A theater with 600 seats sells tickets at $1.20, $1.80, or $2.40 per seat. On Wednesday evening, 1/3 of the...

1. Translate the problem requirements: We have 600 theater seats with tickets at three prices (\(\$1.20\), \(\$1.80\), \(\$2.40\)). We know that \(\frac{1}{3}\) of tickets were sold at \(\$1.80\), total revenue was \(\$1,020\), and we need to find how many tickets were sold at \(\$2.40\).
2. Set up variables for the unknown quantities: Define variables for the number of tickets at \(\$1.20\) and \(\$2.40\) prices, using the given information about \(\$1.80\) tickets.
3. Create equations from the given constraints: Use the fact that total tickets = 600 and total revenue = \(\$1,020\) to establish two equations with our variables.
4. Solve the system systematically: Substitute known values and solve for the number of \(\$2.40\) tickets using algebraic manipulation.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

• The theater has exactly 600 seats, and all 600 tickets were sold
• There are three different ticket prices: \(\$1.20\), \(\$1.80\), and \(\$2.40\)
• We know that \(\frac{1}{3}\) of all tickets sold were at the \(\$1.80\) price
• The total money collected from all 600 tickets was \(\$1,020\)
• We need to find how many tickets were sold at \(\$2.40\)

Since \(\frac{1}{3}\) of the 600 tickets were sold at \(\$1.80\), that means \(\frac{1}{3} \times 600 = 200\) tickets were sold at \(\$1.80\).

This leaves us with \(600 - 200 = 400\) tickets that were sold at either \(\$1.20\) or \(\$2.40\).

Process Skill: TRANSLATE - Converting the word problem into clear mathematical relationships

2. Set up variables for the unknown quantities

Now let's define what we don't know yet:

Let \(\mathrm{x}\) = number of tickets sold at \(\$1.20\)
Let \(\mathrm{y}\) = number of tickets sold at \(\$2.40\)

We already figured out that 200 tickets were sold at \(\$1.80\).

Since we know the total number of tickets and we've accounted for the \(\$1.80\) tickets, we can say:
\(\mathrm{x} + \mathrm{y} = 400\) (the remaining tickets after removing the 200 tickets at \(\$1.80\))

3. Create equations from the given constraints

We have two important constraints to work with:

Constraint 1 - Total number of tickets:
Number of \(\$1.20\) tickets + Number of \(\$1.80\) tickets + Number of \(\$2.40\) tickets = 600
\(\mathrm{x} + 200 + \mathrm{y} = 600\)
Simplifying: \(\mathrm{x} + \mathrm{y} = 400\)

Constraint 2 - Total revenue:
Money from \(\$1.20\) tickets + Money from \(\$1.80\) tickets + Money from \(\$2.40\) tickets = \(\$1,020\)
\((\$1.20 \times \mathrm{x}) + (\$1.80 \times 200) + (\$2.40 \times \mathrm{y}) = \$1,020\)

Let's calculate the money from the \(\$1.80\) tickets: \(\$1.80 \times 200 = \$360\)

So our revenue equation becomes:
\(\$1.20\mathrm{x} + \$360 + \$2.40\mathrm{y} = \$1,020\)
Subtracting \(\$360\) from both sides: \(\$1.20\mathrm{x} + \$2.40\mathrm{y} = \$660\)

4. Solve the system systematically

Now we have two simple equations:
• \(\mathrm{x} + \mathrm{y} = 400\) (total remaining tickets)
• \(1.20\mathrm{x} + 2.40\mathrm{y} = 660\) (remaining revenue)

From the first equation, we can say: \(\mathrm{x} = 400 - \mathrm{y}\)

Substituting this into the second equation:
\(1.20(400 - \mathrm{y}) + 2.40\mathrm{y} = 660\)

Let's expand this step by step:
\(1.20 \times 400 - 1.20\mathrm{y} + 2.40\mathrm{y} = 660\)
\(480 - 1.20\mathrm{y} + 2.40\mathrm{y} = 660\)
\(480 + 1.20\mathrm{y} = 660\)
\(1.20\mathrm{y} = 660 - 480\)
\(1.20\mathrm{y} = 180\)
\(\mathrm{y} = 180 \div 1.20 = 150\)

Therefore, 150 tickets were sold at \(\$2.40\).

Let's verify:
• Tickets at \(\$2.40\): 150
• Tickets at \(\$1.80\): 200
• Tickets at \(\$1.20\): \(400 - 150 = 250\)
• Total tickets: \(150 + 200 + 250 = 600\) ✓
• Total revenue: \((150 \times \$2.40) + (200 \times \$1.80) + (250 \times \$1.20) = \$360 + \$360 + \$300 = \$1,020\) ✓

5. Final Answer

The number of tickets sold at \(\$2.40\) per seat is 150.

This corresponds to answer choice A. 150.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "1/3 of the tickets sold"

Students might think this means \(\frac{1}{3}\) of one of the price categories rather than \(\frac{1}{3}\) of the total 600 tickets. This leads to setting up incorrect equations from the start.

2. Overlooking the constraint that ALL 600 seats were sold

The problem states "from the sale of 600 tickets" but students might miss that this means every single seat was sold, leading them to treat the total number of tickets as an unknown variable instead of a given constraint.

3. Setting up too many variables unnecessarily

Since \(\frac{1}{3}\) of 600 tickets (200 tickets) were sold at \(\$1.80\), students might still create a variable for this known quantity, making their system more complex than needed and increasing chances for errors.

Errors while executing the approach

1. Arithmetic errors when calculating revenue from $1.80 tickets

Students often make mistakes calculating \(\$1.80 \times 200 = \$360\), either getting \(\$300\) or \(\$320\), which throws off the entire revenue equation.

2. Sign errors when rearranging the substitution equation

When substituting \(\mathrm{x} = 400 - \mathrm{y}\) into \(1.20\mathrm{x} + 2.40\mathrm{y} = 660\), students frequently make errors with the negative signs, especially when expanding \(1.20(400 - \mathrm{y})\).

3. Division errors when solving for y

When reaching \(1.20\mathrm{y} = 180\), students sometimes incorrectly calculate \(180 \div 1.20\), getting 120 or 200 instead of the correct answer 150.

Errors while selecting the answer

1. Selecting the number of $1.20 tickets instead of $2.40 tickets

After solving and finding \(\mathrm{y} = 150\) (\(\$2.40\) tickets) and \(\mathrm{x} = 250\) (\(\$1.20\) tickets), students might confuse which variable represents which price and select 250 instead of 150.

No other likely faltering points in this phase.