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30 Apr 2024, 01:31
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C
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Question Stats:
78% (01:59) correct
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If \(n\) is an integer such that \(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\), what is the median of all possible values of \(n\)?
A. -3
B. -2
C. 0
D. 2
E. 3
A. -3
B. -2
C. 0
D. 2
E. 3
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05 May 2024, 12:15
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Official Solution:
If \(n\) is an integer such that \(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\), what is the median of all possible values of \(n\)?
A. -3
B. -2
C. 0
D. 2
E. 3
Given that \(n\) is an integer, for the inequality \(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\) to hold true, \(\frac{1}{n^2-1}\) must be \(\frac{1}{8}\), \(\frac{1}{7}\), \(\frac{1}{6}\), \(\frac{1}{5}\), \(\frac{1}{4}\), or \(\frac{1}{3}\). Hence, \(n^2-1\) must be 8, 7, 6, 5, 4, or 3. Consequently, \(n^2\) must be 9, 8, 7, 6, 5, or 4. However, \(n^2\) cannot be 8, 7, 6, or 5, because in these cases \(n\) won't be an integer. Therefore, \(n^2\) is either 9 or 4, making \(n\) equal to -3, 3, -2, or 2. Therefore, the median of all possible values of \(n\) is the average of the two middle terms, thus \(\frac{-2+2}{2} = 0\).
Answer: C
If \(n\) is an integer such that \(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\), what is the median of all possible values of \(n\)?
A. -3
B. -2
C. 0
D. 2
E. 3
Given that \(n\) is an integer, for the inequality \(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\) to hold true, \(\frac{1}{n^2-1}\) must be \(\frac{1}{8}\), \(\frac{1}{7}\), \(\frac{1}{6}\), \(\frac{1}{5}\), \(\frac{1}{4}\), or \(\frac{1}{3}\). Hence, \(n^2-1\) must be 8, 7, 6, 5, 4, or 3. Consequently, \(n^2\) must be 9, 8, 7, 6, 5, or 4. However, \(n^2\) cannot be 8, 7, 6, or 5, because in these cases \(n\) won't be an integer. Therefore, \(n^2\) is either 9 or 4, making \(n\) equal to -3, 3, -2, or 2. Therefore, the median of all possible values of \(n\) is the average of the two middle terms, thus \(\frac{-2+2}{2} = 0\).
Answer: C
General Discussion
30 Apr 2024, 02:12
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Bunuel
If \(n\) is an integer such that \(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\), what is the median of all possible values of \(n\)?
A. -3
B. -2
C. 0
D. 2
E. 3
A. -3
B. -2
C. 0
D. 2
E. 3
\(\frac{1}{9} < \frac{1}{n^2-1} < \frac{1}{2}\)
Taking reciprocal, we get
\(2 < n^2 - 1 < 9\)
Adding 1 on both sides we get
\(3 < n^2 < 10\)
Therefore, n can take the following values: {-3, -2, 2 , 3}
Median = \(\frac{-2 + 2 }{ 2 }= 0\)
Option C
