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# M10-21

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Math Expert
Joined: 02 Sep 2009
Posts: 50627

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15 Sep 2014, 23:42
1
8
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:17) correct 42% (01:11) wrong based on 114 sessions

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The function $$f$$ is defined by $$f(x) = - \frac{1}{x^2}$$ for all nonzero numbers $$x$$. If $$f(m) = - \frac{1}{16}$$ and $$f(mn) = f(\frac{1}{n})$$, what is the value of $$n^2$$?

A. $$\frac{1}{16}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$2$$
E. $$4$$

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Math Expert
Joined: 02 Sep 2009
Posts: 50627

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15 Sep 2014, 23:42
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Official Solution:

The function $$f$$ is defined by $$f(x) = - \frac{1}{x^2}$$ for all nonzero numbers $$x$$. If $$f(m) = - \frac{1}{16}$$ and $$f(mn) = f(\frac{1}{n})$$, what is the value of $$n^2$$?

A. $$\frac{1}{16}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$2$$
E. $$4$$

Since $$f(x) = - \frac{1}{x^2}$$, then from $$f(m) = - \frac{1}{16}$$ we'll have that $$-\frac{1}{m^2}=-\frac{1}{16}$$, so $$m^2=16$$.

The same way, from $$f(mn) = f(\frac{1}{n})$$ we'll have that $$-\frac{1}{(mn)^2}=-n^2$$, which simplifies to $$n^4=\frac{1}{m^2}$$.

Since $$m^2=16$$, then $$n^4=\frac{1}{m^2}=\frac{1}{16}$$, which gives $$n^2=\frac{1}{4}$$.

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Intern
Joined: 02 Jan 2014
Posts: 39

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28 Oct 2015, 20:00
Hi Bunuel,

Can we now simply solve it like this? Since f(x)=−1/x^2, then from f(m)=−1/16 we'll have that −1/n^4=−1/16, so n^2=4.
As f(n^2) = -1/n^4.

Could you please explain if this is correct?
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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28 Oct 2015, 23:59
Celerma wrote:
Hi Bunuel,

Can we now simply solve it like this? Since f(x)=−1/x^2, then from f(m)=−1/16 we'll have that −1/n^4=−1/16, so n^2=4.
As f(n^2) = -1/n^4.

Could you please explain if this is correct?

There should be m instead of n.
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Joined: 30 Apr 2017
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25 Sep 2017, 23:52
Could you please elaborate on the following：
f(mn)=f(1/n) then we get -1/(mn)^2= - n^2. Don't understand who we got that result
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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25 Sep 2017, 23:56
d975490 wrote:
Could you please elaborate on the following：
f(mn)=f(1/n) then we get -1/(mn)^2= - n^2. Don't understand who we got that result

$$f(x) = - \frac{1}{x^2}$$, hence $$f(mn) = -\frac{1}{(mn)^2}$$ and $$f(\frac{1}{n})=-\frac{1}{(\frac{1}{n})^2}=-n^2$$
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Intern
Joined: 03 May 2014
Posts: 16

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31 Oct 2017, 08:41
Bunuel wrote:
The function $$f$$ is defined by $$f(x) = - \frac{1}{x^2}$$ for all nonzero numbers $$x$$. If $$f(m) = - \frac{1}{16}$$ and $$f(mn) = f(\frac{1}{n})$$, what is the value of $$n^2$$?

A. $$\frac{1}{16}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$2$$
E. $$4$$

What sub-topic/tag does this question fall under? I am looking to solve similar question types both from the GMAT Club and the OG.
Math Expert
Joined: 02 Sep 2009
Posts: 50627

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31 Oct 2017, 08:46
Edofarmer wrote:
Bunuel wrote:
The function $$f$$ is defined by $$f(x) = - \frac{1}{x^2}$$ for all nonzero numbers $$x$$. If $$f(m) = - \frac{1}{16}$$ and $$f(mn) = f(\frac{1}{n})$$, what is the value of $$n^2$$?

A. $$\frac{1}{16}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $$2$$
E. $$4$$

What sub-topic/tag does this question fall under? I am looking to solve similar question types both from the GMAT Club and the OG.

Functions and algebra.

13. Functions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Manager
Joined: 04 Jun 2018
Posts: 61

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14 Sep 2018, 05:41
I think this is a high-quality question and I agree with explanation. I am quite sure that I marked Option B , but it shows that I have marked Option C.
I just hope it is a technical error. If it is kindly look into it,
Intern
Joined: 30 Aug 2017
Posts: 10
GMAT 1: 580 Q45 V26
GMAT 2: 700 Q48 V38

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17 Nov 2018, 07:04
I think this is a high-quality question and I agree with explanation.
Re M10-21 &nbs [#permalink] 17 Nov 2018, 07:04
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# M10-21

Moderators: chetan2u, Bunuel

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