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11 Jul 2024, 07:00
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D
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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
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11 Jul 2024, 08:15
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Bunuel
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
11 Jul 2024, 07:30
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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
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
