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If x and y are non-zero integers. Is

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

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New post 14 Mar 2020, 18:46
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A
B
C
D
E

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Question Stats:

50% (01:28) correct 50% (01:03) wrong based on 34 sessions

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

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

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New post 15 Mar 2020, 05:02
Answer : B
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
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Re: If x and y are non-zero integers. Is  [#permalink]

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New post 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

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

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

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New post 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.

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

If x and y are non-zero integers. Is

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