It is currently 18 Mar 2018, 20:19

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# S95-08

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44298

### Show Tags

16 Sep 2014, 01:49
Expert's post
3
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

59% (01:31) correct 41% (01:26) wrong based on 87 sessions

### HideShow timer Statistics

If $$m$$ denotes a number to the left of 0 on the number line such that $$m^{4}$$ is less than $$\frac{1}{81}$$, then the the reciprocal of $$m^{2}$$ must be

A. less than -3
B. between $$-\frac{1}{3}$$ and 0
C. between 0 and $$\frac{1}{9}$$
D. between $$\frac{1}{9}$$ and 1
E. greater than 9
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 44298

### Show Tags

16 Sep 2014, 01:49
Expert's post
2
This post was
BOOKMARKED
Official Solution:

If $$m$$ denotes a number to the left of 0 on the number line such that $$m^{4}$$ is less than $$\frac{1}{81}$$, then the the reciprocal of $$m^{2}$$ must be

A. less than -3
B. between $$-\frac{1}{3}$$ and 0
C. between 0 and $$\frac{1}{9}$$
D. between $$\frac{1}{9}$$ and 1
E. greater than 9

We must determine which statement correctly the value of, or $$\frac{1}{m^2}$$. We are told that $$m \lt 0$$ and that $$m^{4} \lt \frac{1}{81}$$.

There are two ways to approach this problem. First, it can be solved algebraically. First, take the fourth root of both sides of $$m^{4} \lt \frac{1}{81}$$ remembering that since we are looking for the negative root, it is necessary to flip the sign: $$m \gt -\frac{1}{3}$$. Next, square both sides, remembering once again to flip the sign: $$m^2 \lt \frac{1}{9}$$. Finally, take the reciprocal, remembering that when taking the reciprocal of an inequality in which both sides have the same sign, we must flip the sign. $$m^2$$ and $$\frac{1}{9}$$ are both positive, so we must flip the sign: $$m^2 \lt \frac{1}{9}$$ becomes $$\frac{1}{m^2} \gt 9$$. Thus the square of the reciprocal must be greater than 9.

Choice E is the correct answer.

Alternatively, the problem can be solved by plugging in numbers. We need a negative number which, when raised to the fourth power, is less than $$\frac{1}{81}$$. Since fractions get smaller when they have a larger denominator, we can pick a common fourth power greater than $$3^4$$ for our denominator like $$4^4 = 256$$. Then $$m^4 = \frac{1}{256}$$, so $$m = -\frac{1}{4}$$. In that case, $$m^2 = \frac{1}{16}$$ and the reciprocal of $$m^2$$ is 16. Only choice E describes 16.

_________________
Intern
Joined: 29 Apr 2014
Posts: 26
GMAT 1: 690 Q49 V35
GPA: 3.5

### Show Tags

03 Oct 2014, 08:56
1
KUDOS
1
This post was
BOOKMARKED
Hi Bunuel

I got a similar question on GMATPrep. Got a little confused with reversing the inequality signs. Could really use your help here...

If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and 1/10
E. greater than 10

My solution:

It is given than n<0 and n^2 < 1/100
--> [n] < 1/10 (taking square root both sides)
--> -n < 1/10 (since we are given n<0)
--> n > -1/10 (multiplying both sides by -1 and hence reversing the sign)
--> 1/n > -10 (taking reciprocal on both sides) --> Do I again need to flip the inequality sign here?? If the term on one of the sides is positive (1/n in this case), wouldn't it always be greater than the negative term, even when we take a reciprocal? Then there shouldn't be a need to flip the inequality sign here..?

However, as per the OA (option A), I guess I should be reversing the inequality sign in the last step...Can you help clarify where I'm going wrong?

NB
Math Expert
Joined: 02 Sep 2009
Posts: 44298

### Show Tags

03 Oct 2014, 10:04
Expert's post
1
This post was
BOOKMARKED
nehabhasin wrote:
Hi Bunuel

I got a similar question on GMATPrep. Got a little confused with reversing the inequality signs. Could really use your help here...

If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and 1/10
E. greater than 10

My solution:

It is given than n<0 and n^2 < 1/100
--> [n] < 1/10 (taking square root both sides)
--> -n < 1/10 (since we are given n<0)
--> n > -1/10 (multiplying both sides by -1 and hence reversing the sign)
--> 1/n > -10 (taking reciprocal on both sides) --> Do I again need to flip the inequality sign here?? If the term on one of the sides is positive (1/n in this case), wouldn't it always be greater than the negative term, even when we take a reciprocal? Then there shouldn't be a need to flip the inequality sign here..?

However, as per the OA (option A), I guess I should be reversing the inequality sign in the last step...Can you help clarify where I'm going wrong?

NB

Step by step:

n > -1/10

Multiply by -10 and flip sign: -10n < 1.

Divide by n and flip sign: -10 > 1/n.

Complete solution:
If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and 1/10
E. greater than 10

We have $$n<0$$ and $$n^2<\frac{1}{100}$$

$$n^2<\frac{1}{100}$$ --> $$-\frac{1}{10}<n<\frac{1}{10}$$, but as $$n<0$$ --> $$-\frac{1}{10}<n<0$$.

Multiply the inequality by $$-\frac{10}{n}$$, (note as $$n<0$$, then $$-\frac{10}{n}>0$$, and we don't have to switch signs) --> $$(-\frac{1}{10})*(-\frac{10}{n})<n*(-\frac{10}{n})<0*(-\frac{10}{n})$$ --> so finally we'll get $$\frac{1}{n}<-10<0$$.

Discussed here: if-n-denotes-a-number-to-the-left-of-0-on-the-number-line-91659.html
_________________
Intern
Joined: 05 Aug 2016
Posts: 38

### Show Tags

14 Sep 2016, 11:25
It implies m^2 > 9 .
Intern
Joined: 19 Nov 2017
Posts: 3

### Show Tags

20 Nov 2017, 17:45
Hello!
Please, can you explain this part:
"remembering that since we are looking for the negative root, it is necessary to flip the sign: m>−1/3"
Math Expert
Joined: 02 Sep 2009
Posts: 44298

### Show Tags

20 Nov 2017, 22:56
Fiorella2017 wrote:
Hello!
Please, can you explain this part:
"remembering that since we are looking for the negative root, it is necessary to flip the sign: m>−1/3"

$$m^{4} \lt \frac{1}{81}$$;

Take the fourth root: $$|m| < \frac{1}{3}$$;

$$-\frac{1}{3} < m < \frac{1}{3}$$.

Since we are given that m denotes a number to the left of 0, then m < 0, and thus we get: $$-\frac{1}{3} < m < 0$$.
_________________
Intern
Joined: 19 Nov 2017
Posts: 3

### Show Tags

21 Nov 2017, 13:37
Thank you very much Bunuel
Re: S95-08   [#permalink] 21 Nov 2017, 13:37
Display posts from previous: Sort by

# S95-08

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.