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
$1800
. How many tickets of each type were sold?
Let's denote the number of advance tickets sold as x and the number of same-day tickets sold as y.
According to the problem, we have the following information:
Advance tickets cost $40, and same-day tickets cost $20.
The total number of tickets sold is 55.
The total amount paid for all tickets is $1800.
From these clues, we can set up a system of equations to solve for x and y.
Equation 1: x + y = 55 (since the total number of tickets sold is 55)
Equation 2: 40x + 20y = 1800 (since the total amount paid for all tickets is $1800)
Now, we can solve this system of equations.
Multiplying Equation 1 by 20, we get:
20x + 20y = 1100
Subtracting this equation from Equation 2, we have:
40x + 20y - (20x + 20y) = 1800 - 1100
20x = 700
Dividing both sides by 20, we find:
x = 700 / 20 = 35
Plugging this value of x back into Equation 1:
35 + y = 55
Subtracting 35 from both sides, we get:
y = 55 - 35 = 20
Therefore, 35 advance tickets and 20 same-day tickets were sold.
To solve this problem, we can set up a system of equations.
Let's assume that x represents the number of advance tickets sold, and y represents the number of same-day tickets sold.
Given that there were 55 tickets sold in total, we can write the equation:
x + y = 55 (Equation 1)
The total amount paid for the tickets was $1800. Since advance tickets cost $40 and same-day tickets cost $20, we can write another equation:
40x + 20y = 1800 (Equation 2)
Now we have a system of two equations with two variables. We can solve this system to find the values of x and y, which represent the number of advance tickets and same-day tickets sold, respectively.
To solve this system, we can use any method such as substitution or elimination. Let's use the substitution method.
From Equation 1, we can isolate x as follows:
x = 55 - y
Now substitute this value for x in Equation 2:
40(55 - y) + 20y = 1800
Distribute the 40:
2200 - 40y + 20y = 1800
Combine like terms:
-20y = 1800 - 2200
Simplify:
-20y = -400
Divide by -20:
y = -400 / -20
y = 20
Now substitute this value for y back into Equation 1:
x + 20 = 55
Subtract 20 from both sides:
x = 55 - 20
x = 35
Therefore, 35 advance tickets and 20 same-day tickets were sold.
number of advance --- x
number of same day --- y
x+y = 55
y = 55-x
40x + 20y = 1800
40x + 20(55-x) = 1800
continue
strange price for tickets, usually at the door they are more expensive