GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Feb 2019, 04:26

### 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 February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
• ### FREE Quant Workshop by e-GMAT!

February 24, 2019

February 24, 2019

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

# If the average (arithmetic mean) of 18 consecutive odd integers is 534

Author Message
TAGS:

### Hide Tags

Intern
Joined: 08 Sep 2014
Posts: 25
GMAT Date: 07-11-2015
GPA: 2.9
WE: Account Management (Computer Software)
If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

30 May 2015, 16:47
2
10
00:00

Difficulty:

45% (medium)

Question Stats:

71% (02:06) correct 29% (01:51) wrong based on 216 sessions

### HideShow timer Statistics

If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525
Veritas Prep GMAT Instructor
Joined: 23 Oct 2013
Posts: 144
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

30 May 2015, 18:04
9
4
A very helpful rule to know in arithmetic is the rule that in evenly spaced sets, average = median. By evenly spaced sets I mean any set where the numbers are evenly spaced from one another (they do not have to necessarily be even numbers, or spaced 2 apart, or adhere to any other rule). The sets (-3,-2,-1,0,1,2), (15,20,25), (12,14,16,18,20,22,24), and (-35,-28,-21,-14) are all examples of evenly spaced sets. Because the average will equal the median in these sets, then we quickly know that the median of this set of consecutive odd integer numbers is 534.

There are 18 numbers in the set, and in a set with an even number of terms the median is just the average of the two most median terms (here the 9th and 10th numbers in the set). This means that numbers 9 and 10 in this set are 533 and 535. Because we know that number 9 is 533, we know that the smallest number is 8 odd numbers below this, which means that it is 8*2 = 16 below this (every odd number is every other number). Therefore 533-16 = 517, answer choice A.

I hope this helps!
_________________

Brandon
Veritas Prep | GMAT Instructor

Save \$100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Joined: 21 Feb 2012
Posts: 56
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

30 May 2015, 21:59
3
3
Hello BuggerinOn

I would be most comfortable solving it in this fashion

$$S = \frac{n}{2} [ 2a + (n-1)d ]$$
(you would like to remember this formula which is the sum of an arthimetic progression. also notice number of terms multiplied by average of those numbers is the sum of the set)

$$534*18 = \frac{18}{2} [ 2a + (18-1)2 ]$$

$$1068 = (2a + 34)$$

$$a = 517$$

_________________

Regards
J

Do consider a Kudos if you find the post useful

##### General Discussion
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4391
Location: India
GPA: 3.5
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

27 Oct 2016, 09:53
1
BuggerinOn wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525

Odd numbers can be represented in the form : n , ( n + 2 )............

So , n + ( n + 2 ) + ( n + 4 ) + ( n + 6 )............+ ( n + 34 ) = 18*534

Or, 18n + 306 = 18*534

Or, 18 ( n + 17 ) = 18*534

Or, n + 17 = 534

So, n = 517

Hence, answer will be (A) 517...

_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )

Current Student
Joined: 12 Aug 2015
Posts: 2621
Schools: Boston U '20 (M)
GRE 1: Q169 V154
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

12 Dec 2016, 10:01
3
Here is my solution to this one ->
Let n be the first term
Last term => $$n+17*2=> n+34$$

Now as the series will be in AP.
Mean = Average of the first and the last term
Hence 534 =$$\frac{n+n+34}{2}=> n+17$$
Hence n+17=534=> n=517

Hence A

_________________
Intern
Joined: 17 Jun 2017
Posts: 37
If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

09 Oct 2018, 23:57
Bunuel wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

(A) 517
(B) 518
(C) 519
(D) 521
(E) 525

Sum = n/2[2a+(n-1)d] , a = is the first term , d = is the common difference n=18 as mentioned in the question
Over here
Sum = 18*534 = 9612 , N=18 , A = a ,D=2 (because consecutive odd)
9612 = 18/2(2a+(18-1)2)
9612 = 9(2a + 34)
9612/9 = 2a + 34
1068 = 2a + 34
1068 - 34 = 2a
1034 = 2a
517 = a !

I will go with A
VP
Joined: 31 Oct 2013
Posts: 1162
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

10 Oct 2018, 00:08
1
Bunuel wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

(A) 517
(B) 518
(C) 519
(D) 521
(E) 525

Note: we have total 18 consecutive integers. It means we got average by dividing the sum of 2 integers . Question is which 2 integers?

533 + 535 / 2 = 534.........................a bit common sense is required here.

If total number of integers were 19 , the average would be 10th integers. So, in this case we take care both 9th and 10th integer of the serious.

9th = 533

17-19-21-23-25-27-29-31

517 is the least among the odd integers.

VP
Joined: 09 Mar 2016
Posts: 1284
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

10 Oct 2018, 06:34
BuggerinOn wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525

$$\frac{x}{18}$$= $$534$$

x = 9,532

if each of the numbers were 501 than total would be 501*18 = 9,018

so if i deduct 1 from each of the odd numbers i get 534-18 = 516

so smallest odd number is 517
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4955
Location: United States (CA)
Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534  [#permalink]

### Show Tags

13 Oct 2018, 17:15
BuggerinOn wrote:
If the average (arithmetic mean) of 18 consecutive odd integers is 534, then the least of these integers is

a) 517
b) 518
c) 519
d) 521
e) 525

Since the average of 18 consecutive odd integers is 534, the average of the middle two integers is also 534. Therefore, the middle two integers are 533 and 535. There will be 8 odd integers less than 533 (and 8 more greater than 535). So the smallest of these integers is 533 - 8(2) = 533 - 16 = 517.

Alternate Solution:

If the smallest of these integers is x, then the largest of these integers will be x + 34. Since consecutive odd integers form an evenly spaced set of integers, the average is equal to the average of the smallest and the largest integers; thus:

(x + x + 34)/2 = 534

2x + 32 = 1068

2x = 1034

x = 517

_________________

Scott Woodbury-Stewart
Founder and CEO

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

Re: If the average (arithmetic mean) of 18 consecutive odd integers is 534   [#permalink] 13 Oct 2018, 17:15
Display posts from previous: Sort by