LightTheStage, a stage lighting equipment rental company, charges $x

Official Explanation

Apply the 4-Step Plan

Step 1: Preview the task.

A quick glance at the answer table tells you that this is a quantitative question. Additionally, all answer choices are algebraic expressions in x and y, so you will have to set up some sort of equation and solve for an unknown quantity.

Step 2: Read the prompt.

You have a vendor, a customer, the customer’s budget, and the vendor’s rental prices. Two absolute values are given (the theater’s lighting rental budget and the 4 weeks of the initial rental rate) and two variables for the two rental rates. Circle back to the tasks you have to perform. Note that they are similar but independent of each other. You will have to set up two algebraic expressions, one for each column, and solve the first one for the number of weeks per instrument, and the second one for the number of instruments per week.

Step 3: Proceed to solving, one column at a time.

Column 1: Let W be the maximum number of weeks that 1564 Theatre Group rents each of the 10 instruments. Then \(W – 4\) is the maximum number of weeks per instrument during which 1564 Theatre Group pays $y per instrument (since for the first 4 weeks it pays $x per instrument—and remember, that’s $x in total for each instrument for the first 4 weeks, not $x per week). The total cost per instrument, then, is \(x+y (W-4)\). The theater company is renting 10 instruments, so its total cost is \(10[x+y(W-4)]\) Equate this expression to $2,000 and solve for W:

\(10[x+y(W-4)] = 2000\)

\(x+y(W-4) = 200\)

\(x+yW-4y = 200\)

\(yW=200+4y-x\)

\(W=\frac{200+4y-x}{y}\)

Answer: \(\frac{200+4y-x}{y}\)

Column 2: Follow the same process you did for column 1. If 1564 Theatre Group is renting each instrument for 10 weeks, then it is paying $x for the first four weeks and $y per week for the remaining 6 weeks. Thus, it ispaying \(x + 6y\) in total for each instrument. Let I be the maximum number of instruments the theater company rents. Then, its total cost is \(I(x+6y)\) Equate this expression to $2,000 and solve for I:

\(I(x+6y)=2000\)

\(I=\frac{2000}{(X+6Y)}\)

Answer: \(I=\frac{2000}{(X+6Y)}\)