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Is m+z > 0? (1) m3z > 0 (2) 4zm > 0 [#permalink]
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18 Mar 2007, 16:43
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Is m+z > 0? (1) m3z > 0 (2) 4zm > 0 OPEN DISCUSSION OF THIS QUESTION IS HERE: ismz01m3z024zm106381.html
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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
m3z > 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
4zm > 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 m3z > 0 (A)
o 4zm > 0 (B)
(A) + (B) <=> (m3z) + (4zm) > 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 13382 times ]
Fig2_z inf to m div 3.gif [ 4.32 KiB  Viewed 13377 times ]
Fig3_z sup to m div 4.gif [ 4.08 KiB  Viewed 13373 times ]

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Quote: Both (1) and (2)
We have the system: o m3z > 0 (A) o 4zm > 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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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:
4zm > 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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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
Answer C

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Is m+z > 0? [#permalink]
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15 Jul 2008, 07:55
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Is m+z > 0? (1) m3z > 0 (2) 4zm > 0 OPEN DISCUSSION OF THIS QUESTION IS HERE: ismz01m3z024zm106381.html
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File comment: i hope its understandable....
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Re: DATA SUFF... [#permalink]
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15 Jul 2008, 08:22
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for option C we have to look for overlapping area, for overlapping area, cleary x+y > 0 answer C Edit : i made a mistake in making the drawing, the right one is attached now, Attachment:
DS Q1.JPG [ 12.12 KiB  Viewed 5117 times ]
Last edited by durgesh79 on 15 Jul 2008, 22:08, edited 1 time in total.

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Re: DATA SUFF... [#permalink]
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15 Jul 2008, 08:54
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]
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15 Jul 2008, 09:07
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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]
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15 Jul 2008, 09:09
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]
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15 Jul 2008, 09:18
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]
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15 Jul 2008, 15:45
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arjtryarjtry wrote: is m+z>0? 1.m3z>0 2.4zm>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) 4zm>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]
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16 Jul 2008, 01:18
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arjtryarjtry wrote: is m+z>0? 1.m3z>0 2.4zm>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 ) ==> Answer is (C)

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Re: DATA SUFF... [#permalink]
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Re: DS: Algebra [#permalink]
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01 Feb 2011, 23:33
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 4zm could still be >0 if z=1, m=1,then 4zm = 3 > 0 if z=1, m= 10 so 4zm = 14 > 0......am I missing something?
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Is m+z > 0? (1) m3z > 0 (2) 4zm > 0 [#permalink]
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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 4zm could still be >0 if z=1, m=1,then 4zm = 3 > 0 if z=1, m= 10 so 4zm = 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 > \(m3z+4zm>0\) > \(z>0\), 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. Answer: C. Also discussed here: datasuff67183.html
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Re: DS: Algebra [#permalink]
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03 Feb 2011, 21:45
Hi Bunuel, Z is positive when combining both the statements  understood. Using statement (1)m3z > 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.
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Re: DS: Algebra [#permalink]
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04 Feb 2011, 03:40
puneetj wrote: Hi Bunuel,
Z is positive when combining both the statements  understood. Using statement (1)m3z > 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) m3z>0 and (2) 4zm>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. Hope it's clear.
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Is m+z > 0? 1) m3z >0 2) 4zm >0 According to me, [#permalink]
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14 Oct 2011, 12:47
Is m+z > 0? 1) m3z >0 2) 4zm >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.

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Re: GMAT Prep quant Doubt [#permalink]
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14 Oct 2011, 13:12
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rvind wrote: Is m+z > 0? 1) m3z >0 2) 4zm >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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