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

Video solution from Quant Reasoning starts at 11:40
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Target question: Is m + z > 0?

Statement 1: m - 3z > 0
There are several values of m and z that satisfy statement 1. Here are two:
Case a: m = 10 and z = 1. In this case, the answer to the target question is YES, m + n is greater than zero
Case b: m = 1 and z = -2. In this case, the answer to the target question is NO, m + n is not greater than zero
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 4z - m > 0
There are several values of m and z that satisfy statement 2. Here are two:
Case a: m = 1 and z = 1. In this case, the answer to the target question is YES, m + n is greater than zero
Case b: m = -5 and z = 1. In this case, the answer to the target question is NO, m + n is not greater than zero
Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that m - 3z > 0
Statement 2 tells us that 4z - m > 0

To make things clearer, let's rearrange the terms in the first inequality to get:
-3z + m > 0
4z - m > 0

Since the inequality symbols are facing the SAME direction, we can ADD the inequalities to get: z > 0
Perfect! We now know that z is positive, but we still need information about the value of m.
There are several ways to accomplish this. My approach is to take the system:
-3z + m > 0
4z - m > 0

Multiply both sides of the top inequality by 4, and multiply both sides of the bottom inequality by 3 to get the following equivalent system:
-12z + 4m > 0
12z - 3m > 0
Since the inequality symbols are facing the SAME direction, we can ADD the inequalities to get: m > 0

We now know that m and z are both positive, which means m + z > 0
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

RELATED VIDEO

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.

Bunuel,

is m+z > 0 the same as m/z > -1 ?

if so, then 1. would be m/z > 3, which is SUFF
and 2. would be m/z < 4, which is INSUFF

so I would have said A

Please correct me if I'm wrong - my inequality skills are a bit rusty

No, it's not correct.

When you are writing m/z>-1 from m+z>0 you are actually dividing both parts of inequality by z: never multiply or reduce (divide) inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

So if we knew that z>0 then m+z>0 --> m/z+1>0 and if we knew that z<0 then m+z>0 --> m/z+1<0.

Hope it's clear.

Hello Bunnel,

As you said in the above post..
"(1)+(2) Remember we can add inequalities with the sign in the same direction -->"
what should we have different signs...

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Plus in arbitary values to check is m+z is always greater than 0 or there are other possibilities

st-1 ==> m - 3z >0
m =1 and z = -1 ==> m+z = 0
m=2 and z = -1 ==> m+z >0
not sufficient

st-2 ==> 4z-m>0
m = -1 and z = 0.1 ==> m+z < 0
m=2 and z = 1 ==> m+z >0
not sufficient

combining both

m-3z + 4z -m >0
z > 0.... (P)
from 1, m > 3z and from 2, 4z > m
==> 3z < m < 4z... (Q)
from (P) and (Q) m+Z > 0

Answer C.

I have one question when statements are combined, I got till the part z > 0 now I got confused here since in statement 1 I could prove it m>3z ( m is positive) but didn't how to apply z>0 to the second statement >>> 4z - m > 0

So when you reach that stage you can apply it to any of the statement and conclude its sufficient or you have to conclude using both statement separately?

You got that m is positive with (1), so can stop there. If you combine you get that 4z>m>3z>0.

Hi Bunuel, I have a doubt in this question. it is asked whether m+z>0 this implies that is m>-z?

Statement 1 says: m-3Z>0 then m>3z and 3z is definitely greater than -z then isnt m>-z. Please clarify.

3z is not always greater than -z. It's only true when z is positive, but if z is negative, then 3z < -z.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is m+z > 0

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

In the original condition, there are 2 variables(m,z), which should match with the number of equations. So you need 2 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1)&2), add the 2 equations, which is m-3z+4z+m>0, z>0. Then m>z>0 is derived from m>3z>z, which is m>0. So, m+z>0 is always yes and sufficient. Therefore, the answer is C.


-> For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

Responding to a pm:



There really isn't one suitable method for all inequality questions. The approach depends on the kind of question. Very rarely will I plug in numbers here to find cases where the inequality holds or does not hold.
This is how I will do this question:

"Is m+z > 0?"
Here I think that m and z could both be positive or one of them could be positive with greater absolute value than the other. If both are negative, this doesn't hold. Go on to stmnts.

(1) m-3z > 0

Here, I will try to segregate the variables to get relation between them.
m > 3z
The moment I see this, I naturally go to the number line.

(z positive):

______________________________ 0 ____z_________________3z___________m____________________

OR

(z negative):

__3z_________(m)______________z___(m)__ 0 _______________(m)____________________

Both z and m could be positive, both negative or z negative m positive. Not sufficient.


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

(z positive):
__________________(m)____________ 0 __(m)_______z_____________________(m)_____4z______

OR

(z negative):

______m_______4z____________________z_____ 0 _______________________

Both z and m could be positive, both negative or m negative z positive. Not sufficient.

Both together,

3z < m < 4z

(z positive):

______________________________ 0 ____z_________________3z_____m_______4z___________

OR

(z negative):

___________4z________3z___________________z_____ 0 _______________________

m has to be to the left of 4z but right of 3z. Not possible.

So m and z both must be positive. So m + z > 0

Answer (C)

Thank you so much Karishma. You're a saviour!

Is it possible to visualise such problems rather than drawing on number line? I think that will be prone to errors.

Besides, I have noticed that problem statements concerning number line involves too many possibilities as you explained in detail. For the same reason, I feel these questions are designed to put the test taker in a situation where it consumes more than required time? My gut will always push me to rush on these questions therefore.

Rephrasing the question is vital I guess. Let me know if the only way through such questions is to draw the number line and analyze?

Cheers!

Posted from my mobile device

Most questions can be solved using many different methods.

You can use the graphical approach discussed by walker here: https://gmatclub.com/forum/is-m-z-75657.html

As for too many possibilities on the number line, it seems that way when I draw it out but here is how I see it in my mind.

Stmtn 1:
There are only 2 cases: z is positive (or 0) or z is negative.
If z is to the right of 0, 3z is further to the right of z and m is further to the right of 3z. m has to be positive.
If z is to the left of 0, 3z is further to the left of z but m is anywhere on the right of 3z. m could be negative or positive.
That's all I need to know.

We must determine whether the sum of m and z is greater than zero.

Statement One Alone:

m – 3z > 0

We can manipulate the inequality in statement one to read: m > 3z

However, we still cannot determine whether the sum of m and z is greater than zero. Statement one alone is insufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

4z – m > 0

We can manipulate the inequality in statement two to read: 4z > m

However, we still cannot determine whether the sum of m and z is greater than zero. Statement two alone is insufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using our statements together, we can add together our two inequalities. Rewriting statement 2 as –m + 4z > 0, we have

m – 3z > 0 + (-m +4z > 0)

z > 0

Since we know that z > 0 and that m > 3z (from statement one), m must also be greater than zero. Thus, the sum of m and z is greater than zero.

Answer: C