defect problem kindly help

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Manager

Joined: 19 Sep 2010

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Location: Pune, India

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defect problem kindly help [#permalink]

15 Nov 2010, 08:59

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Question Stats:

11% (00:00) correct

88% (01:26) wrong

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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

not able to understand what question is asking for? kindly help me to solve this one.

will appreciate your help/tips.
cheers,
Sonia saini

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Re: defect problem kindly help [#permalink]

15 Nov 2010, 09:16

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SoniaSaini wrote:

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).
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Manager

Joined: 19 Sep 2010

Posts: 57

Location: Pune, India

Followers: 10

Kudos [?]: 11 [0], given: 9

Re: defect problem kindly help [#permalink]

16 Nov 2010, 07:40

Hey Bunuel,
you're really an awesome person.

thank you very much.

Re: defect problem kindly help [#permalink] 16 Nov 2010, 07:40

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defect problem kindly help

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