The school play sold $550 in tickets one night. The number of $8 adult tickets was 10 less than twice the number of $5 child tickets. How many of each ticket were sold
Answers
Answer:
To solve this question, we first have to know that we are trying to assemble
and solve equations with two variables: $x$ and $y .$
Assuming $x$ is the number of child tickets and $y$ is the number of adult
tickets, we can assemble the equations.
since the number of adult tickets is 10 less than twice the number of child
tickets, we get $y=2 x-10 .$
Additionally, multiplying the number of child tickets by 5 and the number
of adult tickets by $8,$ we get the total value of $x$ and $y$ which appears by this
equation $5 x+8 y=550$ .
Substituting the value of $y$ in the second equation, we get $5 x+8(2 x-10)=$
550 .
Solving this equation, we get that the value of $x$ equals 30 tickets.
$60-10=50$ tickets.
Given : The school play sold $550 in tickets one night.
The number of $8 adult tickets was 10 less than twice the number of
$5 child tickets.
To find : How many of each ticket were sold
Solution:
Let say child ticket sold = x
Adult tickets was 10 less than twice the number of child tickets.
=> Adult ticket sold = 2x - 10
child ticket sold = x
$5 child tickets price
=> Revenue = 5x $
Adult ticket sold = 2x - 10
$8 Adult tickets price
=> Revenue = 8(2x - 10) =16x - 80 $
5x + 16x - 80 = 550
=> 21x = 630
=> x = 30
Child ticket sold = 30
Adult ticket sold = 50
Total ticket sold = 80
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