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What is the value of (m-n)/(m^2-n^2)? 1. (m^2-n^2) = 40 2.

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What is the value of (m-n)/(m^2-n^2)? 1. (m^2-n^2) = 40 2. [#permalink] New post 23 Aug 2004, 22:38
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What is the value of (m-n)/(m^2-n^2)?

1. (m^2-n^2) = 40
2. m+n = 10
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 [#permalink] New post 24 Aug 2004, 01:05
Stem says. Find value of (m-n)/(m^2-n^2). simplify...

(m-n)/ (m+n) (m-n) - - -> 1/(m+n)

Statement A - Does not help...
Statement B - Yes, sufficient.

Hence the answer is B.
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 [#permalink] New post 24 Aug 2004, 09:50
I know what u guys are saying ...

However I am curious about one thing:

In the question it is not mentioned that m is not equal to n
If m=n=5 then the ratio is indefinite! Hence 2 is not suffucient and we need 1 to solve this (just to make sure that m<>n)

Any comments?
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 [#permalink] New post 24 Aug 2004, 13:20
srijay007 wrote:
I know what u guys are saying ...

However I am curious about one thing:

In the question it is not mentioned that m is not equal to n
If m=n=5 then the ratio is indefinite! Hence 2 is not suffucient and we need 1 to solve this (just to make sure that m<>n)

Any comments?


M and N can't be equal because division by 0 would occur. Answer is B.
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 [#permalink] New post 24 Aug 2004, 14:07
sigep thats srijay's point exactly that div by 0 would occur if m = n but nowhere in Q stem does it state m<> n and it usally does in DS problems.

Srijay what is the source of this question and what is the oa? perhaps poor question writing led to ommition of m<>N or maybe it was intentional
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 [#permalink] New post 24 Aug 2004, 19:01
Hi

I do not remmember the source, but ETS does not take the x/0 equation at anytime. As this will lead to infinity (anything by zero is not valid). Hence, the above answer B is valid.

Perhaps, others can put some info they have obtained.
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 [#permalink] New post 24 Aug 2004, 22:04
(m-n)/(m^2-n^2) can be simplified to (m-n)/(m+n)(m-n) = 1/(m+n)

1 gives us the value of (m+n)(m-n) = 40 and nothing more to help us get a value.
So 1. is insufficient.

2. gives us the value of m+n, which we can use to solve.
So 2. is sufficient.

B is the answer.
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 [#permalink] New post 25 Aug 2004, 08:11
i Got B .

equation becomes 1/(m+n)

1 . (m-n)(m+n) = 40

2. m+n = 4

we don't need value of (m-n) .
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 [#permalink] New post 25 Aug 2004, 08:13
thisquestion is from ETS official guide 10 th edition
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Re: DS [#permalink] New post 27 Aug 2004, 10:50
srijay007 wrote:
What is the value of (m-n)/(m^2-n^2)?

1. (m^2-n^2) = 40
2. m+n = 10


I was inclined to go ahead and put down 'B' for answer, but then I thought of testing possible values for m and n in (m^2-n^2) = 40
(m+n)(m-n) = 40
the only values I could come up with are:
(7+3)(7-3)=40

in this case the answer would be D
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 [#permalink] New post 27 Aug 2004, 11:37
I am for C

Lastochka,

We could also have (11+9)*(11-9). So (1) alone won't be sufficient.

Srijay,

I agree.

For values of m and n other than when m=n, (2) alone is sufficient. But when m=n we have the following,

From (2): m-n/(m^2-n^2) = 0/0 (which is indeterminate i.e it can take any value)

So (2) alone is not sufficient. We need (1) to make sure that m is not equal to n.

Rahul,

Could you tell us the exact question number of this problem in OG?
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 [#permalink] New post 27 Aug 2004, 12:35
[quote="shygo"]I am for C

Lastochka,

We could also have (11+9)*(11-9). So (1) alone won't be sufficient.

Good point, :wall
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 [#permalink] New post 30 Aug 2004, 01:10
rahul wrote:
thisquestion is from ETS official guide 10 th edition


Folks,

I also chose C because of the m<>n thing
But OA is B :cry:
I guess the Q is poorly framed;
In most cases Qs clearly mention "m<>n" in cases where divide by zero is a possibility

The source of this Qusetion and Answer is from internet.

Rahul,
If this Q is in OG, could you please check and tell us what the OA is in ETS OG?


Thanks

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 [#permalink] New post 30 Aug 2004, 02:00
I got B as the asnwer as well.


My approach:

M + N = 10

M - N / (M - N)(M + N) = M - N / (M - N)(10) = 1/(1)(10) and therefore
the answer is 1/10

Do you all think this approach is correct?

Regards,

Alex
  [#permalink] 30 Aug 2004, 02:00
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