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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]  22 Aug 2006, 09:15
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

The OA is D but I think it should be C What do you guys think?
Senior Manager
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1) mean 6.5 median 6.5
2) mean 7.16 median 7
3) mean 8 median 8

I and III so D..
Senior Manager
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Jesus that means I have to take a break I divided the first average by 3 instead of 6 Sorry for the dumb question
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Re: Mean Median multiple choice [#permalink]  22 Aug 2006, 11:36
apollo168 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

The OA is D but I think it should be C What do you guys think?

its D.

sum of defects for 5 cars = 4+6+6+7+9+10 = 36

i. sum of defects for 6 cars, if the no. of defects for 6th car is 3 = 3+4+6+7+9+10 = 39
avg. = 39/6 = 6.5
median = (6+7)/2 = 6.5
so mean = median

ii. sum of defects for 6 cars if the no. of defects for the 6th car is 7 = 4+6+7+7+9+10 = 43
avg. = 43/6 = 7.1
median = (7+7)/2 = 7
so mean < median

ii. sum of defects for 6 cars if the no. of defects for the 6th car is 12 = 4+6+7+9+10+12 = 48
avg. = 48/6 = 8
median = (7+9)/2 = 8
so mean = median

So its D.
Re: Mean Median multiple choice   [#permalink] 22 Aug 2006, 11:36
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