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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] New post 24 Sep 2012, 22:14
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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 24 Sep 2012, 23:59, edited 1 time in total.
Edited the question.
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Re: if yz not equal to zero is 0 < y < 1? [#permalink] New post 24 Sep 2012, 23:19
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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] New post 12 Mar 2013, 05:37
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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)

AD (I chose to start with statement 1)
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 :

AD
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 :

AD
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] New post 25 Sep 2012, 00:20
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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
xy-plane-71492.html?hilit=solving%20quadratic#p841486

(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.

Answer: C.

Hope it helps.
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Re: If yz not equal to zero is 0 < y < 1? [#permalink] New post 11 Mar 2013, 10:41
Bunuel wrote:
(1)+(2) Since from (2) y>0, then only one range remains from (1): 0<y<1. Sufficient.

Answer: C.

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] New post 12 Mar 2013, 00:39
Jaisri wrote:
Bunuel wrote:
(1)+(2) Since from (2) y>0, then only one range remains from (1): 0<y<1. Sufficient.

Answer: C.

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,
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Re: If yz not equal to zero is 0 < y < 1? [#permalink] New post 12 Mar 2013, 22:34
Virgilius, Thanks for your detailed explanation.
Re: If yz not equal to zero is 0 < y < 1?   [#permalink] 12 Mar 2013, 22:34
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