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

Hey Bunuel,
you're really an awesome person.

thank you very much.

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)

let's arrange the numbers in ascending order:
4, 6, 7, 9, 10 = sum is 36.

which # if added will result mean=median?
ok, let's test 3:
so, the sum is 36+3=39. we have to divide this by 6 to find the mean, and we have 6.5
let's find the median
3, 4, 6, 7, 9, 10 = we can see that the median is (6+7)/2 so the median is 6.5
ok, so we see that the first one works, and thus we can eliminate B and C.
let's test second one:

new sum is 36+7=43. the average thus would be 43/6, and improper fraction.
new median
4, 6, 7, 7, 9, 10 - so the median is 7. we can see that the median is not equal to the mean. we can thus eliminate E, and we are left with A and D.

let's test the final one:
new sum is 36+12=48. divide by 6 = 8. 8 is the new average.
4, 6, 7, 9, 10, 12 - the new median is (7+9)/2 = 8.
we can see that median=mean, and we can cross A, and select D.

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 these values does the mean number of defects per car for the first six cars equal the median?
I. 3
II. 7
III. 12

Explanation:

Given the number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. Sixth car will have either of 3, 7 or 12 defects.

Let us assume sixth car has x defects

⇒ Mean number of defects = 9+7+10+4+6+x6=36+x6=6+x69+7+10+4+6+x6=36+x6=6+x6 .... (1)


Median of a data can be found out by arranging the terms in ascending order, then finding out the middle term if number of terms is odd and average of the two middle terms if number of terms is even.

Putting x = 3, we get:
Mean = 6.5

Terms arranged in ascending order are 3, 4, 6, 7, 9, 10.
⇒ Median = (6 + 7)/2 = 6.5

Since Mean = Median => x can be 3.

Putting x = 7, we get:
Mean = 6 + 7/6 = 43/6 = 7.16

Terms arranged in ascending order are 4, 6, 7, 7, 9, 10.
⇒ Median = (7 + 7)/2 = 7

Since mean is not equal to median => x cannot be 7.

Putting x = 12, we get:
Mean = 8

Terms arranged in ascending order are 4, 6, 7, 9, 10, 12.
⇒ Median = (7 + 9)/2 = 16/2 = 8

Since mean = median ⇒ x can be 8.

Answer: D.

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.

A quick approach to these kinds of questions is to observe how the shape of the data changes about the median when we introduce a new element to the set. In a symmetric distribution, the median=the mean. So to answer the question, we have to observe which elements make the data symmetric about the median.

First, order the initial set: 4, 6, 7, 9 ,10

I. 3
3, 4, 6, 7, 9 ,10. Observe median will be the average of 6 and 7. The distance between 3 and 4 is 1. The distance between 4 and 6 is 2. Looking to the right of the media, the distance between 7 and 9 is 2. The distance between 9 and 10 is 1. Distribution is symmetric about the median.

II. 7
4, 6, 7, 7, 9, 10.

Distances: +2 +1 (median) +2 +1. NOT SYMMETRIC.

II. 12

4, 6, 7, 9 ,10, 12

Distances: +2 +1 (median) +1 +2 SYMMETRIC

Answer: D

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