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# What is the value of n? 1) n^2 - n = n - n^2 2) -n^2 = n

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Manager
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What is the value of n? 1) n^2 - n = n - n^2 2) -n^2 = n [#permalink]

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13 Apr 2006, 15:49
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What is the value of n?

1) n^2 - n = n - n^2

2) -n^2 = n
Manager
Joined: 09 Feb 2006
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Location: New York, NY
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13 Apr 2006, 16:21
What is the value of n?

1) n^2 - n = n - n^2

2) -n^2 = n

Statement 1: 2n^2 = 2n
n^2 = n

This would be insufficient. because sqrt(n^2) could be positive or negative.

Statement 2: sufficient. This tells us that n = -1.

B
VP
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13 Apr 2006, 17:34
jcgoodchild wrote:
What is the value of n?

1) n^2 - n = n - n^2

2) -n^2 = n

Statement 1: 2n^2 = 2n
n^2 = n

This would be insufficient. because sqrt(n^2) could be positive or negative.

Statement 2: sufficient. This tells us that n = -1.

B

B doesnt tell us that n= -1!!
- n^2 = n (given)
or, n^2 + n =0
or n(n+1) = 0

Thus n = 0 or n =-1! Hence we cannot conclusively prove that n ==-1! (this wud have wored if the question was modifird to say that n was non zero or n was positive)!

Hence E
VP
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13 Apr 2006, 20:55
C.
1) n^2 + n^2 = n + n
2n^2 = 2 n
n^2 = n
n^2 - n = 0
n (n-1) = 0

n = 0 or 1

2) -n^2 = n
n^2 + n= 0
n (n+1) = 0
n = 0, -1

from i and ii, n = 0
VP
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13 Apr 2006, 23:02
Oh Yeah its C.. Tried to solve it mentally, did not notice!!

Its C!
Manager
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16 Apr 2006, 15:53
Professor wrote:
C.
1) n^2 + n^2 = n + n
2n^2 = 2 n
n^2 = n
n^2 - n = 0
n (n-1) = 0

n = 0 or 1

2) -n^2 = n
n^2 + n= 0
n (n+1) = 0
n = 0, -1

from i and ii, n = 0

OA is C.

A question:

In stmnt 1, we get:

n^2 + n^2 = n + n
2n^2 = 2 n

At this step, why can't we just divide by 2n on both sides to get to n = 1 ??? Would appreciate an explanation
Re: DS   [#permalink] 16 Apr 2006, 15:53
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