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# is m+z>0? 1.m-3z>0 2.4z-m>0. is this diag ok,? if

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is m+z>0? 1.m-3z>0 2.4z-m>0. is this diag ok,? if [#permalink]  15 Jul 2008, 06:55
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is m+z>0?
1.m-3z>0
2.4z-m>0.

is this diag ok,? if solved through cartesian method?
the ans isC.
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File comment: i hope its understandable....
X.doc [19 KiB]

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Re: DATA SUFF... [#permalink]  15 Jul 2008, 07:22
for option C we have to look for overlapping area,

for overlapping area, cleary x+y > 0

Edit : i made a mistake in making the drawing, the right one is attached now,

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DS Q1.JPG [ 12.12 KiB | Viewed 1268 times ]

Last edited by durgesh79 on 15 Jul 2008, 21:08, edited 1 time in total.
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Re: DATA SUFF... [#permalink]  15 Jul 2008, 07:54
Durgesh or gmatnub - can you explain this a bit more? I understand you've graphed out the inequalities. One inequalty create the yellow shaded area and the other inequality creates the green shaded area. Can you go into more detail regarding their relationship with each other?
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Re: DATA SUFF... [#permalink]  15 Jul 2008, 08:07
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jallenmorris wrote:
Durgesh or gmatnub - can you explain this a bit more? I understand you've graphed out the inequalities. One inequalty create the yellow shaded area and the other inequality creates the green shaded area. Can you go into more detail regarding their relationship with each other?

its just visulizing the inequalities with two variables....

for example if you have only one variable ... one of the ways of doing such problems is draw anumber line and mar the portion of the number line which falls in that rang

what is the value of x, an integer
1) 2 < x < 8
2) 6 < x < 10

draw a number line
mark the segament between 2 and 8
mark the segament between 6 and 10

the common segamant will give the values of x, which will satisfy both.... as x is an integer, th only possible value is 7

Going back to our question, the idea is to find a target area which is represented by one side of x+y=0

by combinng the two conditions, we can find an area which will always be on one side if x+y=0, no matter what is the value of x and y, this Suff.
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Re: DATA SUFF... [#permalink]  15 Jul 2008, 08:09
so what is the significance of the part on the top right that does not overlap at all?
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Re: DATA SUFF... [#permalink]  15 Jul 2008, 08:18
jallenmorris wrote:
so what is the significance of the part on the top right that does not overlap at all?

In this question no significance actually, that area represents values of x,y which we dont have to consider..... if you are trying to solve the question by plugging values, no value in that area will satisfy (1) or (2)....
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Re: DATA SUFF... [#permalink]  15 Jul 2008, 14:45
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arjtryarjtry wrote:
is m+z>0?
1.m-3z>0
2.4z-m>0.

is this diag ok,? if solved through cartesian method?
the ans isC.

frankly, i dont quite follow the graphing method but algebracially here is how i would do i..
m+z>0?

1)m>3z or m/3 > z

insuff we dont know value of z could be -, + o..dont know insuff

2) 4z-m>0
4z>m
z>m/4 well i dont know anything about m, could be -, +, 0..dont know insuff

togehter

m/4<z<m/3

OK..now m cant be negative since -1/4 IS NOT less than -1/3 ..its GREATer..so the only way know is that M is POSITIVE... its not even 0..since the ineqaulity wont hold..

so right away I know that m+z>0 Sufficient .

C it is..
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Re: DATA SUFF... [#permalink]  15 Jul 2008, 17:10
Great explanation! +1 Thanks for breaking it down into such detailed steps.
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Re: DATA SUFF... [#permalink]  16 Jul 2008, 00:18
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arjtryarjtry wrote:
is m+z>0?
1.m-3z>0
2.4z-m>0.

Just add (1) and (2) and you simply get : z>0

Since (1) tells you m>3z, then m>0 too

Now that you know they are both positive, you directly know that m+z>0 (no calculation )

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Re: DATA SUFF... [#permalink]  31 Jan 2011, 08:46
Das ist fantastisch! The graphic approach to solve inequalities is indeed great!
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Re: DATA SUFF... [#permalink]  31 Jan 2011, 09:05
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Expert's post
I wouldn't recommend graphic approach for this problem, algebraic approach is simpler and fairly straightforward:

Is m+z > 0?

(1) m - 3z > 0. Insufficient on its own.
(2) 4z - m > 0. Insufficient on its own.

(1)+(2) Remember we can add inequalities with the sign in the same direction --> m-3z+4z-m>0 --> z>0, so z is positive. From (1) m>3z=positive, so m is positive too (m is more than some positive number 3z, so it's positive) --> m+z=positive+positive>0. Sufficient.

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Re: DATA SUFF... [#permalink]  31 Jan 2011, 21:56
Which type of questions can most easily be solved by graphic approach ?
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Re: DATA SUFF...   [#permalink] 31 Jan 2011, 21:56
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