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If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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Updated on: 04 Oct 2011, 10:52
Question Stats:
55% (01:44) correct 45% (01:39) wrong based on 47 sessions
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If yz not equal to 0, is 0<y<1 ? (1) \(y< 1/y\) (2) \(y\)=\(z^2\) OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifyznoteq ... 39571.html
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Originally posted by shashankp27 on 02 Oct 2011, 13:59.
Last edited by shashankp27 on 04 Oct 2011, 10:52, edited 1 time in total.



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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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02 Oct 2011, 16:08
shashankp27 wrote: If yz not equal to 0, is 0<y<1 ?
(1) \(y< 1/y\) (2) \(y\)=\(z^2\) My answer is A. Dont find any relevance to 2nd statement.
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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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02 Oct 2011, 21:06
IMO E
Stmt A: the statement satisfies when y is negative or between 0 and 1. Not sufficient Stmt B: Nothing is known about Z. Not Sufficient
Combining the statement also not suffient to conclude.



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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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02 Oct 2011, 21:55
divyakatas wrote: IMO E
Stmt A: the statement satisfies when y is negative or between 0 and 1. Not sufficient Stmt B: Nothing is known about Z. Not Sufficient
Combining the statement also not suffient to conclude. i think from 2nd stmt, we know y is not negative. so it shud be between 0 and 1.



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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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02 Oct 2011, 22:49
1)
y < 1, or 0 < y < 1
NS
2)
y = z^2
y > 0
NS
1+2)
Given y < 1 or 0 < y < 1 and y > 0
then, 0 < y < 1
C



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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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03 Oct 2011, 00:12
(1) \(y<1/y\) \(y^2<1, y>0\) \(y^2>1, y<0\) \(0<y<1, y<1\) So, we could not say whether 0<y<1. (2) \(y=z^2\) This means that y>0 (since yz is not 0), but we do not know whether y>1 or not. (1) and (2) together \(y<1, 0<y<1 and y>0\) \(0<y<1\) So, tho statements together are sufficient. The answer is (C)
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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y =
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26 Nov 2017, 22:34
If yz not equal to zero is 0 < y < 1?\(yz\neq{0}\) means that neither of them is zero. (1) y < 1/y > \(\frac{y^21}{y}<0\) > \(\frac{(y+1)(y1)}{y}<0\) > 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: http://gmatclub.com/forum/x24x94661.html#p731476http://gmatclub.com/forum/inequalitiestrick91482.htmlhttp://gmatclub.com/forum/everythingis ... me#p868863http://gmatclub.com/forum/xyplane7149 ... ic#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. OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifyznoteq ... 39571.html
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Re: If yz not equal to 0, is 0<y<1 ? (1) y< 1/y (2) y = &nbs
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