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List S consists of 10 consecutive odd integers, and list T
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16 Feb 2011, 08:42
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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? A. 2 B. 7 C. 8 D. 12 E. 22
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Re: Arithmetic Statistics
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16 Feb 2011, 08:59




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Re: List S consists of 10 consecutive odd integers, and list T
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23 Jun 2014, 01:53
Picked up numbers: T = 2 , 4 , 6, 8 , 10 (Mean = 6) S = 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (Mean \(= \frac{17+19}{2} = 18\)) Difference = 186 = 12 Answer = D
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Re: Arithmetic Statistics
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16 Feb 2011, 08:56
Sum of evenly spaced progression: \(\frac{n}{2}(2a+(n1)d)\) \(Average = \frac{Sum}{n}\) or \(Average =\frac{1}{2}(2a+(n1)d)\) \(\frac{1}{2}(2a+(101)*2)  \frac{1}{2}(2(a7)+(51)*2)\) \(\frac{1}{2}(2a+(101)*2)  \frac{1}{2}(2(a7)+(51)*2)\) \(a+9(a7+4)\) \(a+9a+74\) \(12\) Ans: "D"
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Re: Arithmetic Statistics
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17 Feb 2011, 04:51
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= 186= 12. Answer D.
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Re: Arithmetic Statistics
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17 Feb 2011, 12:01
Fluke  when u say 2a  what do u mean? shouldnt it be A1 (the first number in the series)? thanks.
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Re: Arithmetic Statistics
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17 Feb 2011, 17:24
Easy by picking numbers, but I didn't know that formula. Thanks.



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17 Feb 2011, 23:46
144144 wrote: Fluke  when u say 2a  what do u mean? shouldnt it be A1 (the first number in the series)?
thanks. You are half right; It should be \(2*a_1\) i.e. twice of the first term. "a" is another way of depicting \(a_1\) The formula for Sum of n numbers in an evenly spaced progression is denoted by: \(\frac{n}{2}(2*a_1+(n1)d)\) If the progression is: 2,4,6. Sum=12 Using formula: \(a_1=2\). First term of the series\(n=3\) Number of elements in the progression(series)\(d=a_2a_1=42=2\) \(Sum=\frac{n}{2}(2a_1+(n1)d)\) \(Sum=\frac{3}{2}(2*2+(31)2)\) \(Sum=\frac{3}{2}*8\) \(Sum=12\)
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Re: Arithmetic Statistics
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Re: Arithmetic Statistics
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18 May 2011, 22:43
(2a + 2a + 8) /2
(2a+7 + 2a + 25)/2
leaves 12 after subtracting.



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Re: Arithmetic Statistics
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19 May 2011, 18:57
I got 12 by solving but i think Bunuel's method is brilliant ( as always )



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Re: List S consists of 10 consecutive odd integers, and list T
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29 Jun 2014, 21:14
PareshGmat wrote: Picked up numbers:
T = 2 , 4 , 6, 8 , 10 (Mean = 6)
S = 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (Mean \(= \frac{17+19}{2} = 18\))
Difference = 186 = 12
Answer = D 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 ST=s1+9t14=t1+7+9t14=12 =>D P/S: Sum of n consecutive integers: n(n+1)/2



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Re: List S consists of 10 consecutive odd integers, and list T
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26 Aug 2015, 10:25
Hi All, This question can be solved by using Number Property Rules or by TESTing VALUES: Here's how TESTing VALUES works: List S: 10 consecutive ODD integers List T: 5 consecutive EVEN integers The least integer is S is 7 more than the least integer in T. Let's say that…. T = {2, 4, 6, 8, 10} Since the least integers in S is 7 MORE than the least integer in T… S = {9, 11, 13, 15, 17, 19, 21, 23, 25, 27} The average of T = 6 The average of S = 18 So, the average of S is 18  6 = 12 more than the average of T. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: List S consists of 10 consecutive odd integers, and list T
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18 Jun 2016, 00:21
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



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Re: List S consists of 10 consecutive odd integers, and list T
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19 Jun 2016, 01:21
Mean of 10 consecutive odd integers is 9 more than the least. Mean of 5 consecutive even integers is 4 more than the least.
9  4 + 7 = 12.



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Re: List S consists of 10 consecutive odd integers, and list T
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20 Jun 2016, 01:02
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



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List S consists of 10 consecutive odd integers, and list T
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23 Jun 2017, 08:52
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 s7 for t, (s+9)(s3)=12 D



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Re: List S consists of 10 consecutive odd integers, and list T
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26 Jun 2017, 16:53
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 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
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28 Jun 2017, 00:05
This is can also by using the concept of evenly spaced numbers. Let Set S be s10, s8, s6.....s+6 10 numbers. Total =(10s10)=> Avg=s1
Set T be e4,e2,e,e+2,e+45 numbers Total=5e => Avg =e
diff of avg is s1e
Other statement for least numbers gives us (s10)=7+(e4) >se=13
Putting it back to se1= 12



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Re: List S consists of 10 consecutive odd integers, and list T
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02 Oct 2018, 03:18
since numbers are evenly spaced so mean = median 7,9,11,13,15......... median = 13+15/2 = 16 0,2,4,6,8 median = 4 164=12




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