Question

A theater has a seating capacity of 750 and charges $2 for children, $4 for students, and $6 for adults. At a certain screening with full attendance, there were half as many adults as children and students combined. The receipts totaled $3300. How many children attended the show? (Let x, y, and z denote the number of children, students, and adults, respectively.)

Answer 1

Answer:

100 children

Step-by-step explanation:

Let x, y, and z denote the number of children, students, and adults, respectively.

A theater has a seating capacity of 750

So, $x+y+z= 750$ ---A

Theater charges $2 for children, $4 for students, and $6 for adults.

Cost for x children = 2x

Cost for y  students = 4y

Cost for y adults = 6z

The receipts totaled $3300

So, $2x+4y+6z=3300$ ---B

Now we are given that  there were half as many adults as children and students combined.

So,$z=\frac{1}{2}(x+y)$ ---C

$2z=x+y$

Substitute the value of x+y in A

$2z+z= 750$

$3z= 750$

$z= 250$

Substitute the value of z in A and B

In A

$x+y+250= 750$

$x+y=500$ ---D

In B

$2x+4y+6(250)=3300$

$2x+4y=1800$ ---E

Solve D and E

Substitute the value of x from D in E

$2(500-y)+4y=1800$

$1000-2y+4y=1800$

$2y=800$

$y=400$

Substitute the value of y in D

$x+400=500$

x=100

Hence 100 children attended the show