02:25.
Last edited by Kinshook on 15 Mar 2020, 21:10, edited 1 time in total.
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Posts: 75
Re: If x and y are non-zero integers. Is  [#permalink]

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15 Mar 2020, 05:02
Statement:1
x>y
3>2,x+y/x-y=5/1>0,yes
-2>-3,x+y/x-y=-5<0,No
Not sufficient
Statement 2
IxI >IyI,then x+y/x-y>0 for x=3,y=2 and x=-3,y=-2.Also for x=-3,y=2
Hence x+y/x-y>0 for all input
Sufficient
Manager
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Posts: 58
Re: If x and y are non-zero integers. Is  [#permalink]

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15 Mar 2020, 05:35
Statement 1: x>y
If both x & y are positive, say x= 8, y = 3
$$\frac{x+y}{x-y}$$ > 0

If both x and y are negative, say x = -2, y = -5
$$\frac{x+y}{x-y}$$
$$\frac{-7}{3}$$ < 0

Insufficient

Statement 2: |x| > |y|
$$x^2$$ > $$y^2$$
$$x^2 - y^2$$ > 0
(x+y)(x-y) > 0
x<-y or x>y
x+y<0 or x-y>0

case 1: x+y<0
x=-2, y=-3: $$\frac{x+y}{x-y}$$ = $$\frac{-5}{1}$$ < 0
x=-4, y=-1 : $$\frac{x+y}{x-y}$$ = $$\frac{-5}{-3}$$ > 0

case 2: x-y>0
x=5, y=2: $$\frac{x+y}{x-y}$$ = $$\frac{7}{3}$$ > 0
x=-5, y=-7: $$\frac{x+y}{x-y}$$ = $$\frac{-12}{2}$$ < 0

Insufficient

Statement 1 & 2: Since x>y,
x=-2, y=-3: $$\frac{x+y}{x-y}$$ = $$\frac{-5}{1}$$ < 0
x=5, y=2: $$\frac{x+y}{x-y}$$ = $$\frac{7}{3}$$ > 0
Insufficient

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Re: If x and y are non-zero integers. Is  [#permalink]

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15 Mar 2020, 06:14
statement 1: x > y
case A: both x & y are negative
$$\frac{x+y}{x-y}<0$$; y is higher magnitude negative value.

case B: both x & y are positive

$$\frac{x+y}{x-y}>0$$

not sufficient

statement 2: |x| > |y|
this tells us that magnitude of x is greater than magnitude of y.

case A: both x & y are negative

$$\frac{x+y}{x-y}>0$$; x is higher magnitude negative value.

case B: both x & y are positive

$$\frac{x+y}{x-y}>0$$

case C: x is negative and y is positive

$$\frac{x+y}{x-y}>0$$

sufficient

Ans: B
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GMAT 1: 600 Q46 V27
Re: If x and y are non-zero integers. Is  [#permalink]

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15 Mar 2020, 06:24
Quote:
If x and y are non-zero integers. Is (x+y)/(x−y)>0?
1) x > y
2) |x| > |y|

Statement 1:

1) x > y

Case 1: -2>-3 Gives us -5 and that's not greater than 0
Case 2: 3>2 Gives us 5 and that's greater than 0

Not Sufficient.

Statement 2: |x| > |y|

This ensures that case 1 is not possible from above. So we are left with case 2

Also checking: x=-3 y =-2 works. We have -5/(-3+2) = 5 greater than 0.

Sufficient.

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Re: If x and y are non-zero integers. Is   [#permalink] 15 Mar 2020, 06:24