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Director  Joined: 27 Dec 2004
Posts: 716
If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 67% (01:10) correct 33% (01:40) wrong based on 441 sessions

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If f is a function defined for all k by $$f(k) = \frac{k^5}{16}$$, what is f(2k) in terms of f(k)?

A. $$\frac{1}{8} f(k)$$

B. $$\frac{5}{8} f(k)$$

C. $$2 f(k)$$

D. $$10 f(k)$$

E. $$32 f(k)$$
Intern  Joined: 09 Jun 2005
Posts: 2
Location: Hong Kong
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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E for me.

F(k) = k^5 / 16

F(2k) = (2k)^5 / 16
= {(2^5) * (k^5)} / 16
= 32 (k^5) / 16
= 32 F(k)
VP  Joined: 18 Nov 2004
Posts: 1155
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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Folaa3 wrote:
If f is a function defined for all k by f(k) = k^5/16, what is f(2k) in terms of f(k)?

A. 1/8 f(k)
B. 5/8 f(k)
C. 2 f(k)
D. 10 f(k)
E. 32 f(k)

E is correct.

f(k) = k^5/16.......f(2k) = (2k)^5/16 ---> 32 k^5/16 ==> 32 f(k)
Director  Joined: 03 Nov 2004
Posts: 653
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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banerjeea_98 wrote:
E is correct.

f(k) = k^5/16.......f(2k) = (2k)^5/16 ---> 32 k^5/16 ==> 32 f(k)

Baner, Can you explain the difference between C & E
VP  Joined: 18 Nov 2004
Posts: 1155
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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banerjeea_98 wrote:
E is correct.

f(k) = k^5/16.......f(2k) = (2k)^5/16 ---> 32 k^5/16 ==> 32 f(k)

Baner, Can you explain the difference between C & E

When a function f(k) is defined as in this case as = k^5/16 , that means that the function is valid for all values of k. So when you are asked f(2k), what u do is simply replace K with 2K. In this case u do this
{(2k)^5}/16 ===> 32 k^/16 ==> 32 f(k)
Intern  Joined: 22 Mar 2011
Posts: 1
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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f(k) = k^5/16
f(2k) = (2k)^5/16 = 32k^5/16 = 32 x (k^5/16) = 32f(k).
So answer is E.
Director  Joined: 01 Feb 2011
Posts: 551
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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f(k) = k^5/16

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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Jasonammex wrote:
If f is the function defined for all k by f(k)=k^5/16, what is f(2k) in terms of f(k)?

A. 1/8 f(k)
B. 5/8 f(k)
C. 2 f(k)
D. 10 f(k)
E. 32 f(k)

I find this the easier with plugging in, instead of trying to solve with all the variables:

Let's assume k = 1, so f(k)=k^5/16 = f(1)=1/16.

The question now asks what is f(2k) in terms of f(k):

f(2k) = 2^5/16 = 32/16 (or 2, but not important to reduce)

Since f(k) = 1/16 and f(2k) = 32/16 f(2k) in terms of f(k) is 32f(k)
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Joined: 04 Sep 2017
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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Can someone explain this to me?

I understand the steps until the very end.

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

Shouldnt it be : 2k^5 = 32*K^5 Instead of just 32 f(k)? Whats happening to the exponent for k?
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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teamryan15 wrote:
Can someone explain this to me?

I understand the steps until the very end.

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

Shouldnt it be : 2k^5 = 32*K^5 Instead of just 32 f(k)? Whats happening to the exponent for k?

Hey teamryan15 ,

Looks like you messed up something there!

Let me explain you:

$$f(2k) = (2k)^5/16$$ = [$$(2)^5 * (k)^5$$]/16 = 2 * $$k^5$$ -- (1)

Now, you know that f(k) = $$k^5$$ / 16

=> $$k^5$$ = 16 f(k) -- (2)

Substitute the value at (2) to (1)

You will get, f(2k) = 32 f(k)

Does that make sense?
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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abhimahna wrote:
teamryan15 wrote:
Can someone explain this to me?

I understand the steps until the very end.

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

Shouldnt it be : 2k^5 = 32*K^5 Instead of just 32 f(k)? Whats happening to the exponent for k?

Hey teamryan15 ,

Looks like you messed up something there!

Let me explain you:

$$f(2k) = (2k)^5/16$$ = [$$(2)^5 * (k)^5$$]/16 = 2 * $$k^5$$ -- (1)

Now, you know that f(k) = $$k^5$$ / 16

=> $$k^5$$ = 16 f(k) -- (2)

Substitute the value at (2) to (1)

You will get, f(2k) = 32 f(k)

Does that make sense?

Yes it does thanks.
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3074
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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Solution:

• $$f(k)= k^5/16$$
• $$f(2k) = (2k)^5/16$$
• $$f(2k) = 32*k^5/16$$
• $$f(2k) = 32* f(k)$$

Hence (E) is correct answer.

_________________ Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)   [#permalink] 21 Mar 2018, 01:00
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