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

Question

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

KarishmaB · Accepted Answer

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.

Bunuel · Answer

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
eq{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).

SoniaSaini · Answer

Hey Bunuel,
 you're really an awesome person.

thank you very much.

yezz · Answer

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)

mvictor · Answer

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.

DHAR · Answer

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.

meenakshimiyer · Answer

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.

czurm · Answer

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

minininini · Answer

Understood the symmetry at the centre but then set formed with 7 would also be symmetrical at the centre....I'm a little confused

egmat · Answer

Hi minininini,

Great instinct to test Karishma's symmetry idea on the third value. The catch is that the set with  7  only  looks  symmetric because the two middle numbers are equal - but symmetry needs the  whole  distribution to mirror, not just the centre.

Let me draw the  7  case the same way Karishma drew the others.

Set:  4, 6, 7, 7, 9, 10 . The centre sits between the two  7 s (at  7 ).

- To the  right  of  7 :  9  and  10  - that's  +2  and  +3  away.
- For symmetry, the left side would need matching points at  -2  and  -3 , i.e.  5  and  4 .
- But the actual left side is  4  and  6  - a  6 , not a  5 .

That single  6  sitting where a  5  should be breaks the mirror. So the set is  not  symmetric, and that's exactly why mean ≠ median here (mean =  43/6  ≈  7.17 , median =  7 ).

Compare with the two that  do  work:

-  3, 4, 6, 7, 9, 10  - distances from the centre:  0.5 ,  2.5 ,  3.5  on  each  side. True mirror. ✓
-  4, 6, 7, 9, 10, 12  - distances:  1 ,  2 ,  4  on  each  side. True mirror. ✓

The takeaway:  equal middle values are not enough. Always check that every point on the left has a partner the same distance out on the right.

Quick drill to lock it in:  which of these is symmetric?

- { 2 ,  3 ,  5 ,  5 ,  7 ,  8 } - reflect the right ( 7 ,  8 ) about  5  - you need  3  and  2  on the left. You have them ✓
- { 2 ,  4 ,  5 ,  5 ,  7 ,  8 } - same middle  5 s, but now the left is { 2 ,  4 } while the mirror wants { 2 ,  3 }. Broken ✗

Same trap as the  7  case: matching middles, mismatched sides.

Answer: D

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards