Is m+z > 0? (1) m-3z > 0 (2) 4z-m > 0 : GMAT Data Sufficiency (DS)
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# Is m+z > 0? (1) m-3z > 0 (2) 4z-m > 0

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Is m+z > 0? (1) m-3z > 0 (2) 4z-m > 0 [#permalink]

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18 Mar 2007, 15:43
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Is m+z > 0?

(1) m-3z > 0
(2) 4z-m > 0

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-m-z-0-1-m-3z-0-2-4z-m-106381.html
[Reveal] Spoiler: OA
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19 Mar 2007, 08:25
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One more approach

From Stmt 1
We have 3Z < M

From Stmt 2
we have M < 4Z

Since we dont know if M and Z are +ve or -ve above are INSUFF individually

Combining
3Z < M < 4Z
Since 3Z < 4Z , Z has to be +ve
so M has to be +ve
and therefore M + Z has to be +ve

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Is m+z > 0? (1) m-3z > 0 (2) 4z-m > 0 [#permalink]

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02 Feb 2011, 01:54
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puneetj wrote:
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?

terp26's reasoning for (1) and (2) is not correct: you cannot write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by $$z$$: never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

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.

Also discussed here: data-suff-67183.html
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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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31 Jan 2011, 09:05
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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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19 Mar 2007, 01:23
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(C) for me

To me, I prefer to draw an XY plan in order to conclude faster .... Let understand m as x and z as y.

m+z > 0 ?
<=> z > -m ? >>> This question ask us if the points (m,z) are above the line z = -m. In the Fig 1, I draw in green the area we are looking for.

From 1
m-3z > 0
<=> z < m/3 >>> It's all points (m,z) below the line z=m/3

Obviously, by looking at the Fig 2, we can conclude that we have points in the green area where z > -m and points in the red one where z < - m.

INSUFF.

From 2
4z-m > 0
<=> z > m/4 >>> It's all points (m,z) above the line z=m/4

Obviously, by looking at the Fig 3, we can conclude that we have points in the green area where z > -m and points in the red one where z < - m.

INSUFF.

Both (1) and (2)
We can conlude with the graph as well, but I prefer here to use some alegbra

We have the system:
o m-3z > 0 (A)
o 4z-m > 0 (B)

(A) + (B) <=> (m-3z) + (4z-m) > 0
<=> z > 0

As well,
(A) <=> m > 3*z
=> m > 3*z > z > 0 as z > 0 then, 3z > z.

Thus,
m>0 and z>0
=> m + z > 0.

SUFF.
Attachments

Fig1_z sup to -m.gif [ 3.99 KiB | Viewed 11778 times ]

Fig2_z inf to m div 3.gif [ 4.32 KiB | Viewed 11775 times ]

Fig3_z sup to m div 4.gif [ 4.08 KiB | Viewed 11772 times ]

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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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15 Jul 2008, 07:22
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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,

Attachment:

DS Q1.JPG [ 12.12 KiB | Viewed 3518 times ]

Last edited by durgesh79 on 15 Jul 2008, 21:08, edited 1 time in total.
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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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14 Oct 2011, 12:12
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rvind wrote:
Is m+z > 0?
1) m-3z >0
2) 4z-m >0

According to me, the answer should be E.

According to GMAT Prep, answer is C. If we consider both the statements, it will end up as 3z<m<4z. How can we confirm that m+z>0 until and unless we know the value of z (+ve or -ve)

Pls help me on this one.

If you consider z negative than equation will be -> 3z>m>4z
consider z=-1, and assume m= - 3.5, these values doesn't fit the above equation. Hence negative value is not be a right for z amd m in this question. hence, C.
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Re: GMAT Prep DS: Is m+z > 0 [#permalink]

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14 Oct 2011, 12:33
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rvind wrote:
Is m+z > 0?
1) m-3z >0
2) 4z-m >0

1)
m-3z>0
m>3z
3z<m----------1
m=-1; z=-1; m+z=-2<0
m=4; z=1; m+z=5>0
Not Sufficient.

2)
4z-m>0
4z>m
OR
m<4z--------------2
m=1; z=1; m+z=2>0
m=-5; z=-1; m+z=-6<0
Not Sufficient.

Putting together:
m-3z+4z-m>0
z>0

From 1 and 2:
We know; 3z<m<4z AND z=+ve; 3z=+ve; m is some +ve greater than 3z.
Sufficient.

Ans: "C"
*******************************

A graph would also be a good way to solve this if someone care to put that down. I know walker is an expert.
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Re: GMAT Prep DS: Is m+z > 0 [#permalink]

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17 Oct 2011, 13:42
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Here is a graph. Two statements define the light green area. For each point of the area m+z>0 is true. So, sufficient.

[Reveal] Spoiler:
Attachment:

graphs_122011.png [ 10.72 KiB | Viewed 1648 times ]

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19 Mar 2007, 03:14
Quote:
Both (1) and (2)

We have the system:
o m-3z > 0 (A)
o 4z-m > 0 (B)

(A) + (B) <m> 0
<z> 0

Fig, this is an excellent way to do this question which I did not follow and took a very long time to solve it. Thanks!!!
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19 Mar 2007, 07:56
How I solved:

Statement 1:

M - 3Z > 0

You can use algebra to get this to look like:

M/Z > 3

This tells us that M or Z could be negative or positive

since to get a positive result M and Z are either both positive or both negative

Since we need to see if M + Z > 0

We need to see if adding the 2 would give us a value above 0

from the statement we get 2 negatives or 2 positives however adding 2 negatives would gives us a positive, adding 2 positives would give us a negative

INSUFF

Statement 2:

4z-m > 0

Use algebra bring us to

M/Z > 4

same thing as statement 1

COMBINED

we have M - 3Z > 0
-M + 4Z > 0

adding both statemetns we are left with

Z > 0

since Z is greater than 0, then M must be greater than as well

since M/Z must be positive according to the statements

M + Z has to be positive as well

C
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15 Jul 2008, 06:55
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Is m+z > 0?

(1) m-3z > 0
(2) 4z-m > 0

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-m-z-0-1-m-3z-0-2-4z-m-106381.html
Attachments

File comment: i hope its understandable....
X.doc [19 KiB]

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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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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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01 Feb 2011, 22:33
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?
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03 Feb 2011, 20:45
Hi Bunuel,

Z is positive when combining both the statements - understood.
Using statement (1)m-3z > 0 you proved that m>3z = positive - understood

How do you prove statement (2) the way you proved statement 1? or should we bother proving it at all?
4z>m which means 4 * some positive number > m. Does that prove anything about m?

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Re: DS: Algebra   [#permalink] 03 Feb 2011, 20:45

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