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Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 [#permalink]
 16 Dec 2010, 10:24
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Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0
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smitakokne wrote: Is m+z > 0
1. m-3z > 0
2. 4z-m > 0
OA : C
Need help in underdstanding how we arrive at C. 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 
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psirus wrote: smitakokne wrote: Is m+z > 0
1. m-3z > 0
2. 4z-m > 0
OA : C
Need help in underdstanding how we arrive at C. 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.
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Thanks Bunuel! That's helps a lot!
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Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m-3z > 0
2. 4z-m > 0
OA : C
Need help in underdstanding how we arrive at C. (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...
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jullysabat wrote: Bunuel wrote: smitakokne wrote: Is m+z > 0
1. m-3z > 0
2. 4z-m > 0
OA : C
Need help in underdstanding how we arrive at C. (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... 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.
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Re: DS Inequality question [#permalink]
26 May 2011, 11:10
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.
Last edited by agdimple333 on 26 May 2011, 12:13, edited 1 time in total.
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Re: DS Inequality question [#permalink]
26 May 2011, 12:07
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.
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Brunel, could you help me to figure out where I am wrong.
From statement 1: M-3Z > 0
Adding 3Z to both sides of inequality
M>3Z
is Z is negative, then 3 times that negative number is definitely less than that number
for example Z = -.25
3Z = -.75 < Z
Z = -1
3Z = - 3 < Z
Zero does not apply because, then both sides would be equal.
Since we its evident that both Z and M are positive Z+M is greater than zero...
Where am I going wrong with my reasoning?
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manishgeorge wrote: Brunel, could you help me to figure out where I am wrong.
From statement 1: M-3Z > 0
Adding 3Z to both sides of inequality
M>3Z
is Z is negative, then 3 times that negative number is definitely less than that number
for example Z = -.25
3Z = -.75 < Z
Z = -1
3Z = - 3 < Z
Zero does not apply because, then both sides would be equal.
Since we its evident that both Z and M are positive Z+M is greater than zero...
Where am I going wrong with my reasoning? m>3z m=0; z=-1; m>3z; m+z=-1<0 m=1; z=0; m>3z; m+z=1>0 m=+1; z=-1; m>3z; m+z=0 Thus, knowing that m>3z is not sufficient to find whether m+z>0 Not Sufficient.
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As all the other answers A and B not sufficient.
I line the inequalities up together
3z<m<4z; only way this holds is if all are positive. so C
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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.
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GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]
21 Nov 2012, 20:03
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.
Last edited by Vips0000 on 21 Nov 2012, 20:30, edited 2 times in total.
Edited the question.placed stems properly
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Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]
21 Nov 2012, 20:34
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 we get, z>0 Thus since m>3z => m is also >0 therefore m+z >0 Sufficient. Ans C it is.
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Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]
21 Nov 2012, 21:06
z must be positive because we know 3z < 4z. Only positive values of z make this statement true. If m is greater than 3z, it must be positive as well. Therefore, m+z is positive. Posted from my mobile device 
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Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0 [#permalink]
22 Nov 2012, 05:55
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. Merging similar questions. Please refer to the solutions above. Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html
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Re: GMATPrep DS Question: Is m+z>0? (1) m-3z>0 (2) 4z-m>0
[#permalink]
22 Nov 2012, 05:55
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