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List S consists of 10 consecutive odd integers, and list T c
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04 Sep 2020, 12:34
04 Sep 2020, 12:34
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Question Stats:
70% (02:28) correct
29% (01:29) wrong based on 41 sessions
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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
(A) 2
(B) 7
(C) 8
(D) 12
(E) 22
Re: List S consists of 10 consecutive odd integers, and list T c
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04 Sep 2020, 12:35
04 Sep 2020, 12:35
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Re: List S consists of 10 consecutive odd integers, and list T c
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13 Sep 2020, 03:01
13 Sep 2020, 03:01
List S: 2n+1,2n+3,2n+5,2n+7,2n+9,2n+11,......
first term in list T must be 7 less than first term in list S
first in list T=2n+1-7=2n-6
list T: 2n-6,2n-4,2n-2,2n,2n+2
average of list S=((2n+1)+........+(2n+19))/10 =(20n+100)/10 = 2n+10
average of list T=((2n-6)+....+(2n+2))/5 = (10n-10)/5 =2n-2
difference is =((2n+10)-(2n-2))=12
so the correct answer is D
if you like the explanation please press kudos please
first term in list T must be 7 less than first term in list S
first in list T=2n+1-7=2n-6
list T: 2n-6,2n-4,2n-2,2n,2n+2
average of list S=((2n+1)+........+(2n+19))/10 =(20n+100)/10 = 2n+10
average of list T=((2n-6)+....+(2n+2))/5 = (10n-10)/5 =2n-2
difference is =((2n+10)-(2n-2))=12
so the correct answer is D
if you like the explanation please press kudos please
