May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 19 Sep 2010
Posts: 55
Location: Pune, India

The number of defects in the first five cars to come through
[#permalink]
Show Tags
15 Nov 2010, 08:59
Question Stats:
56% (02:21) correct 44% (02:14) wrong based on 373 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 55276

Re: defect problem kindly help
[#permalink]
Show Tags
15 Nov 2010, 09:16
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).
_________________



Manager
Joined: 19 Sep 2010
Posts: 55
Location: Pune, India

Re: defect problem kindly help
[#permalink]
Show Tags
16 Nov 2010, 07:40
Hey Bunuel, you're really an awesome person.
thank you very much.



Retired Moderator
Joined: 05 Jul 2006
Posts: 1700

Re: defect problem kindly help
[#permalink]
Show Tags
25 May 2013, 07:33
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)



Board of Directors
Joined: 17 Jul 2014
Posts: 2552
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9234
Location: Pune, India

Re: The number of defects in the first five cars to come through
[#permalink]
Show Tags
23 Dec 2015, 21:55
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. 3467910 The centre is between 6 and 7 and the elements are symmetrical about it. 46791012 The centre is between 7 and 9 and the elements are symmetrical about it.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 15 Dec 2015
Posts: 115
GPA: 4
WE: Information Technology (Computer Software)

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



Manager
Joined: 12 Jan 2019
Posts: 51

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




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






