If f is a function defined for all k by f(k) = k^5/16, what is f(2k)

Question

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)

https://gmatclub.com/forum/download/file.php?id=78080

aowaowb · Accepted Answer

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)

banerjeea_98 · Answer

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

rthothad · Answer

Baner, Can you explain the difference between C & E

banerjeea_98 · Answer

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)

kbdevaraj · Answer

f(k) = k^5/16
f(2k) = (2k)^5/16 = 32k^5/16 = 32 x (k^5/16) = 32f(k).
So answer is E.

Spidy001 · Answer

f(k) = k^5/16

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

Answer is E.

reto · Answer

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)

teamryan15 · Answer

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?

abhimahna · Answer

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?

teamryan15 · Answer

Yes it does thanks.

EgmatQuantExpert · Answer

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.

Answer: E

Basshead · Answer

f(k) = \frac{k^5}{16}

f(2k) = \frac{(2k)^5}{16}

f(2k) = \frac{32 * k^5}{16}

f(2k)  =  2*k^5

Answer = E.  32 f(k)

AnkurGMAT20 · Answer

Bunuel , this one was asked in GMAC official 5 mock(Focus edition). Can you tag it correctly?

Bunuel · Answer

Thank you very much! That realy helps. Updated the tag.

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If f is a function defined for all k by f(k) = k^5/16, what is f(2k)

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If f is a function defined for all k by f(k) = k^5/16, what is f(2k)

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