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

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Retired Moderator
Joined: 25 Feb 2013
Posts: 1214
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Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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03 May 2018, 22:47
Bunuel wrote:
Is m+z > 0

(1) m - 3z > 0. Insufficient on its own.

(2) 4z - m > 0. Insufficient on its own.

$$(m-3z)+(4z-m)>0$$;

$$z>0$$, 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). Therefore, $$m+z=positive+positive>0$$. Sufficient.

For graphic approach refer to: http://gmatclub.com/forum/is-m-z-0-1-m- ... 75657.html

For these types of questions, is there a rule of thumb or a easy way to tell whether the statements alone will be sufficient? I always waste a lot of time on these types of questions.

Note that we have two variables here m & z but each statement offers you only ONE equation. Hence it is not possible to solve. For solving two variables you need at least another equation or a relationship between the two variables (which will be your other equation).

So you can negate the statements simply by looking at it. this will save your time.
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Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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22 Jul 2018, 17:33
Bunuel wrote:
Is m+z > 0

(1) m - 3z > 0. Insufficient on its own.

(2) 4z - m > 0. Insufficient on its own.

$$(m-3z)+(4z-m)>0$$;

$$z>0$$, 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). Therefore, $$m+z=positive+positive>0$$. Sufficient.

For graphic approach refer to: http://gmatclub.com/forum/is-m-z-0-1-m- ... 75657.html

Hi Bunuel..

When we use statement 2, we get that 4z>m... lets say 4(1)>-1 then it is in sufficient; and 4(1)>1 then it is sufficient

Statement 1 satisfies the condition and hence m+z is greater than 0 but statement 2??

Thanks for ur help in advance
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Joined: 19 Aug 2016
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Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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22 Jul 2018, 17:38
Bunuel wrote:
Is m+z > 0

(1) m - 3z > 0. Insufficient on its own.

(2) 4z - m > 0. Insufficient on its own.

$$(m-3z)+(4z-m)>0$$;

$$z>0$$, 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). Therefore, $$m+z=positive+positive>0$$. Sufficient.

For graphic approach refer to: http://gmatclub.com/forum/is-m-z-0-1-m- ... 75657.html

Hi Bunuel...

From statement 2 we know that 4z>m in which case lets say.. 4(1)>1 the statement is suff that m+z>0

But 4(1)>-2 in which case M+z>0 is not true

Thanks For ur help in advance
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Posts: 55150
Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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22 Jul 2018, 20:27
zanaik89 wrote:
Bunuel wrote:
Is m+z > 0

(1) m - 3z > 0. Insufficient on its own.

(2) 4z - m > 0. Insufficient on its own.

$$(m-3z)+(4z-m)>0$$;

$$z>0$$, 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). Therefore, $$m+z=positive+positive>0$$. Sufficient.

For graphic approach refer to: http://gmatclub.com/forum/is-m-z-0-1-m- ... 75657.html

Hi Bunuel...

From statement 2 we know that 4z>m in which case lets say.. 4(1)>1 the statement is suff that m+z>0

But 4(1)>-2 in which case M+z>0 is not true

Thanks For ur help in advance

Yes, (2) is not sufficient because it gives an YES and a NO answers to the question whether m + z > 0 but what is your doubt?
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Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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22 Sep 2018, 16:49
I think except for Bunuel all others are wrong, they simply take the 3z to the other side of the inequality without knowing whether z is positive or negative. Sorry If I am rude, but taking a variable to the other side of the inequality without knowing its sign is just plain wrong. If z were positive there is no change in the inequality sign, but if z were negative we must change the direction of the inequality. Thus the approach of adding two equations seem to be flawless.

So please don't take the z to the other side, or in general do not take any variable to the other side of the inequalilty without knowing it's sign, this affects the equation as a whole. Try and make a rational decision.

You should carefully note that bunuel took the z to the other side, only after "he proved z to be greater than 0". By adding two equations which were already greater than zero individually.
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Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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23 Sep 2018, 02:14
1
sriramsundaram91 wrote:
I think except for Bunuel all others are wrong, they simply take the 3z to the other side of the inequality without knowing whether z is positive or negative. Sorry If I am rude, but taking a variable to the other side of the inequality without knowing its sign is just plain wrong. If z were positive there is no change in the inequality sign, but if z were negative we must change the direction of the inequality. Thus the approach of adding two equations seem to be flawless.

So please don't take the z to the other side, or in general do not take any variable to the other side of the inequalilty without knowing it's sign, this affects the equation as a whole. Try and make a rational decision.

You should carefully note that bunuel took the z to the other side, only after "he proved z to be greater than 0". By adding two equations which were already greater than zero individually.

hi sriramsundaram91

if you are referring to m-3z>0 =>m>3z, then there is nothing wrong in taking to the other side of inequality. We do not need to know the sign of the variable here to subtract or add.
But if you are multiplying or dividing the inequality with a variable, then you need to know the sign of the variable because in this case sign of inequality will change based on the nature of the variable.
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Joined: 05 Mar 2019
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Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0  [#permalink]

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10 Apr 2019, 16:34
Is there a list of similar questions to practice?
Re: Is m + z > 0 (1) m - 3z > 0 (2) 4z - m > 0   [#permalink] 10 Apr 2019, 16:34

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