Let's call the number of students each van and bus can carry "x".
For High School A:
- Number of vans: 5
- Number of buses: 12
- Total number of students: 584
We can set up the following equation:
5x + 12x = 584
Simplifying:
17x = 584
x = 34.35
Since we can't have a fraction of a student, we can round up to the nearest whole number. Therefore, each van and bus at High School A can carry 35 students.
Now let's do the same for High School B:
- Number of vans: 13
- Number of buses: 13
- Total number of students: 754
We can set up the following equation:
13x + 13x = 754
Simplifying:
26x = 754
x = 28.92
Again, we'll round up to the nearest whole number. Therefore, each van and bus at High School B can carry 29 students.
Write and solve a system of equations to determine how many students each van can carry and how many students each bus can carry.
The senior class at High School A and High School B planned separate trips to the state fair. The senior class at High School A rented and filled 5 vans and 12 buses with 584 students. The senior class at High School B rented and filled 13 vans and 13 buses with 754 students. Each van held the same number of students as did each bus. How many students can each van and bus carry?
3 answers
help
AAAaannndd the bot gets it wrong yet again!
5v+12b = 584
13v+13b = 754
or, getting rid of the big numbers,
5v+12b = 584
v+b = 58
now a simple substitution gives
5v+12(58-v) = 584
so v=16 and b=42
5v+12b = 584
13v+13b = 754
or, getting rid of the big numbers,
5v+12b = 584
v+b = 58
now a simple substitution gives
5v+12(58-v) = 584
so v=16 and b=42