A city manager made the graph below to represent the number of passengers that city buses can carry, where the number of passeng
ers is represented by x, and the number of buses by y. If all city buses carry the same number of passengers, what is the least number of buses needed to carry 710 passengers?
2 answers:
In my research to help you, I found the figure that comes with the question. It is in the file attached now.
Also, I found the set of choices from which you have to select the right answer. Thi is it:
5
10
11
12
From the graph you have this information:
1 bus: 1 to 70 passengers
2 buses: 71 to 140 passengers
3 buses: 141 to 210 passengers
4 buses: 211 to 280 passengers
5 buses: 281 to 350 passengers
That means that each bus add an addtional capacity to carry 70 passengers.
So, to increase the capacity from 350 to 710 passengers you need this additional amount of buses:
[710 - 350] / 70 = 360 / 70 = 5,1 buses which must be rounded to the next integer, this is 6 additional buses, over 5 buses (which carry 350 passengers).
The total amount is 5 + 6 = 11 buses.
Answer: the least number of buses needed to carry 710 passengers is 11.
Also, I found the set of choices from which you have to select the right answer. Thi is it:
5
10
11
12
From the graph you have this information:
1 bus: 1 to 70 passengers
2 buses: 71 to 140 passengers
3 buses: 141 to 210 passengers
4 buses: 211 to 280 passengers
5 buses: 281 to 350 passengers
That means that each bus add an addtional capacity to carry 70 passengers.
So, to increase the capacity from 350 to 710 passengers you need this additional amount of buses:
[710 - 350] / 70 = 360 / 70 = 5,1 buses which must be rounded to the next integer, this is 6 additional buses, over 5 buses (which carry 350 passengers).
The total amount is 5 + 6 = 11 buses.
Answer: the least number of buses needed to carry 710 passengers is 11.
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