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# If mn is different from 0, what is the ratio of m to n^2?

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Manager
Joined: 03 Aug 2005
Posts: 134
If mn is different from 0, what is the ratio of m to n^2? [#permalink]

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03 Nov 2005, 01:50
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If mn is different from 0, what is the ratio of m to n^2?

(1) The ratio of m^2 to 1 is 7/5

(2) The ratio of m^2 to n is 7/5
Manager
Joined: 21 Sep 2005
Posts: 232

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03 Nov 2005, 08:59
Ya...with both, I get m/n2 = + or - root(7/5) so E.
Manager
Joined: 17 Sep 2005
Posts: 72
Location: California

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03 Nov 2005, 11:35
E for me too same reason as mohish.

However, I did not have to use the fact that mn not= 0. Not sure what the trap that is?
Manager
Joined: 03 Aug 2005
Posts: 134

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03 Nov 2005, 13:03
E for me too.

OA is C but I think it is wrong.
SVP
Joined: 28 May 2005
Posts: 1707
Location: Dhaka

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03 Nov 2005, 13:28
I got C.

A.The ratio of m^2 to 1 is 7/5 means m^2 /1 = 7/5 we can't solveve ratio of m to n^2

B. The ratio of m^2 to n is 7/5 means m^2/n =7/5 we can't solveve ratio of m to n^2

But if we combine the 2 equation, we get m^2/n =m^2 /1

Since mn is not equal to zero, we can get n=1

so if n = 1 , we can get the value of m which is equal to sqrt(5/7)

now we can calculte the ration of m to n^2 from here.
_________________

hey ya......

Director
Joined: 24 Oct 2005
Posts: 572
Location: NYC

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03 Nov 2005, 13:53
nakib77 wrote:
I got C.

A.The ratio of m^2 to 1 is 7/5 means m^2 /1 = 7/5 we can't solveve ratio of m to n^2

B. The ratio of m^2 to n is 7/5 means m^2/n =7/5 we can't solveve ratio of m to n^2

But if we combine the 2 equation, we get m^2/n =m^2 /1

Since mn is not equal to zero, we can get n=1

so if n = 1 , we can get the value of m which is equal to sqrt(5/7)

now we can calculte the ration of m to n^2 from here.

but when n=1.. m can be +sqrt(7/5)
or m can be -sqrt(7/5)
??
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Success is my only option, failure is not -- Eminem

Manager
Joined: 11 Jul 2005
Posts: 86
Location: New York

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03 Nov 2005, 14:31
According to a Kaplan Coursebook that I have, Unless the 'square root sign' is explicitly given in the question, both positive and negative roots have to be taken into consideration.

But here I feel the answer is C. you donot have to take the roots. The first statement says m^2/1=7/5 .The second statement says m^2/n=7/5.

From St:1 we get M^2= 7/5. Though it is obvious, I am just explaining. Substitute m^2 in statement 2 ,we get n=1. tThere is no need to find the square root.

Manager
Joined: 11 Jul 2005
Posts: 86
Location: New York

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03 Nov 2005, 14:33
According to a Kaplan Coursebook that I have, Unless the 'square root sign' is explicitly given in the question, both positive and negative roots have to be taken into consideration.

But here I feel the answer is C. you donot have to take the roots. The first statement says m^2/1=7/5 .The second statement says m^2/n=7/5.

From St:1 we get M^2= 7/5. Though it is obvious, I am just explaining. Substitute m^2 in statement 2 ,we get n=1. tThere is no need to find the square root.

Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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03 Nov 2005, 15:58
this is E...

m^2 can be +-m...so we dont know...m/n

while we agree that n^2 is always positive...that cannot be said about m..
Manager
Joined: 11 Jul 2005
Posts: 86
Location: New York

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03 Nov 2005, 16:09
fresinha12 wrote:
this is E...

m^2 can be +-m...so we dont know...m/n

while we agree that n^2 is always positive...that cannot be said about m..

Am I missing something here? how can m^2 be positive /negative??????I understand m can be +/-

you donot have to find m here. We don't need m/n, we need m^2/n ;Just substitute for m^2 in m^2/n
Director
Joined: 21 Aug 2005
Posts: 789

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03 Nov 2005, 16:16
amy_v wrote:
fresinha12 wrote:
this is E...

m^2 can be +-m...so we dont know...m/n

while we agree that n^2 is always positive...that cannot be said about m..

Am I missing something here? how can m^2 be positive /negative??????I understand m can be +/-

you donot have to find m here. We don't need m/n, we need m^2/n ;Just substitute for m^2 in m^2/n

Manager
Joined: 11 Jul 2005
Posts: 86
Location: New York

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03 Nov 2005, 16:19
gsr wrote:
amy_v wrote:
fresinha12 wrote:
this is E...

m^2 can be +-m...so we dont know...m/n

while we agree that n^2 is always positive...that cannot be said about m..

Am I missing something here? how can m^2 be positive /negative??????I understand m can be +/-

you donot have to find m here. We don't need m/n, we need m^2/n ;Just substitute for m^2 in m^2/n

you are right!!!!!!!!!!!!!!! silly me!!!!!!!!!!!!!!
SVP
Joined: 24 Sep 2005
Posts: 1885

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03 Nov 2005, 18:19
amy_v wrote:
According to a Kaplan Coursebook that I have, Unless the 'square root sign' is explicitly given in the question, both positive and negative roots have to be taken into consideration.

But here I feel the answer is C. you donot have to take the roots. The first statement says m^2/1=7/5 .The second statement says m^2/n=7/5.

From St:1 we get M^2= 7/5. Though it is obvious, I am just explaining. Substitute m^2 in statement 2 ,we get n=1. tThere is no need to find the square root. So the answer is C.

yeah, but the question asks you to find m/n^2 ...finally, you still have to calculate sqrt(m^2)!
Senior Manager
Joined: 07 Jul 2005
Posts: 402

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03 Nov 2005, 23:52
(1) Insufficient

(2) Insufficietn

(1/2) Sufficient. Combine (1) and (2) and first solve for n. Then solve for m^2. Finally divide n with m^2.
Manager
Joined: 03 Aug 2005
Posts: 134

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04 Nov 2005, 00:24
Question says m/n^2 and m can be +/- sqrt(7/5)
Senior Manager
Joined: 15 Apr 2005
Posts: 415
Location: India, Chennai

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04 Nov 2005, 02:14
jdtomatito wrote:
If mn is different from 0, what is the ratio of m to n^2?

(1) The ratio of m^2 to 1 is 7/5

(2) The ratio of m^2 to n is 7/5

From 1 we get m^2 = 7/5 or m = sqrt(7/5)

From 2 we get m^2/n = 7/5

Combining both we get m^2 = 7/5 and n = 1
=> m = sqrt(7/5) and n^2 = 1
m:n^2 = sqrt(7/5)
Hence C
Manager
Joined: 03 Aug 2005
Posts: 134

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04 Nov 2005, 05:16
Why do not you consider m= - sqrt(7/5)?

If m^2 = 7/5, m = sqrt(7/5) or m = -sqrt (7/5).
04 Nov 2005, 05:16
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