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13 Mar 2025, 04:36
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Maximum: 12
Minimum: 4
At toy store, five types of model cars are available, priced as follows: $15, $25, $35, $45, and $55.
A group of customers visits the store and makes purchases with the following conditions:
- Each customer can buy at least 1 car and at most 3 cars.
- A customer cannot buy more than one of each car type.
- On average, each model car purchased costs $35.
In addition, the group’s total spending is $420.
Based on the above information, select for Maximum the maximum number of possible people in the group and for Minimum the minimum number of possible people in the group.
A group of customers visits the store and makes purchases with the following conditions:
- Each customer can buy at least 1 car and at most 3 cars.
- A customer cannot buy more than one of each car type.
- On average, each model car purchased costs $35.
In addition, the group’s total spending is $420.
Based on the above information, select for Maximum the maximum number of possible people in the group and for Minimum the minimum number of possible people in the group.
| Maximum | Minimum | |
| 3 | ||
| 4 | ||
| 5 | ||
| 10 | ||
| 11 | ||
| 12 |
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Official Answer
Maximum: 12
Minimum: 4
13 Mar 2025, 04:36
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Official Solution:
The question asks us to find the limits to the number of possible people in the group.
Let’s first analyze the possible car combinations people could’ve bought:
• 1 model car: $35. The cost is $35.
• 2 model cars: $25 and $45; $15 and $55. The cost is $70.
• 3 model cars: $15, $35, and $55; $25, $35, and $45. The cost is $105.
For there to be as many people as possible, everyone needs to spend as little as possible - which is buying 1 model car for $35. Likewise, for there to be as few people as possible, everyone needs to spend as much as possible - which is buying 3 model cars for $105.
Then, the maximum number of people in the group is equal to \(\frac{420}{35} = 12\) and the minimum number of people in the group is equal to \(\frac{420}{105} = 4\).
Correct answer:
Maximum "12"
Minimum "4"
Bunuel
The question asks us to find the limits to the number of possible people in the group.
Let’s first analyze the possible car combinations people could’ve bought:
• 1 model car: $35. The cost is $35.
• 2 model cars: $25 and $45; $15 and $55. The cost is $70.
• 3 model cars: $15, $35, and $55; $25, $35, and $45. The cost is $105.
For there to be as many people as possible, everyone needs to spend as little as possible - which is buying 1 model car for $35. Likewise, for there to be as few people as possible, everyone needs to spend as much as possible - which is buying 3 model cars for $105.
Then, the maximum number of people in the group is equal to \(\frac{420}{35} = 12\) and the minimum number of people in the group is equal to \(\frac{420}{105} = 4\).
Correct answer:
Maximum "12"
Minimum "4"
04 May 2025, 22:57
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When we say Maximum number of people
so we can assume that
420/35 = 12
Minimum
55+35+15 = 105
so 3
why 4
so we can assume that
420/35 = 12
Minimum
55+35+15 = 105
so 3
why 4
Bunuel
