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(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.

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.

(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.

Answer: C.

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...

(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.

Answer: C.

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 \(5>1\) --> \(3-5<4-1\). _________________

Re: DS Inequality question [#permalink]
26 May 2011, 11:07

1

This post received KUDOS

Wow - Thank you agdimple333. That was quick and clear!

I was combining both to get that 3z < m < 4z but I did not add them to get that z > 0 as well. Now that I think again if m lies between 3z and 4z, then by this equation alone z has to be GT 0 because there is NO z < 0 that will ever satisfy this equation. Thanks again. _________________

I'd like you to conform my solution to this problem. I got this right but not sure if I solved it correctly.

Stm 1) m -3z > 0

m > 3z --- > since both m and n can be negative or positive, it is insufficient.

Stm 2) 4z - m >0

4z > m ------ > again, since both m and n can be negative or positive, it is insufficient.

Together ) m > 3z and m < 4z ---- > 3z < m < 4z ---> if either m or n is negative, this inequality is impossible. Hence, this stm tells us that both m and n are positive. In other words, m + n > 0 --------- > Sufficient.

Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]
21 Nov 2012, 19:34

arvindbhat1887 wrote:

Is m+z>0?

(1) m > 3z (2) m < 4z.

The answer is given as (C) Both statements together. I don't understand how. Can someone please explain? I think the answer is (E) Not sufficient with both (1) tells you that m > 3z (2) tells you that m < 4z. Either of the two cases taken individually is not sufficient. So Rule out (A), (B), (D) Therefore, 3z<m<4z when you combine the two. Now z can take both -ve and +ve values. So, m + z can be either -ve or +ve depending on the value of z. Hence, (E). Please help.

Stem 1: m>3z or m-3z >0 doesnt tell us anything about m+z. Not sufficient.

Stem 2: m<4z or m-4z <0 or 4z-m >0 tells us nothing again. Not sufficient.

combining, we get that 3z <m <4z also, if we add both equations, m-3z >0 4z-m >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.

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? _________________

(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.

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. _________________

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