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moneyman: Master | Next Rank: 500 Posts; Posts: 468; Joined: Sat Mar 03, 2007 10:17 pm; Thanked: 5 times

by moneyman » Mon Mar 31, 2008 2:43 am

Is m+z>0 ??

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

Fromt this I deduced that 3z<m<4z and the expression can be positive as well as negative right??

Pls help!!

Ans is C

Maxx

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Source: Beat The GMAT — Data Sufficiency | Mon Mar 31, 2008 2:43 am

camitava: Legendary Member; Posts: 645; Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members; Location: India; Thanked: 34 times

Re: TestPrep

by camitava » Mon Mar 31, 2008 2:48 am

moneyman wrote:Is m+z>0 ??

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

Fromt this I deduced that 3z<m<4z> 0 or z > 0
If z > 0, m > 3z i.e. m > 0
So m + z > 0

So go for C. Got me, Maxx?

Correct me If I am wrong

Regards,

Amitava

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Mon Mar 31, 2008 4:06 am

Though I have not started working on inequalities, so it is hard for me to understand the above solution posted by Camitava.

However, there is also another way to solve this problem without having much knowledge about inequalities. Here it goes:-

Statement 1: m - 3z > 0 => m > 3z
Statement 2: 4Z - m > 0 => 4z > m

From both statement as you deduce that 4z > m > 3z

Now lets check the solution with negative values to prove that m + z < 0

Lets take z = -1 (simplest one)
i.e -4 > m > -3
which is incorrect as -4 can't greater than -3.

Now lets check with some negative fractional value.
Lets take -1/12 (divisible by both 3 and 4)

therefore equation becomes, -0.33 > m > -0.25, which again incorrect, as -0.33 cannot be greater than -0.25

So, it means z cannot be negative. So, if z is not negative, then m also cannot be negative.

Therefore adding 2 positive number yields a positive number.
i.e. m + z > 0

There C is the answer

Last edited by codesnooker on Mon Mar 31, 2008 5:27 am, edited 3 times in total.

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

Re: TestPrep

by codesnooker » Mon Mar 31, 2008 4:09 am

camitava wrote:
From this I deduced that 3z<m<4z> 0 or z > 0

Hi Camitava, Can you explain here how you have taken z > 0 ?

I didn't understand this point.

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camitava: Legendary Member; Posts: 645; Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members; Location: India; Thanked: 34 times

Re: TestPrep

by camitava » Mon Mar 31, 2008 4:49 am

codesnooker wrote:
camitava wrote:
From this I deduced that 3z<m<4z> 0 or z > 0
Hi Camitava, Can you explain here how you have taken z > 0 ?

I didn't understand this point.

Sorry! I really don;t know how did I mess up while posting my last post.

I meant to say that - add the two equs : m-3z>0
and 4z-m>0.
So m - 3z + 4z - m > 0
or z > 0
From this m > 3z or m > 0
So m + z > 0

Got me, Codesnooker?

Correct me If I am wrong

Regards,

Amitava

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camitava: Legendary Member; Posts: 645; Joined: Wed Sep 05, 2007 4:37 amFollowed by: 5 members; Location: India; Thanked: 34 times

by camitava » Mon Mar 31, 2008 4:51 am

Ohhh! Got it now... Last time I forgot to uncheck the option - Disable HTML in this post or may be I messed up in some other way ...

Correct me If I am wrong

Regards,

Amitava

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Mon Mar 31, 2008 4:52 am

Sure, thanks

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codesnooker: Legendary Member; Posts: 543; Joined: Fri Jan 18, 2008 1:01 am; Thanked: 43 times; GMAT Score:580

by codesnooker » Mon Mar 31, 2008 5:29 am

Yes HTML code is ruining the answers in case of questions of inequalities. Disable them, before posting answers or questions of inequalities

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Mon Mar 31, 2008 9:02 am

Just as a general tip, you can turn html off under your profile and then just unclick the box if you ever decide you need it.

As a quick point on this discussion (if it's even needed anymore), Codesnooker got to:

From both statement as you deduce that 4z > m > 3z

From here, we can see that:

4z > 3z

which can only be true if z is positive.

And, since m > 3z, this means that m is also positive.

So, since both z and m are positive, it must be true that:

m + z

is also positive.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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cjiang16: Senior | Next Rank: 100 Posts; Posts: 71; Joined: Sun Apr 06, 2008 2:24 pm; Thanked: 1 times

by cjiang16 » Sat Apr 12, 2008 4:41 pm

C is the answer.

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