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03 Dec 2024, 03:11
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Maximum: 60
Minimum: 20
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (02:21) correct
47%
(02:38)
wrong
based on 624
sessions
History
Date
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Not Attempted Yet
A certain number of people visited an amusement park on a specific day, which featured three rides: the Ferris Wheel, Classic Wooden Rollercoaster, and Merry-Go-Round, costing $1, $2, and $3 per ride, respectively.
Each person spent an average of $2 per ride, and the total money spent was $120. Every person took at least one ride and did not repeat any ride.
Based on this information, select for Maximum the greatest possible number of visitors and for Minimum the smallest possible number of visitors who could have visited the park.
Each person spent an average of $2 per ride, and the total money spent was $120. Every person took at least one ride and did not repeat any ride.
Based on this information, select for Maximum the greatest possible number of visitors and for Minimum the smallest possible number of visitors who could have visited the park.
| Maximum | Minimum | |
| 10 | ||
| 15 | ||
| 20 | ||
| 30 | ||
| 60 | ||
| 120 |
ShowHide Answer
Official Answer
Maximum: 60
Minimum: 20
04 Jan 2025, 06:15
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Bunuel
Official Solution:
Each person spending an average of $2 per ride implies the following three scenariois:
• A person could have taken all three rides ($1 + $2 + $3 = $6 total), resulting in an average of $2 per ride.
• A person could have taken the $1 and $3 rides ($1 + $3 = $4 total), again averaging $2 per ride.
• A person could have taken only the $2 ride.
Since the total spending was $120, the minimum number of people occurs when the maximum spending per person ($6) is achieved. This results in 120/6 = 20 people, meaning all 20 people took each of the three rides once. Conversely, the maximum number of people occurs when the minimum spending per person ($2) is achieved. This results in 120/2 = 60 people, meaning all 60 people took only the $2 ride once.
Correct answer:
Maximum "60"
Minimum "20"
Attachment:
GMAT-Club-Forum-5ggs3tsk.png [ 9.54 KiB | Viewed 2219 times ]
General Discussion
03 Dec 2024, 12:17
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1. First let's analyze the problem. "Each person spent an average of $2 per ride" means that each person did one of the following combinations: $1 and $3 rides; $2 ride; $1, $2, and $3 rides.
2. The total amount of money spent is $120. The smallest number of possible visitors happens when everyone spends as much as possible. In this case, it would be when everyone goes on $1, $2, and $3 rides together - or spending $6 each. Then, the number of visitors would be \(\frac{120}{6} = 20\).
3. Now the largest number of possible visitors happens when everyone spends as little as possible. In this case, it would be when everyone goes on $2 rides - or spending $2 each. Then, the number of visitors would be \(\frac{120}{2} = 60\).
4. Our answer will be: Maximum - 60 and Minimum - 20.
2. The total amount of money spent is $120. The smallest number of possible visitors happens when everyone spends as much as possible. In this case, it would be when everyone goes on $1, $2, and $3 rides together - or spending $6 each. Then, the number of visitors would be \(\frac{120}{6} = 20\).
3. Now the largest number of possible visitors happens when everyone spends as little as possible. In this case, it would be when everyone goes on $2 rides - or spending $2 each. Then, the number of visitors would be \(\frac{120}{2} = 60\).
4. Our answer will be: Maximum - 60 and Minimum - 20.
