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

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If mn is different from 0, what is the ratio of m to n^2? [#permalink] New post 03 Nov 2005, 00:50
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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
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 [#permalink] New post 03 Nov 2005, 07:59
Ya...with both, I get m/n2 = + or - root(7/5) so E.
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 [#permalink] New post 03 Nov 2005, 10:35
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?
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 [#permalink] New post 03 Nov 2005, 12:03
E for me too.

OA is C but I think it is wrong.
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 [#permalink] New post 03 Nov 2005, 12:28
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.
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 [#permalink] New post 03 Nov 2005, 12:53
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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 [#permalink] New post 03 Nov 2005, 13:31
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.

So the answer is C.
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 [#permalink] New post 03 Nov 2005, 13:33
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.

So the answer is C.
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 [#permalink] New post 03 Nov 2005, 14:58
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..
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 [#permalink] New post 03 Nov 2005, 15:09
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
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 [#permalink] New post 03 Nov 2005, 15:16
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


Question asks for m/n^2
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 [#permalink] New post 03 Nov 2005, 15:19
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


Question asks for m/n^2



you are right!!!!!!!!!!!!!!! silly me!!!!!!!!!!!!!!
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 [#permalink] New post 03 Nov 2005, 17:19
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)!
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 [#permalink] New post 03 Nov 2005, 22:52
(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.
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 [#permalink] New post 03 Nov 2005, 23:24
Question says m/n^2 and m can be +/- sqrt(7/5)
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Re: DS number properties [#permalink] New post 04 Nov 2005, 01:14
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
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 [#permalink] New post 04 Nov 2005, 04:16
Why do not you consider m= - sqrt(7/5)?

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