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The number of defects in the first five cars to come through [#permalink]
15 Nov 2010, 07:59

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Difficulty:

65% (hard)

Question Stats:

52% (02:35) correct
48% (01:38) wrong based on 82 sessions

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

Re: defect problem kindly help [#permalink]
15 Nov 2010, 08:16

1

This post received KUDOS

Expert's post

SoniaSaini 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

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

will appreciate your help/tips. cheers, Sonia saini

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 \(mean={median}\).

Re: defect problem kindly help [#permalink]
25 May 2013, 06:33

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

Intuitively each answer choice except for 7 , together with the given forms the union of 2 AP that has the same difference (d =3) and same number of terms.

3,4,6,7,9,10 = {3,6,9} U {4,7,10}

4,6,7,9,10,12 = { 4,7,10} U {6,9,12}

mean and median of such union is equal (symmetric distribution) and is equivalent to the average of both sets median (means)

gmatclubot

Re: defect problem kindly help
[#permalink]
25 May 2013, 06:33

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