At a carnival, food tickets cost $2 each and ride tickets cost $3 each. A total of $1,240 was collected at the carnival. The number of food tickets sold was 10 less than twice the number of ride tickets sold.

Question

The system of equations represents x, the number of food tickets sold, and y, the number of ride tickets sold.

2x + 3y = 1240

x = 2y – 10

How many of each type of ticket were sold?

180 food tickets and 293 ride tickets
180 food tickets and 350 ride tickets
293 food tickets and 180 ride tickets
350 food tickets and 180 ride tickets

Answer 1

To solve the problem, we start with the system of equations provided:

1. \( 2x + 3y = 1240 \)
2. \( x = 2y - 10 \)

We can substitute the expression for \( x \) from the second equation into the first equation.

Substituting (2) into (1):

\[ 2(2y - 10) + 3y = 1240 \]

Expanding this gives:

\[ 4y - 20 + 3y = 1240 \]

Combining like terms:

\[ 7y - 20 = 1240 \]

Adding 20 to both sides:

\[ 7y = 1240 + 20 \]

\[ 7y = 1260 \]

Dividing by 7:

\[ y = \frac{1260}{7} = 180 \]

Now that we have \( y \), we can find \( x \) using equation (2):

\[ x = 2y - 10 = 2(180) - 10 = 360 - 10 = 350 \]

Thus, the solution is:

• \( x = 350 \) (food tickets sold)
• \( y = 180 \) (ride tickets sold)

Double checking:

1. Total revenue from tickets: \[ 2x + 3y = 2(350) + 3(180) = 700 + 540 = 1240 \] This matches the total revenue collected.

2. Condition check: \[ x = 2y - 10 \implies 350 = 2(180) - 10 \implies 350 = 360 - 10 \implies 350 = 350 \] This also holds true.

Therefore, the final answer is:

350 food tickets and 180 ride tickets.