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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?

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

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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

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

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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)
_________________

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.
_________________

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
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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

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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.

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

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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+ 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