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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

n^2 < 1/100 -1/10 < n < 1/10 n is negative so forget right side of equation multiply both sides by 1/n 1/n * -1/10 < n * 1/n flip signs as n is negative 1/10n > 1 note 1/n * -1/10 = +1/10n now multiply both sides by 10 10 * 1/10n < 1 * 10 Don't flip signs 1/n < 10

Answer is A! Please explain your answers.

Last edited by study on 25 Nov 2011, 22:59, edited 3 times in total.

n^2 < 1/100 -1/10 < n < 1/10 n is negative so forget right side of equation multiply both sides by 1/n 1/n * -1/10 < n * 1/n -1/10n < 1 now multiply both sides by -10 -10 * -1/10n < 1 * -10 Flip signs 1/n > -10

Answer E!

First, that is not what E says (E says that 1/n is greater than positive 10, not negative 10), so that might have suggested that your answer wasn't quite right. I've highlighted your mistake in red. When you multiply on both sides of the inequality by 1/n, you must reverse the inequality, because 1/n is negative.
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Thanks, Ian. I edited the post. That was a typo. Would you please look at the post now and correct my mistake. Thanks.

In your edited post, the negative sign vanished - it still needs to be there. You have:

-1/10 < n

Now if we multiply by 1/n on both sides, we must reverse the inequality, since 1/n is negative:

-1/10n > 1

Now we can multiply by 10 on both sides:

-1/n > 10

Now we can multiply by -1 on both sides, reversing the inequality since we are multiplying by a negative:

1/n < -10

Note that you can also do this problem very quickly by finding any suitable number for n (say -1/100) and working out the reciprocal of n.
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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Yes, the negative sign vanished, cuz negative * negative = positive. So a negative 1/n * negative 1/10 = + 1/10n And that is exactly the part I don't understand. Why would you retain a negative after multiplying a negative number by another negative number? Would you please explain To illustrate -2 * -5 = 10. Not -10 so why would -1/n * -1/10 = -1/10n?

If you multiply, say, -1 by x, the result is -x. It makes no difference if x is positive or negative. If x is negative, then -x is a *positive* number, even if it might look negative because of the negative sign in front. There is a second negative sign 'hidden' inside of 'x'.

That's the issue with the step you took in your edited post. When you multiply 1/n by -1/10, the result is *always* equal to -1/10n. It makes no difference at all if n is positive or negative. If n is negative, then -1/10n is a positive number, because you have two negatives in the fraction, one in the numerator and one in the denominator (since n is negative).

I'd strongly suggest you review this part of algebra, because it's fundamental in many GMAT questions.
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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

if n=4 for example 4>-1/10 and 1/4>-10. why to flip sign?

Welcome to GMAT Club. Below is an answer to your question.

Since n denotes a number to the left of 0 on the number then n is negative, so it can not be 4 as you assumed.

Next, when you multiply (or divide) an inequality by a negative number you must flip the sign of the inequality.

So, for \(n>-\frac{1}{10}\) --> multiply by negative -10 and flip the sign: \(-10n<1\) --> divide by negative \(n\) and flip the sign again: \(-10>\frac{1}{n}\).

Hi Bunuel Can it be right if go like the following way?

given (-n)^2<1/100 =>n^2<1/00 =>n<1/10

Now Q ask us to find reciprocal of n So take the reciprocal on both side simply we get 1/n<10 Ans

No. Given: n < 0 \(n^2 < (1/100)\) Note that you will not use -n because n is negative. n already includes the negative sign. When you take square root, you get |n| < 1/10 (and not n < 1/10). This means -1/10 < n < 1/10. But since n < 0, -1/10 < n < 0.

We need to find the value of reciprocal i.e. 1/n.

n > -1/10 1/n < -10 Note that in an inequality, if both sides of the inequality have the same sign (positive or negative), the sign of inequality (<, >) flips when you take the reciprocal. Here both sides are negative so sign flips. (Or do what Bunuel did: multiply by -10/n.) So reciprocal is less than -10.

Re: If n denotes a number to the left of 0 on the number line [#permalink]

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03 Oct 2014, 19:39

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.

Hi Bunnel ,

I am weak with inequalities and flipping of signs , where could i get theory for it / additional similar practice questions ?

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.

Hi Bunnel ,

I am weak with inequalities and flipping of signs , where could i get theory for it / additional similar practice questions ?

Re: If n denotes a number to the left of 0 on the number line [#permalink]

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14 Feb 2016, 09:25

These types of problems can be easier to do visually :

Step 1: range of n^2 < 1 /100 | --------- 0 <--------.01 -----|

Step 2: range of \sqrt{n^2} = \sqrt{1/100} (question states only to the left of 0) : | ---------(-1/10)-------->0 ---------|

* note to see if the range of n is going towards 0 or away test a number of n^2 and see what the sq root gives you i.e. \sqrt{1/10000} = +/- \frac{1}{100} which is smaller.

Step 3: inverse of the range n | <--------- (-10) ---- 0 ----------|

Re: If n denotes a number to the left of 0 on the number line [#permalink]

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28 Mar 2017, 10:57

I approached it this way: n is number to the left from 0 n^2<1/100 question asks 1/n-? so from the data we know that -1/10<n<1/10 let's flip so -10>1/n>10 As we know that n is to the left from 0, then 1/n<-10 Answer is A

Re: If n denotes a number to the left of 0 on the number line [#permalink]

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08 Jul 2017, 20:13

Bunuel 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

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

Answer: A.

Bunuel, why do you Multiply \(-\frac{1}{10}<n<0\) by \(-\frac{10}{n}\)?

Can you not just take the reciprocal of \(-\frac{1}{10}<n<0\) to give you: \(-10<\frac{1}{n}<0\)?

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

Answer: A.

Bunuel, why do you Multiply \(-\frac{1}{10}<n<0\) by \(-\frac{10}{n}\)?

Can you not just take the reciprocal of \(-\frac{1}{10}<n<0\) to give you: \(-10<\frac{1}{n}<0\)?

As you can see you are not getting correct inequality, so you cannot do that way.

-10 < -5 < -1 but -1/10 < -1/5 < -1 is not correct.
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Re: If n denotes a number to the left of 0 on the number line [#permalink]

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10 Jul 2017, 19:39

Math mods and others - Please check my approach:

Given: n is less than 0, implying n is negative.

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

Square rooting both sides:

n< -\(\frac{1}{10}\) AND n < \(\frac{1}{10}\)

Reciprocating both the inequalities, \(\frac{1}{n}\) <-10 AND \(\frac{1}{n}\) <10

Since, \(\frac{1}{n}\) MUST be less than -10 and 10 at all times, the option that always satisfies this is A.

Is this a right approach? I always falter when removing the squares on both sides of the equation/inequality and I want to make sure my approach is correct so that I can register it in my noggin.

Re: If n denotes a number to the left of 0 on the number line [#permalink]

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26 Aug 2017, 08:41

experts, please correct me if i'm wrong:

isn't it easy to just take a glance here and eliminate A/C? - You're told the SQUARE of n = fraction meaning n = fraction --> The reciprocal of n must therefore b an integer. Elim B, C, D straight away

Between A & E, you know that n = negative, so its reciprocal cannot be positive. Elim E.