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# What is the value of y?

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Manager
Joined: 04 Oct 2011
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Location: India
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What is the value of y? [#permalink]

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04 Dec 2012, 23:23
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Question Stats:

59% (01:50) correct 41% (01:08) wrong based on 62 sessions

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What is the value of y?

(1) y^3+2y=y+2y^2

(2) y^2=y
[Reveal] Spoiler: OA

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Manager
Joined: 04 Oct 2011
Posts: 224
Location: India
GMAT 1: 440 Q33 V13
GMAT 2: 0 Q0 V0
GPA: 3
Followers: 0

Kudos [?]: 39 [0], given: 44

Re: What is the value of y? [#permalink]

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04 Dec 2012, 23:30
shanmugamgsn wrote:
What is the value of $$y$$?

(1)$$y^3+2y=y+2y^2$$

(2)$$y^2=y$$

My choice was A..
I did this way can any one correct me if im wrong??

(2) INSUFFICIENT :
$$y^2-y=0$$
$$y(y-1)$$
So $$y=0$$ or $$y=1$$

(1) i felt this is SUFFICIENT:
$$y^3+2y=y+2y^2$$
divide by y on both sides
$$y^2+2=1+2y$$
$$y^2-2y+1=0$$
$$(y-1)^2=0$$
So solution will be $$y=1$$

Is there any mistakes in my calculation?
If y this soln is wrong???
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Re: What is the value of y? [#permalink]

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05 Dec 2012, 00:42
shanmugamgsn wrote:
shanmugamgsn wrote:
What is the value of $$y$$?

(1)$$y^3+2y=y+2y^2$$

(2)$$y^2=y$$

My choice was A..
I did this way can any one correct me if im wrong??

(2) INSUFFICIENT :
$$y^2-y=0$$
$$y(y-1)$$
So $$y=0$$ or $$y=1$$

(1) i felt this is SUFFICIENT:
$$y^3+2y=y+2y^2$$
divide by y on both sides
$$y^2+2=1+2y$$
$$y^2-2y+1=0$$
$$(y-1)^2=0$$
So solution will be $$y=1$$

Is there any mistakes in my calculation?
If y this soln is wrong???

At first glance of the equation given in Statement 1, you should be able to see that 0 is a solution to the equation. Every expression contains a y term and hence 0 will satisfy the equation. Your first step is dividing by y. Since y is an unknown variable, that step cannot be allowed. As the famous mems of the internet go No one except Chuck Norris can divide by 0. ...
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Re: What is the value of y? [#permalink]

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05 Dec 2012, 02:02
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Expert's post
What is the value of y?

(1) y^3+2y=y+2y^2 --> $$y^3-2y^2+y=0$$ --> $$y(y^2-2y+1)=0$$ --> $$y(y-1)^2=0$$ --> $$y=0$$ or $$y=1$$. Not sufficient.

(2) y^2=y --> $$y(y-1)=0$$ --> $$y=0$$ or $$y=1$$. Not sufficient.

(1)+(2) $$y=0$$ or $$y=1$$. Not sufficient.

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Kudos [?]: 68591 [0], given: 9815

Re: What is the value of y? [#permalink]

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05 Dec 2012, 02:04
Expert's post
shanmugamgsn wrote:
shanmugamgsn wrote:
What is the value of $$y$$?

(1)$$y^3+2y=y+2y^2$$

(2)$$y^2=y$$

My choice was A..
I did this way can any one correct me if im wrong??

(2) INSUFFICIENT :
$$y^2-y=0$$
$$y(y-1)$$
So $$y=0$$ or $$y=1$$

(1) i felt this is SUFFICIENT:
$$y^3+2y=y+2y^2$$
divide by y on both sides
$$y^2+2=1+2y$$
$$y^2-2y+1=0$$
$$(y-1)^2=0$$
So solution will be $$y=1$$

Is there any mistakes in my calculation?
If y this soln is wrong???

Never reduce an equation by a variable (or expression with a variable), if you are not certain that the variable (or the expression with a variable) doesn't equal to zero. We can not divide by zero.

So, if you divide (reduce) y^3+2y=y+2y^2 by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y=1 AND y=0 satisfy the equation).

Hope it's clear.
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What is the value of y? [#permalink]

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06 May 2014, 06:48
What is the value of y?
(1) y^3+2y=y+2y^2

(2) y2=y

official ans. is E. But I got A. Please someone explain issue in following logic

y^3+2y = y+2y^2

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

now cancel y on both sides

y^2+2 = 1+2y
Y^2-2y+1 = 0

solving this I will get (y-1) = 0 so y =1

or we can see b^2-4ac = 0 so it has one root.

Thanks
Math Expert
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Posts: 32610
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Kudos [?]: 68591 [0], given: 9815

Re: What is the value of y? [#permalink]

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06 May 2014, 08:22
Expert's post
PathFinder007 wrote:
What is the value of y?
(1) y^3+2y=y+2y^2

(2) y2=y

official ans. is E. But I got A. Please someone explain issue in following logic

y^3+2y = y+2y^2

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

now cancel y on both sides

y^2+2 = 1+2y
Y^2-2y+1 = 0

solving this I will get (y-1) = 0 so y =1

or we can see b^2-4ac = 0 so it has one root.

Thanks

Merging similar topics. Please refer to the discussion above.

Hope it helps.
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Re: What is the value of y?   [#permalink] 06 May 2014, 08:22
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