Answer:
There were 40 advanced tickets and 15 same-day tickets sold
Step-by-step 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
Answer:
Advanced tickets = 40
Same-day tickets = 15
Step-by-step 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 [tex]A[/tex] to be the number of advance tickets and [tex]S[/tex] to be the number of same-day tickets, we can write the following equations:
[tex]A+S=55[/tex] --- (1)
[tex]40A+20S=1900[/tex] --- (2)
Simplifying equation (2):
[tex]5(8A+4S)=1900[/tex]
[tex]8A+4S=380[/tex] --- (3)
Now solving for A and S:
Substituting A from equation 1, [tex]A=55-S[/tex] in (3):
[tex] 8(55-S)+4S=380[/tex]
[tex]440-8S+4S=380[/tex]
[tex] 4S=60[/tex]
S = 15
Substituting this value of S in (1):
[tex]A+15=55[/tex]
[tex]A=55-15[/tex]
A = 40
