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:

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

Problem Solving Question: 70 Category:Arithmetic Statistics Page: 70 Difficulty: 600

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

30 Jan 2014, 01:57

2

This post received KUDOS

Expert's post

7

This post was BOOKMARKED

SOLUTION

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: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

30 Jan 2014, 02:54

1

This post received KUDOS

List S consists of 10 consecutive odd integers, and list T consists of 5 consecutive even integers. If the least integer in Sis 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

Sol: Let List T has the following members : 2,4,6,8 and 10 Then S has : 9,11,13,15,17,19,21,23,25,27

Now If we find the average of List T is 6 and average of List S is (19+17)/2 =18 So Ans is 12.

Suppose if we S also had 5 members and all the other condition remains same then Average of S would have been 13 and diferecne between the 2 would be 7 cause when the same number is added/subtracted from a given set then the average of the new set increases or decreases by the same number

So ans is D.

Average difficulty level of 650 is okay _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

30 Jan 2014, 03:18

1

This post received KUDOS

We could do this by taking value for the lists List T=-4,-2,0,2,4.Mean=0 List S=3,5,7,...21=>Mean=12;(21+3)/2 (S has started from 3 as -4+7=3) Difference=12 Ans.D

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

01 Feb 2014, 08:59

Expert's post

SOLUTION

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: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

28 May 2014, 02:47

Since the least no. in S is 7 greater than the least no. in T, lets assume S starts at 7 so T will start at 0. For S mean will be the average of 5th and 6th no.: {7, 9, 11, 13, 15, 17....} = (15+17)/2 = 16 For T mean will be the 3rd no. {0, 2, 4...} = 4 Answer=16-4=12 D! _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

15 Jun 2015, 04:12

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. _________________

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

24 Jan 2016, 23:49

Bunuel wrote:

SOLUTION

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;

The difference will be (x + 9) - (x - 3) = 12.

Answer: D.

Hi Bunel,

I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3;"

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

25 Jan 2016, 00:32

Expert's post

amanlalwani wrote:

Bunuel wrote:

SOLUTION

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;

The difference will be (x + 9) - (x - 3) = 12.

Answer: D.

Hi Bunel,

I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3;"

Hi, there are 10 consecutive odd numbers , means each number is 2 more than the previous number... if the least number here is x, the next number will be x+2, third will be x+2*2... and so on till 10th term= x+9*2.. also we can find this through arithmetic progression.. Nth term = first term + (N-1)d, d is the constant difference between two consecutive numbers..

2ND part.. "the mean of T will simply be the median or the third term: mean = (x - 7) + 2*2 = x - 3 in the second set, there are only five consecutive numbers so the median=mean=the central number, which is third number here.. the least integer in s is 7 less than T, so it will become x-7... the third term here will be (x-7) + 2*2..same as nthterm above _________________

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

Show Tags

25 Jan 2016, 02:20

1

This post received KUDOS

Expert's post

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

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

Since S is the list consisting of 10 consecutive odd integers we can put S={s, s + 2, s + 4, ...., s + 18}, where s is the least odd integer of S. So the average of S is (10*s + 2+4+....+18)/10=(10*s + 90)/10= s+9.

Similarly we may put T={t, t+2, ..., t+8}, where t is the least even integer of T. So the average of T is (5*t + 2+ 4+ ....+8)/5 = t+4. s+9-(t+4)=s-t+5=7+5=12. So the answer is 12. ---> (D). _________________

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?

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

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 (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...