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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
The number of defects in the first five cars to come through a new pro
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25 Nov 2021, 04:07
25 Nov 2021, 04:07
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Question Stats:
33% (01:34) correct
66% (02:35) wrong based on 6 sessions
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The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
I. 3
II. 7
III. 12
A. I only
B. II only
C. III only
D. I and III only
E. I, II, and III
Re: The number of defects in the first five cars to come through a new pro
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26 Nov 2021, 06:07
26 Nov 2021, 06:07
Expert Reply
Basically we have a set with 6 terms: {4, 6, 7, 9, 10, x}. The question asks if \(x\) is either 3, 7, or 12 then for which values of \(x\) the mean of the set equals to the median (note that \(mean=\frac{4+6+7+9+10+x}{6}=\frac{36+x}{6}\) and the median will be the average of two middle terms, so it depends on the value of \(x\)).
If \(x=3\) then \(mean=\frac{36+3}{6}=6.5\) and \(median=\frac{6+7}{2}=6.5\), so \(mean={median}\);
If \(x=7\) then \(mean=\frac{36+7}{6}=\frac{43}{6}\) and \(median=\frac{7+7}{2}=7\), so \(mean\neq{median}\);
If \(x=12\) then \(mean=\frac{36+12}{6}=8\) and \(median=\frac{7+9}{2}=8\), so \(mean={median}\).
Answer: D (I and III only).
If \(x=3\) then \(mean=\frac{36+3}{6}=6.5\) and \(median=\frac{6+7}{2}=6.5\), so \(mean={median}\);
If \(x=7\) then \(mean=\frac{36+7}{6}=\frac{43}{6}\) and \(median=\frac{7+7}{2}=7\), so \(mean\neq{median}\);
If \(x=12\) then \(mean=\frac{36+12}{6}=8\) and \(median=\frac{7+9}{2}=8\), so \(mean={median}\).
Answer: D (I and III only).
gmatclubot
Re: The number of defects in the first five cars to come through a new pro [#permalink]
26 Nov 2021, 06:07
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