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If m and n are integers, and m/n = 20n/(m-n), what is the

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If m and n are integers, and m/n = 20n/(m-n), what is the [#permalink] New post 30 Oct 2008, 22:03
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If m and n are integers, and m/n = 20n/(m-n), what is the ratio of m/n?

1) 1:2
2) 2:1
3) 3:1
4) 4:1
5) 5:1
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Re: PS questions [#permalink] New post 30 Oct 2008, 22:14
4:1

Equation m/n = 20n/(m-n) reduces to
20n^2 -mn - m^2 = 0
Solving this will lead to values
m = -5n
or
m = 4n

Looking at the gine options
4:1 is the answer
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Re: PS questions [#permalink] New post 30 Oct 2008, 22:25
mbajingle wrote:
4:1

Equation m/n = 20n/(m-n) reduces to
20n^2 -mn - m^2 = 0
Solving this will lead to values
m = -5n
or
m = 4n

Looking at the gine options
4:1 is the answer


I think, you meant m = 5n or m = -4n and hence 5:1 should be the answer.
The equation above should be 20n^2 +mn - m^2 = 0
From the equation above, (5n-m)(4n+m) = 0
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Re: PS questions [#permalink] New post 30 Oct 2008, 22:35
scthakur wrote:
The equation above should be 20n^2 +mn - m^2 = 0
From the equation above, (5n-m)(4n+m) = 0


how do we factor such equations 20n^2 +mn - m^2 = 0? I got stuck on this step....
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Re: PS questions [#permalink] New post 30 Oct 2008, 22:37
amitdgr wrote:
scthakur wrote:
The equation above should be 20n^2 +mn - m^2 = 0
From the equation above, (5n-m)(4n+m) = 0


how do we factor such equations 20n^2 +mn - m^2 = 0? I got stuck on this step....


Two solutions of quadratic equation of form
a(x^2) + bx + c = 0, where a, b and c are constants,are

x = (-b + sqrt(b^2 -4ac))/2a
x = (-b - sqrt(b^2 - 4ac))/2a
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Re: PS questions [#permalink] New post 30 Oct 2008, 22:42
mbajingle wrote:
amitdgr wrote:
scthakur wrote:
The equation above should be 20n^2 +mn - m^2 = 0
From the equation above, (5n-m)(4n+m) = 0


how do we factor such equations 20n^2 +mn - m^2 = 0? I got stuck on this step....


Two solutions of quadratic equation of form
a(x^2) + bx + c = 0, where a, b and c are constants,are

x = (-b + sqrt(b^2 -4ac))/2a
x = (-b - sqrt(b^2 - 4ac))/2a


so here a = 20, b = m and c = -m² ???
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Re: PS questions [#permalink] New post 30 Oct 2008, 22:52
m/n=20n/m-n = m/1=20/m-n

m^2-mn=20 = m(m-n)=20 so 5(5-1)=20 m/n =5:1
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Re: PS questions [#permalink] New post 31 Oct 2008, 01:00
bigtreezl wrote:
m/n=20n/m-n = m/1=20/m-n
m^2-mn=20 = m(m-n)=20 so 5(5-1)=20 m/n =5:1



Bigtreez .... where did "n" go ??
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Re: PS questions   [#permalink] 31 Oct 2008, 01:00
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