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Is 0 < y < 1 ?

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Is 0 < y < 1 ?  [#permalink]

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New post 14 Sep 2012, 17:48
5
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

68% (01:04) correct 32% (01:10) wrong based on 474 sessions

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Re: Is 0 < y < 1 ?  [#permalink]

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New post 14 Sep 2012, 18:39
3
Is 0 < y < 1 ?

(1) 0 < \sqrt{y} < 1

(2) y^2=1/4

Stat1
0 < \sqrt{y} < 1

Square root and square of any number between 0 and 1 is always between 0 and 1
So, 0 < y < 1
So, SUFFICIENT

Stat2
y^2=1/4
y = +- 1/2
So, NOT SUFFICIENT

So, Answer will be A. Hope it helps!
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Re: Is 0 < y < 1 ?  [#permalink]

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New post 15 Sep 2012, 01:16
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Re: Is 0 < y < 1 ?  [#permalink]

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Re: Is 0 < y < 1 ?  [#permalink]

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Re: Is 0 < y < 1 ?  [#permalink]

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New post 21 Jun 2017, 14:28
Is 0 < y < 1 ?

(1) \(0 < \sqrt{y} < 1\)

From above we understand that y is positive as we cannot take a square root of a negative number. ======> Sufficient

(2) y^2=1/4[/quote]

y can be +1/4 or -1/4 =====> Not Sufficient.

Hence, Answer is A
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Re: Is 0 < y < 1 ?  [#permalink]

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New post 21 Jun 2017, 21:10
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If √y is a number between 0 and 1, then y (which is square of √y) will also lie between 0 and 1. We can check with any value that is between 0 and 1, its square also cannot exceed 1, rather its square will be even lesser than the original value. So statement 1 is sufficient.

If y^2 = 1/4, then y can take both positive (1/2) and negative (-1/2) values. So statement 2 is insufficient.

Hence A answer
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Re: Is 0 < y < 1 ?  [#permalink]

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New post 22 Jun 2017, 07:29
Solution:

Statement 1: the value of y has to be between 0 and 1. As sqrt of negative no's is not defined.Sufficient.
Statement 2: y can be +[1][/2] or -[1][/2]. Insufficient.

Therefore the answer is Option A.
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Re: Is 0 < y < 1 ?  [#permalink]

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New post 22 Jun 2017, 09:07
carcass wrote:
Is 0 < y < 1 ?

(1) \(0 < \sqrt{y} < 1\)
(2) y^2=1/4


St1: \(0 < \sqrt{y} < 1\)
Lets assume \(\sqrt{y}\)=0.1
y=0.01 which is in the range 0 < y < 1 Suff

St2: \(y^2=1/4\)
\(y=\frac{1}{2}
y=\frac{-1}{2}\) NS

Option A
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Is 0 < y < 1 ?  [#permalink]

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New post 23 Jun 2017, 11:09
carcass wrote:
Is 0 < y < 1 ?

(1) \(0 < \sqrt{y} < 1\)

(2) y^2=1/4


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There is 1 variable and 0 equation. Thus the answer is D most likely.

Condition 1)
Squaring all sides of the inequalities \(0 < \sqrt{y} < 1\), we have \(0 < y < 1\).
Thus this is sufficient.

Condition 2)
From \(y^2 = \frac{1}{4}\), we have \(y = \frac{1}{2}\) or \(y = -\frac{1}{2}\).
The answer is not unique. Hence this it not sufficient.

Therefore A is the answer.


For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is 0 < y < 1 ?  [#permalink]

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New post 24 Jun 2017, 04:40
IMO A
From 1 we have 0<√y<1 if we square it we have 0<y<1 hence sufficient
From 2 we have y^2=1/4 so y=-1/2,y=1.2 hence insufficient
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Re: Is 0 < y < 1 ?  [#permalink]

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