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# x/m) (m^2 + n^2 + k^2) = xm+yn+zk? 1. z/k = x/m 2. x/m = y/n

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Senior Manager
Joined: 10 Mar 2008
Posts: 341
x/m) (m^2 + n^2 + k^2) = xm+yn+zk? 1. z/k = x/m 2. x/m = y/n [#permalink]

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11 Sep 2008, 12:55
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(x/m) (m^2 + n^2 + k^2) = xm+yn+zk?

1. z/k = x/m

2. x/m = y/n

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SVP
Joined: 07 Nov 2007
Posts: 1756
Location: New York

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11 Sep 2008, 13:30
vksunder wrote:
(x/m) (m^2 + n^2 + k^2) = xm+yn+zk?

1. z/k = x/m

2. x/m = y/n

when
z/k = x/m=y/n

(x/m) (m^2 + n^2 + k^2) = (x/m)*m^2 +(y/n)*n^2++(z/k)*k^2 = xm+yn+zk

C.
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VP
Joined: 17 Jun 2008
Posts: 1322

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11 Sep 2008, 19:08
vksunder wrote:
(x/m) (m^2 + n^2 + k^2) = xm+yn+zk?

1. z/k = x/m

2. x/m = y/n

solving the equation i question we get
(n^2)( x/m - y/n)+(k^2)(-z/k +x/m) => requires both (1) and (2)
to be equal to 0

IMO C

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Its Now Or Never

Re: DS: Equations   [#permalink] 11 Sep 2008, 19:08
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