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29 Jul 2024, 02:39
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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
29 Jul 2024, 02:39
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Official Solution:
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
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
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
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
01 Aug 2024, 08:56
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If we consider Homer and Merge as 2 extras then the equation would be 12x + 2 <= 100.
This would again give us x = 8. But the total number of people would be 98.
In this case the maximum number of people ahead of Homer can be 56.
How is this answer not correct?
@Bunuel
This would again give us x = 8. But the total number of people would be 98.
In this case the maximum number of people ahead of Homer can be 56.
How is this answer not correct?
@Bunuel
