Last visit was: 04 Jun 2026, 10:26 It is currently 04 Jun 2026, 10:26
Home
Messages
MBA Fair
Tests
Schools
Events
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ (Hard)|   Min-Max Problems|   Word Problems|         
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 Jun 2026
Posts: 111,096
Own Kudos:
818,803
 [36]
Given Kudos: 106,641
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 111,096
Kudos: 818,803
 [36]
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 11 Jul 2024, 07:00
11
Kudos
Add Kudos
25
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

32% (02:26) correct 68% (02:57) wrong based on 480 sessions

History

Date
Time
Result
Not Attempted Yet
­Homer and Marge are in a ticket line containing fewer than 100 people, and Homer is ahead of Marge in the line. The ratio of the number of people between them to the number of people behind Homer to the number of people ahead of Marge is 2 to 5 to 9. What is the maximum number of people that could be ahead of Homer?­

A. 40
B. 48
C. 54
D. 55
E. 56­

­
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

­
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 Jun 2026
Posts: 111,096
Own Kudos:
818,803
 [9]
Given Kudos: 106,641
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 111,096
Kudos: 818,803
 [9]
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 11 Jul 2024, 08:15
5
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Bunuel
­Homer and Marge are in a ticket line containing fewer than 100 people, and Homer is ahead of Marge in the line. The ratio of the number of people between them to the number of people behind Homer to the number of people ahead of Marge is 2 to 5 to 9. What is the maximum number of people that could be ahead of Homer?­

A. 40
B. 48
C. 54
D. 55
E. 56
Show more
­
­

GMAT Club Official Explanation:



Given:

  • 2x people between Homer and Marge
  • 5x people behind Homer (including Marge)
  • 9x people ahead of Marge (including Homer)­

The total number of people in the line can be expressed as: 9x + 5x - 2x = 12x. Since this should be fewer than 100 people, we get 12x < 100, which gives x < 8.something. Since x must be a positive integer, the maximum value for x is 8, making the maximum number of people in the line equal to 12x = 96. Thus, the maximum number of people behind Homer would be 5x = 40. Therefore, Homer is 96 - 40 = 56th in the line.

Therefore, the maximum number of people that could be ahead of Homer is 55.

Answer: D.­
General Discussion
User avatar
Fantasy_app
User avatar
Joined: 17 Jun 2024
Last visit: 14 Aug 2024
Posts: 16
Own Kudos:
20
 [1]
Given Kudos: 9
Posts: 16
Kudos: 20
 [1]
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 11 Jul 2024, 07:30
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
­Let people in front of Homer be 'a', between homer and Marge be 'b' and behind Marge is 'c'

Also let people behind homer be B and ahead of marge be A. 

So, we know b:B:A = 2:5:9
Let b=2k, B=5k and A=9k
then we know, a= A-(Homer+b) = 9k -(1+2k) = 7k-1
and c= B-(b+Marge) => c= 5k-(2k+1) =>   c=3k-1

Now we also know a+Homer+b+Marge+c<100 = > k<100/12
So, to maximize a we need maximum k possible which is k=8

THerefore, a= 7k-1 = 7*8-1 = 55

Answer: Option D
User avatar
meet459
User avatar
Goizueta / Scheller Moderator
Joined: 03 Jun 2024
Last visit: 29 May 2025
Posts: 85
Own Kudos:
96
 [2]
Given Kudos: 32
Location: India
Schools: Ross '27 (S) Haas '27 (S) Scheller '27 (A) Goizueta '27 (I) McCombs '27 Jones '27 (II) Kenan-Flagler '27 (I)
GMAT Focus 1: 675 Q90 V81 DI79
GPA: 8.55
Products:
My Reviews
Schools: Ross '27 (S) Haas '27 (S) Scheller '27 (A) Goizueta '27 (I) McCombs '27 Jones '27 (II) Kenan-Flagler '27 (I)
GMAT Focus 1: 675 Q90 V81 DI79
Posts: 85
Kudos: 96
 [2]
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 11 Jul 2024, 07:37
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
­Homer and Marge are in a ticket line containing fewer than 100 people, and Homer is ahead of Marge in the line. The ratio of the number of people between them to the number of people behind Homer to the number of people ahead of Marge is 2 to 5 to 9. What is the maximum number of people that could be ahead of Homer?­

Let us assume this is the line:  (x) ---- Homer ------ (y) ------ Marge ----- (z)

Where,
x represents the number of people ahead of Homer
y represents the number of people between Homer and Marge
z represents the number of people behind Marge

According to the question, it is given that,

y : y+z+1 : x+y+1 = 2 : 5 : 9

And x + y + z + 2 < 100

Using the ratio given to us, we can say that

y / (y+z+1) = 2/5     which gives us z = (3y-2)/2

and

y / (x+y+1) = 2/9    which gives us x = (7y-2)/2

Replacing x and z in the above equation, we get

(7y-2)/2 + y + (3y-2)/2 < 100

which can be simplified as     12y < 200    -> 3y < 50, Hence the max value of y can be 16

So replacing that value to get the value of x (which is the ask of the question), we get x = 55

Hence the anwer is (D) 55
User avatar
Oppenheimer1945
User avatar
Joined: 16 Jul 2019
Last visit: 04 Jun 2026
Posts: 770
Own Kudos:
670
 [2]
Given Kudos: 236
Location: India
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
GPA: 7.81
Schools: INSEAD '26 (D) ISB '27 (D) IIMB '26 (A) IIMA '26 (A) HEC '26 (D) LBS '26 IESE '27 (D) Fuqua '27 (WL) Tepper '27 (WL)
GMAT Focus 1: 645 Q90 V76 DI80
Posts: 770
Kudos: 670
 [2]
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 11 Jul 2024, 08:30
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
9x+(5x-2x)<100
12x<100
x_max=8

Gives no of people behind H: 5x=40

total people 12*8=96
No of people ahead of H (including H)=96-40=56
No of people ahead of H=56-1=55­
Attachments

image_2024-07-11_210021052.png
image_2024-07-11_210021052.png [ 11.48 KiB | Viewed 3879 times ]

User avatar
said.tojiboev
User avatar
PhD Forum Moderator
Joined: 04 Oct 2018
Last visit: 15 Apr 2026
Posts: 65
Own Kudos:
60
 [1]
Given Kudos: 10
Location: Uzbekistan
Concentration: Strategy, General Management
Schools: Stanford '27
GPA: 4.49
WE:Project Management (Education)
Schools: Stanford '27
Posts: 65
Kudos: 60
 [1]
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 11 Jul 2024, 22:23
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
­Homer and Marge are in a ticket line containing fewer than 100 people, and Homer is ahead of Marge in the line. The ratio of the number of people between them to the number of people behind Homer to the number of people ahead of Marge is 2 to 5 to 9. What is the maximum number of people that could be ahead of Homer?­

A. 40
B. 48
C. 54
D. 55
E. 56­

­
 


This question was provided by GMAT Club
for the GMAT Club Olympics Competition

Win over $30,000 in prizes such as Courses, Tests, Private Tutoring, and more

 

­
Show more
­
Show Spoiler
Attachment:
GMAT-Club-Forum-mb9fmjwj.png
GMAT-Club-Forum-mb9fmjwj.png [ 45.57 KiB | Viewed 2866 times ]
User avatar
Legallyblond
User avatar
Joined: 14 May 2024
Last visit: 04 Jun 2026
Posts: 80
Own Kudos:
7
Given Kudos: 1,984
Location: India
Posts: 80
Kudos: 7
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 12 Jul 2024, 08:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

Bunuel
­Homer and Marge are in a ticket line containing fewer than 100 people, and Homer is ahead of Marge in the line. The ratio of the number of people between them to the number of people behind Homer to the number of people ahead of Marge is 2 to 5 to 9. What is the maximum number of people that could be ahead of Homer?­

A. 40
B. 48
C. 54
D. 55
E. 56
­
­

GMAT Club Official Explanation:



Given:


  • 2x people between Homer and Marge
  • 5x people behind Homer (including Marge)
  • 9x people ahead of Marge (including Homer)­

The total number of people in the line can be expressed as: 9x + 5x - 2x = 12x. Since this should be fewer than 100 people, we get 12x < 100, which gives x < 8.something. Since x must be a positive integer, the maximum value for x is 8, making the maximum number of people in the line equal to 12x = 96. Thus, the maximum number of people behind Homer would be 5x = 40. Therefore, Homer is 96 - 40 = 56th in the line.

Therefore, the maximum number of people that could be ahead of Homer is 55.

Answer: D.­
Show more
­Hello Bunuel,

I just wanna know why did you subtarct 2x from 9x and 5x
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 Jun 2026
Posts: 111,096
Own Kudos:
818,803
Given Kudos: 106,641
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 111,096
Kudos: 818,803
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 12 Jul 2024, 09:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sakshijjw

Bunuel

Bunuel
­Homer and Marge are in a ticket line containing fewer than 100 people, and Homer is ahead of Marge in the line. The ratio of the number of people between them to the number of people behind Homer to the number of people ahead of Marge is 2 to 5 to 9. What is the maximum number of people that could be ahead of Homer?­

A. 40
B. 48
C. 54
D. 55
E. 56
­
­

GMAT Club Official Explanation:



Given:



  • 2x people between Homer and Marge
  • 5x people behind Homer (including Marge)
  • 9x people ahead of Marge (including Homer)­

The total number of people in the line can be expressed as: 9x + 5x - 2x = 12x. Since this should be fewer than 100 people, we get 12x < 100, which gives x < 8.something. Since x must be a positive integer, the maximum value for x is 8, making the maximum number of people in the line equal to 12x = 96. Thus, the maximum number of people behind Homer would be 5x = 40. Therefore, Homer is 96 - 40 = 56th in the line.

Therefore, the maximum number of people that could be ahead of Homer is 55.

Answer: D.­
­Hello Bunuel,

I just wanna know why did you subtarct 2x from 9x and 5x
Show more
­
The 2x people between Homer and Marge are subtracted because they are counted in both groups: the 5x people behind Homer and the 9x people ahead of Marge. Subtracting 2x avoids double-counting these individuals.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 39,179
Own Kudos:
1,128
Posts: 39,179
Kudos: 1,128
GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line 25 Jul 2025, 09:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderator:
Bunuel
Math Expert
111096 posts