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The number of defects in the first five cars to come through

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The number of defects in the first five cars to come through [#permalink] New post 11 Nov 2008, 14:00
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-number-of-defects-in-the-first-five-cars-to-come-through-104862.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Jan 2014, 04:01, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: defects [#permalink] New post 11 Nov 2008, 18:36
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Jcpenny wrote:
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


D

Depending on the number of defects in 6th car,
3,4,6,7,9,10 (median 13/2 =6.5) OR
4,6,7,7,9,10 (median 14/2 = 7) OR
4,6,7,9,10,12 (median 16/2 = 8)

To find mean we can use the sum of numbers given (= 36) and add either 3, 7, or 12 to it and then divide by 6 (number of cars).

(36 + 3) /6 = 39/6 = 6.5 (== median above)
(36 + 7) /6 = 43/6 = 7.1 (does not equal median above)
(36 + 12) /6 = 48/6 = 8 (== median above)

Thus I and III are valid => D
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Re: The number of defects in the first five cars to come through [#permalink] New post 10 Jan 2014, 20:40
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Re: The number of defects in the first five cars to come through [#permalink] New post 11 Jan 2014, 04:03
Expert's post
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

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).

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-number-of-defects-in-the-first-five-cars-to-come-through-104862.html
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Re: The number of defects in the first five cars to come through   [#permalink] 11 Jan 2014, 04:03
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