DS-Prob 1

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DS-Prob 1

by er.twi.fb » Sat Jul 31, 2010 11:16 pm
Is m+z>0?


1>m>3z

2>m<4z


OA-C

My Solution is as follows.

1-Not Suff
2-Not Suff

1+2> 3z<m<4z


let z=5, then 15<m<20 and suppose m =16,so m+z=5+16>0
let z=-5, then -15<m<-20 and suppose m=-16,so m+z=-5-16<0

Not Sufficient..

So I go for "E" but OA is "C". Any suggestions??
Source: — Data Sufficiency |

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by kvcpk » Sun Aug 01, 2010 12:47 am
Is m+z>0?
m>3z
put m = 1, z= 0
1+0 >0.. YES
put m = -1, z = -2
-1-2<0.. NO
INSUFF

m<4z
put m = 0, z=1
0+1>0.. YES
put m= -10, z = -1
-10-1 <0.. NO
INSUFF

Combining:
m>3z
m<4z
3z<m<4z
4z<m+z<5z
4z<5z implies that z>0
m>3z
implies that m>0
Hence m+z>0
SUFF

pick C

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by er.twi.fb » Sun Aug 01, 2010 12:51 am
Great. I got it now. Just a simple trick. I should have cracked that :-(

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by kvcpk » Sun Aug 01, 2010 1:21 am
er.twi.fb wrote:Great. I got it now. Just a simple trick. I should have cracked that :-(
Questions look tricky.. What is the source?

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by selango » Sun Aug 01, 2010 8:41 am
m+z>0?

stmt1,

m>3z

No info abt m and z signs.

Insuff

stmt2,

m<4z

No info abt m and z signs.

Insuff

Combining 1 and 2,

m>3z --A

m<4z or -m>4z--B

A+B-->7z>0 or Z>0

3z<m<4z

Since Z>,m must surely positive and m>0

Hence m+z>0
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by selango » Sun Aug 01, 2010 8:42 am
I think this is one of GMATPREP problem..rt?
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by lunarpower » Tue Aug 03, 2010 4:00 am
to the original poster:
your error is here --
er.twi.fb wrote:let z=-5, then -15<m<-20 and suppose m=-16,so m+z=-5-16<0
"-15 < m < -20" is impossible; a number can't be both greater than -15 and less than -20 at once.
if this is not obvious, the obfuscation comes from the fact that the numbers are negative -- many people don't really have the proper intuition regarding negative numbers. however, the statement -15 < m < -20 is a lot like the statement 10 < m < 5; both are contradictory to themselves.

in fact, the statement 3z < m < 4z, by itself, proves that m and z are both positive, since "3z < 4z" is a false statement unless z is positive (and once you've established that z this positive, it follows that m must also be positive, since it's sandwiched between two positive values).
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by lunarpower » Tue Aug 03, 2010 4:31 am
by the way, this problem is from GMATPREP, and it has not been reproduced faithfully.

the original of the problem is:

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

note that this original form makes it much easier to combine the statements: just add them together!

once you've figured out that the two individual statements are insufficient, add them together:
m - 3z > 0
4z - m > 0
--------------
z > 0

once you figure out that z > 0, then realize that m > 3z, so m is also positive.
therefore m and z are also positive, so m + z > 0.
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by lunarpower » Tue Aug 03, 2010 4:36 am
lunarpower wrote:the original of the problem is:

Is m + z > 0 ?
(1) m - 3z > 0
(2) 4z - m > 0
just for fun, here's an awesome (but useless, in terms of takeaways/studying) solution!

* first, eliminate statements 1 and 2 individually

* then, some magic:
multiply statement (1) by 5, and multiply statement (2) by 4
then add them!
you get:
5m - 15z > 0
16z - 4m > 0
-----------------
m + z > 0

i will now return to providing solutions that are actually useful :)
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