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Joined: 16 May 2014
Posts: 590
Given Kudos: 0
GRE 1: Q165 V161
GRE Math Challenge #56-If mn does not equal to zero
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28 Feb 2015, 02:31
28 Feb 2015, 02:31
Expert Reply
00:00
Question Stats:
64% (01:35) correct
35% (01:34) wrong based on 113 sessions
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If mn does not equal to zero, and \(m^2n^2 + mn = 12\), then m could be:
I. \(-\frac{4}{n}\)
II.\(\frac{2}{n}\)
III.\(\frac{3}{n}\)
A. I only
B. II only
C. III only
D. I & II only
E. I & III only
I. \(-\frac{4}{n}\)
II.\(\frac{2}{n}\)
III.\(\frac{3}{n}\)
A. I only
B. II only
C. III only
D. I & II only
E. I & III only
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #56
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16 May 2015, 15:34
16 May 2015, 15:34
2
soumya1989 wrote:
If mn does not equal to zero, and \(m^2n^2 + mn = 12\), then m could be:
I. \(-4/n\)
II.\(2/n\)
III.\(3/n\)
A. I only
B. II only
C. III only
D. I & II only
E. I & III only
I. \(-4/n\)
II.\(2/n\)
III.\(3/n\)
A. I only
B. II only
C. III only
D. I & II only
E. I & III only
Take m²n² + mn = 12 and rewrite as m²n² + mn - 12 = 0
Factor: (mn + 4)(mn - 3) = 0
So, we have two possibilities.
case a: mn + 4 = 0
This means mn = -4, which means m = -4/n
This matches option I
case b: mn - 3 = 0
This means mn = 4, which means m = 3/n
This matches option III
Answer:
Show: ::
E
Cheers,
Brent
