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List S consists of 10 consecutive odd integers, and list T c [#permalink]
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The Official Guide For GMAT® Quantitative Review, 2ND EditionList 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 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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SOLUTIONList 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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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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30 Jan 2014, 02:54
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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. Average difficulty level of 650 is okay
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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30 Jan 2014, 03:18
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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]
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30 Jan 2014, 10:17
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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 186=12
answer is D



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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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01 Feb 2014, 08:59
SOLUTIONList 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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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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28 May 2014, 02:47
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=164=12 D!
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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24 Jan 2016, 23:49
Bunuel wrote: 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. Hi Bunel, I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3;"



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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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25 Jan 2016, 00:32
amanlalwani wrote: Bunuel wrote: 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. Hi Bunel, I could not understand how x+9*2 is the final term and similarly "the mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3;" Hi, there are 10 consecutive odd numbers , means each number is 2 more than the previous number... if the least number here is x, the next number will be x+2, third will be x+2*2... and so on till 10th term= x+9*2.. also we can find this through arithmetic progression.. Nth term = first term + (N1)d, d is the constant difference between two consecutive numbers.. 2ND part.. "the mean of T will simply be the median or the third term: mean = (x  7) + 2*2 = x  3in the second set, there are only five consecutive numbers so the median=mean=the central number, which is third number here.. the least integer in s is 7 less than T, so it will become x7... the third term here will be (x7) + 2*2.. same as nthterm above
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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25 Jan 2016, 02:20
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. 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 Since S is the list consisting of 10 consecutive odd integers we can put S={s, s + 2, s + 4, ...., s + 18}, where s is the least odd integer of S. So the average of S is (10*s + 2+4+....+18)/10=(10*s + 90)/10= s+9. Similarly we may put T={t, t+2, ..., t+8}, where t is the least even integer of T. So the average of T is (5*t + 2+ 4+ ....+8)/5 = t+4. s+9(t+4)=st+5=7+5=12. So the answer is 12. > (D).
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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22 Apr 2016, 12:07
can someone please post a couple more problems like this one? involving counting techniques for odds\evens, evenly spaced sets etc? thank you!



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List S consists of 10 consecutive odd integers, and list T c [#permalink]
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Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList 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 Let the set of numbers be S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 } T = { 2 , 4 , 6 , 8 , 10 } Sum of the set S = 180 Mean of set T = 18 Sum of the set T = 30 Mean of set T = 6 So, The arithmetic mean of set S is 12 more than the mean of set T
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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17 Nov 2016, 15:16
General formula for odd numbers is 2n + 1 and even 2n
Assume that 2 is the least in the even set then 2+7 = 9 has to be the first in the odd set.
So 2n + 1 = 9 gives n = 4 so the index n of the 10th value for the odd set is (4 + 10)  1 = 13 AND THE magical 1 occurs because the formula has "zero based" indexing. Hence value n for the 10th is 13 AND NOT 14.
Therefore 2(13) + 1 = 27. This means that for the ODD set min = 9 and max = 27 so mean = 18
The mean of the EVEN set is the median which is equal to 6 so the difference is 12 and Correct answer D



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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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29 Nov 2016, 16:43
Bunuel 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
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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11 Dec 2016, 11:48
S = 10 consecutive interegers T = 5 c. integers
S>7t
If the least integer of T is 1, than the least integer of S is 8.
The largest number of T is thus: 1+2x4=9, 9+1=10/2 = 5, so mean is 5 And for S is 8+2x9= 26, 26+8, 34/2=17, so mean is 17
The difference is 175, which is 12. Hence answer D



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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]
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07 May 2017, 03:24
Abhishek009 wrote: Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionList 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 Let the set of numbers be S = { 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 } T = { 2 , 4 , 6 , 8 , 10 } Sum of the set S = 180 Mean of set T = 18 Sum of the set T = 30 Mean of set T = 6 So, The arithmetic mean of set S is 12 more than the mean of set T guess thats the easiest approach. However, we could also start with 0 and recognise that we are dealing with an evenly spaced set, hence median = mean
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