I think this is the solution.....

is 1 sufficient?
Plug in variables

Say m=1 z=-2, the answer is NO
Saym=100 and z=1, then YES

So 1 is insufficent

2
Say z=1, m=6, YES
Say z=-4, m=2, NO

Not sufficient

combined sufficent?

If you want to flip an inequality, multiply the whole side by -1
So, m-3z > 0 as is m-4z>0

m>3z, or m>4z. Insert variables -1 and -3 or 4 and 1. Different solutions. Therefore

E, no solution

It should be 'C'.

From 1: m-3z>0 => m>3z

both m and z could be positive or negative to satusfy this.
example : let m=7; z=2 so,m>3z ; here m+z>0

let m=-5 and z=-2 so, m>3z but here m+z<0
so not sufficient

From 2: 4z-m<0=>4z<m
let z=2, m=9 so 4z<m ; here m+n>0
let z=-2 amd m=-3; here 4z<m; but m+n <0, not sufficient

Now combining both :

from (1) m-3z>0
from (2) 4z-m<0=>m-4z>0
solving (1) and (2) z>0 => m>0 (since m>3z)
So, m+n >0; sufficient.


Amit

Hi Amit,

Just co-relating to the solution you have presented:

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

there are cases when

-->M Z
1>+ +
2> - -
3>+ -
what Im trying to say is that there are values which satisfy all the above cases of M&Z based on the 2 inequalities stated above(for eg: substitute m=5,z=1for op1,m=-1,z=-1 for op2,m=4,z=-1for op3). And for these options there can be different options of vaues for m+Z i.e. it can + for
option 1,- for op 2, not sure for op3 - so then how can both be suffiecient to arrive at an answer ....Could you please elaborate for my understanding ??[/u]

Hi Amit,

Just co-relating to the solution you have presented:

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

there are cases when

-->M Z
1>+ +
2> - -
3>+ -
what Im trying to say is that there are values which satisfy all the above cases of M&Z based on the 2 inequalities stated above(for eg: substitute m=5,z=1for op1,m=-1,z=-1 for op2,m=4,z=-1for op3). And for these options there can be different options of vaues for m+Z i.e. it can + for
option 1,- for op 2, not sure for op3 - so then how can both be suffiecient to arrive at an answer ....Could you please elaborate for my understanding ??[/u]

raptor84 wrote:IS m+z>0

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


OA C

hmm...i'm also getting E.

we know it's not A or B so it's either C or E.

option 1: m=5, z=1
Satisfies statement 1 & 2 and 5+1 = 6 > 0

option 2: m= -4, z= -2
1) -4 - 3(-2) = 2 > 0
2) 4(-2) - -4 = -4 < 0

m + z > 0 ====> -4 + - 2 = -6 < 0

it's giving conflicting results so shouldn't the answer be E?

Anyone?

the ans should be E
m=2 z = -3
m =4 z = -3

These two values satisfies 1 & 2 and will give you both negative and positive value for m+z...
Hence E

raptor84 wrote:IS m+z>0

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


OA C

I go with C..

here it is..

Adding statement 1 and 2 we get Z > 0 and if Z > 0
and if M-3z>0, it implies that M >0

thus hence m+z > 0

Hope it helps.. do let me know if u have any doubts..

@Sudhir3127 can u please comment on Kshin78's post. Thanks!

Suyog wrote:@Sudhir3127 can u please comment on Kshin78's post. Thanks!

When 'm' and 'z' are assumed; one time with +ve sign and other time with -ve sign they give no definite solution whether (m+z) >0 or not.

At times m+z>0 and at times (m+z) is <0.

When we combine the two (given) statements, we get that z is definitely > 0 and since m>3z => m is +ve. (i.e. 'z' and 'm' get their unique signs as being both positive when they both satisfy stem 1 & stem2).

When we combine two statements, we generally try to get some definite value/s or sign/s out of it to see whether it gives a definite ans always.
And that is what we get here , both "m" & "z" are positive and m>z, both satisfying two statements at the same time.

Hope this justifies.


Amit

IS m+z>0

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

1) m>3z
For z<0:
m+z can be +ve as well as -ve
Insufficient

2) m>4z
Again For z<0:
m+z can be +ve as well as -ve
Insufficient

Combining the 2 conditions we have
m>3z & m>4z.
These conditions will be true simultaneously only when
m>4z.
Which essentially gives us the second condition.
So insufficient.

I'll go with answer E.

Any comments?

Thanks

amitansu wrote:

Suyog wrote:@Sudhir3127 can u please comment on Kshin78's post. Thanks!

When 'm' and 'z' are assumed; one time with +ve sign and other time with -ve sign they give no definite solution whether (m+z) >0 or not.

At times m+z>0 and at times (m+z) is <0.

That's precisely what I'm trying to say, if with (I) & (II), we are getting different answer then how come its 'C'

When we combine two statements, we generally try to get some definite value/s or sign/s out of it to see whether it gives a definite ans always.
And that is what we get here , both "m" & "z" are positive and m>z, both satisfying two statements at the same time.

I guess m=5, z=1 are definite values

Which satisfies (I) & (II) both and it says the question asked is true.

Also, m= -4, z= -2 are again definite values

which again satisfies (I) & (II) both but it says the question asked is false.

I still don't agree...Sorry..

Someone needs to convince me....

Suyog wrote:

amitansu wrote:

Suyog wrote:@Sudhir3127 can u please comment on Kshin78's post. Thanks!

When 'm' and 'z' are assumed; one time with +ve sign and other time with -ve sign they give no definite solution whether (m+z) >0 or not.

At times m+z>0 and at times (m+z) is <0.

That's precisely what I'm trying to say, if with (I) & (II), we are getting different answer then how come its 'C'

When we combine two statements, we generally try to get some definite value/s or sign/s out of it to see whether it gives a definite ans always.
And that is what we get here , both "m" & "z" are positive and m>z, both satisfying two statements at the same time.

I guess m=5, z=1 are definite values

Which satisfies (I) & (II) both and it says the question asked is true.

Also, m= -4, z= -2 are again definite values

which again satisfies (I) & (II) both but it says the question asked is false.

I still don't agree...Sorry..

Someone needs to convince me....

Well, combining both statements we get both 'z' and 'm' as positive.
So we can't assume them negative.

amitansu wrote:

Suyog wrote:

amitansu wrote:

Suyog wrote:@Sudhir3127 can u please comment on Kshin78's post. Thanks!

When 'm' and 'z' are assumed; one time with +ve sign and other time with -ve sign they give no definite solution whether (m+z) >0 or not.

At times m+z>0 and at times (m+z) is <0.

That's precisely what I'm trying to say, if with (I) & (II), we are getting different answer then how come its 'C'

When we combine two statements, we generally try to get some definite value/s or sign/s out of it to see whether it gives a definite ans always.
And that is what we get here , both "m" & "z" are positive and m>z, both satisfying two statements at the same time.

I guess m=5, z=1 are definite values

Which satisfies (I) & (II) both and it says the question asked is true.

Also, m= -4, z= -2 are again definite values

which again satisfies (I) & (II) both but it says the question asked is false.

I still don't agree...Sorry..

Someone needs to convince me....

Well, combining both statements we get both 'z' and 'm' as positive.
So we can't assume them negative.

Why u r considering 'z' & 'm' as positive???
Consider
m= -4, z= -2

Now solve...

1) m-3z>0
-4 + 6 = 2, which is greater than 0

2) 4z-m<0
-8 +4 = -4, which is less than 0

doesn't value for m & z both satisfy 1 as well as 2

Now would u say that combining both statements we get both 'z' and 'm' as negative.
So we can't assume them positive.?????

moreover que doesn't say that m and v are positive...
I'm not assuming anything.....

[/quote]Why u r considering 'z' & 'm' as positive???
Consider
m= -4, z= -2

Now solve...

1) m-3z>0
-4 + 6 = 2, which is greater than 0

2) 4z-m<0
-8 +4 = -4, which is less than 0

doesn't value for m & z both satisfy 1 as well as 2

Now would u say that combining both statements we get both 'z' and 'm' as negative.
So we can't assume them positive.?????

moreover que doesn't say that m and v are positive...
I'm not assuming anything.....[/quote]


i'm so with you Suyog... i can't seem to figure it out...

the argument here is that m> 0 because z > 0. ok, true if z > 0, but WHY??? We can't assume that z > 0 when there are negative values that holds true for both equations.