M+Z>0

[saege]

How to solve this Q?

[rhymes_with_luck]

once you rule out 1 and 2 independently and combine them,

what you get is

1) 4z>m>3z

2) by adding two inequalities

z>0

hence m>0

and m+z>0

C

[saege]

z>0

How come z>0

[rhymes_with_luck]

add ineq 1 and ineq 2

[sumanal]

From what i could make out..

1. says m>3z.. in which case either of m or z or both can be negative.

same is the case with statement 2.

If u consider both the statements..

m>3z but m<4z
this can happen only when both the variables are positve because when z is negative for ex: -3 ..there exists no solution for m which will be
>-9 but <-12 so for the inequality to hold true.. z should be positve n therefore m also positve.
and hence the addition of 2 positive numbers positive

Hope that gave a better insight.

[gmattester]

Can someone plz. help me understand this problem
If m-3z>0
Then m>3z only if m and z are positive.
And m<3z if m and z are negative
So 1st is insufficient

2) Similarly for 2nd
4z>m only if m and z are positive
Else 4z<m

Now combining 1 and 2 we get
3z<m<4z

Now m and z can be positive as well as negative
So how we reached conclusion that m+z>0

Also I am not sure this is right way to solve this problem

[Ian Stewart]

Can someone plz. help me understand this problem
If m-3z>0
Then m>3z only if m and z are positive.
And m<3z if m and z are negative

If m-3z > 0, then m > 3z. It does not matter whether m and z are positive or negative. We're only adding 3z to both sides of the inequality here, and you never need to worry about reversing an inequality when you add or subtract the same quantity on both sides.

I expect you're thinking about multiplying (or dividing) on both sides of an inequality- if you multiply or divide by a negative, you must reverse the inequality.

[parallel_chase]

Now combining 1 and 2 we get
3z<m<4z

Now m and z can be positive as well as negative
So how we reached conclusion that m+z>0

Also I am not sure this is right way to solve this problem

if you combine I & II
m-3z>0
4z-m>0

If you add these two inequalities

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

if z > 0, then m will also be greater than 0 because m>3z

Hence m+z>0, answer is C.

Always remember the following rule for inequalities.

1. Inequalities can only be added to each other, they cannot be subtracted.

2. When adding two inequalities, the signs of both the inequalities should be in the same direction.

Like we did in this case.

Hope this helps. Let me know if you still have any problems.

[Suyog]

https://www.beatthegmat.com/gprep-1-expl ... 16244.html

[gmattester]

Can someone plz. help me understand this problem
If m-3z>0
Then m>3z only if m and z are positive.
And m<3z if m and z are negative

If m-3z > 0, then m > 3z. It does not matter whether m and z are positive or negative. We're only adding 3z to both sides of the inequality here, and you never need to worry about reversing an inequality when you add or subtract the same quantity on both sides.

I expect you're thinking about multiplying (or dividing) on both sides of an inequality- if you multiply or divide by a negative, you must reverse the inequality.

Thanks Ian. I read one of your post where it was x/y>1 and you explained this reversing an inequality thing. So I applied same concept here but forgot this rule is only true for multiplication and division. Thanks for clearing that.

[gmattester]

Now combining 1 and 2 we get
3z<m<4z

Now m and z can be positive as well as negative
So how we reached conclusion that m+z>0

Also I am not sure this is right way to solve this problem

if you combine I & II
m-3z>0
4z-m>0

If you add these two inequalities

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

if z > 0, then m will also be greater than 0 because m>3z

Hence m+z>0, answer is C.

Always remember the following rule for inequalities.

1. Inequalities can only be added to each other, they cannot be subtracted.

2. When adding two inequalities, the signs of both the inequalities should be in the same direction.

Like we did in this case.

Hope this helps. Let me know if you still have any problems.

Thanks chase.........
Now it is more clear to me.

[codesnooker]

https://www.beatthegmat.com/gprep-1-expl ... 16244.html

Suyog, the link that you have given, in that post, the question is wrongly posted. Check again the inequalities sign for the second statement.

And yes answer (C) is correct answer.