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If f(x) = k(x - k) and k is a constant, what is the value

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If f(x) = k(x - k) and k is a constant, what is the value [#permalink]

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If f(x) = k(x - k) and k is a constant, what is the value of f(4) - f(3), in terms of k?

(A) 1
(B) k
(C) 7k - 1
(D) k^2 + k
(E) k^2 - k

GH-04.12.13 | OE to follow
[Reveal] Spoiler: OA
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Re: If f(x) = k(x - k) and k is a constant, what is the value [#permalink]

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New post 24 Oct 2013, 10:38
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avohden wrote:
If f(x) = k(x - k) and k is a constant, what is the value of f(4) - f(3), in terms of k?

(A) 1
(B) k
(C) 7k - 1
(D) k^2 + k
(E) k^2 - k

GH-04.12.13 | OE to follow


f(x)= k(x-k).
f(4)= k(4-k).
f(3)=k(3-k).

f(4)-f(3)= [k(4-k)-k(3-k)].. taking the common k out.

= k[4-k-3+k]---- +k and -k cancelled out.
=k[1]

So Answer is B.
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Status: 1,750 Q's attempted and counting
Affiliations: University of Florida
Joined: 09 Jul 2013
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Location: United States (FL)
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GMAT 1: 600 Q45 V29
GMAT 2: 590 Q35 V35
GMAT 3: 570 Q42 V28
GMAT 4: 610 Q44 V30
GPA: 3.45
WE: Accounting (Accounting)
Followers: 25

Kudos [?]: 645 [0], given: 630

Re: If f(x) = k(x - k) and k is a constant, what is the value [#permalink]

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New post 27 Oct 2013, 10:40
Official Explanation

Answer: B
First, distribute f(x). k(x - k) = kx - k^2. Now evaluate f(4) and f(3):

f(4) = k(4) - k^2 = 4k - k^2

f(3) = 3k - k^2

Now subtract: (4k - k^2) - (3k - k^2) ---> = 4k - k^2 - 3k + k^2 ---> = 4k - 3k = k, choice (B).
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Re: If f(x) = k(x - k) and k is a constant, what is the value [#permalink]

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New post 14 Nov 2014, 22:48
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Re: If f(x) = k(x - k) and k is a constant, what is the value   [#permalink] 14 Nov 2014, 22:48
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