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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.

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

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23 Jun 2014, 01:53

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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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26 Aug 2015, 07:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

gmatclubot

Re: List S consists of 10 consecutive odd integers, and list T
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20 Jun 2016, 02:02

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