(x-y)/ (x+y) < 1

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(x-y)/ (x+y) < 1

by gkumar » Wed Oct 21, 2009 6:34 am
I couldn't find this problem in the search results. I am confused by the below GPrep question:
If x is not equal to y, is (x-y)/ (x+y) < 1?
1) x>0
2) y<0

I rephrased the question to be the following:
Case 1: x+y>0
Cross multiply to get: x-y < x+y
-y < y
0 < 2y
0 < y

Case 2: x-y<0
Cross multiply and switch inequality sign: x-y > x+y
-y > y
0 > 2y
0 > y

So the rephrased question is asking: "Is y>0 or y<0?"

Looking at (1), X is not involved in the rephrase that contains only y, so it is irrelevant and hence insufficient.

Looking at (2), y<0 is given and answers the rephrased question of "is y<0" and the answer is yes. So I assumed B.

[spoiler]But the official answer is E. I realize that I could plug in numbers to get this answer, but isn't that time consuming? Why did my approach not work?[/spoiler]
Source: — Data Sufficiency |

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by xcusemeplz2009 » Thu Oct 22, 2009 2:36 am
IMO E is the correct ans

as it is not given that x and y are int.

in fractional value and whole no we get diff results even on combining bth.
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by Talkativetree » Thu Oct 22, 2009 8:19 pm
If x and y were integers, I believe the answer would be b, but x and y could also be fractions, so that would chance the answer to insufficient, because we could end up with a fraction over a fraction, which could potentially be greater than 0.

I thought it was B at first as well, then remembered that.

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by grockit_jake » Fri Oct 23, 2009 11:56 am
The reason cross multiplying doesn't work is if you multiply an inequality by a negative number, then the sign switches. So if you don't know whether (x+y) is positive or negative, then you can't know the direction of the inequality, which is obviously very important.

While the signs of x and are clearly important in this and other questions like it, the key here is recognizing that the absolute values of x and y are not given. So if x and y are opposite signs, we don't know if their sum is +/-/0 since that depends on their magnitude.
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by panda8989 » Fri Oct 23, 2009 1:45 pm
I think your approach has an error in these steps

Case 1: x+y>0
Cross multiply to get: x-y < x+y
-y < y
0 < 2y <----- this should be -1 < 1 or -1 > 1 depending on the sign of y
0 < y

Case 2: x-y<0
Cross multiply and switch inequality sign: x-y > x+y
-y > y
0 > 2y <----- same as in case (1)
0 > y

When I try to solve these types of questions, I check to see if the inequality sign will change when you cross multiply...

(x-y)/(y+x) < 1

(x-y) < (y+x) if (y+x) > 0
(x-y) > (y+x) if (y+x) < 0

even if both (1) and (2) are true, it cannot determine the sign of (y+x) since we don't know if |x| > |y| or |x| < |y|.

Not sure if this is a valid approach.. but that's how I got to (E)... Hope it helps!

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by agganitk » Sat Jul 31, 2010 2:16 am
Can some one explain to me where I am wrong?

if X+Y >0 ==> X> -Y
THEN
X-Y > X + Y
Y<0
bcoz X>-Y ==> X>0
hence for [X>0, Y<0] (X-Y)/(X+Y)>1 [CASE 1]
-------------------------------------------
if X+Y <0 ==> X< -Y
THEN
X-Y < X + Y
Y>0
bcoz X<-Y ==> X<0
hence for [X<0,Y>0] (X-Y)/(X+Y)>1 [CASE 2]

-----------------------------------------------

statement 1 and 2 tell us X>0 and Y <0
therefore using case one, we can say (X-Y)/(X+Y) >1


ANS SHOULD BE C. BUT OA IS E

?????????????????

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by Asif » Sun Aug 01, 2010 2:12 am
Answer is E.

x = 1, Y = -1

solve for (x-y)/(x+y)= 2/0 which is undefined, so the answer is E.

Singly both the statements are not sufficient as well.