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Question 1111087: The Enormous State University History Department offers three courses—Ancient, Medieval, and Modern History—and the chairperson is trying to decide how many sections of each to offer this semester. The department is allowed to offer 44 sections total, there are 4,900 students who would like to take a course, and there are 58 professors to teach them. Sections of Ancient History have 100 students each, sections of Medieval History hold 50 students each, and sections of Modern History have 200 students each. Modern History sections are taught by a team of two professors, while Ancient and Medieval History need only one professor per section. How many sections of each course should the chair schedule in order to offer all the sections that they are allowed, accommodate all of the students, and give one teaching assignment to each professor? (Assume each section has the maximum number of students allowed.)
Ancient History
sections
Medieval History
sections
Modern History
sections
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let A = number of sections of ancient history
let B = number of sections of medieval history
let C = number of sections of modern history
you get 3 equations that need to be solved simultaneously.
they are:
A + B + C = 44 (first equation)
100A + 50B + 200C = 4900 (second equation)
A + B + 2C = 58 (third equation)
first equation is number of sections
second equation is number of students
third equation is number of teachers
subtract the first equation from the third equation to get:
A + B + 2C = 58 minus (A + B + C = 44) results in C = 14
since you know the value of C, rewrite your 3 original equations with the value of C being evaluated in each equation.
your equations become:
A + B + 14 = 44 (first equation)
100A + 50B + 2800 = 4900 (second equation)
A + B + 28 = 58 (third equation)
subtract the constants on the left side of the equal sign from both sides of the equations to get:
A + B = 30 (first equation)
100A + 50B = 2100 (second equation)
A + B = 30 (third equation)
subtract the second equation from 100 times the third equation to get:
100A + 100B = 3000 minus (100A + 50B = 2100) results in 50B = 900
solve for B to get B = 18
you now have:
C = 14
B = 18
replace B and C in the first equation to get:
A + B + C = 44 becomes A + 18 + 14 = 44
solve for A to get A = 12
you now have:
A = 12
B = 18
C = 14
that's your solution.
number of sections is A + B + C which becomes 12 + 18 + 14 which is equal to 44
number of students in each course is 100A + 50B + 200C = 1200 + 900 + 2800 which is equal to 4900
number of teachers in each course is A + B + 2C which becomes 12 + 18 + 28 which is equal to 58.
12 sections in ancient history have 1200 students and 12 teachers.
that's 1 teacher per course and 100 students per section.
18 sections in medieval history have 900 students and 18 teachers.
that's 1 teacher per course and 50 students per section.
14 sections in modern history have 2800 students and 28 teachers.
that's 2 teachers per course and 200 students per section.
answer to your questions is:
12 sections of ancient history and 18 sections of medieval history and 14 sections of modern history.
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