Inequality: A fundamental doubt

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Inequality: A fundamental doubt

by Uri » Tue Apr 07, 2009 7:33 am
Image

I simplified the question as:
(x - y) > (x + y)
i.e., I need to know whether y < 0.
This information is provided in (2). But (B) is not the answer.

Can you please tell me where I have made the mistake? Have I overlooked some rule in of inequality while simplifying?

OA: [spoiler](E)[/spoiler]
Last edited by Uri on Tue Apr 07, 2009 11:27 am, edited 1 time in total.
Source: — Data Sufficiency |

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by mike22629 » Tue Apr 07, 2009 8:01 am
Not positive, but this is what I think.

Obviously we agree A is inefficient.

With B,

If x = -4 and y = -1, then x + y = -5, thus you are multiplying both sides of equation by a negative number and must change the direction of the sign.

Thus,

x-y < x + y

Someone correct me if I am wrong plz.

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Re: Inequality: A fundamental doubt

by sacx » Tue Apr 07, 2009 8:49 am
Uri wrote:Image

I simplified the question as:
(x - y) > (x + y)
i.e., I need to know whether y < 0.
This information is provided in (2). But (B) is not the answer.

Can you please tell me where I have made the mistake? Have I overlooked some rule in of inequality while simplifying?

OA: [spoiler](E)[/spoiler]
You can cross- multiply if and only if we know it is positive. In this question we dot know the sign - if it was negative then the inequality sign would change as well

Stmt I

x = 5, y = 10

the fraction is <1

x = 5, y = -1

the fraction is >1

Insufficient

Stmt II

x = 5, y = -1

the fraction is >1

x = 10, y = -10

the fraction is not defined

Insufficient

Stmt I and II

x = 10, y = -10

the fraction is not defined

x = 5, y = -1

the fraction is >1

[spoiler]Insufficient

E[/spoiler]
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Re: Inequality: A fundamental doubt

by Musiq » Tue Apr 07, 2009 9:43 am
Uri wrote:Image

I simplified the question as:
(x - y) > (x + y)
i.e., I need to know whether y < 0.
This information is provided in (2). But (B) is not the answer.

Can you please tell me where I have made the mistake? Have I overlooked some rule in of inequality while simplifying?

OA: [spoiler](E)[/spoiler]
Hi Uri,

My solution is hard to see, but if you can train yourself in Number Properties to think along similar lines, this will really really help you on test day.

Thought Process:
On a GMAT question, when you see inequalities that contain variables in conjunction with Division or Multiplication, the first thing you have to ask yourself is "What do I know about the SIGN of the variable?"

Q: Is (x-y) / (x+y) > 1?

We see variables, we see inequalities and we see division....RED ALERT.

Immediately focus your energies on trying to reframe the question.....this part is hard to see, but with practice will become easier.

Restated Question:
Is Absolute Value of (x-y) > Absolute Value of (x+y)?

The absolute value reframing of the question will allow you to multiply...since Absolute value cannot be negative.

Within 10 to 15 seconds you should eliminate answer choices A / B and D. We need information about both variables to answer the question....but each statement gives us only one variable.

It's between C & E now.

Since we dotn have the value of the variables we really cannot say Yes or NO with any conviction.

Leaves us with E as the answer.
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by Uri » Tue Apr 07, 2009 11:26 am
thanks, buddies! i too was suspicious about the sign when dividing or cross multiplying in inequality problems. you all have done perfectly what i was looking for.

musiq, you have given a totally new method for me. i'll try this with some more problems. hope this method will work fine in all inequality problems involving variables and multiplication/ division.

thanks once again.

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by vpr » Tue Apr 07, 2009 11:31 am
Since you dont know the sign of (x+y), you cant multiply inequality by x+y to arrive at x-y>x+y. For example consider inequality 8>5. When you multiply this inequality by a positive number say 5, sign doesnt change (40>25) but when you multiply it by a negative number say -5, sign reverses (-40<-25). Point is you cant not multiply inequlity by a number whose sign you dont know.

Lets solve the original problem

x-y/x+y >1

x-y/x+y -1 > 0

x-y-x-y/x+y > 0

-2y/x+y > 0

or

y/x+y < 0

to have this one you can must have y and x+y opposite sign. So correct answer is E

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by cubicle_bound_misfit » Tue Apr 07, 2009 11:32 am
is (x-y)/(x+1) > 1?

==> is x-y/x+y -1 >0?

=> is -2y/x+y >0

=> is 2y/x+y < 0?

stmt 1. x>0 now for the expression to be true 2y and x+y shud be of opposite sign.

from x>0 if |x| > |Y| x<0 and y <0 this does not hold trur
x<0 and y>0 holds true. INSUFF

from stmt 2 Y < 0

again if x>0 Y< 0 but |Y| > |X| this does not hold

together, x>0 Y< 0 still teh expression can be positive.

Hence E.
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by maihuna » Tue Apr 14, 2009 10:15 am
After simplifying the inequality I will get:

2y/x+y <0

just by knowing the sign of x and y we cant conclude, for example:

y = -1, x = 3 it works
y = -4, x=3 it doesnt

so E.