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

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16 Feb 2011, 09:42

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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?

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;

If you didnt know the formula for evenly spaced sets, you can pick numbers and solve this question very easily.

Pick first number of set T=2. Since number of elements = 5, the mean will be the middle term ie the 3rd term, which will be 6.

From the question, you can infer that first number of the set T, will be 2+7= 9. Since this set consists of even number of terms,ie 10, the mean will be the average of the middle 2 terms, ie 5th and 6th term, which are 17 and 19 respectively and their average will be 18= mean of the set.

Therefore, the difference between the mean of two sets= 18-6= 12.

Answer D.
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PareshGmat's solution is the quickest way. Anyway, here is mine: T=(t1+t2+t3+t4+t5)/5 =[t1+(t1+1*2)+(t1+2*2)+(t1+3*2)+(t1+4*2)]/5 =[5t1+2*(1+2+3+4)]/5 =[5t1+2*(4*(4+1)/2)]/5 =[5t1+20]/5=t1+4 S=(s1+...+s10)/10 =[s1+(s1+1*2)+...+(s1+9*2)]/10 =[10s1+2*(1+...+9)]/10 =[10s1+2*(9*(9+1)/2)]/10 =[10s1+90]/10=s1+9 We have s1=t1+7 S-T=s1+9-t1-4=t1+7+9-t1-4=12 =>D P/S: Sum of n consecutive integers: n(n+1)/2

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

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18 Jun 2016, 01:21

the idea is to remember the rule for consecutive numbers whether even or odd that it will have median=mean=average(1st term+last term)....then for S and T series have a difference of 7 of first digit of series when creating hypothetical digits for series....this will result in exact solution

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

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20 Jun 2016, 02:02

when we read consecutive odd/even integers...that implies mean(average)=median=average(1st term+last term).....make series S and T with hypothetical numbers according to the condition given....on setting up the numbers.....find A.M according to the method described above leading to the right solution

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

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23 Jun 2017, 09:52

Baten80 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

let s=least integer of S t=least integer of T mean of S=s+range/2=s+18/2=s+9 mean of T=t+range/2=t+8/2=t+4 substituting s-7 for t, (s+9)-(s-3)=12 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

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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