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

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Senior Manager
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Is m + z > 0? (1) m - 3z > 0 (2) 4z - m > 0 [#permalink]

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New post 02 Aug 2008, 07:59
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A
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C
D
E

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Is m + z > 0?

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

Kudos [?]: 67 [0], given: 0

Manager
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Joined: 27 May 2008
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Kudos [?]: 45 [0], given: 0

Re: DS m=z> 0? [#permalink]

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New post 02 Aug 2008, 08:48
Is m + z > 0?

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


here there is no clue that m and z are integers or positive or negative.

So let approach the problem,
if m and n, both are negative then letz take -1 and -2
so -1 +6 >0
but -1+ -2 <0 but still sufficient to say.

lets take m = 4 and z = 1
we have 4 - 3 >0
but if u substitiute in m+n 5>0
So there is a contradiction here So statement A is INSUFFICIENT

Similarly for statement 2 as well.

i may be wrong, Answer might be D as well.

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Intern
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Re: DS m=z> 0? [#permalink]

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New post 02 Aug 2008, 09:41
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m-3z>0 from sattement I, therefore m>3z

if we were to rephrease statement II then we get m-4z < 0

from both the statments 3z < m < 4z . hence for any value of z < 0 the condition does not satisfy the equation.

therefore z>0 and m>0 is the only solution, implying that m+z>0.

choice C

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Director
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Re: DS m=z> 0? [#permalink]

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New post 02 Aug 2008, 09:50
refer same question here p493511?t=67183

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Re: DS m=z> 0? [#permalink]

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New post 02 Aug 2008, 22:45
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I think it can be solved with both together

With both statements, 4z>m>3z.

Now plot a number line, with z, 2z, 3z and 4z. Considering both statements, m lies between 3z and 4z. Also, since 4z>3z, z is +ve and all numbers lie on right of zero.

now, since m > 3z, m+z is at least > 3z+z, which means at least 4z and z being +ve, its > 0.

Kudos [?]: 2 [1], given: 0

Re: DS m=z> 0?   [#permalink] 02 Aug 2008, 22:45
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Is m + z > 0? (1) m - 3z > 0 (2) 4z - m > 0

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