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If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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If $$x = \frac{1}{2}$$, is y equal to 1?

(1) $$y^2(x + 1/2) = 1$$

(2) $$y(2x - 1) = 2x - y$$

why is GMAT so biased on sqr(y) = 1. I mean, why sqroot(y) not equal to 1, instead it is + or - 1? for GMAT \sqrt{1} = 1

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Originally posted by pradeepss on 09 Oct 2013, 20:05.
Last edited by Bunuel on 05 Jun 2019, 05:23, edited 3 times in total.
Renamed the topic and edited the question.
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Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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if X = 1/2, is Y equal to 1?

1. Y(X+1/2) = 1
2. Y(2X-1) = 2X - Y

why is GMAT so biased on sqr(y) = 1. I mean, why sqr(y) not equal to 1, instead it is + or - 1? for GMAT \sqrt{1} = 1

Are you sure of the OA. And is the 1st statement $$Y*(X+\frac{1}{2}) = 1$$??

Because if that is so then it is sufficient. We can substitute value of X = 1/2 in this and we certainly get Y = 1. I don't believe there is any trick here? Can someone kick in here?

And by the way about your query on $$\sqrt{Y^2}$$ you can visit the following link where Bunuel has explained it very nicely.

http://gmatclub.com/forum/confusion-on-square-roots-127807.html

For this question my Answer would be D.

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Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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Shameek,

You have the correct expression for 1.
I read in MGMAT guides as follow:
if square of Y = 1, then Y = + or - 1. GMAT approach is, if Y = 1 then square of Y = 1 and if Y = -1 the square if Y is still 1.

Hope that helps.
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Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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Shameek,

You have the correct expression for 1.
I read in MGMAT guides as follow:
if square of Y = 1, then Y = + or - 1. GMAT approach is, if Y = 1 then square of Y = 1 and if Y = -1 the square if Y is still 1.

Hope that helps.

First of all I would like to know where does square of Y comes into picture in this question. If you open the 1st statement's expression you get

$$X*Y + (\frac{Y}{2}) = 1$$

which will be $$\frac{Y}{2} + \frac{Y}{2} = 1$$

i.e. Y = 1 - Hence Sufficient

For the second statement :-
Y*(2X-1) = 2X - Y

2XY - Y = 2X - Y

i.e. 2XY - 2X = 0

i.e. 2X(Y-1) = 0

either X = 0 or Y = 1

Given that $$X = \frac{1}{2}$$ we are left with Y = 1

And yes $$\sqrt{Y^2} = |Y|$$; and therefore 2 solutions for Y.

But why are you concerned with it for this question? I don't get that.

Also according to me the answer should be "D" not "B".
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Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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1
If x = 1/2, is y equal to 1?

(1) y^2(x + 1/2) = 1
(2) y(2x - 1) = 2x - y

why is GMAT so biased on sqr(y) = 1. I mean, why sqroot(y) not equal to 1, instead it is + or - 1? for GMAT \sqrt{1} = 1

NOTE: edited the first statement.

If x = 1/2, is y equal to 1?

(1) y^2(x + 1/2) = 1 --> substitute: $$y^2=1$$ --> $$y=-1$$ or $$y=1$$. Not sufficient.

(2) y(2x - 1) = 2x - y --> substitute: $$0=1-y$$ --> $$y=1$$. Sufficient.

Hope it's clear.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rules 3 and 5. Thank you.
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Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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2
1
If x = 1/2, is y equal to 1?

(1) y^2(x + 1/2) = 1
(2) y(2x - 1) = 2x - y

We are given that x = ½ and we must determine whether y = 1.

Statement One Alone:

y^2(x + ½) = 1

We can substitute ½ for x in the equation:

y^2(x + ½) = 1

y^2(1/2+ 1/2) = 1

y^2(1) = 1

y^2 = 1

y = 1 or y = -1

Statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

y(2x – 1) = 2x – y

We can substitute ½ for x in the equation:

y(2x – 1) = 2x – y

y(2(1/2) – 1) = 2(1/2) – y

y(1- 1) = 1 – y

y(0) = 1 – y

0 = 1 – y

y = 1

Statement two alone is sufficient to answer the question.

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Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x  [#permalink]

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_________________ Re: If x = 1/2, is y equal to 1? (1) y^2(x + 1/2) = 1 (2) y(2x - 1) = 2x   [#permalink] 11 Jun 2019, 21:20
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