A community college offers courses in Applied Algebra, Busin

Salvatore Boone

Salvatore Boone

Answered question

2021-12-21

A community college offers courses in Applied Algebra, Business Calculus, and Statistics. Each section of Applied Algebra has 50 students and earns the college $20,000 in revenue. Each section of Business Calculus has 20 students and earns the college $50,000 in revenue. Each section of Statistics has 30 students and earns the college $60.000 in revenue. Assuming the college wishes to offer a total of 7 sections, accommodate 230 students, and bring in $320,000 in revenue, how many sections of each course should they offer?
? sections of Applied Algebra
? sections of Business Calculus
? sections of Statistics

Answer & Explanation

rodclassique4r

rodclassique4r

Beginner2021-12-22Added 37 answers

Step 1
let x ,y and z be the number of sections of Applied Algebra, Business Calculus and Statistics.
Total number of sections is 7, so x+y+z=7 ... (1)
Total number of students is 230, so
50x+20y+30z=230
5x+2y+3z=23 ...(2)
Total revenue is 320,000, so
20,000x+50,000y+60,000z=320,000
2x+5y+6z=32 ...(3)
Step 2
Subtract equation (2) from 5 times equation (1)
3y+2z=12 ...(4)
Subtract twice the equation (1) from equation (3)
3y+4z=18 ...(5)
Subtract equation (4) from equation (5)
2z=6
z=3
From equation (4)
3y+2z=12
3y+2(3)=12
3y+6=12
3y=6
y=2
From equation (1)
x+y+z=7
x+2+3=7
x=2
Step 3
Ans: The number of sections of Applied Algebra, Business Calculus and Statistics are 2,2 and 3 respectively.
yotaniwc

yotaniwc

Beginner2021-12-23Added 34 answers

Step 1
Assume number of sections in applied algebra, Business calculus and Statistics be a, b and s respectively.
Step 2
Total number of sections =7
a+b+s=7 ...(1)
Total number of students =230
50a+20b+30s=230
5a+2b+3s=23 ...(2)
Total revenue =320,000
20,000a+50,000b+60,000s=320,000
2a+5b+6s=32 ...(3)
Step 3
Substitute a=7bs into equation (2).
5(7bs)+2b+2s=23
355b5s+2b+3s=23
3b2s=12
3b+2s=12 ...(4)
Substitute a=7bs into equation (3).
2(7bs)+5b+6s=32
142b2s+5b+6s=32
3b+4s=18 ...(5)
Step 4
Subtract equation (4) from equation (5).
(3b+4s)(3b+2s)=1812
2s=6
s=3
Substitute s=3 into the equation (4).
3b+2(3)=12
3b+6=12
3b=6
b=2
Substitute s=3 and b=2 into the equation (1).
a+2+3=7
a=2
Step 5
Answer:
2 sections of applied algebra
2 sections of business calculus
3 sections of statistics
nick1337

nick1337

Expert2021-12-28Added 777 answers

Step 1
Given information
Applied Algebra has 40 student and revenue is $70,000
Business Calculus has 30 student and revenue is $40,000
Statistics has 20 student and revenue is $50,000
And total number of section is 11 and total revenue is $550,000
We have to find the sections of each courses.
Let the number of section of Applied Algebra, Business Calculus and Statistics is x,y and z respectively.
According to given condition
Since, total number of sections is 11
Hence, x+y+z=11 ...(1)
And total number of student is 300
NSK
Hence, 40x+30y+20z=300
And total revenue is $550,000
Hence, 70,000x+40,000y+50,000z=550,000 ...(3)
Step 2
NSK
Subtract equation from equation (2), we get
20x+10y=80 ...(4)
Now, subtract equation 50,000×(1) from equation (3), we get
20,000x10,000y=0 ...(5)

Solve equation (5)
20,000x10,000y=0
20,000x=10,000y
x=10,00020,000y
x=y2
Step 3
Now, substitute x=y2 in equation (4)
20×y2+10y=80
10y+10y=80
20y=80
y=8020
y=4
Hence, y=4
Substitute y=4 in x=y2
x=42
=2
Hence, x=
Now, substitute x=2 and y=4 in equation (1)
2+4+z=11
6+z=11
z=116
z=5
Hence, z=5
Step 4
Therefore,
2 section of Applied Algebra
4 section of Business Calculus
5 section of Statistic

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?