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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

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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
30 Jan 2014, 00:57

1

This post received KUDOS

Expert's post

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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! _________________

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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
15 Jun 2015, 03:12

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