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# M21-16

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:10
1
2
00:00

Difficulty:

35% (medium)

Question Stats:

59% (00:48) correct 41% (00:44) wrong based on 152 sessions

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What is the value of $$y$$?

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

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

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16 Sep 2014, 00:10
1
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.

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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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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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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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06 Sep 2016, 08:14
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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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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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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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]

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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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24 Nov 2017, 06:00
+1 for E.
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24 Apr 2018, 10:29
I think this is a high-quality question and I agree with explanation. I think this is a good quality question
Re M21-16 &nbs [#permalink] 24 Apr 2018, 10:29
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# M21-16

Moderators: chetan2u, Bunuel

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