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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 14580 times ]

Fig2_z inf to m div 3.gif [ 4.32 KiB | Viewed 14576 times ]

Fig3_z sup to m div 4.gif [ 4.08 KiB | Viewed 14571 times ]

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?
_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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.

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)....

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..

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.

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?
_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

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.

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?

Please confirm.
_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

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?

Please confirm.

On the GMAT, two data sufficiency statements always provide TRUE information. So: (1) m-3z>0 and (2) 4z-m>0 are given to be true, you shouldn't prove them. You should check whether m+z>0 is true, which is done when we take these two (true) statements together: by adding them we get that z>0 and then looking on (1) with this info we get that m>0 too.

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)

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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