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M21-16

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M21-16 [#permalink]

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New post 16 Sep 2014, 00:10
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  35% (medium)

Question Stats:

58% (00:42) correct 42% (00:40) wrong based on 110 sessions

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Re M21-16 [#permalink]

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New post 16 Sep 2014, 00:10
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Official Solution:


(1) \(y^3+2y=y+2y^2\). Rearrange and factor out \(y\) to get \(y(y^2-2y+1)=0\), which is the same as \(y(y-1)^2=0\), so \(y=0\) or \(y=1\). Not sufficient.

(2) \(y^2=y\). The same two solutions: \(y=0\) or \(y=1\). Not sufficient.

(1)+(2) Nothing new. Not sufficient.


Answer: E
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Re: M21-16 [#permalink]

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New post 10 Dec 2014, 00:55
Is it based on the logic : Don't divide the variable by the same variable ? Because statement 1 can be reduced to Y = 1, if we divide both the sides by y.

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Re: M21-16 [#permalink]

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New post 10 Dec 2014, 03:08
Ram1987 wrote:
Is it based on the logic : Don't divide the variable by the same variable ? Because statement 1 can be reduced to Y = 1, if we divide both the sides by y.


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 y=0 satisfies the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.

Check more tips on Algebra here: algebra-tips-and-hints-175003.html

Hope it helps.
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Re M21-16 [#permalink]

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New post 06 Sep 2016, 08:10
I don't agree with the explanation. Hi ,
Here we got two values of Y. So what s wrong with choice c

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Re: M21-16 [#permalink]

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New post 06 Sep 2016, 08:14
PadmajaH wrote:
I don't agree with the explanation. Hi ,
Here we got two values of Y. So what s wrong with choice c


Official Guide:

In data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.
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M21-16 [#permalink]

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New post 09 Apr 2017, 05:56
1. y3-y = 2y2 - 2y
y(y+1)(y-1) = 2y(y-1)
y+1 = 2
y = 1 ......sufficient

2. y2 = y
y2-y = 0
y(y-1) = 0
So, y=0, y=1 ........not sufficient

Ans : option A

**FYI - Consider the nos. coming after the variables as powers

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Re: M21-16 [#permalink]

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New post 10 Apr 2017, 09:19
Hi Bunuel

1. y3-y = 2y2 - 2y
y(y+1)(y-1) = 2y(y-1)
y+1 = 2
y = 1 ......sufficient

2. y2 = y
y2-y = 0
y(y-1) = 0
So, y=0, y=1 ........not sufficient

Ans : option A

**FYI - Consider the nos. coming after the variables as powers[/quote]

Pls let me know your take on this.

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New post 10 Apr 2017, 09:45
bibinphilips wrote:
Hi Bunuel

1. y3-y = 2y2 - 2y
y(y+1)(y-1) = 2y(y-1)
y+1 = 2
y = 1 ......sufficient

2. y2 = y
y2-y = 0
y(y-1) = 0
So, y=0, y=1 ........not sufficient

Ans : option A

**FYI - Consider the nos. coming after the variables as powers


Pls let me know your take on this.[/quote]

Please read the whole discussion before posting a question: https://gmatclub.com/forum/m21-184270.html#p1453874
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Re: M21-16 [#permalink]

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New post 22 May 2017, 02:08
Bunuel wrote:
bibinphilips wrote:
Hi Bunuel

1. y3-y = 2y2 - 2y
y(y+1)(y-1) = 2y(y-1)
y+1 = 2
y = 1 ......sufficient

2. y2 = y
y2-y = 0
y(y-1) = 0
So, y=0, y=1 ........not sufficient

Ans : option A

**FYI - Consider the nos. coming after the variables as powers


Pls let me know your take on this.




BUT HOW AM I WRONG? THE METHOD SEEMS PRETTY SOLID.

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Re: M21-16 [#permalink]

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New post 24 Nov 2017, 06:00
+1 for E.
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Re: M21-16   [#permalink] 24 Nov 2017, 06:00
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