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]
30 Jan 2014, 00:57
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]
30 Jan 2014, 01:54
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]
30 Jan 2014, 02:18
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]
01 Feb 2014, 07:59
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]
28 May 2014, 01:47
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]
15 Jun 2015, 03:12
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]
24 Jan 2016, 22:49
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]
24 Jan 2016, 23:32
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]
25 Jan 2016, 01:20
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). _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Ninety-five percent of the Full-Time Class of 2015 received an offer by three months post-graduation, as reported today by Kellogg’s Career Management Center(CMC). Kellogg also saw an increase...
By Dean Nordhielm So you just got into business school. Congrats! At this point you still have months before you actually begin classes. That seems like a lot of...