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# (n-x)-(n-y)-(n-z)-(n-k)=?

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Intern
Joined: 23 Sep 2007
Posts: 35

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09 Nov 2008, 09:40
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Hello guys...try this one..

(n-x)-(n-y)-(n-z)-(n-k)=?
1). X, y, z, and k are consecutive integers
2). The average of x, y, z, and k is n.
SVP
Joined: 17 Jun 2008
Posts: 1529

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09 Nov 2008, 10:21
1
KUDOS
singaks wrote:
Hello guys...try this one..

(n-x)-(n-y)-(n-z)-(n-k)=?
1). X, y, z, and k are consecutive integers
2). The average of x, y, z, and k is n.

C.

x + x+1 + x+2 + x+3 = 4n
or, 4x + 6 = 4n
or, n-x = 6/4 = 3/2
n-y = n-(x+1) = (n-x) - 1 = 1/2
n-z = n-y-1 = -1/2
n-k = n-z-1 = -3/2

Hence, (n-x)-(n-y)-(n-z)-(n-k)= 3/2 -1/2 + 1/2 + 3/2 = 3.
Re: Solve me..   [#permalink] 09 Nov 2008, 10:21
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# (n-x)-(n-y)-(n-z)-(n-k)=?

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