Revenue Practice Problems
Set 1: Introductino to Revenue Word Problems
To review business applications of quadratic equations, go to Revenue Help
Practice
Let x represent the number of increases/decreases in the price.

The owner of an apartment complex knows he can rent all 50 apartments when the monthly rent is $400. He thinks that for each $25 increase in the rent, he will lose two tenants.
P = _____________
Q = _____________
R = _____________

A grocery store sells 4000 gallons of milk per week when the price is $2.80 per gallon. Customer research indicates that for each $0.10 decrease in the price, 200 more gallons of milk will be sold.
P = _____________
Q = _____________
R = _____________

A movie theater’s concession stand sells an average of 500 buckets of popcorn each weekend when the price is $4 per bucket. The manager knows from experience that for every $0.05 decrease in the price, 20 more buckets of popcorn will be sold each weekend.
P = _____________
Q = _____________
R = _____________

An automobile repair shop performs 40 oil changes per day when the price is $30. Industry research indicates that the shop will lose 5 customers for each $2 increase in the price.
P = _____________
Q = _____________
R = _____________

A fast food restaurant sells an average of 250 orders of onion rings each week when the price is $1.50 per order. The manager believes that for each $0.05 decrease in the price, 10 more orders will be sold.
P = _____________
Q = _____________
R = _____________

A shoe store sells a certain athletic shoe for $40 per pair. The store averages sales of 80 pairs each week. The store owner’s past experience leads him to believe that for each $2 increase in the price of the shoe, one less pair would be sold each week.
P = _____________
Q = _____________
R = _____________
Solutions

P = 400 + 25 x
Q = 50 – 2 x
R = (400 + 25 x )(50 – 2 x )

P = 2.80 – 0.10 x
Q = 4000 + 200 x
R = (2.80 – 0.10 x )(4000 + 200 x )

P = 4 – 0.05 x
Q = 500 + 20 x
R = (4 – 0.05 x )(500 + 20 x )

P = 30 + 2 x
Q = 40 – 5 x
R = (30 + 2 x )(40 – 5 x )

P = 1.50 – 0.05 x
Q = 250 + 10 x
R = (1.50 – 0.05 x )(250 + 10 x )

P = 40 + 2 x
Q = 80 – 1 x
R = (40 + 2 x )(80 – x )
Set 2: Solving the Revenue Word Problem
To review the revenue formula, go to Revenue Help
Practice
 The owner of an apartment complex knows he can rent all 50 apartments when the monthly rent is $400. He thinks that for each $25 increase in the rent, he will lose two tenants. What should the rent be for the revenue to be $20,400?
 A grocery store sells 4000 gallons of milk per week when the price is $2.80 per gallon. Customer research indicates that for each $0.10 decrease in the price, 200 more gallons of milk will be sold. What does the price need to be so that weekly milk sales reach $11,475?
 A movie theater’s concession stand sells an average of 500 buckets of popcorn each weekend when the price is $4 per bucket. The manager knows from experience that for every $0.05 decrease in the price, 20 more buckets of popcorn will be sold each weekend. What should the price be so that $2450 worth of popcorn is sold? How many buckets will be sold at this price (these prices)?
 An automobile repair shop performs 40 oil changes per day when the price is $30. Industry research indicates that the shop will lose 5 customers for each $2 increase in the price. What would the shop have to charge in order for the daily revenue from oil changes to be $1120? How many oil changes will the shop perform each day?
 A fast food restaurant sells an average of 250 orders of onion rings each week when the price is $1.50 per order. The manager believes that for each $0.05 decrease in the price, 10 more orders are sold. If the manager wants $378 weekly revenue from onion ring sales, what should she charge for onion rings?
 A shoe store sells a certain athletic shoe for $40 per pair. The store averages sales of 80 pairs each week. The store owner’s past experience leads him to believe that for each $2 increase in the price of the shoe, one less pair would be sold each week. What price would result in $3648 weekly sales?
Solutions
If x = 1, the rent should be 400 + 25(1) = $425. If x = 8, the rent should be 400 + 25(8) = $600.
If x = 2.50, the price should be 2.80 – 0.10(2.5) = $2.55. If x = 5.5, the price should be 2.80 – 0.10(5.50) = $2.25.

P = 4 – 0.05 x Q = 500 + 20 x R = (4 – 0.05 x )(500 + 20 x )
2450 = (4 – 0.05 x )(500 + 20 x )
2450 = 2000 + 55 x – x ^{2}
x ^{2} – 55 x + 450 = 0
( x – 45)( x − 10) = 0
If x = 45, the price should be 4 – 0.05(45) = $1.75 and 500 + 20(45) = 1400 buckets would be sold. If x = 10, the price should be 4 – 0.05(10) = $3.50 and 500 + 20(10) = 700 buckets would be sold.
The price should be 30 + 2(1) = $32. There would be 40 – 5(1) = 35 oil changes performed each day.

P = 1.50 – 0.05 x Q = 250 + 10 x
R = (1.50 – 0.05 x )(250 + 10 x )
378 = (1.50 – 0.05 x )(250 + 10 x )
378 = 375 + 2.5 x – 0.5 x ^{2}
0.5 x ^{2} – 2.5 x + 3 = 0
2(0.5 x ^{2} – 2.5 x + 3) = 2(0)
x ^{2} – 5 x + 6 = 0
( x – 2)( x – 3) = 0
If x = 2, the price should be 1.50 – 0.05(2) = $1.40. If x = 3, the price should be 1.50 – 0.05(3) = $1.35.
( x = 56 is not likely to be a solution—the price would be $152!) If the price of the shoes are 40 + 2(4) = $48 per pair, the revenue will be $3648.
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From Algebra Demystified: A SelfTeaching Guide. Copyright © 2003 by The McGrawHill Companies, Inc. All Rights Reserved.