 Salvatore Boone

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 Statistics rodclassique4r

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)
$3y+2z=12$
$3y+2\left(3\right)=12$
$3y+6=12$
$3y=6$
$y=2$
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

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 $a=7-b-s$ into equation (2).
$5\left(7-b-s\right)+2b+2s=23$
$35-5b-5s+2b+3s=23$
$-3b-2s=-12$
$3b+2s=12$ ...(4)
Substitute $a=7-b-s$ into equation (3).
$2\left(7-b-s\right)+5b+6s=32$
$14-2b-2s+5b+6s=32$
$3b+4s=18$ ...(5)
Step 4
Subtract equation (4) from equation (5).
$\left(3b+4s\right)-\left(3b+2s\right)=18-12$
$2s=6$
$s=3$
Substitute $s=3$ into the equation (4).
$3b+2\left(3\right)=12$
$3b+6=12$
$3b=6$
$b=2$
Substitute into the equation (1).
$a+2+3=7$
$a=2$
Step 5
2 sections of applied algebra
3 sections of statistics nick1337

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 $50,000×\left(1\right)$ from equation (3), we get
$20,000x-10,000y=0$ ...(5)

Solve equation (5)
$⇒20,000x-10,000y=0$
$⇒20,000x=10,000y$
$⇒x=\frac{10,000}{20,000}y$
$⇒x=\frac{y}{2}$
Step 3
Now, substitute $x=\frac{y}{2}$ in equation (4)
$20×\frac{y}{2}+10y=80$
$10y+10y=80$
$20y=80$
$y=\frac{80}{20}$
y=4
Hence, y=4
Substitute
$x=\frac{4}{2}$
=2
Hence, x=
Now, substitute x=2 and y=4 in equation (1)
$⇒2+4+z=11$
$⇒6+z=11$
$⇒z=11-6$
$⇒z=5$
Hence, z=5
Step 4
Therefore,
2 section of Applied Algebra