A charity sold 60 tickets for a benefit concert: some were premium seats, and the rest were standard seats. The price of a standard seat was \2,250. Revenue from premium seats exceeded revenue from standard seats by \$150. What was the price, in dollars, of one premium seat?

Solution available with step-by-step explanation

SAT Systems of Two Linear Equations Question 3 | Free SAT Prep | Aniko

00:00

Question 3·Hard·Systems of Two Linear Equations in Two Variables

0 / 140

A charity sold 60 tickets for a benefit concert: some were premium seats, and the rest were standard seats.

The price of a standard seat was $5 less than the price of a premium seat.
Total ticket sales equaled $2,250.
Revenue from premium seats exceeded revenue from standard seats by $150.

What was the price, in dollars, of one premium seat?

Your answer

(Express the answer as an integer)

When a problem gives you a total and a difference between two quantities, add and subtract the equations to find each quantity directly. Then connect those values to the unknowns you need to find.

Hints

Click me!

Start with the revenues

You know the total revenue and how much more premium brought in than standard. Can you find each revenue separately before worrying about prices?

Connect revenue to tickets

Once you know each revenue, think about: revenue = price × quantity. How can you express the number of each ticket type in terms of the unknown premium price?

Use the total ticket count

You have 60 tickets total. Write an equation using your expressions for premium and standard ticket counts.

Desmos Guide

Graph the ticket-count equation

Enter y = 1200/x + 1050/(x - 5) and y = 60. Find where the curves intersect with $x > 5$ . The $x$ -coordinate is the premium price.

Step-by-step Explanation

Find each revenue using sum and difference

Let $R_p$ = premium revenue and $R_s$ = standard revenue.

We know:

$R_p + R_s = 2250$ (total revenue)
$R_p - R_s = 150$ (premium exceeds standard)

Add the equations: $2R_p = 2400$ , so $R_p = 1200$ .

Substitute into $R_p + R_s = 2250$ : we get $1200 + R_s = 2250$ , so $R_s = 1050$ .

Set up the ticket-count equation

Let $p$ be the premium price. Then the standard price is $p - 5$ .

Number of premium tickets: $\dfrac{1200}{p}$
Number of standard tickets: $\dfrac{1050}{p - 5}$

Since 60 tickets were sold:

\dfrac{1200}{p} + \dfrac{1050}{p - 5} = 60

Solve for the premium price

Try $p = 40$ :

Premium tickets: $\dfrac{1200}{40} = 30$
Standard tickets: $\dfrac{1050}{35} = 30$
Total: $30 + 30 = 60$ ✓

Since $p = 40$ works and gives a positive standard price of $35, the premium seat price is $40.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation