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

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Is m+z > 0? (1) m-3z > 0 (2) 4z-m > 0 [#permalink] New post 18 Mar 2007, 15:43
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Is m+z > 0?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-m-z-0-1-m-3z-0-2-4z-m-106381.html
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 [#permalink] New post 19 Mar 2007, 01:23
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(C) for me :)

To me, I prefer to draw an XY plan in order to conclude faster :).... Let understand m as x and z as y.

m+z > 0 ?
<=> z > -m ? >>> This question ask us if the points (m,z) are above the line z = -m. In the Fig 1, I draw in green the area we are looking for.

From 1
m-3z > 0
<=> z < m/3 >>> It's all points (m,z) below the line z=m/3

Obviously, by looking at the Fig 2, we can conclude that we have points in the green area where z > -m and points in the red one where z < - m.

INSUFF.

From 2
4z-m > 0
<=> z > m/4 >>> It's all points (m,z) above the line z=m/4

Obviously, by looking at the Fig 3, we can conclude that we have points in the green area where z > -m and points in the red one where z < - m.

INSUFF.

Both (1) and (2)
We can conlude with the graph as well, but I prefer here to use some alegbra

We have the system:
o m-3z > 0 (A)
o 4z-m > 0 (B)

(A) + (B) <=> (m-3z) + (4z-m) > 0
<=> z > 0

As well,
(A) <=> m > 3*z
=> m > 3*z > z > 0 as z > 0 then, 3z > z.

Thus,
m>0 and z>0
=> m + z > 0.

SUFF.
Attachments

Fig1_z sup to -m.gif
Fig1_z sup to -m.gif [ 3.99 KiB | Viewed 4463 times ]

Fig2_z inf to m div 3.gif
Fig2_z inf to m div 3.gif [ 4.32 KiB | Viewed 4463 times ]

Fig3_z sup to m div 4.gif
Fig3_z sup to m div 4.gif [ 4.08 KiB | Viewed 4459 times ]

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 [#permalink] New post 19 Mar 2007, 03:14
Quote:
Both (1) and (2)

We have the system:
o m-3z > 0 (A)
o 4z-m > 0 (B)

(A) + (B) <m> 0
<z> 0


Fig, this is an excellent way to do this question which I did not follow and took a very long time to solve it. Thanks!!!
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 [#permalink] New post 19 Mar 2007, 07:56
How I solved:



Statement 1:

M - 3Z > 0

You can use algebra to get this to look like:

M/Z > 3

This tells us that M or Z could be negative or positive

since to get a positive result M and Z are either both positive or both negative


Since we need to see if M + Z > 0

We need to see if adding the 2 would give us a value above 0

from the statement we get 2 negatives or 2 positives however adding 2 negatives would gives us a positive, adding 2 positives would give us a negative

INSUFF

Statement 2:

4z-m > 0

Use algebra bring us to

M/Z > 4

same thing as statement 1


COMBINED

we have M - 3Z > 0
-M + 4Z > 0

adding both statemetns we are left with

Z > 0

since Z is greater than 0, then M must be greater than as well

since M/Z must be positive according to the statements

M + Z has to be positive as well


C
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 [#permalink] New post 19 Mar 2007, 08:25
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One more approach :lol:

From Stmt 1
We have 3Z < M

From Stmt 2
we have M < 4Z

Since we dont know if M and Z are +ve or -ve above are INSUFF individually

Combining
3Z < M < 4Z
Since 3Z < 4Z , Z has to be +ve
so M has to be +ve
and therefore M + Z has to be +ve

Answer C
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 [#permalink] New post 19 Mar 2007, 15:17
Hi Kyatin!

Yours is also a good approach. Thanks! :)
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Re: DS: Algebra [#permalink] New post 01 Feb 2011, 22:33
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?
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Re: DS: Algebra [#permalink] New post 02 Feb 2011, 01:54
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puneetj wrote:
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?


terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by z: never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 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.

Also discussed here: data-suff-67183.html
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Re: DS: Algebra [#permalink] New post 03 Feb 2011, 20:45
Hi Bunuel,

Z is positive when combining both the statements - understood.
Using statement (1)m-3z > 0 you proved that m>3z = positive - understood

How do you prove statement (2) the way you proved statement 1? or should we bother proving it at all?
4z>m which means 4 * some positive number > m. Does that prove anything about m?

Please confirm.
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Re: DS: Algebra [#permalink] New post 04 Feb 2011, 02:40
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puneetj wrote:
Hi Bunuel,

Z is positive when combining both the statements - understood.
Using statement (1)m-3z > 0 you proved that m>3z = positive - understood

How do you prove statement (2) the way you proved statement 1? or should we bother proving it at all?
4z>m which means 4 * some positive number > m. Does that prove anything about m?

Please confirm.


On the GMAT, two data sufficiency statements always provide TRUE information. So: (1) m-3z>0 and (2) 4z-m>0 are given to be true, you shouldn't prove them. You should check whether m+z>0 is true, which is done when we take these two (true) statements together: by adding them we get that z>0 and then looking on (1) with this info we get that m>0 too.

Hope it's clear.
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Re: DS: Algebra [#permalink] New post 05 Feb 2011, 19:53
Yes clear. Thanks Bunuel :)
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Re: DS: Algebra [#permalink] New post 11 May 2013, 07:10
Bunuel wrote:
puneetj wrote:
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?


terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by z: never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 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.

Also discussed here: data-suff-67183.html


Hi BUnuel
I agree to your solution and even i approached the problem in the same fashion. But i have got a doubt ...I understand that whn we combine both the statements we get....Z is positive and when we substitute in the equation m>3z...we end up saying that m must also be greater than 0. On the contrary when we substitute any positive value for z in statement 2 ; we end up getting both negative and positive value for m....
Since we have 4z-m>0 ie m<4z
Now if we take say z=1....m<4 that means we can have negative values too for z.....

Need your help...

Archit
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Re: DS: Algebra [#permalink] New post 12 May 2013, 01:45
Expert's post
Archit143 wrote:
Bunuel wrote:
puneetj wrote:
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?


terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by z: never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 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.

Also discussed here: data-suff-67183.html


Hi BUnuel
I agree to your solution and even i approached the problem in the same fashion. But i have got a doubt ...I understand that whn we combine both the statements we get....Z is positive and when we substitute in the equation m>3z...we end up saying that m must also be greater than 0. On the contrary when we substitute any positive value for z in statement 2 ; we end up getting both negative and positive value for m....
Since we have 4z-m>0 ie m<4z
Now if we take say z=1....m<4 that means we can have negative values too for z.....

Need your help...

Archit


Yes, from z>0 and m<4z we cannot get that m must be positive and that's exactly why we use m>3z to get that.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-m-z-0-1-m-3z-0-2-4z-m-106381.html

In case of any questions please post in that thread.
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Re: DS: Algebra [#permalink] New post 27 Apr 2014, 18:33
Bunuel wrote:
puneetj wrote:
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?


terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by z: never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 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.

Also discussed here: data-suff-67183.html


Bunuel, can you explain why (1) and (2) are immediately insufficient? I didn't quite understand your reasoning.
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Re: DS: Algebra [#permalink] New post 28 Apr 2014, 02:19
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TooLong150 wrote:
Bunuel wrote:
puneetj wrote:
why does M/Z have to be greater than 4 according to the 2nd statement by terp26? Z could be =1 and m could be = -10 and 4z-m could still be >0
if z=1, m=1,then 4z-m = 3 > 0
if z=1, m= -10 so 4z-m = 14 > 0......am I missing something?


terp26's reasoning for (1) and (2) is not correct: you can not write m/z>3 from m>3z in (1) or 4>m/z from 4z>m in (2) (or 4<m/z).

What terp26 is actually doing when writing m/z>3 from m>3z is dividing both parts of the inequality by z: never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So, m/z>3 would be correct in case z>0 but in case z<0 it'll be m/z<3.

Is m+z > 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.

Also discussed here: data-suff-67183.html


Bunuel, can you explain why (1) and (2) are immediately insufficient? I didn't quite understand your reasoning.


Plug values:

Is m+z > 0?

(1) m - 3z > 0. If m=1 and z=0, then the answer is YES but if m=1 and z=-1, then the answer is No.

(2) 4z - m > 0. If m=0 and z=1, then the answer is YES but if m=-1 and z=1, then the answer is No.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-m-z-0-1-m-3z-0-2-4z-m-106381.html
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: DS: Algebra   [#permalink] 28 Apr 2014, 02:19
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3 Experts publish their posts in the topic Is m+z > 0? 1) m-3z >0 2) 4z-m >0 According to me, rvind 8 14 Oct 2011, 11:47
29 Experts publish their posts in the topic Is m+z > 0 (1) m-3z > 0 (2) 4z-m > 0 smitakokne 29 16 Dec 2010, 09:24
Is m+z > 0? 1) m - 3z > 0 2) 4z - m > 0 uzonwagba 2 26 Jul 2009, 21:43
PFA is m+z >0 1: m -3z >0 2: 4z -m >0 vr4indian 2 05 Oct 2008, 02:39
Is m+z > 0? (1) m -3 z > 0. (2) 4z - m > 0 kapilnegi 4 07 Sep 2008, 06:31
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