It is currently 19 Sep 2017, 21:09

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 41601

Kudos [?]: 124053 [0], given: 12070

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

### Show Tags

19 Jul 2017, 23:35
00:00

Difficulty:

35% (medium)

Question Stats:

78% (01:28) correct 22% (01:31) wrong based on 74 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 124053 [0], given: 12070

Intern
Joined: 19 May 2017
Posts: 2

Kudos [?]: 0 [0], given: 271

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

### Show Tags

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

Kudos [?]: 0 [0], given: 271

BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 1319

Kudos [?]: 518 [0], given: 16

Location: India
WE: Sales (Retail)
Re: Set S consists of consecutive positive integers. If the lowest integer [#permalink]

### Show Tags

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)
_________________

Stay hungry, Stay foolish

Kudos [?]: 518 [0], given: 16

Intern
Joined: 19 Jan 2014
Posts: 36

Kudos [?]: 1 [0], given: 42

Schools: Melbourne '20
GMAT 1: 670 Q45 V37
Re: Set S consists of consecutive positive integers. If the lowest integer [#permalink]

### Show Tags

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)

Kudos [?]: 1 [0], given: 42

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7604

Kudos [?]: 16876 [0], given: 230

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

### Show Tags

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 16876 [0], given: 230

Manager
Joined: 06 Nov 2016
Posts: 91

Kudos [?]: 17 [0], given: 15

GMAT 1: 710 Q50 V36
Re: Set S consists of consecutive positive integers. If the lowest integer [#permalink]

### Show Tags

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

+1 kudos if u like the post

Kudos [?]: 17 [0], given: 15

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1414

Kudos [?]: 754 [1], given: 5

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

### Show Tags

25 Jul 2017, 11:46
1
KUDOS
Expert's post
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

_________________

Jeffery Miller

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

Kudos [?]: 754 [1], given: 5

Intern
Joined: 15 Jul 2017
Posts: 7

Kudos [?]: 1 [0], given: 116

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

### Show Tags

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

Kudos [?]: 1 [0], given: 116

Set S consists of consecutive positive integers. If the lowest integer   [#permalink] 20 Aug 2017, 08:40
Similar topics Replies Last post
Similar
Topics:
44 What is the lowest positive integer that is divisible by 19 12 Jun 2017, 16:00
3 Set M consists of 50 consecutive integers. If the range of the negativ 4 02 Mar 2017, 18:20
1 Set Q consists of 6 consecutive even integers beginning with -4, and S 1 24 Jan 2017, 09:59
39 A set of consecutive positive integers beginning with 1 is 13 22 Jul 2017, 02:26
8 Set A consists of 2 positive integers, and set B consists 5 20 Mar 2017, 03:21
Display posts from previous: Sort by