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List S consists of 10 consecutive odd integers, and list T c

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List S consists of 10 consecutive odd integers, and list T c [#permalink] New post 30 Jan 2014, 00:57
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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] New post 30 Jan 2014, 00:57
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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;

The difference will be (x + 9) - (x - 3) = 12.

Answer: D.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink] New post 30 Jan 2014, 01:54
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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 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.

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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink] New post 30 Jan 2014, 02:18
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
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink] New post 30 Jan 2014, 09:17
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Easy one.
Let us consider two set S and T.
1) T is the even consecutive set and S is odd consecutive set .
2) Least value of T +7=Least value of S

So if least value of T is 2 then least value of S is 9.

Its a series of even and odd consecutive integer.
So T 5th term= 2+4*2=10 ... Mean=(10+2)/2=6
Similarly S 10 th term = 9+2*9=27.... Mean=(27+9)/2=18

Difference is 18-6=12

answer is D
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink] New post 01 Feb 2014, 07:59
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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;

The difference will be (x + 9) - (x - 3) = 12.

Answer: D.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink] New post 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] 28 May 2014, 01:47
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