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 350,000 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:

List S consists of 10 consecutive odd integers, and list T [#permalink]
16 Feb 2011, 08:42

2

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

66% (02:43) correct
34% (01:50) wrong based on 127 sessions

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?

Re: Arithmetic Statistics [#permalink]
16 Feb 2011, 08:59

6

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Baten80 wrote:

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in S is 7 more than the least integer in T, how much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?

a) 2 b) 7 c) 8 d) 12 e) 22

For any evenly spaced set median=mean=the average of the first and the last terms.

So the mean of S will be the average of the first and the last terms: mean=(x+x+9*2)/2=x+9, where x is the first term; The mean of T will simply be the median or the third term: mean=(x-7)+2*2=x-3;

Re: Arithmetic Statistics [#permalink]
17 Feb 2011, 04:51

2

This post received KUDOS

If you didnt know the formula for evenly spaced sets, you can pick numbers and solve this question very easily.

Pick first number of set T=2. Since number of elements = 5, the mean will be the middle term ie the 3rd term, which will be 6.

From the question, you can infer that first number of the set T, will be 2+7= 9. Since this set consists of even number of terms,ie 10, the mean will be the average of the middle 2 terms, ie 5th and 6th term, which are 17 and 19 respectively and their average will be 18= mean of the set.

Therefore, the difference between the mean of two sets= 18-6= 12.

Re: List S consists of 10 consecutive odd integers, and list T [#permalink]
23 Jun 2014, 00:53

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

PareshGmat's solution is the quickest way. Anyway, here is mine: T=(t1+t2+t3+t4+t5)/5 =[t1+(t1+1*2)+(t1+2*2)+(t1+3*2)+(t1+4*2)]/5 =[5t1+2*(1+2+3+4)]/5 =[5t1+2*(4*(4+1)/2)]/5 =[5t1+20]/5=t1+4 S=(s1+...+s10)/10 =[s1+(s1+1*2)+...+(s1+9*2)]/10 =[10s1+2*(1+...+9)]/10 =[10s1+2*(9*(9+1)/2)]/10 =[10s1+90]/10=s1+9 We have s1=t1+7 S-T=s1+9-t1-4=t1+7+9-t1-4=12 =>D P/S: Sum of n consecutive integers: n(n+1)/2

gmatclubot

Re: List S consists of 10 consecutive odd integers, and list T
[#permalink]
29 Jun 2014, 21:14

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

I had an interesting conversation with a friend this morning, and I realized I need to add a last word on the series of posts on my application process. Five key words:...