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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]

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New post 15 Nov 2010, 08:59
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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
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Re: defect problem kindly help  [#permalink]

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New post 15 Nov 2010, 09:16
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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 \(mean={median}\).

Answer: D (I and III only).
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Re: defect problem kindly help  [#permalink]

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New post 16 Nov 2010, 07:40
Hey Bunuel,
you're really an awesome person.

thank you very much.
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Re: defect problem kindly help  [#permalink]

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New post 25 May 2013, 07:33
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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

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)
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Re: The number of defects in the first five cars to come through  [#permalink]

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New post 23 Dec 2015, 19:20
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.
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Re: The number of defects in the first five cars to come through  [#permalink]

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New post 23 Dec 2015, 21:55
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yezz 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

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)


Another intuitive way to see that mean will be equal to median is to imagine them on a number line. Both sets (with 3 and with 12) are symmetrical about the centre and hence mean = median.


------3-4--6-7--9-10-----
The centre is between 6 and 7 and the elements are symmetrical about it.

-------4--6-7--9-10--12-------
The centre is between 7 and 9 and the elements are symmetrical about it.
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Re: The number of defects in the first five cars to come through  [#permalink]

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New post 26 Jul 2017, 12:44
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.
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Re: The number of defects in the first five cars to come through  [#permalink]

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New post 12 Jan 2019, 22:36
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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Re: The number of defects in the first five cars to come through   [#permalink] 12 Jan 2019, 22:36
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