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

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

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25 May 2009, 06:52
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Is m + z > 0 ?

(1) m - 3z > 0

(2) 4z - m > 0

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Senior Manager
Joined: 08 Jan 2009
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25 May 2009, 07:08
stmt 1 :

m-3z >0 ---1
so m > 3z
m > z. Not sufficient

stmt 2 :
4z-m >0 ---2
Not sufficient.

So combining adding 1 and 2
z >0 anf from one m >z so m and z are +ve
so m+z >0. So sufficient.Ans. C

Walker graphical methods are good for such problems.

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Manager
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26 May 2009, 16:47
Hello...
Can you plz explain this in more detail. I am trying to understand the graphical method and I am not able to solve this using the graphical method.
Your help wud be deeply appreciated

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27 May 2009, 09:15
1st statement
m - 3z > 0
=> m > 3z

2nd statement

4z -m > 0
=> m < 4z

Now combining 1st & 2nd statement
z > 0

So

m + z > 0

Hence Both statement combinedly justify the question.

C)

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Manager
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27 May 2009, 22:00
Can someone pls. explain the graphical approach.

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Intern
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27 May 2009, 22:11
Is m + z > 0

(1) m-3z > 0,
we can rewrite this as m > 3z ---- (i)

from the question, we know m + z > 0,
reframing this, we get m > z ------ (ii)

This is not sufficient for concluding.

(2) 4z - m > 0
rewriting this as m - 4z < 0 => m < 4z ----(iii)
from (iii), we cannot identify (ii)

combining (1) and (2)
m-3z > 0
4z-m < 0
---------
z > 0
---------

from (ii) we know that m > z,
so we can conclude m + z > 0

Thanks.

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Intern
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30 May 2009, 10:31
mdfrahim wrote:
Can someone pls. explain the graphical approach.

graphic-approach-to-problems-with-inequalities-68037.html

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30 May 2009, 14:43
Let me try to stitch together what happens when we look at the two statements together. From S1, we got that $$m>3z$$. From S2, we got that $$m<4z$$, right?

Thus, we can combine those two facts to say $$3z<m<4z$$. From there, extract that $$3z<4z$$.

Option 1) From here, simply subtract $$3z$$ from both sides; we explicitly see that $$z>0$$.

Option 2) This is where the magic happens. Try to divide both sides by $$z$$. Whoa, panic! You don't divide both sides of an inequality by a variable; are you crazy??! ...except that we end up with $$3<4$$, which is, uh, true. Therefore, it was ok to divide by z, which tells us that z is positive.

If you have a copy of it, OG12 DS#11 illustrates a similar problem, using Option 1.

Now, at this point. Recall from S1 that $$m>3z$$, and since $$z$$ is positive, clearly $$3z$$ is positive, thus $$m$$ is positive. Finally, since $$m$$ and $$z$$ are both positive, clearly their sum is positive. We now have sufficiency.

I noticed an error in someone's explanation earlier; they concluded that if $$m>3z$$, then it must also be true that $$m>z$$. This would only hold if we were guaranteed that $$z$$ were positive, so that we can chain that $$m>3z>z$$ and then get $$m>z$$. However, if $$z$$ is negative, then we can't be sure. For example, let $$m=-2$$ and $$z=-1$$. Thus, $$-2 = m > 3z = -3$$, but it's not true that $$-2 = m > z = -1$$.

Furthermore, to be perfectly honest, it took me a bit more to prove to myself that either statements 1 or 2 were insufficient! I personally do not declare INS until I either realize that some information is OBVIOUSLY missing (OG12 DS#15 S1, for example), or until I see proof through examples that the QUESTION can be answered multiple ways.

For example, in S1, to confirm insufficiency, I said let $$m=10$$, and $$z=3$$. This satisfies the given condition ($$10-3*3>0$$), and then the answer to the question that is asked is YES ($$10+3>0$$).

Then, I tried another one; I said what if z is negative? Ok, let's try $$m=10$$, and $$z=-10$$. The condition still holds ($$10+3*(-10)>0$$), but now the answer to the question is NO ($$10+(-10)>0$$).

Now, I safely declare insufficient. Similarly, the test cases $$(m,z)=(1,1)$$ and $$(m,z)=(-1000,1)$$ establish that statement 2 is insufficient.

As someone who has an extremely rigorous formal math background, I don't move forward until I get a positive signal that something is insufficient. I know that "insufficient" and "insufficient for me!" don't mean the same thing

I always have the question in mind "what does insufficiency look like?"

Here's an example I just made up:
Quote:
If x and y are integers, what is the ratio of x to y?
1) 3x + 5y = 20

The easiest way to prove that this is insufficient is just to say, well, what if $$(x,y)$$ are $$(5,1)$$? $$15+5=20$$, check, then the ratio is $$5/1=5$$. But what if $$(x,y)$$ are $$(0,4)$$? $$0+20=20$$, check, then the ratio is $$0/4=0$$. Different. Insufficient. You can't argue that.

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Re: inequalities   [#permalink] 30 May 2009, 14:43
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