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GRE Prep Club Team Member
Joined: 20 Feb 2017
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Three straight metal rods have an average (arithmetic mean) length of
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12 Dec 2022, 06:08
12 Dec 2022, 06:08
Expert Reply
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Question Stats:
80% (01:35) correct
20% (00:52) wrong based on 5 sessions
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Three straight metal rods have an average (arithmetic mean) length of 77 inches and the shortest rod has a length of 65 inches. What is the maximum possible value of the median length, in inches, of the three rods?
A. 71
B. 77
C. 80
D. 83
E. 89
A. 71
B. 77
C. 80
D. 83
E. 89
Re: Three straight metal rods have an average (arithmetic mean) length of
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14 Dec 2022, 04:35
14 Dec 2022, 04:35
Expert Reply
Say the lengths of the rods in ascending order are \(x_1\), \(x_2\), and \(x_3\), where \(x_1\leq{x_2}\leq{x_3}\).
The median of a set with odd number of terms is just the middle term, when arranged in ascending/descending order, hence the median is \(x_2\).
Given that \(x_1+x_2+x_3=3*77\) --> \(65+x_2+x_3=3*77\) --> \(x_2+x_3=166\). We need to maximize \(x_2=median\), so we need to minimize \(x_3\).
The minimum value of \(x_3\) is \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).
Answer: D.
The median of a set with odd number of terms is just the middle term, when arranged in ascending/descending order, hence the median is \(x_2\).
Given that \(x_1+x_2+x_3=3*77\) --> \(65+x_2+x_3=3*77\) --> \(x_2+x_3=166\). We need to maximize \(x_2=median\), so we need to minimize \(x_3\).
The minimum value of \(x_3\) is \(x_2\) --> \(x_2+x_2=166\) --> \(x_2=median=83\).
Answer: D.
gmatclubot
Re: Three straight metal rods have an average (arithmetic mean) length of [#permalink]
14 Dec 2022, 04:35
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