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?

Question ID
1651110

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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.

3170255

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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.

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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