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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
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A. -5
B. -4.5
C. -4
D. -3.5
E. -3
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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
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General Discussion
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
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
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
8Harshitsharma
Joined: 25 Oct 2017
Last visit: 06 Jul 2025
Posts: 125
GMAT Focus 1: 655 Q87 V80 DI80 
GMAT 1: 690 Q49 V35
GRE 1: Q165 V160
GRE 2: Q170 V162
GPA: 9.25
Bunuel
Well, another inequality problem EZ, right? WRONG. Or well, you might be wrong, so be careful out there when solving such questions they have nasty traps sometimes, that are easy to miss.
Let's go, let's flip 'em all to make it easy, right? Not quite.
We must check first of all, that the given fraction of n is only positive or it may give negative values as well.
We see that possibility, (1-n) is negative when n>1 and positive when n<1.
So flipsy-doodle can be done only when that term is positive i.e. when n<1. But, upon paying close attention to the problem, we see that it is indirectly implied that n is indeed less than 1, because the given limits are both positive fractions.
\(\frac{1}{3} > \frac{1}{1 - n} > \frac{1}{7}\)
Flip the fraction and reverse the signs,
\(3 < 1- n < 7\)
\( -2 > n > -6\)
So, n can have -3, -4, -5 as its values. Since the values are in an AP, Mean = -4.
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