GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 27 Jun 2019, 01:27

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

M10-21

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55804
M10-21  [#permalink]

Show Tags

16 Sep 2014, 00:42
1
9
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:24) correct 40% (01:11) wrong based on 123 sessions

HideShow timer Statistics

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. $$4$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 55804
Re M10-21  [#permalink]

Show Tags

16 Sep 2014, 00:42
1
2
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 $$n^2=\frac{1}{4}$$.

Answer: B
_________________
Intern
Joined: 02 Jan 2014
Posts: 38
Re: M10-21  [#permalink]

Show Tags

28 Oct 2015, 21:00
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: 55804
Re: M10-21  [#permalink]

Show Tags

29 Oct 2015, 00:59
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.
_________________
Intern
Joined: 30 Apr 2017
Posts: 4
Re M10-21  [#permalink]

Show Tags

26 Sep 2017, 00:52
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: 55804
Re: M10-21  [#permalink]

Show Tags

26 Sep 2017, 00:56
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 $$f(\frac{1}{n})=-\frac{1}{(\frac{1}{n})^2}=-n^2$$
_________________
Intern
Joined: 03 May 2014
Posts: 16
Re: M10-21  [#permalink]

Show Tags

31 Oct 2017, 09:41
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: 55804
Re: M10-21  [#permalink]

Show Tags

31 Oct 2017, 09:46
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.
_________________
Manager
Joined: 04 Jun 2018
Posts: 157
GMAT 1: 610 Q48 V25
GMAT 2: 690 Q50 V32
GMAT 3: 710 Q50 V36
Re M10-21  [#permalink]

Show Tags

14 Sep 2018, 06:41
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
GMAT 3: 710 Q49 V39
Re M10-21  [#permalink]

Show Tags

17 Nov 2018, 08:04
I think this is a high-quality question and I agree with explanation.
Re M10-21   [#permalink] 17 Nov 2018, 08:04
Display posts from previous: Sort by

M10-21

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne