A theater has a seating capacity of 900 and charges $5 for children, $7 for students, and $9 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $6100. How many children attended the show?
5 c + 7 s + 9 a = 6100
c + s + a = 900
a = .5 (c+s) or 2a - c - s = 0
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add two and three
3 a = 900
so
a = 300
c+s = 600 so s = 600 - c
3 c + 7 s + 2700 = 6100
3 c + 7(600-c) = 3400
3 c + 4200 - 7 c = 3400
4 c = 800
c = 200
a = adults
c = childrens
s = students
With a full audience:
a + c + s = 900
There were half as many adults as children and students combined mean:
a = ( s + c ) / 2
Replace this in equation:
a + c + s = 900
( s + c ) / 2 + c + s = 900 Multiply both sides by 2
s + c + 2 c + 2 s = 1,800
3 c + 3 s = 1,800 Divide both sides by 3
c + s = 600
The receipts totaled $6,100 mean:
$9 * adults + $7 * students + $5 * chidrens= $6,100
$9 * a + $7 * s + $5 * c = $6,100
9 a + 7 s + 5 c = 6,100
Replace a = ( s + c ) / 2 in this equation.
9 * ( s + c ) / 2 + 7 s + 5 c = 6,100 Multiply both sides by 2
9 ( s + c ) + 2 * 7 s + 2 * 5 c = 2 * 6,100
9 s + 9 c + 14 s + 10 c = 12,200
23 s + 19 c = 12,200
Now you must solve 2 equations with 2 unknow:
c + s = 600
23 s + 19 c = 12,200
Try that.
The solutions are : c = 400 , s = 200
400 childrens
200 students
Proof:
a = ( s + c ) / 2
a = ( 400 + 200 ) / 2 = 600 / 2 = 300
300 adults
$9 * a + $ 7 * s + $5 * c = $6,100
$9 * 300 + $ 7 * 200 + $5 * 400 = $6,100
$2,700 + $1,400 + $2,000 = $6,100
$6,100 = $6,100
A theater is presenting a program on drinking and driving for students and their parents or other responsible adults. The proceeds will be donated to a local alcohol information center. Admission is $ 10.00 for adults and $ 5.00 for students. However, this situation has two constraints: The theater can hold no more than 210 people and for every two adults, there must be at least one student. How many adults and students should attend to raise the maximum amount of money?
To raise the maximum amount of money,
nothing adults and
nothing students should attend.
To solve this problem, we can set up a system of equations based on the given information.
Let's assume the number of children attending the show is C, the number of students is S, and the number of adults is A. We are also given some information based on the problem:
1. The total seating capacity of the theater is 900, so the total number of people attending the show is C + S + A = 900.
2. The number of adults is half the sum of the number of children and students, so A = (C + S)/2.
3. The total receipts from the show amount to $6100, so we can write an equation based on the prices and quantities of each ticket: 5C + 7S + 9A = 6100.
Now let's substitute the value of A from equation 2 into equation 3 to form a single equation with two variables (C and S):
5C + 7S + 9((C + S)/2) = 6100
Simplifying the equation:
5C + 7S + 9C/2 + 9S/2 = 6100
Multiplying through by 2 to get rid of the fractions:
10C + 14S + 9C + 9S = 12200
Combining like terms:
19C + 23S = 12200
Now we have a system of two equations:
C + S + A = 900 (equation 1)
19C + 23S = 12200 (equation 2)
We can solve this system using various methods such as substitution or elimination. Let's use the substitution method.
Using equation 1, we can express A in terms of C and S:
A = 900 - C - S
Now substitute this expression for A in equation 2:
19C + 23S = 12200
A = (C + S)/2
(900 - C - S) = (C + S)/2
Multiply through by 2 to eliminate the fractions:
2(900 - C - S) = C + S
1800 - 2C - 2S = C + S
Rearrange the equation:
3C + 3S = 1800
Now we have a system of two equations:
19C + 23S = 12200 (equation 2)
3C + 3S = 1800 (equation 3)
Solving this system of equations will give us the values of C and S, which represent the number of children and students attending the show, respectively.