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# n-x) + (n-y) + (n-c) + (n-k) What is the value of the

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Director
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n-x) + (n-y) + (n-c) + (n-k) What is the value of the [#permalink]

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18 Mar 2006, 21:30
(n-x) + (n-y) + (n-c) + (n-k)
What is the value of the expression above?
(1) The average (arithmetic mean) of x, y, c, and k is n.
(2) x, y, c, and k are consecutive integers.
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Re: old paper test q? [#permalink]

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18 Mar 2006, 21:32
A. the value of this expression is 0.
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18 Mar 2006, 21:40
oa is A

cant beleive i overlooked this stupid q
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19 Mar 2006, 08:24
Can someone explain this?
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22 Mar 2006, 00:59
(n-x) + (n-y) + (n-c) + (n-k) = n-x+n-y+n-c+n-k
= 4n -(x+y+c+k)............(i)

now if (1) The average (arithmetic mean) of x, y, c, and k is n.
means (x+y+c+k)/4 = n
means (x+y+c+k) = 4n

substitute this in (i)

4n - 4n
=0

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07 May 2013, 14:17
Angela780 wrote:
Can someone explain this?

Can you explain this problem for me, I'm not understanding it.
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07 May 2013, 19:04
Quote:
(n-x) + (n-y) + (n-c) + (n-k)
What is the value of the expression above?
(1) The average (arithmetic mean) of x, y, c, and k is n.
(2) x, y, c, and k are consecutive integers.

laythesmack23 wrote:
Angela780 wrote:
Can someone explain this?

Can you explain this problem for me, I'm not understanding it.

It asks for the value of (n-x) + (n-y) + (n-c) + (n-k) .......or 4n-(x+y+c+k) .....so that means we need to know the value of n

& x+y+c+k .....

statement:: 1 says ..... The average (arithmetic mean) of x, y, c, and k is n..... that means ....

$$\frac{x+y+c+k}{4}= n$$ ... Theforefore, 4n = x+y+c+k ...

when plugging in the value in the expression given .... 4n-4n .. Therefore, 0.

The value of above expression is 0. Sufficient.

Statement :: 2 says x, y, c, and k are consecutive integers. ..lets say for ex. the consecutive integers x,y,c,k are 1,2,3,4, respectively...

therefore, we have (n-1)(n-2)(n-3)(n-4)..... or 4n-10 .. so here we need to know the value of n in order to know the value of the

expression above........ so, Insufficient.

Hence, A ............

Hope it Helps !! & let me know if there is any problem ..............
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Re: Re:   [#permalink] 07 May 2013, 19:04
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