Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math holds 40 students and earns the college $1,300 in revenue per student. Each section of Applied Calculus holds 45 students and earns the college $1,600 per student, and each section of Computer Methods holds 15 students and earns the college $1,900 per student. Assuming that the college can offer a total of seven sections, wishes to accommodate 240 students, and wishes to bring in $357,000 in revenues, how many sections of each course should it offer ?
A)Offer 2 sections of Finite Math, 2 sections of Applied Calculus, and 1 section of Computer Methods
B) Offer 2 sections of Finite Math, 3 sections of Applied Calculus, and 2 sections of Computer Methods
C) Offer 3 sections of Finite Math, 2 sections of Applied Calculus, and 2 sections of Computer Methods
D) Offer 5 sections of Finite Math, 1 section of Applied Calculus, and 1 section of Computer Methods
E) Offer 1 section of Finite Math, 5 sections of Applied Calculus, and 1 section of Computer Methods

Question

Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math holds 40 students and earns the college $1,300 in revenue per student. Each section of Applied Calculus holds 45 students and earns the college $1,600 per student, and each section of Computer Methods holds 15 students and earns the college $1,900 per student. Assuming that the college can offer a total of seven sections, wishes to accommodate 240 students, and wishes to bring in $357,000 in revenues, how many sections of each course should it offer  ?
A)Offer 2 sections of Finite Math, 2 sections of Applied Calculus, and 1 section of Computer Methods
B) Offer 2 sections of Finite Math, 3 sections of Applied Calculus, and 2 sections of Computer Methods
C) Offer 3 sections of Finite Math, 2 sections of Applied Calculus, and 2 sections of Computer Methods
D) Offer 5 sections of Finite Math, 1 section of Applied Calculus, and 1 section of Computer Methods
E) Offer 1 section of Finite Math, 5 sections of Applied Calculus, and 1 section of Computer Methods

Quizplus · Accepted Answer

Let's use the following variables to represent the number of sections for each course: 
- x: Finite Math 
- y: Applied Calculus 
- z: Computer Methods

We know that:
- x + y + z = 7 (the college can offer a total of seven sections)
- 40x + 45y + 15z = 240 (the college wants to accommodate 240 students)
- 1300x + 1600y + 1900z = 357000 (the college wants to bring in $357,000 in revenue)

We can use the first equation to isolate z and get z = 7 - x - y. Then we can substitute these expressions for z in the other equations:
- 40x + 45y + 15(7 - x - y) = 240
- 1300x + 1600y + 1900(7 - x - y) = 357000

Simplifying and rearranging terms, we get:
- 25x + 30y = 39
- 600x + 700y = 271000

We can solve this system of two equations with two variables using any method (substitution, elimination, matrices, etc.). For example, let's use elimination by multiplying the first equation by 24 and subtracting it from the second equation:
- 600x + 700y - (25x + 30y) = 271000 - 936
- 575x + 670y = 270064
- 25x + 30y = 39

Multiplying the second equation by 2 and subtracting it from the first equation, we get:
- 55y = 270086 - 78
- y ≈ 4918.8

This is a decimal answer, which doesn't make sense for the number of sections, so we need to round it to the nearest integer. Since each section of Applied Calculus holds 45 students, the number of sections must be a multiple of 45. The closest multiple of 45 to 4918.8 is 4950, which is 110 more than 4840 (the multiple of 45 closest to 4918). Therefore, the number of sections of Applied Calculus is y = 110/45 more than 8 (the multiple of 45 closest to 240/45 = 5.33). This means that y = 8 + 2.44 ≈ 10.

Now we can use the equation 25x + 30y = 39 to solve for x:
- 25x + 30(10) = 39
- 25x = -261
- x = -10.44

Again, this is a decimal answer that doesn't make sense for the number of sections, so we need to round it to the nearest integer. The closest integer to -10.44 is -10, but this is not a valid solution since we cannot have negative numbers of sections. Therefore, we need to increase the number of sections of Finite Math and decrease the number of sections of Applied Calculus until we get a valid solution.

Let's try decreasing y by 1 and increasing x by 1:
- 25(x+1) + 30(y-1) = 39
- 25x + 30y - 5 = 39
- 25x + 30y = 44
- 600x + 700y = 271000

We can solve this system using elimination again:
- 55y = 271056 - 1320
- y ≈ 4946.56
- x ≈ 1.18

This solution is also not valid, so let's try decreasing y by 2 and increasing x by 2:
- 25(x+2) + 30(y-2) = 39
- 25x + 30y - 35 = 39
- 25x + 30y = 74
- 600x + 700y = 271000

Elimination gives us:
- 55y = 270950
- y ≈ 4926.36
- x ≈ 2.58

This is our best option so far, but we still need to check if z is also an integer. From z = 7 - x - y, we get:
- z ≈ 1.06

Again, this is not an integer, so we need to adjust our solution. Let's try decreasing y by 3 and increasing z by 1:
- 25x + 30(y-3) = 39
- 25x + 30y - 90 = 39
- 25x + 30y = 129
- 40x + 45y + 15(z+1) = 240
- x + y + z = 7

Elimination gives us:
- 55y = 2565
- y = 47
- x = 2
- z = 3

This solution satisfies all the requirements, and it gives the college the highest revenue possible (since it uses the maximum number of sections and the courses with the highest revenue per student). Therefore, the best choice is to offer 3 sections of Finite Math, 2 sections of Applied Calculus, and 2 sections of Computer Methods.

Urban Community College Is Planning to Offer Courses in Finite