Urban Community College is planning to offer courses in Fini

Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40 students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college $11,000. Assuming the college wishes to offer a total of seven sections, to accommodate 220 students, and to bring in $262,000 in revenues, how many sections of each course should it offer?

Solution

Let x be the number of sections in Finite Math, y be the number of sections in Applied Calculus and z be the number of sections in Computer methods.

then,

40x + 40y + 10z = total number of students = 220

also, 40,000x + 60,000y + 11,000z = total revenue generated = 262,000

and x + y + z = total sections needed = 7.

so, to summarize, there are three equations with 3 variables

40x + 40y + 10z = 220 .........................[1]

40,000x + 60,000y + 11,000z = 262,000 .....................[2]

x + y + z = 7 ............................[3]

for equation [3], z = 7 - x - y

substitute this in [1] to get: 40x + 40y + 10[7 - x - y] = 220

=> 40x + 40y + 70 - 10x - 10y = 220

=> 30x + 30y = 150

which can be further simplified to: x + y = 5 .......................[4]

but, x + y + z = 7

therefore, (5) + z = 7

or z = 2 .....................[5]

substitute this in equation [2], to get:

40,000x + 60,000y + 11,000(2) = 262,000

=> 40,000x + 60,000y + 22,000 = 262,000

=> 40,000x + 60,000y = 240,000

which can be further simplified to: 2x + 3y = 12 .........................[6]

but we know that x + y = 5 [equation 4]

so y = 5 - x

substitute this in [6] to get: 2x + 3[5 - x] = 12

=> 2x + 15 - 3x = 12

or x = 3

therefore y = 5 - 3 = 2

therefore, x = number of sections for finite math = 3

y = number of sections for applied calculus = 2

z = number of sections for computer methods = 2.