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b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink]
28 Jan 2016, 13:24
Question Stats:
60% (00:50) correct
40% (00:48) wrong based on 30 sessions
b − 3, b − 1, b + 2, b + 3, b + 4 The median of the ﬁve terms listed above is 5, where b is a constant. What is the average (arithmetic mean) of the ﬁve terms? A 3 B 4 C 5 D 6 E 7 Practice Questions Question: 11 Page: 468 Difficulty: easy/medium
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Founder
Joined: 18 Apr 2015
Posts: 6874
Followers: 114
Kudos [?]:
1332
[0], given: 6293

Re: b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink]
28 Jan 2016, 13:29
SolutionThe median is the middle number in a set of odd numbers. As such, we do have from stem that \(b+2=5\) so \(b=3\) It follows that the values of the five terms are \(\frac{0+2+5+6+7}{5}=\frac{20}{5}=4\) The answer is \(B\)
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Joined: 08 Mar 2018
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Re: b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink]
27 Mar 2018, 08:51
1
This post received KUDOS
the answer key in the original question is wrong. Correct answer should be B (4)



Founder
Joined: 18 Apr 2015
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Followers: 114
Kudos [?]:
1332
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Re: b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink]
27 Mar 2018, 09:42
Thank you. We had a mismatch when updated the questions with the new timer. Regards
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Re: b − 3, b − 1, b + 2, b + 3, b + 4
[#permalink]
27 Mar 2018, 09:42





