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by btgyes » Thu Feb 24, 2011 4:22 am
after solving both .

3Z < M < 4Z

now let say M = 3.5 Z,
now putting in main eq ie is M + Z > 0

ie M + 3.5 Z > 0
or 4.5 Z > 0

can't say because Z still can be negative.

So i found and E

but OA is different

plz help..

ta
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Source: — Data Sufficiency |
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by maihuna » Thu Feb 24, 2011 4:30 am
Aha so u donot got the expert only reply so reposting is't.

I think someone/may be u recently posted this Q.

C is a clear ans:

Adding both statements it gives : z>0

1 says : m>3z => m>0

COmbining m>0 as z>0

so m+z>0

It is a clear C, what is the confusion?
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by Anurag@Gurome » Thu Feb 24, 2011 4:48 am
(1) m - 3z > 0 or m > 3z. From here we cannot say whether m + z > 0.
So, (1) is NOT SUFFICIENT.

(2) 4z - m > 0 or 4z > m. Again we cannot say whether m + z > 0.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), 3z < m < 4z

If z = -1, then the above inequality becomes -3 < m < -4, but that is not true as -3 > -4. Hence z can never be negative as 3z < m < 4z turns false in that case.
This implies z has to be positive and since m lies between 3z and 4z, which are positive so m will also be positive. Hence m + z > 0 is true.

The correct answer is C.
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by maihuna » Thu Feb 24, 2011 4:52 am
Anurag,
you can simply add the two stmts that gives z>0

No need to cumbersome arithmetic, preety st forward.
Anurag@Gurome wrote:(1) m - 3z > 0 or m > 3z. From here we cannot say whether m + z > 0.
So, (1) is NOT SUFFICIENT.

(2) 4z - m > 0 or 4z > m. Again we cannot say whether m + z > 0.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), 3z < m < 4z

If z = -1, then the above inequality becomes -3 < m < -4, but that is not true as -3 > -4. Hence z can never be negative as 3z < m < 4z turns false in that case.
This implies z has to be positive and since m lies between 3z and 4z, which are positive so m will also be positive. Hence m + z > 0 is true.

The correct answer is C.
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by btgyes » Thu Feb 24, 2011 6:00 am
Thanks Guys..

You are brilliant...!
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