Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

15 Nov 2010, 08:59

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

52% (02:42) correct
48% (01:51) wrong based on 146 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

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 \(mean={median}\).

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)

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.

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

Show Tags

23 Dec 2015, 21:55

1

This post received KUDOS

Expert's post

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. _________________

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I may have spoken to over 50+ Said applicants over the course of my year, through various channels. I’ve been assigned as mentor to two incoming students. A...