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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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New post 25 May 2009, 06:52
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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

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Re: inequalities [#permalink]

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New post 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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Re: inequalities [#permalink]

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New post 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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Re: inequalities [#permalink]

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New post 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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Re: inequalities [#permalink]

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

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Re: inequalities [#permalink]

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New post 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

This gives us Answer C.

Thanks.

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Re: inequalities [#permalink]

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New post 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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Re: inequalities [#permalink]

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New post 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 :P

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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