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# Are x and y both positive? 1) 2x - 2y = 1 2) x/y

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Manager
Joined: 17 Dec 2008
Posts: 171
Are x and y both positive? 1) 2x - 2y = 1 2) x/y [#permalink]

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03 Mar 2009, 17:20
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Are x and y both positive?
$$1) 2x - 2y = 1$$
$$2) x/y > 1$$

I found the answer plugging numbers and checking every condition again.
Is there a mathematical way to solve/check these conditions.
Manager
Joined: 27 Feb 2009
Posts: 59

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03 Mar 2009, 17:52
C

b, x/y > 1
x is -ve, y is -ve
x is +ve, y is +ve not sufficient

a, x-y = 1/2

x = 1, y = 1/2
x=-1/2, y = -1
or x = 1/4, y = -1/4 not sufficient

taken to-geather, x should be +ve and y should be +ve to have x/y>1 hence C
Manager
Joined: 17 Dec 2008
Posts: 171

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03 Mar 2009, 18:07
Thanks but is punching in 2 value with fractions sufficient to judge these type of questions?
You tried x=1, y=1/2 AND x=-1, y=-1/2.
Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

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03 Mar 2009, 20:48
I am trying to follow Walker's approach on nailing down the Inequalities under DS. It sure can be solved under 2 min mark with 100% accuracy, provided you are comfortable drawing the graphs from the Inequalities. So attached is my approach graphically.
Attachments

Are x and y both positive.doc [117 KiB]

Manager
Joined: 27 Feb 2009
Posts: 59

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03 Mar 2009, 21:04
what is walkers explanation..i m new to this forum..can u send me the file? I waste a lot of time in plugging numbers

Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

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03 Mar 2009, 21:36
well......Walker is one of the forum members just like us. But his Quant approaches are wonderful. I have attached the document on how I solved the DS problem. But I wanted to give credit to Walker for his approach on this.

Search the forum on Inequalities. You might get various approaches on how to solve them. Of them I picked this approach.

p502749-graphic-approach-to-problems-with-inequalities?t=68037&hilit=+Inequalities#p502749

Even though it is time consuming to understand the concepts on how to draw graphs from the Linear (In)equalities , once you get it right, you should be comfortable solving the DS questions.
Manager
Joined: 27 May 2008
Posts: 200

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04 Mar 2009, 05:27
mrsmarthi wrote:
well......Walker is one of the forum members just like us. But his Quant approaches are wonderful. I have attached the document on how I solved the DS problem. But I wanted to give credit to Walker for his approach on this.

Search the forum on Inequalities. You might get various approaches on how to solve them. Of them I picked this approach.

p502749-graphic-approach-to-problems-with-inequalities?t=68037&hilit=+Inequalities#p502749

Even though it is time consuming to understand the concepts on how to draw graphs from the Linear (In)equalities , once you get it right, you should be comfortable solving the DS questions.

hi mrsmarthi, you explanation is good. but please highlight the intersection or proof you are trying to prove in the diagram, so that the diagram would be more expressive.
Walker - please suggest us the name of the software you are using to draw the graph.
Intern
Joined: 03 Mar 2009
Posts: 48

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04 Mar 2009, 05:39
Be sure to check out my first post about graphing inequalities quickly without doing any math!! :D
t76255-quick-way-to-graph-inequalities

Here's a website where you can graph your inequalities. Just save them as a picture after it's generated.
http://www.hostsrv.com/webmab/app1/MSP/ ... s&s3=basic
Manager
Joined: 19 Aug 2006
Posts: 238

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04 Mar 2009, 09:59
Let me take a stab.

1. 2x-2y-=1
2x = 1 + 2y
x = (1+2y)/2=0.5 +y
not suff.

2. x/y>1
multiply by y => x>y
not suff.

combining them both, it's sufficient:
x=0.5+y and x>y => 0.5+y>y
this makes sense only if y<0

So, it's C.
Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

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04 Mar 2009, 11:09
2
KUDOS
peraspera wrote:
Let me take a stab.

1. 2x-2y-=1
2x = 1 + 2y
x = (1+2y)/2=0.5 +y
not suff.

2. x/y>1
multiply by y => x>y
not suff.

combining them both, it's sufficient:
x=0.5+y and x>y => 0.5+y>y
this makes sense only if y<0

So, it's C.

X/y > 1 ==> x > y This is is not true in Ineqalities. You cannot multiple denominator value on both sides.

Here is an ex Let x = -3 and y = -2 ==? -3 / -2 > 1. This does not mean that -3 > -2. So your approach on Second clue is incorrect. I think you need to revisit the logic.
Manager
Joined: 19 Aug 2006
Posts: 238

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04 Mar 2009, 13:03
mrsmarthi wrote:
peraspera wrote:
Let me take a stab.

1. 2x-2y-=1
2x = 1 + 2y
x = (1+2y)/2=0.5 +y
not suff.

2. x/y>1
multiply by y => x>y
not suff.

combining them both, it's sufficient:
x=0.5+y and x>y => 0.5+y>y
this makes sense only if y<0

So, it's C.

X/y > 1 ==> x > y This is is not true in Ineqalities. You cannot multiple denominator value on both sides.

Here is an ex Let x = -3 and y = -2 ==? -3 / -2 > 1. This does not mean that -3 > -2. So your approach on Second clue is incorrect. I think you need to revisit the logic.

Thank you so much, I can't believe I still make this stupid mistake.
You are correct, I cannot do that.

Here's the (2) and the solution, then:

2. x/y>1, so then x and y have the same signs, either -, or +.
Furthermore, x>y if their signs are positive and x<y if their signs are negative.
insuff.

Let's combine (1) and (2).

if x and y are negative, then -x=0.5+(-y) or -x+y=0.5 - this can never be possible, because from (2) we know that x<y if their signs are negative. E.g. if x=-3, and y=-2, then -3+2 cannot equal 0.5.

if x and y are positive, then x=0.5+y or x-y=0.5 - this is possible, because from (2) we know that x>y if their signs are positive.

So, C is sufficient, x and y are both positive.
I hope it makes sense.
Manager
Joined: 17 Dec 2008
Posts: 171

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04 Mar 2009, 16:29
This is a great approach. Thanks all. Kudos to mrsmarthi.
Can this be applied to inequalities with absolute values? How would things change?

Few questions though:

In the example provided (walker's post)
p502749-graphic-approach-to-problems-with-inequalities?t=68037&hilit=+Inequalities#p502749
The condition (case 1) and (case 2) covers a subset of the solution.

1.) Is a subset of the solution a SUFFICIENT condition? Or should I make sure that ALL points of the condition case statement meets the original statement? In his example there is only a subset and looks like it is SUFFICIENT.
Manager
Joined: 05 Jan 2009
Posts: 78

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04 Mar 2009, 17:18
Hi mrsmarthi ,
Could you please explain how will you draw x/y >1 using your technique?
using same technique how are you getting solution space between x-axis and the line x/y>1?..your equation is x-y>0; according to your technique, solution space will be right side of the line x-y=0;
Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

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04 Mar 2009, 20:40
Was surprised that I got couple of questions to be answered. So thought I have a separate thread to document my preparation on solving Inequalities in DS. Here is the link.

http://gmatclub.com/forum/t76273-how-to-crack-inequality-questions-in-ds

Hope it might solve the questions being asked in this thread. If not do let me know and I will try to answer them.
Re: DS: Number properties   [#permalink] 04 Mar 2009, 20:40
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