Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1311:00 AM EDT
-12:30 PM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
03 Dec 2024, 03:11
Kudos
Bookmarks
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% (02:26) correct
47%
(02:34)
wrong
based on 496
sessions
History
Date
Time
Result
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
Kudos
Bookmarks
Bunuel
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.
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 658 times ]
General Discussion
03 Dec 2024, 12:17
Kudos
Bookmarks
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.
