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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
S is the set of all prime numbers that satisfy the inequality
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19 Feb 2022, 23:26
19 Feb 2022, 23:26
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S is the set of all prime numbers that satisfy the inequality \(\frac{1}{|10-x|}\leq\frac{5}{x^2}\)
What is the value of the numerator of the fraction that is equal to the average (arithmetic mean) of S expressed in its most reduced terms?
A. 8
B. 9
C. 10
D. 11
E. 12
What is the value of the numerator of the fraction that is equal to the average (arithmetic mean) of S expressed in its most reduced terms?
A. 8
B. 9
C. 10
D. 11
E. 12
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: S is the set of all prime numbers that satisfy the inequality
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28 Feb 2022, 21:58
28 Feb 2022, 21:58
Expert Reply
Explanation
\(\frac{1}{|10 - x|} \leq \frac{5}{x^2}\) can be written as \(\frac{x^2}{|10 - x|} \leq 5\)
When x = 2, we have \(\frac{2^2}{|10 - 2|} = \frac{4}{8}\), which is \(\leq\) 5
When x = 3, we have \(\frac{3^2}{|10 - 7|} = \frac{9}{7}\), which is \(\leq\) 5
When x = 5, we have \(\frac{5^2}{|10 - 5|} = \frac{25}{5}\), which is \(\leq\) 5
Any other prime number greater than 5 will have a fractional value greater than 5.
S = {2, 3, 5}
The arithmetic mean = \(\frac{2 + 3 + 5}{3} = \frac{10}{3}\). The numerator is 10.
Option C
\(\frac{1}{|10 - x|} \leq \frac{5}{x^2}\) can be written as \(\frac{x^2}{|10 - x|} \leq 5\)
When x = 2, we have \(\frac{2^2}{|10 - 2|} = \frac{4}{8}\), which is \(\leq\) 5
When x = 3, we have \(\frac{3^2}{|10 - 7|} = \frac{9}{7}\), which is \(\leq\) 5
When x = 5, we have \(\frac{5^2}{|10 - 5|} = \frac{25}{5}\), which is \(\leq\) 5
Any other prime number greater than 5 will have a fractional value greater than 5.
S = {2, 3, 5}
The arithmetic mean = \(\frac{2 + 3 + 5}{3} = \frac{10}{3}\). The numerator is 10.
Option C
Re: S is the set of all prime numbers that satisfy the inequality
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07 Nov 2023, 20:24
07 Nov 2023, 20:24
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