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Set S consists of consecutive positive integers. If the lowest integer
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19 Jul 2017, 23:35
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Set S consists of consecutive positive integers. If the lowest integer in Set S is 1 and the sum of the integers of Set S is 231, what is the median of Set S? A. 10.5 B. 11 C. 12 D. 13 E. 22
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Set S consists of consecutive positive integers. If the lowest integer
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20 Jul 2017, 01:20
sum of n consecutive integers = n* (n+1) /2 so, n * (n+1) / 2 = 231 n * (n+1) = 462 implying, n = 21 so the numbers are 1,2,3 .... 21 median is the middle number
Hence answer is 11, option B



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Re: Set S consists of consecutive positive integers. If the lowest integer
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20 Jul 2017, 02:05
Formula used : Sum of n consecutive positive integers = \(\frac{n(n+1)}{2}\)
Since the lowest number is 1, and the sum of integers are 231 If there are n consecutive positive integers which are added to get the sum, \(\frac{n(n+1)}{2} = 231\) \(n(n+1) = 462\) Solving for n, n=21. For numbers starting 1 to 21, the median is 11(Option B)
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Re: Set S consists of consecutive positive integers. If the lowest integer
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20 Jul 2017, 02:16
Given Sum = 231 and Least integer in the set = 1
Sum of n consecutive positive integers is given by the formula = (n(n+1))/2
(n(n+1))/2=231 => n(n+1)=462 => n(n+1)=21*22 => n=21
for a set of consecutive positive integers mean = median.
mean of the set = (last term+first term)/2 => mean=(1+21)/2 => mean = 11
Hence, median of the set = 11. (Option B)



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Re: Set S consists of consecutive positive integers. If the lowest integer
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20 Jul 2017, 02:42
Bunuel wrote: Set S consists of consecutive positive integers. If the lowest integer in Set S is 1 and the sum of the integers of Set S is 231, what is the median of Set S?
A. 10.5 B. 11 C. 12 D. 13 E. 22 The only tricky part of this question is solving for n when we say n*(n+1)/2 = 231 Note that do not try to multiply 231 with 2. Instead try to break down 231 into factors. 231 = 11 * 21 When you multiply 11 with 2, you will get 22 i.e. consecutive numbers 21 and 22. So you know that n*(n+1) = 21*22 n = 21 Then you know that median is 11.
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Re: Set S consists of consecutive positive integers. If the lowest integer
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22 Jul 2017, 15:14
if a1 = 1 and consecutive positive integers is the sequence
sum is n(n+1)/2 = 231
n^2 + n  462 = 0 solving n = 21 or n = 22 n can't be negative as it is a number of something
n = 21
hence median for odd number of sample set is n+1/2 = 11th value also 11 as it is consecutive integers set from 1.
Option B
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Re: Set S consists of consecutive positive integers. If the lowest integer
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25 Jul 2017, 11:46
Bunuel wrote: Set S consists of consecutive positive integers. If the lowest integer in Set S is 1 and the sum of the integers of Set S is 231, what is the median of Set S?
A. 10.5 B. 11 C. 12 D. 13 E. 22 We can use the following formula: sum = average x quantity. We can let the largest value in the set = n. Thus: average = (n + 1)/2 quantity = n  1 + 1 = n So, we have: sum = (n + 1)/2 x n 231 = (n^2 + n)/2 462 = n^2 + n n^2 + n  462 = 0 (n + 22)(n  21) = 0 n = 22 or n = 21 Since n is positive, it must be 21. Finally, we can determine the median (which, in this case, is equal to the average): median = (21 + 1)/2 = 11 Answer: B
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Set S consists of consecutive positive integers. If the lowest integer
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20 Aug 2017, 08:40
Bunuel wrote: Set S consists of consecutive positive integers. If the lowest integer in Set S is 1 and the sum of the integers of Set S is 231, what is the median of Set S?
A. 10.5 B. 11 C. 12 D. 13 E. 22 The answers above are better. But, I am posting this method anyway. This method, for timing purposes, is viable, if you know that the sum of first 10 natural numbers is 55. That is, (10 + 11)/2 = 110/2 = 55 We are still far away from the total of 231. Let's try another 10 numbers. Applying previous logic, sum of numbers 11 through 20, is just, 11 = 1 + 10 12 = 2 + 10 . . 19 = 9 + 10 20 = 10 + 10 55 + (10 * 10) = 155 Now, 55 + 155 = 210 We are almost there. You can see that adding 21 to 210 gives 231. We have our set, which is, 1 through 21. The median is, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 11 [ 12, 13, 14, 15, 16, 17, 18, 19, 20, 21



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Re: Set S consists of consecutive positive integers. If the lowest integer
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30 Sep 2017, 04:42
Bunuel wrote: Set S consists of consecutive positive integers. If the lowest integer in Set S is 1 and the sum of the integers of Set S is 231, what is the median of Set S?
A. 10.5 B. 11 C. 12 D. 13 E. 22 Sum of consecutive integers= Median x No. of terms 231= median x no. of terms Sum of consecutive is odd then last term must be odd and median must be odd also. then C and E out. also consecutive odd integers then A is also out. No. of terms= Last + First 1 = last Odd term+ 11= last odd term 231=3x7x11/ Last odd term = median, in the ans B or D B satisfies the condition above B is answer



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Re: Set S consists of consecutive positive integers. If the lowest integer
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23 Apr 2018, 07:35
+1 for option B. Note that (n)*(n+1)/2=231. Now split 231 as 11*21. The only trick here is in splitting 231 as a product of 11 and 21.
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Re: Set S consists of consecutive positive integers. If the lowest integer
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