
Wyatt B. answered 06/03/21
ACT Prep, High School Math and Physics Tutor
Hi Nina!
The best way to solve these questions is to assign assign a variable to each piece we're looking for. First let's figure out how many buses and vans we need. Let's call Vans = V and Buses = B. We don't need to assign teachers and students a variable because we know there are 491 total and in this case the problem doesn't differentiate between them.
So the statements made about Vans and Buses "There are 5 more buses than vans." gives us the equation:
B = V + 5
and the statement about the cost, which we can call C, gives us:
C = B*$1200 + V*$700
We know that Buses can hold 52 students/teachers, and vans hold 25. The total number of students and teachers we need to carry are at minimum 491. From this, we can write the inequality:
B*52 + V*25 ≥ 491
This is a two variable equation, which has infinite solutions, but with the nice equation we came up with earlier we can substitute it in for B to get an equation with only one variable:
(V + 5)*52 + V*25 ≥ 491
Which we can simplify:
V*52 + 260 + V*25 ≥ 491
V*77 + 260 ≥ 491
V*77 ≥ 231
V ≥ 231/77
V ≥ 3
If we had gotten an answer like V ≥ 3.2 , since we can't have a fraction of a van, we would have needed to round up to 4. Since we got a nice round 3, though, we don't need to do any rounding and we can carry on. Now that we know we have 3 vans, we know that we have 8 buses because of our earlier equation relating B and V. Now we can calculate the cost, our cost equation from the price of buses and vans provided:
C = V*700 + B*1200
And we can plug in our numbers and get the following:
C = 3*700 + 8*1200 = 11,700
So our answer is $11,700.