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

Notice that 3z < m < 4z implies that z is positive (3z < 4z --> 0 < z), so you should not use negative numbers for z. Positive z on the other hand implies that m is positive too. So, m + z = positive + positive = positive.

Hope it helps.

Thank you so much for replying this quickly!

I understood. Thanks a lot Bunuel. You are a saviour!

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 2 variables: Let the original condition in a DS question contain 2 variables . In other words, there are two fewer equations than variables. Now, we know that each condition (1) and (2) would usually give us an equation, however, since we need 2 equations to match the numbers of variables and equations in the original condition, the unequal number of equations and variables should logically give us an answer C.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find whether m + z > 0 .


Second and the third step of Variable Approach: From the original condition, we have 2 variables (m and z).To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation each, C would most likely be the answer.

Let’s take a look at both conditions together .

Condition(1) tells us that m - 3z > 0 or m > 3z .

Condition(2) tells us that 4z - m > 0.

=> 4z - m > 0 or m < 4z

Adding both inequalities: m - 3z + 4z - m > 0 => z > 0 [Hence, z is positive]

Combining both: 3z < m < 4z

When z > 0 then m + z will also be greater than '0' - is m +z > 0 - YES

Since the answer is unique YES , both conditions together are sufficient by CMT 1.


So, C is the correct answer.

Answer: C

The question pertains to the sign of an addition of terms. This should give us a clue that if we have to combine the inequalities, we need to add the inequalities.

From statement I alone, m – 3z > 0.
Therefore, m > 3z

If z = -1 and m = 1, m > 3z but m + z = 0. The main question can be answered with a NO.
If z = -1 and m = 2, m > 3z but m + z > 0. The main question can be answered with a YES.
Statement I alone is insufficient. Answer options A and D can be eliminated.

From statement II alone, 4z – m > 0.
Therefore, 4z > m or m < 4z.

If z = 0, m = -1, m <4z but m + z < 0. The main question can be answered with a NO
If z = 1, m = 1, m < 4z but m + z > 0. The main question can be answered with a YES.
Statement II alone is insufficient. Answer option B can be eliminated.

Combining statements I and II, we have the following:
From statement I alone, m > 3z; from statement II alone, m < 4z. Therefore, 3z < m < 4z.

Adding the two inequalities given, m – 3z + 4z – m > 0, we can see that z > 0. Therefore, m has to be positive.
Since both m and z are positive, m + z > 0.
The combination of statements is sufficient. Answer option E can be eliminated.

The correct answer option is C.

Hope that helps!
Aravind B T

Statement 1: m>3z
m=1; z=-1;
m-3z>0; sure
m+z = 0; So, no

m=8, z=1;
m>3z? yes
m+z>0? yes

So, Statement 1 is insufficient

Statement 2: 4z>m
z=1; m=2
4z>m? Yes
m+z>0? Yes

z=1, m=-2
4z>m? Yes
m+z>0? No

So, Statement 2 is insufficient

Combining both:
(1) m>3z
(2)m<4z

Multiply (1) with -1: -m<-3z
Add (1) to (2)
0<z
i.e. z>0
Put this in original (1)
m>0
Therefore, m+z> 0

Therefore, (C)

We are going to Tahiti

My 2 cents on this:

For me, the best way to solve this was to realize that both statements alone were not giving any answer. Goal is to find whether m + z > 0

Jumping ahead to St 1 and 2 together, By adding both statements:

m - 3z> 0
-m + 4z> 0
-------------
z>0

Therefore, if m - 3z >0

m>3z

m> 3 (z, which is some positive number)

hence m is positive and m+z > 0

C

I did the same mistake

Can you please explain how did you draw these diagrams?