-
5 Day FREE TrialStudy Smarter, Not Harder
Available with Beat the GMAT members only code
MORE DETAILS
-
Free Trial & Practice ExamBEAT THE GMAT EXCLUSIVE
Available with Beat the GMAT members only code
MORE DETAILS
-
Free Veritas GMAT ClassExperience Lesson 1 Live Free
Available with Beat the GMAT members only code
MORE DETAILS
-
5-Day Free Trial5-day free, full-access trial TTP Quant
Available with Beat the GMAT members only code
MORE DETAILS
-
Free Practice Test & ReviewHow would you score if you took the GMAT
Available with Beat the GMAT members only code
MORE DETAILS
-
FREE GMAT ExamKnow how you'd score today for $0
Available with Beat the GMAT members only code
MORE DETAILS
-
Award-winning private GMAT tutoringRegister now and save up to $200
Available with Beat the GMAT members only code
MORE DETAILS
-
Get 300+ Practice Questions25 Video lessons and 6 Webinars for FREE
Available with Beat the GMAT members only code
MORE DETAILS
-
-
How do you solve this!
This topic has
3 expert replies and 2 member replies
How do you solve this!
Tue Jun 24, 2014 11:22 am
- +1 Upvote Post
- Quote
- Flag
GMAT/MBA Expert
Rich.C@EMPOWERgmat.com
Elite Legendary Member
- Joined
- 23 Jun 2013
- Posted:
- 9439 messages
- Followed by:
- 480 members
- Upvotes:
- 2867
- GMAT Score:
- 800
Tue Jun 24, 2014 11:39 am
Hi AkiB,
This question can be solved by using Number Property Rules or by TESTing VALUES:
Here's how TESTing VALUES works:
List S: 10 consecutive ODD integers
List T: 5 consecutive EVEN integers
The least integer is S is 7 more than the least integer in T.
Let's say that….
T = {2, 4, 6, 8, 10}
Since the least integers in S is 7 MORE than the least integer in T…
S = {9, 11, 13, 15, 17, 19, 21, 23, 25, 27}
The average of T = 6
The average of S = 18
So, the average of S is 18 - 6 = 12 more than the average of T.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
This question can be solved by using Number Property Rules or by TESTing VALUES:
Here's how TESTing VALUES works:
List S: 10 consecutive ODD integers
List T: 5 consecutive EVEN integers
The least integer is S is 7 more than the least integer in T.
Let's say that….
T = {2, 4, 6, 8, 10}
Since the least integers in S is 7 MORE than the least integer in T…
S = {9, 11, 13, 15, 17, 19, 21, 23, 25, 27}
The average of T = 6
The average of S = 18
So, the average of S is 18 - 6 = 12 more than the average of T.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
- +1 Upvote Post
- Quote
- Flag
GMAT/MBA Expert
Brent@GMATPrepNow
GMAT Instructor
- Joined
- 08 Dec 2008
- Posted:
- 11516 messages
- Followed by:
- 1229 members
- Upvotes:
- 5254
- GMAT Score:
- 770
Tue Jun 24, 2014 11:41 am
AkiB wrote:
GIVEN:
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
a) 2
b) 7
c) 8
d) 12
e) 22
Set S: 10 consecutive ODD integers
Set T: 5 consecutive EVEN integer
Let's create some sets that satisfy the given conditions.
...the least integer in S is 7 more than the least integer in T.
How about:
T: {0,.......}
S: {7,.......}
This satisfies the given condition
Now complete the sets:
T: {0, 2, 4, 6, 8}
S: {7, 9, 11, 13,..., 25}
How much greater is the average (arithmetic mean) of the integers in S than the average of the integers in T?
SHORTCUT: In ANY set where the numbers are equally spaced, the average = (smallest term + biggest term)/2 [aka the average of the smallest and biggest values]
So, average of set T = (0 + 8)/2 = 4
Average of set S = (7 + 25)/2 = 16
Difference = 16 - 4 = 12 = D
Cheers,
Brent
_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide
Check out the online reviews of our course
Come see all of our free resources
- +1 Upvote Post
- Quote
- Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
Wed Jun 25, 2014 5:44 am
- +1 Upvote Post
- Quote
- Flag
Top Member
GMATinsight
Legendary Member
- Joined
- 10 May 2014
- Posted:
- 1050 messages
- Followed by:
- 23 members
- Upvotes:
- 205
Wed Jun 25, 2014 8:52 am
- +1 Upvote Post
- Quote
- Flag
GMAT/MBA Expert
Jeff@TargetTestPrep
GMAT Instructor
- Joined
- 09 Apr 2015
- Posted:
- 1013 messages
- Followed by:
- 13 members
- Upvotes:
- 39
Thu Dec 07, 2017 7:08 am
AkiB wrote:
We can let x = the least integer in T. Thus, T contains the following integers: x, x + 2, x + 4, x + 6, and x + 8.
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
a) 2
b) 7
c) 8
d) 12
e) 22
Since the least integer in S is 7 more than the least integer in T, x + 7 = the least integer in S, and so S has the following integers: x + 7, x + 9, x + 11, x + 13, x + 15, x + 17, x + 19, x + 21, x + 23, and x + 25.
Since each list is an evenly spaced set, the average of each list is the respective median. Since the median of the integers in T is x + 4, and the median of integers in S is [(x +15) + (x + 17)]/2 = (2x + 32)/2 = x + 16, the averages of the integers in T and S are x + 4 and x +16, respectively.
Therefore, the average of list S is (x + 16) - (x + 4) = 12 more than the average of list T.
Answer: D
_________________
Jeffrey Miller Head of GMAT Instruction
- +1 Upvote Post
- Quote
- Flag
Top First Responders* |
|||
|---|---|---|---|
| 1 |  |
GMATGuruNY | 67 first replies |
| 2 |  |
Rich.C@EMPOWERgma... | 44 first replies |
| 3 |  |
Brent@GMATPrepNow | 40 first replies |
| 4 |  |
Jay@ManhattanReview | 25 first replies |
| 5 |  |
Terry@ThePrinceto... | 10 first replies |
| * Only counts replies to topics started in last 30 days | |||
| See More Top Beat The GMAT Members | |||
Most Active Experts |
|||
|---|---|---|---|
| 1 | |
GMATGuruNY
The Princeton Review Teacher |
132 posts |
| 2 | |
Rich.C@EMPOWERgma...
EMPOWERgmat |
112 posts |
| 3 |  |
Jeff@TargetTestPrep
Target Test Prep |
95 posts |
| 4 |  |
Scott@TargetTestPrep
Target Test Prep |
92 posts |
| 5 |  |
Max@Math Revolution
Math Revolution |
91 posts |
| See More Top Beat The GMAT Experts | |||
