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

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Intern
Joined: 12 Jul 2006
Posts: 8
If m and n are integers, and m/n = 20n/(m-n), what is the [#permalink]

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30 Oct 2008, 23:03
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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
Intern
Joined: 30 Oct 2008
Posts: 30

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30 Oct 2008, 23:14
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
SVP
Joined: 17 Jun 2008
Posts: 1504

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30 Oct 2008, 23:25
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

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
VP
Joined: 30 Jun 2008
Posts: 1019

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30 Oct 2008, 23:35
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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"You have to find it. No one else can find it for you." - Bjorn Borg

Intern
Joined: 30 Oct 2008
Posts: 30

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30 Oct 2008, 23:37
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
VP
Joined: 30 Jun 2008
Posts: 1019

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30 Oct 2008, 23:42
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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Director
Joined: 23 May 2008
Posts: 757

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30 Oct 2008, 23:52
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
VP
Joined: 30 Jun 2008
Posts: 1019

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31 Oct 2008, 02:00
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, 02:00
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