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b − 3, b − 1, b + 2, b + 3, b + 4

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b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink] New post 28 Jan 2016, 13:24
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Question Stats:

58% (00:50) correct 41% (00:51) wrong based on 41 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
[Reveal] Spoiler: OA

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Founder
Founder
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Joined: 18 Apr 2015
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Kudos [?]: 2012 [0], given: 8049

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Re: b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink] New post 28 Jan 2016, 13:29
Expert's post
Solution

The 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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Re: b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink] New post 27 Mar 2018, 08:51
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the answer key in the original question is wrong. Correct answer should be B (4)
Founder
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Re: b − 3, b − 1, b + 2, b + 3, b + 4 [#permalink] New post 27 Mar 2018, 09:42
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Thank you. We had a mismatch when updated the questions with the new timer.

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Re: b − 3, b − 1, b + 2, b + 3, b + 4   [#permalink] 27 Mar 2018, 09:42
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b − 3, b − 1, b + 2, b + 3, b + 4

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