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# If yz not equal to zero is 0 < y < 1?

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If yz not equal to zero is 0 < y < 1? [#permalink]

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24 Sep 2012, 23:14
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If yz not equal to zero is 0 < y < 1?

(1) y < 1/y

(2) y = z^2
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2012, 00:59, edited 1 time in total.
Edited the question.
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Re: if yz not equal to zero is 0 < y < 1? [#permalink]

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25 Sep 2012, 00:19
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harikris wrote:

if yz not equal to zero is 0 < y < 1?

1)y < 1/y

2) y=z^2

Statement 1 is true for values between 0 and 1, but is also true for any negative number less than -1 (you can plug in, say, -2 to confirm). So Statement 1 is not sufficient. Statement 2 is clearly not sufficient since we have no information about z besides the fact that it's nonzero.

Taking the Statements together, from Statement 2, y must be positive, since it is equal to a nonzero square. So looking at Statement 1, we can multiply both sides by y, since y is positive, without worrying about whether to reverse the inequality. We then find that y^2 < 1, and if y is positive, that means 0 < y < 1. So the two statements together are sufficient and the answer is C.
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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25 Sep 2012, 01:20
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If yz not equal to zero is 0 < y < 1?

$$yz\neq{0}$$ means that neither of them is zero.

(1) y < 1/y --> $$\frac{y^2-1}{y}<0$$ --> $$\frac{(y+1)(y-1)}{y}<0$$ --> roots are -1, 0, and 1 --> this gives us 4 ranges: $$y<-1$$, $$-1<y<0$$, $$0<y<1$$, and $$y>1$$. Now, test some extreme value: for example if $$y$$ is very large number then the whole expression is positive. Here comes the trick: since in the fourth range, when $$y>1$$, the expression is positive, then in third range it'll be negative, in the second positive, and in the first range it'l be negative again: -+-+. Thus, the ranges when the expression is negative are: $$y<-1$$ and $$0<y<1$$. Not sufficient.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

(2) y = z^2. Since, we know that $$z\neq{0}$$, then $$z^2>0$$, hence $$y=z^2>0$$ Clearly insufficient.

(1)+(2) Since from (2) $$y>0$$, then only one range remains from (1): $$0<y<1$$. Sufficient.

Hope it helps.
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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12 Mar 2013, 06:37
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To be honest, I'm a huge fan of DS so forgive me in advance for being too giddy ! :p

So let's get cracking !

Let's start by understanding the problem : We're told that y*z is not equal to zero. This should automatically give you the following information : neither y nor z are equal to 0 !!. Then we're asked to answer by yes or no (consider the question stem : "Is... ?") whether $$0<y<1$$

Put your answer choice grid up : (Something I've inherited from MGMAT prep)

BCE

Statement 1 : $$y <\frac{1}{y}$$

Since y isn't equal to 0, the inequality is sound. However since we don't know the sign of y (whether y is positive or negative) we can't multiply each side of the inequality by y since there's the risk that the inequality is flipped. So statement 1 is insufficient.

The grid should look like this by now :

BCE

Statement 2 : $$y = z^2$$

The only information we can gather from this statement is that y is positive. This will be particularily useful when we'll combine both statements. But up to this point, statement 2 is insufficient.

Our grid should look like this :

BCE

Let's combine statements 1 and 2.

From statement 2, we know that y is positive. Therefore, we can multiply the inequality in statement 1 by y without risking to have the inequality flipped. From statement 1 we have, $$y^2<1$$. Applying the square root to this inequality yields $$y<1$$.

Therefore, we can clearly say that $$0<y<1$$. So the final answer is C ! (Oh and don't forget to cross off the E in your grid as well ).

Hope that helped
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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11 Mar 2013, 11:41
Bunuel wrote:
(1)+(2) Since from (2) $$y>0$$, then only one range remains from (1): $$0<y<1$$. Sufficient.

Hope it helps.

Hi Bunuel,
Won't the intersecting range between Stm I and II be both $$0<y<1$$ and y>1 (as y>1 from Stm I and y>0 from Stm II becomes y>1)? Please help me understand this.

Thanks.
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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12 Mar 2013, 01:39
Jaisri wrote:
Bunuel wrote:
(1)+(2) Since from (2) $$y>0$$, then only one range remains from (1): $$0<y<1$$. Sufficient.

Hope it helps.

Hi Bunuel,
Won't the intersecting range between Stm I and II be both $$0<y<1$$ and y>1 (as y>1 from Stm I and y>0 from Stm II becomes y>1)? Please help me understand this.

Thanks.

Never mind. Y>1 is already ruled out because y is positive in that region. My mistake.

Thanks,
Jaisri
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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12 Mar 2013, 23:34
Virgilius, Thanks for your detailed explanation.
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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07 Jul 2015, 05:19
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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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07 Jul 2015, 14:27
harikris wrote:
If yz not equal to zero is 0 < y < 1?

(1) y < 1/y

(2) y = z^2

Given : If yz not equal to zero

Question : Is 0 < y < 1?

Statement 1: y < 1/y

Case 1: y = 1/2 i.e. Answer to the question is YES
Case 1: y = -2 i.e. Answer to the question is NO
NOT SUFFICIENT

Statement 2: y = z^2

i.e. y is positive as the square of any NON-Zero number z will always be positive but the range of values of y is still unknown hence
NOT SUFFICIENT

Combining the two statements
y is positive and y < 1/y
i.e. y^2 < 1
i.e. 0 < y < 1
SUFFICIENT

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Re: If yz not equal to zero is 0 < y < 1? [#permalink]

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26 Jul 2016, 06:56
Hello from the GMAT Club BumpBot!

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Re: If yz not equal to zero is 0 < y < 1?   [#permalink] 26 Jul 2016, 06:56
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