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CARS

by phelps » Wed May 04, 2011 11:00 am

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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

CORRECT ANSWER
D
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by pemdas » Wed May 04, 2011 11:33 am
by arranging the set {4,6,7,9,10} we know that mean=median when the values are consecutive; so must be {4,five,6,7,eight,9,10} or {three,4,6,7,9,10,twelve,thirteen} where 3,4 and 6,7 and 9,10 and 12,13 are considered as clued, One unit expressions -> 7, 13, 19, 25 (the difference is 6, so consecutive properties). Thirteen is out BUT we must keep it in mind as possible number in this set. Option d 3 and 12
phelps 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

CORRECT ANSWER
D
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by Tani » Wed May 04, 2011 12:03 pm
The sum of the first five cars is 36.

If the sixth car has 3, the sum of the first 6 is 39, which makes the mean 6.5
arranging them in order 3,4,6,7,9,10 - the median is the average of the two middle numbers = 6.5

"I" works. Rule out B and C

Since III appears twice we test it next.
the new sum = 36+12 = 48 average = 8.
again arranging 4,6,7,9,10, 13 = the median is the average of 7 and 9 = 8
"III" works - rule out A.

We still have to check II.
New sum = 45. Average = 7.5
arranging 4,6,7,7,9,10 - the median is 7. II doesn't work, the answer is D
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by meenakshimiyer » Sat Jan 12, 2019 9:26 pm
The sum of defects in the first five cars is 9 + 7 + 10 + 4 + 6 = 36. With six cars, the median will be the average of third and fourth ranked defects, when arranged in ascending order. Since number of defects is always an integer, hence the median must either be an integer or integer plus 0.5.
Now if there are X defects in the sixth car, then the mean is obtained as
M = (36 + X)/6
Since 36 is already divisible by 6, hence to satisfy the median condition, X must either be a multiple of 6 or a multiple of 3. From the given options, 7 does not satisfy the condition, hence it is out. We now need to check for both 3 and 12.
With X = 3, the mean is M = 39/6 = 6.5; and the values arranged in ascending order are {3, 4, 6, 7, 9, 10} for which the median is (6 + 7)/2 = 6.5 = M. So it matches for I.
With X = 12, the mean is M = 48/6 = 8; and the values arranged in ascending order are {4, 6, 7, 9, 10, 12} for which the median is (7 + 9)/2 = 8 = M. So it matches for III.
Both I and III match, hence D.
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by [email protected] » Sun Jan 13, 2019 10:50 am
Hi All,

We're told that the number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively and that the sixth car through the production line has either 3, 7, or 12 defects. We're asked which of the following three values does the MEAN number of defects per car for the first six cars equal the MEDIAN. This question comes down to organizing the data (to find the Median) and using the Average Formula.

I. 3 defects

IF... the 6th car has 3 defects, then the six numbers would be: 3, 4, 6, 7, 9 and 10
The Median would be (6+7)/2 = 6.5
The Average would be (3+4+6+7+9+10)/6 = 39/6 = 6.5
The Average and the Median ARE equal.
Eliminate Answers B and C.

II. 7 defects

IF... the 6th car has 7 defects, then the six numbers would be: 4, 6, 7, 7, 9 and 10
The Median would be (7+7)/2 = 7
The Average would be (4+6+7+7+9+10)/6 = 43/6 = 7 1/6
The Average and the Median are NOT equal.
Eliminate Answer E

III. 12 defects

IF... the 6th car has 12 defects, then the six numbers would be: 4, 6, 7, 9, 10 and 12
The Median would be (7+9)/2 = 8
The Average would be (4+6+7+9+10+12)/6 = 48/6 = 8
The Average and the Median ARE equal.
Eliminate Answer A.

Final Answer: D

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Rich
Contact Rich at [email protected]
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by Scott@TargetTestPrep » Mon Jan 14, 2019 5:54 pm
phelps 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
Let's check each number in the given Roman numerals.

I. 3

The average is (3 + 4 + 6 + 7 + 9 + 10)/6 = 6.5.

The median is (6 + 7)/2 = 13/2 = 6.5.

We see that I is correct.

II. 7

The average is (4 + 6 + 7 + 7 + 9 + 10)/6 ≈ 7.17.

The median is (7 + 7)/2 = 14/2 = 7.

We see that II is not correct.

III. 12

The average is (4 + 6 + 7 + 9 + 10 + 12)/6 = 8.

The median is (7 + 9)/2 = 16/2 = 8.

We see that III is correct.

Answer: D

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