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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If n denotes a number to the left of 0 on the number line [#permalink]
29 Dec 2009, 07:50

14

This post received KUDOS

Expert's post

14

This post was BOOKMARKED

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 \(\frac{1}{n}<-10<0\).

Re: Inequalities and reciprocals [#permalink]
21 Mar 2010, 10:24

nifoui wrote:

If n denotes a number to the left of the 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

now lets say p = 1/n There can be two equations n < 0 => 1/p <0 since 1 cannot be negative this implies p< 0 and n^2 < 1/100 => 1/p^2 < 1/100 =>p^2>100 => (p -10)(p+10) > 0 => either p <-10 or p >10 since p < 0 from first equation we can deduce that p < -10 hence answer is A _________________

Re: Inequalities and reciprocals [#permalink]
21 Mar 2010, 10:28

nifoui wrote:

If n denotes a number to the left of the 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

The number of left of the 0 on the number line. We know that n is negative. The square of n is less than 1/100 so, -1/10 < n < 0. Choose a number between -1/10 and 0, for example -1/20. The reciprocal of that is -20. Thus the answer is A. _________________

Re: Inequalities and reciprocals [#permalink]
21 Mar 2010, 17:32

1

This post received KUDOS

A good approach for these kind of problems is always a Trail&Error/Substitution.

In this case since the square must be less than \(\frac{1}{100}\) consider any number in denominator more than 100. lets say 121 that will give us \(\frac{1}{121}\).

Since n is -ve the square root of this will be \(\frac{-1}{11}\). and the reciprocal of this number is -11.

Now Go back to the options and compare the one that satisfy the answer.

If there is only one option then that's your answer. If there are two try to eliminate one by reason.(can also try a different fraction.)

Re: PS Number line [#permalink]
23 Mar 2010, 05:41

1

This post received KUDOS

Expert's post

hardnstrong wrote:

i dont get your point bangolarian we have n^2 < 100 so -1/10 < n < 1/10 how you got n < - 1/10

Can somebody please explain if n > -1/10. does that mean its reciprocal is 1/n < -10 . Do we change less than and greater than sign with reciprocal of any interger?

Refer to my previous post:

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 \(n<0\) expression \(-\frac{10}{n}>0\), 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 \(\frac{1}{n}<-10<0\). OR \(\frac{1}{n}<-10\).

Re: PS Number line [#permalink]
24 Mar 2010, 00:07

Bunuel wrote:

hardnstrong wrote:

i dont get your point bangolarian we have n^2 < 100 so -1/10 < n < 1/10 how you got n < - 1/10

Can somebody please explain if n > -1/10. does that mean its reciprocal is 1/n < -10 . Do we change less than and greater than sign with reciprocal of any interger?

Refer to my previous post:

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 \(n<0\) expression \(-\frac{10}{n}>0\), 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 \(\frac{1}{n}<-10<0\). OR \(\frac{1}{n}<-10\).

Answer: A.

Hope it helps.

Bunuel, Thanks so much for your explanation

As i am weak in this type of problems so just need clarification on 1 more thing We have \(n<0\) and \(n^2<\frac{1}{100}\)

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

because n < 0 and from above we know n >-1/10 so instead of multiplying by -10/n can we straight so it like this -> if n>-1/10 , therefore 10>-1/n which means -10>1/n

Re: PS Number line [#permalink]
12 May 2010, 09:55

4

This post received KUDOS

2

This post was BOOKMARKED

Bunuel wrote:

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 \(n<0\) expression \(-\frac{10}{n}>0\), 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 \(\frac{1}{n}<-10<0\).

Answer: A.

Perhaps it's easier to remember this regarding switching of the signs...

When a and b are both positive or both negative:

* If a < b then 1/a > 1/b * If a > b then 1/a < 1/b

However when either a or b is negative (but not both) the direction stays the same:

* If a < b then 1/a < 1/b * If a > b then 1/a > 1/b

Take the above rules into consideration.... n^2 < 1/100

|n| < 1/10

but as n<0, therefore

-n < 1/10 or we can write as n > -1/10

( think of the above mentioned rules regarding the signs of the inequality, and as we know that n is also negative and right side of the inequality is also negative, thus taking the reciprocal will invert the sign of the inequality )

Re: if n denotes a number to the left of 0 [#permalink]
02 Nov 2010, 05:17

Expert's post

1

This post was BOOKMARKED

anilnandyala wrote:

if n denotes a number to the left of 0 on the number line such that the square of n is less then 1/100 then reciprocal of n must be? a less then -10 b B/w -1 & -1/10 c B/w -1/10 & 0 d B/w 0 & 1/10 e greater then 10

Since the number is to the left of 0, it must be a negative number. The squares pf -1/10 will be 1/100. If the square has to be less than 1/100, the number,n, must be something like -1/15 to get 1/225 etc. So numbers lying between -1/10 and 0 will have squares less than 1/100. But C is not the answer since we need to find reciprocal of n. Reciprocal of -1/10 is -10, of -1/15 is -15. Therefore, we see that numbers less than -10 are reciprocals of n and hence our answer is (A). _________________

Re: Number properties [#permalink]
21 Aug 2011, 06:49

1

This post received KUDOS

If n denotes a number to the left of 0 - means that n is negative square of n is less than 1/100 - means n is less than -1/10 then the reciprocal of n must be 1/(1/10) = less than -10 _________________

Show Thanks to fellow members with Kudos its shows your appreciation and its free

Re: Gmatprep number line PS [#permalink]
03 Sep 2011, 00:39

2

This post received KUDOS

800_gal wrote:

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

kkalyan wrote:

so when n^2 < 1/100 then n<-1/10 It means when we take square root on both side we have to change the sign? Can anyone explain the logic?

\(n^2<\frac{1}{100}\)

\(n^2<\frac{1}{(10)^2}\)

Taking square root both sides:

\(\sqrt{n^2}<\sqrt{\frac{1}{(10)^2}}\)

\(\hspace{3} \sqrt{n^2}=|n|\)

\(\therefore \hspace{3} |n|<\frac{1}{10}\)

\(|x|<y\) is equivalent to \(-y<x<+y\)

Similarly, \(-\frac{1}{10}<n<\frac{1}{10}\)

From the stem we know, \(n<0\)

So, \(-\frac{1}{10}<n<0\)

Write those separately: \(-\frac{1}{10}<n\) Flip RHS and LHS and change the symbol to ">" \(n>-\frac{1}{10}\) Multiply both sides by -10: \(-10*n<1\) Divide each side by n(n is -ve so signs flip again) \(-10>\frac{1}{n}\)

Flip RHS and LHS and change the symbol to "<" \(\frac{1}{n}<-10\)--------1

Also, \(n<0\)-----------2

In 1 and 2; 1 is more limiting \(\frac{1}{n}<-10\)

Ans: "A"

************************************************************** Note: Few of these steps could be avoided, but I just wanted to make clear how the inequality symbols and polarity signs change. _________________

I am not panicking. Nope, Not at all. But I am beginning to wonder what I was thinking when I decided to work full-time and plan my cross-continent relocation...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...