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M21-16

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Sachin9
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M21-16 [#permalink] New post   10 Feb 2013, 04:27
1 gives us
y=0 or y=1

2 gives us the same.

Had 2 resulted in y=0 or y=2, would the answer have been C? cause y=0 is common in both statements ?

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Re: M21-16 [#permalink] New post   10 Feb 2013, 04:35
Expert Reply
Sachin9 wrote:
1 gives us
y=0 or y=1

2 gives us the same.

Had 2 resulted in y=0 or y=2, would the answer have been C? cause y=0 is common in both statements ?


Please, DO NOT post pictures of questions. Rule #6 here: rules-for-posting-please-read-this-before-posting-133935.html

What is the value of \(y\)?

(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.

As for your question: if from (2) we had that y=0 or y=2, then the answer would be C: y=0.
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Sachin9
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Joined: 22 Jul 2012
Status:Gonna rock this time!!!
Posts: 356
Own Kudos [?]: 166 [0]
Given Kudos: 562
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE:Information Technology (Computer Software)
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Re: M21-16 [#permalink] New post   10 Feb 2013, 04:58
Bunuel wrote:
Sachin9 wrote:
1 gives us
y=0 or y=1

2 gives us the same.

Had 2 resulted in y=0 or y=2, would the answer have been C? cause y=0 is common in both statements ?


Please, DO NOT post pictures of questions. Rule #6 here: rules-for-posting-please-read-this-before-posting-133935.html

What is the value of \(y\)?

(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.

As for your question: if from (2) we had that y=0 or y=2, then the answer would be C: y=0.


Sorry Sir, Will adhere to rule 6 going forward.

Thanks for the reply..
Had 2 resulted in y=3 or y=4, answer would have been E ?
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Bunuel
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Re: M21-16 [#permalink] New post   10 Feb 2013, 05:04
1
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Expert Reply
Sachin9 wrote:
Bunuel wrote:
Sachin9 wrote:
1 gives us
y=0 or y=1

2 gives us the same.

Had 2 resulted in y=0 or y=2, would the answer have been C? cause y=0 is common in both statements ?


Please, DO NOT post pictures of questions. Rule #6 here: rules-for-posting-please-read-this-before-posting-133935.html

What is the value of \(y\)?

(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.

As for your question: if from (2) we had that y=0 or y=2, then the answer would be C: y=0.


Sorry Sir, Will adhere to rule 6 going forward.

Thanks for the reply..
Had 2 resulted in y=3 or y=4, answer would have been E ?


No. In this case the question would be flawed: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

Thus the case when from (1) we have that y=0 or y=1 and from (2) we have that y=3 or y=4, is not possible, since in this case the statements clearly contradict each other.

Hope it's clear.

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Re: M21-16 [#permalink]
10 Feb 2013, 05:04
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Bunuel
Math Expert
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