Question

Suppose that there are two types of tickets to a show: advance and same-day. Advance tickets cost $40

and same-day tickets cost $20
. For one performance, there were
55 tickets sold in all, and the total amount paid for them was
$1900. How many tickets of each type were sold?

Answer 1

Advanced tickets = 40

Same-day tickets = 15

Explanation:

We know that the advance tickets cost $40 and same-day ticket costs $20.

Also, there were 55 tickets sold and the total amount paid for them was $1900.

Assuming
`A` to be the number of advance tickets and
`S` to be the number of same-day tickets, we can write the following equations:

`A+S=55` --- (1)

`40A+20S=1900` --- (2)

Simplifying equation (2):

`5(8A+4S)=1900`

`8A+4S=380` --- (3)

Now solving for A and S:

Substituting A from equation 1,
`A=55-S` in (3):

`8(55-S)+4S=380`

`440-8S+4S=380`

`4S=60`

S = 15

Substituting this value of S in (1):

`A+15=55`

`A=55-15`

A = 40

Answer 2

There were 40 advanced tickets and 15 same-day tickets sold

Explanation:

* Lets change the story problem to equations and solve them

- Let x is the number of the advance tickets

- Let y is the number of the same-day tickets

∵ There were 55 tickets sold

∴ x + y = 55 ⇒ (1)

- The cost of the advance ticket is $40

- The cost of the same-day tickets is $20

- The total amount paid for them was $1900

∴ 40x + 20y = 1900 ⇒ (2)

* Now lets solve the two equations (1) and (2) to find the number

of tickets of each type were sold

- By using substitution method

- From equation (1) ⇒ y = 55 - x

- Substitute the value of y in equation (2)

∴ 40x + 20(55 - x) = 1900 ⇒ open the bracket

∴ 40x + 1100 - 20x = 1900 ⇒ collect the like term

∴ 20x + 1100 = 1900 ⇒ subtract 1100 from both sides

∴ 20x = 800 ⇒ divide both sides by 20

∴ x = 40

- Substitute this value of x in the equation of y

∴ y = 55 - 40 = 15

* There were 40 advance tickets and 15 same-day tickets sold