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# Is m + z > 0?

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Manager
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Is m + z > 0? [#permalink]

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11 Feb 2009, 17:20
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Question Stats:

60% (00:57) correct 40% (01:35) wrong based on 217 sessions

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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

Last edited by Bunuel on 30 Aug 2013, 04:36, edited 1 time in total.

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11 Feb 2009, 20:57
3
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I will go with C.

From clue 1, m-3z > 0 here m, z can be positive or negative. So cannot say whether m + z > 0. Hence Insufficient.

From clue 2, 4z-m > 0, here m, z can be positive or negative. So cannot say whether m + z > 0. Hence Insufficient.

Cosider both the clues.

m-3z>0--->Ine1
4z-m>0 --->ine2

Add both the inequalities, z > 0.
m-3z > 0 ==> m > 3z. Since Z is positive m also should be positive.

So we can say that m+z > o. Hence sufficient.

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12 Feb 2009, 03:38
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It is also possible to solve it by drawing:

1. Let's draw our statement using two colors: red (statement is false) and green (statement is true). We will have two color cake:
Attachment:

t75657_1.png [ 6.27 KiB | Viewed 11395 times ]

2. Now, cut a bit of the cake by first condition:
Attachment:

t75657_2.png [ 9.33 KiB | Viewed 11392 times ]

3. cut a bit of the cake by second condition:
Attachment:

t75657_3.png [ 8.99 KiB | Viewed 11384 times ]

4. Now, use both conditions:
Attachment:

t75657_4.png [ 10.54 KiB | Viewed 11390 times ]

So, C is obvious after drawing as the statement is only true.
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27 Dec 2009, 16:33
2
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Q : m + z > 0 ?

1. m-3z > 0

m+z > 4z

For this to be sufficient, z > 0 should be known, but we don't know that, hence insufficient.

2. 4z-m > 0

m+z > 5/4 m OR 4z > m

For this to be sufficient, m > 0 should be known, but we don't know that, hence insufficient.

Combining 1 & 2 :

m+z > 4z > m

m+z-m > 4z-m > m-m

z > 4z-m > 0. This solves our question in 1. Hence sufficient. => C

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Is m + z > 0? [#permalink]

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27 Apr 2010, 14:41
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Is m + z > 0?

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

[Reveal] Spoiler:
I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

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Re: Question on GMAT PREP DS [#permalink]

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27 Apr 2010, 15:01
2
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CharmWithSubstance wrote:
The question asks is m + z > 0?

1. m - 3z > 0
2. 4z - m > 0

I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

U r right here, the sign of m depends on sign of Z.
let say Z is +1 , then m> 3z and m< 4z, which simply means that m>3 and m<4, and m is positive.

Second case
let say Z is -1 , then m> 3(-1) and m< 4(-1),
which translates to m> -3 and m< -4, and m is negative. But if we draw these regions on a number line, there is no overlap, these regions do not give any solution.

m> -3 : m can be -2,1-,0,1....
m< -4 : m can be -5,-6,-7.....

SO effectively only first case z= +1 is valid case and so both z and m are positive..!!

I must say, it's not an easy catch..!!

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Re: Question on GMAT PREP DS [#permalink]

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27 Apr 2010, 15:09
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CharmWithSubstance wrote:
The question asks is m + z > 0?

1. m - 3z > 0
2. 4z - m > 0

I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

(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 same sign --> $$m-3z+4z-m>0$$ --> $$z>0$$, so $$z$$ is positive. From (1) $$m>3z=positive$$, so $$m$$ is positive too --> $$positive+positive>0$$. Sufficient.

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Re: Question on GMAT PREP DS [#permalink]

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27 Apr 2010, 15:11
Both great explanations. Thank you!

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Re: Inequality from GMAT PREP [#permalink]

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26 Jul 2010, 00:02
mn2010 wrote:
Is m+z > 0

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

Thanks

1) m-3z>0 If z=1 m=4 m+z>0 whereas if z=-4 m=1 m+z<0. Hence Statement 1alone is not sufficient
2) 4z-m>0 If m=4 z=2 m+z>0 whereas if m=-4 and z=2 m+z<0. Hence statement 2 alone is not sufficient

m-3z >0
4z-m>0
--------
z>0 Since m>3z m is also greater than 0 and Hence m+z>0 . So C
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Re: Question on GMAT PREP DS [#permalink]

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26 Jul 2010, 05:04
thanks crack 700 for the solution, I was struggling between C and E.

Thanks bunuel for merging the topic, I was trying to find this Q on the forum before I posted it but Couldn't find it.

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22 Aug 2010, 21:26
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Hi,

(1) m - 3z > 0

m > 3z

Let z be 100 and m be 400. Here, m+z > 0

Let z be -100 and me be -200. Here, m+z > 0

(2) 4z - m > 0

m < 4z

Let z be 100 and m be 300. Here, m+z > 0

Let z be -100 and me be -500. Here, m+z < 0

Combine 1 and 2:

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

Combining, we get 3z < m < 4z. This is possible ONLY when both z and m are +ve.

Hence, C.

Hope this helps.
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29 Aug 2013, 22:06
Can you explain after drawing graphs how to interpret them.?

I am able to draw all but how to interpret while taking two lines at a time, I am not getting it.Please advise.

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30 Aug 2013, 04:38
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ygdrasil24 wrote:
Can you explain after drawing graphs how to interpret them.?

I am able to draw all but how to interpret while taking two lines at a time, I am not getting it.Please advise.

Explained here: graphic-approach-to-problems-with-inequalities-68037-80.html

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-m-z-0-1-m-3z-0-2-4z-m-106381.html
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Re: Is m + z > 0? [#permalink]

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06 Feb 2016, 13:21
Bunuel wrote:
CharmWithSubstance wrote:
The question asks is m + z > 0?

1. m - 3z > 0
2. 4z - m > 0

I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

(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 same sign --> $$m-3z+4z-m>0$$ --> $$z>0$$, so $$z$$ is positive. From (1) $$m>3z=positive$$, so $$m$$ is positive too --> $$positive+positive>0$$. Sufficient.

Please help with last part (statement 1+ statement 2). I understand that if z>0 then 3z<m means that m>0 (since posiive number can't less than negative. But this is not true for condition 4z>m where m can be + or -. What i am missing?
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Re: Is m + z > 0? [#permalink]

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07 Mar 2016, 21:37
CharmWithSubstance wrote:
Is m + z > 0?

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

[Reveal] Spoiler:
I chose E, but correct answer was C. Can anyone explain?

From statement 1, m > 3z. From statement 2, 4z > m. So m is between 3z and 4z. Then m + z can have values between 4z and 5z (not inclusive). So shouldn't the sign of this depend on z? If z is negative, m + z is negative, and if z is positive, m + z is positive. So we can't determine whether m+z is positive or negative.

What am i missing?

Thanks!

Adding 1 and 2 we get => z>0
and m>3z=> m>0 too .
hence C
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Re: Is m + z > 0? [#permalink]

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13 Mar 2016, 06:43
Rule used here => WE CAN ADD TWO OR MORE INEQUALITIES if and only if They are in the same direction is Same < or > symbol .
Never Subtract ,multiply or divide them as it will get messed up..
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Re: Is m + z > 0?   [#permalink] 13 Mar 2016, 06:43
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