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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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Bumping for review and further discussion.
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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
Here we can square all the parts of the inequality as all are >0 hence A is sufficient
but statement 2 will give 2 values as 1/2 and -1/2 do we cant say anything from here.
Hence A
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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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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 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
Expert Reply
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 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
carcass wrote:
Is 0 < y < 1 ?

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

(2) y^2=1/4


This is how i solved

Target question is 0 < y < 1 ?

St 1

(1) 0<y√<1

So we learnt that whenever a no is between 0 and 1 the root of a no is also between o and 1 eg : let the no be 1/4 the root is 1 / 2
or the root for 0.25 is 0.5 notice both are between 0 and 1

Sufficient

St 2

y^2 = 1 /4 we know when the power of any no is even it will convert the value into a positive no.... in this case, y^2 = 1/4 so Y could be both 1/2 or - 1/2 so we cant answer the target question with certainty

Ans = A
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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
Is 0 < y < 1 ?

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

We can square all 3 numbers =

0 < y < 1. SUFFICIENT.

(2) \(y^2=1/4\)
y = \(\frac{1}{2}, \frac{-1}{2}\)

INSUFFICIENT.
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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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Re: Is 0 < y < 1 ? (1) 0 < y^(1/2) < 1 (2) y^2 = 1/4 [#permalink]
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