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Register now and save up to $200 Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code ## How do you solve this! tagged by: AkiB This topic has 3 expert replies and 2 member replies AkiB Junior | Next Rank: 30 Posts Joined 27 Apr 2013 Posted: 16 messages #### How do you solve this! Tue Jun 24, 2014 11:22 am 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 OA D ### GMAT/MBA Expert Jeff@TargetTestPrep GMAT Instructor Joined 09 Apr 2015 Posted: 647 messages Followed by: 9 members Upvotes: 39 Thu Dec 07, 2017 7:08 am AkiB 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 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. 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 GMATinsight Legendary Member Joined 10 May 2014 Posted: 1007 messages Followed by: 21 members Upvotes: 205 Wed Jun 25, 2014 8:52 am Another easy way is as follows in Arithmetic Progression, the median is always equal to the Mean i.e. Average Therefore if least of 5 even integer is x then least of 10 odd integers is x+7 Therefore mean of 5 even integer is x+4 (x, x+2, x+4,...) then mean of 10 odd integers is x+7+9 [average of (x+7)+8 and (x+7)+10] The difference is = (x+16) - (x+4) = 12 ANSWER _________________ Bhoopendra Singh & Sushma Jha - Founder "GMATinsight" Testimonials e-mail: info@GMATinsight.com I Mobile: +91-9999687183 / +91-9891333772 To register for One-on-One FREE ONLINE DEMO Class Call/e-mail One-On-One Private tutoring fee - US$40 per hour & for FULL COURSE (38 LIVE Sessions)-US\$1000

shrivats Junior | Next Rank: 30 Posts
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Wed Jun 25, 2014 5:44 am
a bit long, but we can solve this algebraically...

T= x, x+2, x+4, x+6, x+8

sum of all elements of T
= x + x+2 + x+4 + x+6 + x+8
= 5x + 2(1+2+3+4)
= 5x + 20
A.M (T) = (5x+20)/5 = x+4

S = x+7,x+9,x+11,x+13 .... , x+25 ( since the least element of S is 7 more than x )

sum of all elements in S
(x+7)+ (x+9) + (x+11) + .... + (x+25)
7*10 + {x + (x+2) + (x+4) + (x+6) .... + (x+18)}
70 + 10x + 2 ( 1+2+3 .... +9)
70 + 10x + 2*45 = 10x + 160

A.M (S) = (10x+160)/10 = x+16

x+16 - (x+4) = 12

### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Tue Jun 24, 2014 11:41 am
AkiB 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
GIVEN:
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.
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

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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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 thatâ€¦.
T = {2, 4, 6, 8, 10}
Since the least integers in S is 7 MORE than the least integer in Tâ€¦
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.

GMAT assassins aren't born, they're made,
Rich

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