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25 Nov 2023, 02:46
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Number of weeks per instrument: \(\frac{200+4y-x}{y}\)
Number of instruments per week: \(\frac{2000}{x+6y}\)
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
67% (03:22) correct
33%
(03:25)
wrong
based on 302
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LightTheStage, a stage lighting equipment rental company, charges $x for the first four weeks that a lighting instrument is rented, and $y per week for each week after that. The company 1564 Theatre Group has budgeted $2,000 to spend on the rental of lighting instruments from LightTheStage for its upcoming production.
In the table, indicate which expression corresponds to the maximum number of weeks per instrument for which 1564 Theatre Group can rent 10 instruments given its budget, as well as which expression corresponds to the maximum number of instruments 1564 Theatre Group can rent for a total of 10 weeks per instrument given its budget. Make only two selections, one in each column.
In the table, indicate which expression corresponds to the maximum number of weeks per instrument for which 1564 Theatre Group can rent 10 instruments given its budget, as well as which expression corresponds to the maximum number of instruments 1564 Theatre Group can rent for a total of 10 weeks per instrument given its budget. Make only two selections, one in each column.
| Number of weeks per instrument | Number of instruments per week | Expression |
| \(\frac{1000}{2x+3y}\) | ||
| \(\frac{2000}{x+6y}\) | ||
| \(\frac{1000}{x+6y}\) | ||
| \(\frac{4(50+y-x)}{y}\) | ||
| \(\frac{200+4y-x}{y}\) | ||
| \(\frac{200+4y+x}{y}\) |
ShowHide Answer
Official Answer
Number of weeks per instrument: \(\frac{200+4y-x}{y}\)
Number of instruments per week: \(\frac{2000}{x+6y}\)
26 Nov 2023, 05:44
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Premises:
x$ for first 4 weeks
y$ for remaining weeks
Maximum budget=2000$
First Scenario:
Instrument rented is constant => 10
Variable is number of weeks. Assume it to be z
Thus the expression will be:
10x+(z-4)10y=2000
\(z=\frac{200-x+4y}{y}\)
Second Scenario:
Number of weeks rented is constant => 10
Variable is number of instruments. Assume it to be z
Thus the expression will be:
zx+6zy=2000
\(z=\frac{2000}{x+6y}\)
x$ for first 4 weeks
y$ for remaining weeks
Maximum budget=2000$
First Scenario:
Instrument rented is constant => 10
Variable is number of weeks. Assume it to be z
Thus the expression will be:
10x+(z-4)10y=2000
\(z=\frac{200-x+4y}{y}\)
Second Scenario:
Number of weeks rented is constant => 10
Variable is number of instruments. Assume it to be z
Thus the expression will be:
zx+6zy=2000
\(z=\frac{2000}{x+6y}\)
28 Nov 2023, 09:39
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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: \(I=\frac{2000}{(X+6Y)}\)
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: \(I=\frac{2000}{(X+6Y)}\)
