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30 Apr 2024, 01:31
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Difficulty:
15%
(low)
Question Stats:
86% (01:34) correct
14%
(01:23)
wrong
based on 91
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If \(n\) is an integer such that \(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\), what is the average (arithmetic mean) of all possible values of \(n\)?
A. -5
B. -4.5
C. -4
D. -3.5
E. -3
A. -5
B. -4.5
C. -4
D. -3.5
E. -3
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05 May 2024, 14:15
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Official Solution:
If \(n\) is an integer such that \(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\), what is the average (arithmetic mean) of all possible values of \(n\)?
A. -5
B. -4.5
C. -4
D. -3.5
E. -3
Given that \(n\) is an integer, for the inequality \(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\) to hold, \(\frac{1}{1 - n}\) must be \(\frac{1}{4}\), \(\frac{1}{5}\), or \(\frac{1}{6}\). Hence, solving for \(n\) yields \(n = -3\), \(n = -4\), and \(n = -5\). Therefore, the average of all possible values of \(n\) is \(-4\).
Answer: C
If \(n\) is an integer such that \(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\), what is the average (arithmetic mean) of all possible values of \(n\)?
A. -5
B. -4.5
C. -4
D. -3.5
E. -3
Given that \(n\) is an integer, for the inequality \(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\) to hold, \(\frac{1}{1 - n}\) must be \(\frac{1}{4}\), \(\frac{1}{5}\), or \(\frac{1}{6}\). Hence, solving for \(n\) yields \(n = -3\), \(n = -4\), and \(n = -5\). Therefore, the average of all possible values of \(n\) is \(-4\).
Answer: C
General Discussion
30 Apr 2024, 02:08
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Bunuel
\(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\)
\(\frac{1}{7} < \frac{1}{1 - n} < \frac{1}{3}\)
Taking reciprocal of the terms
\(3 < 1 - n < 7\)
Subtracting 1 on both sides, we get
\( 2 < - n < 6\)
Multiplying by -1
\(-6 < n < -2\)
Hence, n ={ -5, -4 ,-3}
As the number is in arithmetic progression,
Mean = Median
Hence, average = -4
Option C
