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S95-08

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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

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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.


Answer: E
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Re: S95-08 [#permalink]

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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?

Thanks in advance!

NB

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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?

Thanks in advance!

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

Answer: A.

Discussed here: if-n-denotes-a-number-to-the-left-of-0-on-the-number-line-91659.html
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New post 14 Sep 2016, 10:25
It implies m^2 > 9 .

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New post 20 Nov 2017, 16:45
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"

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New post 20 Nov 2017, 21:56
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: \(-\frac{1}{3} < m < 0\).
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Collection of Questions:
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Re: S95-08 [#permalink]

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New post 21 Nov 2017, 12:37
Thank you very much Bunuel

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Re: S95-08   [#permalink] 21 Nov 2017, 12:37
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