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OG 11 - x+y problem

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OG 11 - x+y problem

by maxim730 » Mon Feb 05, 2007 6:19 am
From OG 11

Problem 139 or 140:

If x != -y, is (x-y)/(x+y) > 1? ( != means not equal)

1) x>0
2) y<0

I don't understand their solution. OA when a couple of people reply.

P.S.

is it ok for me to rewrite the inequality as x-y > x+y?


MY REASONING:

Here is my reasoning. Please critique and help me understand what I'm doing wrong!

1) X>0

x = 1, y = 2

(1-2)/(1+2) > 1 NO

x = 3 y = -1

(3-(-1))/(3+(-1)) > 1
4/2 > 1 YES

so INSUFFICIENT

2) y < 0


x = 3 y = -1

(3-(-1))/(3+(-1)) > 1
4/2 > 1 YES

x = -2, y = -1

(-2-(-)1)/(-2+(-1)) > 1
-1/-3 > 1 NO

so INSUFFICIENT

Both 1 and 2)

I rewrote the inequality as x-y > x+y

x>0 and y<0

x = 2 y = -3

2-(-3) > 2+(-3)
5 > -1 YES

x = 4 y = -2

4+2 > 2 YES

So, why not C? What am i doing wrong?!?!
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by banona » Mon Feb 05, 2007 1:00 pm
Hi,
you reasoning is missing an important thing which the following :
in order to rewrite ( x-y)/(x+y)> 1 in the following form ( x-y)>(x+y),your multiplying both sides of the inequality by ( x+y), in which case you need to now the sign ( x+y) in order to keep or to inverse the inequality;
In your case, you supposed that it's positive number;
back to the DS question, we are given to information about x>0, which is insuffisant,
Knowing that y<0; then we ca conclude that:
x-y is a STRICT postive number and that ( x-y > x+y) , dividig all by (x-y), we find that ( x+y)/(x-y)< 1; One can conclude that by inversing the inequality we could find that ( x-y)/(x+y) > 1; but we can not do that unless we know the sign of ( x+y); consequently
E is the answer;

to be more confident pick the following numbers with x >0 and y<0

you can consider : x = 10 and y = -1 ( this works)
but if x =1 and y = -2 ( this doesnt work)


I hope this could help

ANY COMMENT

Banona
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by diebeatsthegmat » Mon May 10, 2010 11:51 pm
banona wrote:Hi,
you reasoning is missing an important thing which the following :
in order to rewrite ( x-y)/(x+y)> 1 in the following form ( x-y)>(x+y),your multiplying both sides of the inequality by ( x+y), in which case you need to now the sign ( x+y) in order to keep or to inverse the inequality;
In your case, you supposed that it's positive number;
back to the DS question, we are given to information about x>0, which is insuffisant,
Knowing that y<0; then we ca conclude that:
x-y is a STRICT postive number and that ( x-y > x+y) , dividig all by (x-y), we find that ( x+y)/(x-y)< 1; One can conclude that by inversing the inequality we could find that ( x-y)/(x+y) > 1; but we can not do that unless we know the sign of ( x+y); consequently
E is the answer;

to be more confident pick the following numbers with x >0 and y<0

you can consider : x = 10 and y = -1 ( this works)
but if x =1 and y = -2 ( this doesnt work)


I hope this could help

ANY COMMENT

Banona
hi,
i think i'd like to solve this in different way.
x#-y , (x-y)/(x+y)>1
=> x-y>x+y => 2y<0 => y<0
thus, i think only 2 is enough to answer the question.
how do you guys think? please post the answer
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by raviki8208 » Tue May 11, 2010 12:24 am
maxim730 wrote:From OG 11

Problem 139 or 140:

If x != -y, is (x-y)/(x+y) > 1? ( != means not equal)

1) x>0
2) y<0

I don't understand their solution. OA when a couple of people reply.

P.S.

is it ok for me to rewrite the inequality as x-y > x+y?


MY REASONING:

Here is my reasoning. Please critique and help me understand what I'm doing wrong!

1) X>0

x = 1, y = 2

(1-2)/(1+2) > 1 NO

x = 3 y = -1

(3-(-1))/(3+(-1)) > 1
4/2 > 1 YES

so INSUFFICIENT

2) y < 0


x = 3 y = -1

(3-(-1))/(3+(-1)) > 1
4/2 > 1 YES

x = -2, y = -1

(-2-(-)1)/(-2+(-1)) > 1
-1/-3 > 1 NO

so INSUFFICIENT

Both 1 and 2)

I rewrote the inequality as x-y > x+y

x>0 and y<0

x = 2 y = -3

2-(-3) > 2+(-3)
5 > -1 YES

x = 4 y = -2

4+2 > 2 YES

So, why not C? What am i doing wrong?!?!
Hi,

We should not rewrite the inequality as x-y > x+y?

It depends on the sign of the denominator.
a/b > 1 => a > b if b is +ve
a/b > 1 => a < b if b is -ve

see an example below,
3/2 > 1 => 3 > 2
(-3)/(-2) > 1 => -3 < -2


MY reasoning,

Given, x+y != 0


1) x > 0

case1: if x+y > 0 & y > 0, x-y can be -ve so (x-y)/(x+y) can be < 0 NOT SATISFIED
case2: if x+y > 0 & y < 0, x-y always > x+y
case3: if x+y < 0 & y > 0, NOT possible
case4: if x+y < 0 & y < 0, x-y is +ve and x+y is -ve => (x-y)/(x+y) < 0 NOT SATISFIED

2) y < 0

case4 above is not satisfied so, no need to go further

using both, case4 above is not satisfied so, no need to go further

so E
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by Nipun Ravi » Tue May 11, 2010 12:57 am
Maxim730: You were doing absolutely right until you tested the last condition (both statement 1 and 2 to be true).

You already tested a YES case with x>0 and y<0, however if you consider x=1 and y=-5, still x>0 and y<0 and x!=-y.

In this case (x-y)/(x+y)=(1-(-5))/(1+(-5))=(1+5)/(1-5)=6/(-4)=-(3/2) which is less than 1, so you arrive at a NO situation.

Thus both 1 and 2 together are also not consistent.

Thus the answer is E.
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