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by AkiB » 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
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by [email protected] » 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

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by Brent@GMATPrepNow » 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.
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 = [spoiler]12 = D[/spoiler]

Cheers,
Brent
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by shrivats » 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
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by GMATinsight » 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
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by Jeff@TargetTestPrep » 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
[email protected]

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