GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line

Question

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

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

Bunuel · Accepted Answer

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.

Fantasy_app · Answer

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

meet459 · Answer

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

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

Oppenheimer1945 · Answer

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

said.tojiboev · Answer

https://gmatclub.com/forum/download/file.php?id=82492 GMAT-Club-Forum-mb9fmjwj.png

Legallyblond · Answer

Hello Bunuel,

I just wanna know why did you subtarct 2x from 9x and 5x

Bunuel · Answer

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.

bumpbot · Answer

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GMAT Club Olympics 2024 (Day 4): Homer and Marge are in a ticket line

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