Question

last tuesday, regal cinemas sold a total of 8500 movie tickets. proceeds totaled 64000. tickets can be bought in one of the 3 ways: a matinee admission costs $5, student admission is $6 all day, and regular admissions are $8.50. How many of each type of ticket was sold if twice as many students tickets were sold as matinee tickets?

Answer 1

Number of student ticket is $571

Number of matinee ticket is $1142

Number of regular ticket is $6,787

Explanation:

Let number of student ticket be x

Let number of matinee ticket be 2x

Number of regular ticket be 8500 - (x + 2x)

= 8500 - 3x

Total cost is:

$( 5x + ( 6 x 2x) + (8500 - 3x) x 8)

= $( 5x + 12x + 68000 + 24x)

= $( 68000 - 7x)

From the condition given

68000 - 7x = 64000

7x = 4000

x = 4000 / 7

x = 571.43

x = 571

Number of student ticket is $571

Number of matinee ticket is 2 x 571 = $1142

Number of regular ticket = 8500 - (3 x 571)

= 8500 - 1713

= $6,787

Answer 2

Let's assume

number of matinee admission tickets =x

number of student admission tickets =y

number of regular admission tickets =z

First equation:

regal cinemas sold a total of 8500 movie tickets.

so, we get

`x+y+z=8500`

Second equation:

a matinee admission costs $5,

student admission is $6 all day,

and regular admissions are $8.50

proceeds totalled 64000

we will get

`5x+6y+8.50z=64000`

Third equation:

twice as many students tickets were sold as matinee tickets

`-2x+y=0`

so, we got system of equations as

`x+y+z=8500`

`5x+6y+8.50z=64000`

`-2x+y=0`

we can solve for x , y and z

and we get

`x=971`

`y=1941`

`z=5588`

number of matinee admission tickets =971

number of student admission tickets =1941

number of regular admission tickets =5588...........Answer