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

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

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(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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New post 18 Mar 2006, 22:32
A. the value of this expression is 0.

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New post 18 Mar 2006, 22:40
oa is A

cant beleive i overlooked this stupid q :x

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New post 19 Mar 2006, 09:24
Can someone explain this?
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New post 22 Mar 2006, 01:59
(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

:-D

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New post 07 May 2013, 15:17
Angela780 wrote:
Can someone explain this?



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

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New post 07 May 2013, 20:04
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: n-x) + (n-y) + (n-c) + (n-k) What is the value of the [#permalink]

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New post 03 May 2017, 08:21
Statement 1: (x+y+c+k)/4 = n
or x+y+c+k = 4n = n+n+n+n. Now take all the terms from left hand side to right hand side
It gives (n-x)+(n-y)+(n-c)+(n-k) = 0. Sufficient.

Statement 2: It doesn't tell us anything about n. OR even about the differences between n and any of the four variables x,y,c or k. Insufficient.

Hence A.

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

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New post 03 May 2017, 10:04
joemama142000 wrote:
(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: n-x) + (n-y) + (n-c) + (n-k) What is the value of the   [#permalink] 03 May 2017, 10:04
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