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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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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]

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30 Jan 2014, 02:54

5

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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

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]

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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=16-4=12 D!
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Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

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15 Jun 2015, 04:12

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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;"

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 + (N-1)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 - 3 in 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 x-7... the third term here will be (x-7) + 2*2..same as nthterm above _________________

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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

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)=s-t+5=7+5=12. So the answer is 12. ---> (D).
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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?

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

Re: List S consists of 10 consecutive odd integers, and list T c [#permalink]

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29 Nov 2016, 16:43

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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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