A school district needs to arrange transportation for at least 360 students for a field trip. The district can use large buses or small buses. The large buses hold 60 students each and have an operating expense of $300 for the day. The small buses hold 45 students each and have an operating expense of $200 for the day. The district only has chaperones for, at most, seven buses. What is the minimum cost for the transportation for the field trip?. A.$1,400. B. $1,600. C.$1,700. D.$1,800

This can be answered using a system of equations. Let us list the equations that we know:

To find the minimum cost, we use the equation: 300x + 200y = N

Where N is the minimum cost for the transportation; x = number of large buses; y = number of small buses.

To find the values for x and y, we use the following 2 equations:

60x + 45y = 360

x + y = 7

x = 7 - y

60 (7 - y) + 45y = 360

420 - 60y + 45y = 360

-15y = - 60

y = 4

x + 4 = 7

x = 3

There will be 3 large buses and 4 small buses. Calculating the minimum cost:

300 (3) + 4 (200) = 1700

Therefore, the minimum cost for transportation will be $1,700.