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Set S consists of consecutive positive integers. If the lowest integer

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Set S consists of consecutive positive integers. If the lowest integer  [#permalink]

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New post 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  [#permalink]

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New post 20 Jul 2017, 01:20
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 20 Jul 2017, 02:42
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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  [#permalink]

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New post 22 Jul 2017, 15:14
1
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

+1 kudos if u like the post
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Re: Set S consists of consecutive positive integers. If the lowest integer  [#permalink]

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New post 25 Jul 2017, 11:46
1
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  [#permalink]

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New post 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  [#permalink]

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New post 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+ 1-1= 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  [#permalink]

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New post 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 &nbs [#permalink] 23 Apr 2018, 07:35
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