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16 Sep 2014, 01:49
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:51) correct
30%
(02:03)
wrong
based on 83
sessions
History
Date
Time
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Not Attempted Yet
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
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
16 Sep 2014, 01:49
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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
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
03 Oct 2014, 08:56
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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
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
