Word problems for algebra

Find below a wide variety of hard word problems in algebra. Most tricky and tough algebra word problems are covered here.  If you can solve these, you can probably solve any algebra problems.Teachers! Feel free to select from this list and give them to your students to see if they have mastered how to solve tough algebra problems.Find out below how you can print these problems. You can also purchase a solution if needed.

1. The
cost of petrol rises by 2 cents a liter. last week a man bought 20
liters at the old price. This week he bought 10 liters at the new
price. Altogether, the petrol costs $9.20. What was the old price for
1 liter?

2. Teachers
divided students into groups of 3. Each group of 3 wrote a report
that had 9 pictures in it. The students used 585 pictures altogether.
How many students were there in all?

3. Vera
and Vikki are sisters. Vera is 4 years old and Vikki is 13 years old.
What age will each sister be when Vikki is twice as old as Vera?

4. A
can do a work in 14 days and working together A and B can do the same
work in 10 days. In what time can B alone do the work?

5. 7
workers can make 210 pairs of cup in 6 days. How many workers are required
to make 450 pairs of cup in 10 days?

6. Ten
years ago the ratio between the ages of Mohan and Suman was 3:5. 11
years hence it will be 11:16. What is the present age of Mohan?

7. The
ratio of girls to boys in class is 9 to 7 and there are 80 students in
the class. How many girls are in the class?

8.  One
ounce of solution X contains only ingredients a and b in a ratio of 2:3.
One ounce of solution Y contains only ingredients a and b in a ratio of
1:2. If solution Z is created by mixing solutions X and Y in a ratio of
3:11, then 2520
ounces of solution Z contains how many ounces of a?

9.  This
week Bob puts gas in his truck
when the tank was about half empty.
Five days later, bob puts gas again when the tank was about three
fourths full. If Bob Bought 24 gallons of gas, how many gallons does
the tank hold?

10. A
commercial
airplane flying with a speed of 700 mi/h is detected 1000 miles
away with a radar. Half an hour later an interceptor plane flying with
a speed of 800 mi/h is dispatched. How long will it take the interceptor
plane to meet with the other plane?

11. There
are 40 pigs and chickens in a farmyard. Joseph counted 100 legs in
all. How many pigs and how many chickens are there?

12. The
top of a box is a rectangle with a perimeter of 72 inches. If the box
is 8 inches high, what dimensions will give the maximum volume?

13. You
are raising money for a charity. Someone made a fixed donation of
500. Then, you require each participant to make a pledge of 25
dollars. What is the minimum amount of money raised if there are 224 participants.

14. The
sum of two positive numbers is 4 and the sum of their squares
is 28.
What are the two numbers?

15. Flying
against the jet stream, a jet travels 1880 mi in 4 hours. Flying with
the jet stream, the same jet travels 5820 mi in 6 hours. What is the
rate of the jet in still air and what is the rate of the jet stream?

16. Jenna
and her friend, Khalil, are having a contest to see who can save the
most money. Jenna has already saved $110 and every week she saves an
additional $20. Khalil has already saved $80 and every week he saves an
additional $25. Let x represent the number of weeks and y represent the
total amount of money saved. Determine in how many weeks Jenna and Khalil
will have the same amount of money.

17. The
sum of three consecutive terms of a geometric sequence is 104 and their
product is 13824.find the terms.

18. The
sum of the first and last of four consecutive odd integers is 52. What
are the four integers?

19. A
health club charges a one-time initiation fee and
a monthly fee. John paid
100 dollars for 2 months of membership. However, Peter paid 200 for 6
months of membership. How much will Sylvia pay for 1 year of
membership?

20. The
sum of two positive numbers is 4 and the sum of their cubes is 28. What
is the product of the two numbers?

21. A
man selling computer parts realizes that when he sells 16 computer parts,
his earning is $1700. When he sells 56 computer parts, his earning
is $4300. What will the earning be if the man sells 30 computers parts?

22. A
man has 15 coins in his pockets. These coins are dimes and quarters
that add
up 2.4 dollars. How many quarters and how many dimes does the man have?

23. The
lengths of the sides of a triangle are in the ratio 4:3:5. Find
the lengths
of the sides if the perimeter is 18 inches.

24. The
ratio of base to height of a equilateral
triangle is 3:4. If the area of
the triangle is 6, what is the perimeter of the triangle?

25. The
percent return rate of a growth fund, income fund, and money market are
10%, 7%, and 5% respectively. Suppose you have 3200 to invest and you want
to put twice as much in the growth fund as in the money market to maximize
your return. How should you invest to get a return of 250 dollars
in 1 year?

26. A
shark was caught whose tail weighted 200 pounds. The head of the
shark weighted
as much as its tail plus half its body. Its body weighted as much
as its head and tail. What is the weight of the shark?

27. The
square root of a
number plus two is the same as the
number. What
is the
number?

28. Suppose
you have a coupon worth 6 dollars off any item at a mall. You go to
a store at the mall that offers a 20% discount. What do you need to
do to
save the most money?

29. Suppose
your grades on three math exams are 80, 93, and 91. What grade do you
need on your next exam to have at least a 90 average on the four exams?

30. Peter
has a photograph that is 5 inches wide and 6 inches long. She enlarged
each side by the same amount. By how much was the photograph enlarged
if the new area is 182 square inches?

31. The
cost to produce a book is 1200 to get started plus 9
dollars per book. The
book sells for 15 dollars each. How many books must be sold to make a profit?

32. Store
A sells CDs
for 2 dollars each if you pay a one-time
fee of 104
dollars. Store B offers 12 free CDs and charges 10 dollars for each additional
CD. How many CD must you buy so it will cost the same under both
plans?

33. When
4 is added to two numbers, the ratio is 5:6. When 4 is subtracted
from the two numbers, the ratio is 1:2. Find the two numbers.

34. A
store owner wants to sell 200 pounds of pistachios and walnuts mixed
together. Walnuts cost 4 dollars per pound and pistachio cost 6
dollars per
pounds. How many pounds of each type of nuts should be mixed if the
store owner will charge 5 dollars for the mixture?

35. A
cereal box manufacturer makes 32-ounce boxes of cereal. In a perfect
world, the box will be 32-ounce every time is made. However, since the
world is not perfect, they allow a difference of 0.06 ounce. Find the
range of acceptable size for the cereal box.

36. A
man weighing 600 kg has been losing 3.12% of his weight each month
with some
heavy exercises and eating the right food. What will the man weigh after
20 months?

37. An
object is thrown into the air at a height of 60 feet. After 1 second and
2 second, the object is 88 feet and 84 feet in the air respectively. What
is the initial speed of the object?

38. A
transit is 200 feet from the base of a building. There is man
standing on top of the building. The angles of elevation from the top
and bottom of the man are 45 degrees and 44 degrees. What is the
height of the man?

39. A
lemonade consists of 6% of lemon juice and a strawberry juice
consists of 15% pure fruit juice. How much of each kind should be
mixed together to get a 4 Liters of a 10% concentration of fruit
juice?

40. Ellen
can wash her car in 60 minutes. Her older sister Sarah can do the
same job in 45 minutes. How long will it take if they wash the car
together?

41. A
plane flies 500 mi/h. The plane can travel 1100 miles with the wind
in the same amount of time as it travels 900 miles against the wind.
What is the speed of the wind?

42. A
company produces boxes that are 5 feet long, 4 feet wide, and 3 feet
high. The company wants to increase each dimension by the same amount
so that the new volume is twice as big. How much is the increase in
dimension?

43. James
invested half of his money in land, a tenth in stock, and a twentieth
in saving bonds. Then, he put the remaining 21000 in a CD. How much
money did James saved or invested?

44. The
motherboards for a desktop computer can be manufactured for 50
dollars each.
The development cost is 250000. The first 20 motherboard are samples and
will not be sold. How many salable motherboards will yield an average cost
of 6325 dollars?

45. How
much of a 70% orange juice drink must be mixed with 44 gallons of a
20% orange juice drink to obtain a mixture that is 50% orange juice?

46. A
company sells nuts in bulk quantities. When bought in bulk, peanuts
sell for $1.20 per pound, almonds for $ 2.20 per pound, and cashews
for $3.20 per pound. Suppose a specialty shop wants a mixture of 280
pounds that will cost $2.59 per pound. Find the number of pounds of
each type of nut if the sum of the number of pounds of almonds and
cashews is three times the number of pounds of peanuts. Round your
answers to the nearest pound.

47. A
Basketball player has successfully made 36 of his last 48 free
throws. Find
the number of consecutive free throws the player needs to increase his
success rate to 80%.

48. John
can wash cars 3 times as fast as his son Erick. Working together,
they need to wash 30 cars in 6 hours. How many hours will it takes
each of them working alone?

49. In
a college, about 36% of student are under 20 years old and 15% are
over 40 years old. What is the probability that a student chosen at
random is under 20 years old or over 40 years?

50. Twice
a number plus the square root of the number is twelve minus the
square root of the number.

51. The
light intensity, I , of a light bulb varies inversely as the square
of the
distance from the bulb. A a distance of 3 meters from the bulb, I = 1.5 W/m^2 .
What is the light intensity at a distance of 2 meters from the bulb?

52. The
lengths of two sides of a triangle are 2 and 6. find the range of
values for the possible lengths of the third side.

53. Find
three consecutive integers such that one half of their sum is between
15 and 21.

54. After
you open a book, you notice that the
product of the
two page numbers on
the facing pages is 650. What are the two page numbers?

55. Suppose
you start with a number. You multiply the number by 3, add 7, divide
by ½, subtract 5, and then divide by 12. The result is 5. What
number did you start with?

56. You
have 156 feet of fencing to enclose a rectangular garden. You want
the garden to be 5 times as long as it is wide. Find the dimensions
of the garden.

57. The
amount of water a dripping faucet wastes water varies directly with
the amount of time the faucet drips. If the faucet drips 2 cups of
water every 6 minutes, find out how long it will take the faucet to
drip 10.6465 liters of water.

58. A
washer costs 25% more than a dryer. If the store clerk gave a 10% discount
for the dryer and a 20% discount for the washer, how much is the
washer before the discount if you paid 1900 dollars.

59. Baking
a tray of blueberry muffins takes 4 cups of milk and 3 cups of wheat
flour. A tray of pumpkin muffins takes 2 cups of milk and 3 cups of
wheat flour. A baker has 16 cups of milk and 15 cups of wheat flour.
You make 3 dollars profit per tray of blueberry muffins and 2 dollars
profit per tray of pumpkin muffins. How many trays of each type of
muffins should you make to maximize profit?

60. A
company found that -2p + 1000 models the number of TVs sold per month
where p can be set as low as 200 or as high as 300. How can the
company maximize the revenue?

More hard word problems in algebra

61. Your
friends say that he has $2.40 in equal numbers of quarters, dimes,
and nickels. How many of each coin does he have?

62. I
am a two-digit number whose digit in the tenth place is 1 less than
twice the digit in the ones place. When the digit in the tenth place
is divided by the digit in the ones place, The quotient is 1 and the
remainder is 4. What number am I?

63. A
two-digit number is formed by randomly selecting from the digits 2,
4, 5,
and 7 without replacement. What is the probability that a two-digit number
contains a 2 or a 7?

64. Suppose
you interview 30 females and 20 males at your school to find out who
among them are using an electric toothbrush. Your survey revealed
that only 2 males use an electric toothbrush while 6 females use it.
What is the probability that a respondent did not use an electric
toothbrush given that the respondent is a female?

65. An
employer pays 15 dollars per hour plus an extra 5 dollars per hour
for every hour worked beyond 8 hours up to a maximum daily wage of
220 dollars. Find
a piecewise function that models this situation.

66. Divide
me by 7, the remainder is 5. Divide me by 3, the remainder is 1 and
my quotient is 2 less than 3 times my previous quotient. What number
am I?

67. A
company making luggage have these requirements to follow. The length is
15 inches greater than the depth and the sum of length, width, and
depth may not exceed 50 inches. What is the maximum value for the
depth if the manufacturer will only use whole numbers?

68. To
make an open box, a man cuts equal squares from each corner of a
sheet of metal that is 12 inches wide and 16 inches long. Find an
expression for the volume in terms of x.

69. Ten
candidates are running for president, vice-president, and secretary
in the students government. You may vote for as many as 3 candidates.
In how many ways can you vote for 3 or fewer candidates?

70. The
half-life of a medication prescribed by a doctor is 6 hours. How
many mg of this medication is left after 78 hours if the doctor prescribed
100 mg?

71. Suppose
you roll a red number cube and a yellow number cube. Find
P(red 2, yellow 2) and the probability to get any matching pairs of
numbers.

72. A
movie theater in a small town usually open its doors 3 days in a row
and then closes the next day for maintenance. Another movie theater 3
miles away open 4 days in a row and then closes the next day for the
same reason. Suppose both movie theaters are closed today and today
is Wednesday, when is the next time they will both be closed again on
the same day?

73. An
investor invests 5000 dollars at 10% and the rest at 5%. How much was
invested at 5% if the yield is one-fifth of the amount invested at at
10%?

74. 20000
students took a standardized math test. The scores on the test are
normally distributed, with a mean score of 85 and a standard
deviation of 5. About how many students scored between 90 and 95?

75. A
satellite, located 2400 km above Earth’s surface, is in circular
orbit around the earth. If it takes the satellite 3 hours to complete
1 orbit, how far is the satellite after 1 hour?

76. In
a group of 10 people, what is the probability that at least two
people in
the group have the same birthday?

77. During
a fundraising for cancer at a gala, everybody shakes hands with
everyone else in the room before the event and after the event is
finished. If
n people attended the gala, how many different handshakes occur?

78. Two
cubes have side lengths that are equal to 2x and 4x. How many times greater
than the surface of the small cube is the surface area of the large
cube?

79. Suppose
you have a job in a restaurant that pays $8 per hour. You also have
a job at Walmart that pays $10 per hour. You want to earn at least
200 per week. However, you want to work no more than 25 hours per
week . Show 3 different ways you could work at each job.

80. Two
company offer tutoring services. Company A realizes that when they
tutor for 3 hours, they make 45 dollars. When they tutor for 7 hours
they make 105 hours. Company B realizes that when they tutor for 2
hours, they make 34 dollars. When they tutor for 6 hours they make
102 hours. Assuming that the number of hours students sign for
tutoring is the same for both company, which company will generate
more revenue?

81. You
want to fence a rectangular area for kids in the backyard. To save on
fences, you will use the back of your house as one of the four sides.
Find the possible dimensions if the house is 60 feet wide and you
want to use at least 160 feet of fencing.

82. When
a number in increased by 20%, the result is the same when it is
decreased by 10% plus 12. What is the number?

83. The
average of three numbers is 47. The biggest number is five more than
twice the smallest. The range is 35. What are the three numbers?

84. The
percent of increase of a number from its original amount to 36 is
80%. What
is the original amount of the number?

85. When
Peter drives to work, he averages 45 miles per hour because of
traffic. On the way back home, he averages 60 miles per hour because
traffic is not as bad. The total travel time is 2 hours. How far is
Peter’s house from work?

86. An
advertising company takes 20% from all revenue that it generates for
its affiliates. If the affiliates were paid 15200 dollars this month,
how much revenue
did the advertising company generate this month?

87. A
company’s revenue can be modeled with a quadratic equation. The
company noticed that when they sell either 2 or or 12 items, the
revenue is 0. How much is the revenue when they sell 20 items?

88. A
ball bounced 4 times, reaching three-fourths of its previous height
with each bounce. After the fourth bounce, the ball reached a height
of 25 cm. How high was the ball when it was dropped?

89. A
rental company charges 40 dollars per day plus $0.30 per mile. You
rent a car and drop it off 4 days later. How many miles did you drive
the car if you paid 325.5 dollars which included a 5% sales tax?

90. Two
students leave school at the same time and travel in opposite
directions along the same road. One walk at a rate of 3 mi/h. The
other bikes at a rate of 8.5 mi/h. After how long will they be 23
miles part?

91. Brown
has the same number of brothers as sisters. His sister Sylvia as
twice as many brothers as sisters. How many children are in the
family?

92. Ethan
has the same number of male classmates as female classmates. His
classmate Olivia has three-fourths as many female classmates as male
classmates. How many students are in the class?

93. Noah
wants to share a certain amount of money with 10 people. However, at
the last minute, he is thinking about decreasing the amount by 20 so
he can keep 20 for himself and share the money with only 5 people.
How much money is Noah trying to share if each person still gets the
same amount?

94. The
square root of me plus the square root of me is me. Who am I?

95. A
cash drawer contains 160 bills, all 10s and 50s. If the total value
of the 10s and 50s is $1,760. How many of each type of bill are in
the drawer?

96. You
want to make 28 grams of protein snack mix made with peanuts and
granola. Peanuts contain 7 grams of protein per ounce and granola
contain 3 grams of protein per ounce. How many ounces of granola
should you use for 1 ounce of peanuts?

97. The
length of a rectangular prism is quadrupled, the width is doubled,
and the height is cut in half. If V is the volume of the rectangular
prism before the modification, express the volume after the
modification in terms of V.

98. A
car rental has CD players in 85% of its cars. The CD players are
randomly distributed throughout the fleet of cars. If a person rents
4 cars, what is the probability that at least 3 of them will have CD
players?

99. Jacob’s
hourly wage is 4 times as much as Noah. When Jacob got a raise of 2
dollars, Noah accepted a new position that pays him 2 dollars less
per hour. Jacob now earns 5 times as much money as Noah. How much
money do they make per hour after Jacob got the raise?

100.

You
own a catering business that makes specialty cakes. Your company has
decided to create three types of cakes. To create these cakes, it
takes a team that consists of a decorator, a baker, and a design
consultant. Cake A takes the decorator 9 hours, the baker 6 hours,
and the design consultant 1 hour to complete. Cake B takes the
decorator 10 hours, the baker 4 hours, and the design consultant 2
hours. Cake C takes the decorator 12 hours, the baker 4 hour, and the
design consultant 1 hour. Without hiring additional employees, there
are 398 decorator hours available, 164 baker hours available, and 58
design consultant hours available. How many of each type of cake can
be created?

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REAL WORLD PROBLEMS:
How to Write Equations Based on Algebra Word Problems

I know that you often sit in class and wonder, «Why am I forced to learn about equations, Algebra and variables?»

But… trust me, there are real situations where you will use your
knowledge of Algebra and solving equations to solve a problem that is
not school related. And… if you can’t, you’re going to wish that you
remembered how.

It might be a time when you are trying to figure out how much you
should get paid for a job, or even more important, if you were paid
enough for a job that you’ve done. It could also be a time when you are
trying to figure out if you were over charged for a bill.

This is important stuff — when it comes time to spend YOUR money —
you are going to want to make sure that you are getting paid enough and
not spending more than you have to.

Ok… let’s put all this newly learned knowledge to work.

Click here if you need to review how to solve equations.

There are a few rules to remember when writing Algebra equations:

Let’s look at an example of an algebra word problem.

Example 1: Algebra Word Problems

---

Linda was selling tickets for the school play.  She sold 10 more adult tickets than children
tickets and she sold twice as many senior tickets as children tickets.

1. Let x represent the number of children’s tickets sold.
2. Write an expression to represent the number of adult tickets sold.
3. Write an expression to represent the number of senior tickets sold.
4. Adult tickets cost $5, children’s tickets cost $2, and senior tickets cost $3. Linda made $700. Write an equation to represent the total ticket sales.
5. How many children’s tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?
   

---

As you can see, this problem is massive!  There are 5 questions to answer with many expressions to write. 

Solution

---

A few notes about this problem

---

1. In this problem, the variable was defined for
you.  Let x represent the number of
children’s tickets sold tells what x stands for in this problem.  If this had not been done for you, you might
have written it like this:

        Let x = the number of children’s tickets sold

2. For the first expression, I knew that 10 more adult
tickets were sold.  Since more means add,
my expression was x +10.  Since the
direction asked for an expression, I don’t need an equal sign.  An equation is written with an equal sign and
an expression is without an equal sign. 
At this point we don’t know the total number of tickets.

3. For the second expression, I knew that my key words,
twice as many meant two times as many. 
So my expression was 2x.

4. 
We know that to find the total
price we have to multiply the price of each ticket by the number of
tickets.  Take note that since x + 10 is the quantity of adult tickets, you must put
it in parentheses!  So, when you multiply
by the price of $5 you have to distribute the 5.

5.  Once I solve
for x, I know the number of children’s tickets and I can take my expressions
that I wrote for #1 and substitute 50 for x to figure out how many adult and
senior tickets were sold.

Example 2 — Identifying a Constant

---

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20.

1. Write an equation that models this situation.

2. How many minutes were charged on this bill?

---

Solution

---

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20.

1. Write an equation that models this situation.

Notes For Example 2

---

• $12.95 is a monthly rate. Since this is a set fee for each
  month, I know that this is a constant. The rate does not change;
  therefore, it is not associated with a variable.
• $0.25 per
  minute per call requires a variable because the total amount will change
  based on the number of minutes. Therefore, we use the expression 0.25m
• You must solve the equation to determine the value for m, which is the number of minutes charged.

The last example is a word problem that requires an equation with variables on both sides.

Example 3 — Equations with Variables on Both Sides

---

You have $60 and your sister has $120. You are saving $7 per week and your sister is saving $5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve.

Solution

---

Notes for Example 3

---

• $60 and $120 are constants because this is the amount of money that they each have to begin with. This amount does not change.
• $7 per week and $5 per week are rates. They key word «per» in this situation means to multiply.
• The key word «same» in this problem means that I am going to set my two expressions equal to each other.
• When we set the two expressions equal, we now have an equation with variables on both sides.
• After solving the equation, you find that x = 30, which means that after 30 weeks, you and your sister will have the same amount of money.

I’m hoping that these three examples will help you as you solve real world problems in Algebra!

1. Home

>

2. Solving Equations

>

3. Algebra Word Problems

In algebra word problems worksheets, we will talk about algebra, which is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. They represent quantities without fixed values, known as variables. Solving algebraic word problems requires us to combine our ability to create and solve equations. Translating verbal descriptions into algebraic expressions is an essential initial step in solving word problems.

Benefits of Algebra Word Problems Worksheets

We use algebra in our everyday life without even realising it. Solving algebraic word problems worksheets help kids relate and understand the relevance of algebra in the real world. Algebra finds its way while cooking, measuring ingredients, sports, finance, professional advancement etc. They help in logical thinking and help students to break down a problem and then find its solution.

Download Algebra Word Problems Worksheet PDFs

These math worksheets should be practiced regularly and are free to download in PDF formats.

☛ Check Grade wise Algebra Word Problems Worksheets

• 1st Grade Algebra Worksheets
• Grade 3 Algebra Worksheets
• Algebra Worksheets for Grade 4
• 5th Grade Algebra Worksheets
• Grade 6 Algebra Worksheets
• Algebra Worksheets for Grade 7
• 8th Grade Algebra Worksheets

English to Math Translation	Rate/Distance Word Problem
Unit Rate Problem	Profit Word Problem
“Find the Numbers” Word Problem	Converting Repeating Decimal to Fraction Word Problem
Simple Interest Rate Problem	Inequality Word Problems
Percent Word Problem	Integer Function Problem
Percent Increase Word Problem	Direct, Inverse, Joint and Combined Variation Word Problems (in that section)
Ratio/Proportion Word Problems	Work Word Problems (in Rational Functions and Equations)
Weighted Average Word Problem	Systems of Equations Word Problems (in that section)
Consecutive Integer Word Problem	More Word Problems using Rational Functions (in Rational Functions section)
Age Word Problem	Absolute Value Word Problem (in Absolute Values section)
Money (Coins) Word Problem	Composition of Functions Word Problems (in Advanced Functions section)
Mixture Word Problem, Percent Mixture Word Problem

The algebra word problems here only involve one variable; later we’ll work on some that involve more than one, such as here in the Systems of Linear Equations and Word Problems section.

English to Math Translation for Word Problems

Doing word problems is almost like learning a new language like Spanish or French; you can basically translate word-for-word from English to Math, and here are some translations:

English	Math		English	Math
is, yields, will be	(=)		(p) percent	(displaystyle frac{p}{{100}}), or move decimal left 2 places
what number, how much (look at question)	“(x)”  (or any variable)		half, twice	(displaystyle frac{n}{2},,,,2n)
in addition to, added to, increased by	(+)		consecutive numbers	Let (n=) first number, (n+1=) second number, (n+2=) third number…
sum of (x) and (y)	(x+y)		odd/even consecutive numbers	Let (n=) first number, (n+2=) second number, (n+4=) third number… Note: Even if you are looking for odd consecutive numbers, use (n, n+2, n+4, …).
difference of (x) and (y)	(x-y)		average of (x,y) and (z) (and so on)	(displaystyle frac{{x+y+z+…}}{{text{(how many},,text{numbers},,text{on},,text{top)}}})
product of (x) and (y)	(xtimes y)		(x) per (y), (x) to (y), (x) over (y), (x) part of (y)	(xdiv y) or (displaystyle frac{x}{y}) Example: number of girls to total people can be represented by (displaystyle frac{{text{girls}}}{{text{total}}}).
quotient of (x) and (y)	(displaystyle xdiv y,,,,,text{or },,,frac{x}{y})		(x) per (y), as in (x) “for every” (y) Other examples of multiplication: each hour, every hour, an hour	Multiplication, or (xtimes y). Example: if you drive 50 miles per hour, how many miles will you drive in 5 hours:  250 miles.
opposite of (x)	(-x)		(y) increased by (x%)	(displaystyle y+left( {ytimes frac{x}{{100}}} right))
ratio of (x) to (y)	(displaystyle xdiv y,,,,,text{or},,,,frac{x}{y})		(y) decreased by (x%)	(displaystyle y-left( {ytimes frac{x}{{100}}} right))
a number (n) less 3	(n-3)		(y) is at least (or no less than) (x)	(yge x)
a number (n) less than 3	(3-n)		(y) is at most (or no more than) (x)	(yle x)
a number (n) reduced by 3	(n-3)		(y) is between (x) and (z)	(xle yle z) (inclusive)    (x<y<z)  (exclusive)
of	times

Remember these important things:

• If you’re wondering what the variable (or unknown) should be when working on a word problem, look at what the problem is asking. This is usually what your variable is!
• If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!
• If the problem asks for a Unit Rate, you want the ratio of the (y)-value (typically the dollar amount) to the (x)-value, when the (x)-value is 1.This is basically the slope of the linear functions. You may see feet per second, miles per hour, or amount per unit; these are all unit rates.

Here are problems that use some of the translations above. We’ll get to more difficult algebra word problems later.

Unit Rate Problem:

Problem: You buy 5 pounds of apples for $3.75. What is the unit rate of a pound of apples?

Solution: To get the unit rate, we want the amount for one pound of apples; this is when “(x)” (apples) is 1. We can set up a ratio: (displaystyle frac{{3.75}}{5}=frac{x}{1};,,,x=$.75). We also could have just divided the amount spent by the number of pounds, which makes sense.

(Another cool way to figure out what to do is to think of a simpler problem. If we buy 20 pounds of apples for $10, we “feel” the unit rate is $.50 per pound. And the math agrees with us!)

“Find the Numbers” Word Problems:

Problem: The sum of two numbers is 18. Twice the smaller number decreased by 3 equals the larger number. What are the two numbers?   

Solution: We always have to define a variable, and we can look at what they are asking. The problem is asking for both the numbers, so we can make “(n)” the smaller number, and “(18-n)” the larger.
Do you see why we did this? The way I figured this out is to pretend the smaller is 10. (This isn’t necessarily the answer to the problem!) I knew the sum of the two numbers had to be 18, so I knew to take 10 and subtract it from 18 to get the other number. It is easier to think of real numbers, instead of variables when you’re coming up with the expressions.

Translate the English into math and solve:

Problem/Math	Notes
The sum of two numbers is 18. Twice the smaller number decreased by 3 equals the larger number. What are the two numbers? (displaystyle begin{align}2n-3&=18-n\2n-3,,(+n)&=18-n,,(+n)\3n,-3&=18\,,3n&=21\,frac{{3n}}{3}&=frac{{21}}{3}\n&=7,,,,,,text{(smaller number)}\18-7&=11,,,,text{(larger number)}end{align})	Let (n=) the smaller number, and (18-n=) the larger number. The translation is pretty straight forward; note that we didn’t have distribute the 2 since the problem only calls for twice the smaller number, and then we subtract 3. Check our work: The sum of 7 and 11 is 18.   √ Twice the smaller ((2times 7)) decreased by 3 would be (14-3=11).   √

Another Problem:

Problem: If the product of a number and –7 is reduced by 3, the resulting number is 33 less than twice the opposite of that number. What is the number?

Solution: Define a variable, and look at what they are asking. The problem is asking for a number, so make that (n). Translate word-for-word, and remember that the “opposite” of a number just means to make it negative if it’s positive or positive if it’s negative. Thus, we can just put a negative sign in front of the variable.

If you’re not sure if you should multiply, add, or subtract, try “real numbers” to see what you should do. For example, “8 reduced by 3” is 5, so for the “reduce by 3” part, we subtract 3. Also, “33 less than 133” is 100, so for the “33 less than”, we subtract 33 at the end:

Problem/Math

Notes

If the product of a number and –7 is reduced by 3, the resulting number is 33 less than twice the opposite of that number. What is the number? (displaystyle begin{align}left( {-7} right)n-3&=2left( {-n} right)-33\,,,,,-7n-3&=-2n-33\-7n-3,,left( {+7n+33} right)&=-2n-33,,left( {+7n+33} right)\30&=5n\,frac{{30}}{5}&=frac{{5n}}{5}\n&=6end{align})

The opposite of a number means we basically just multiply the number by –1 (in our case, put a negative in front of it). Translate English to math, and, after solving, get 6 as our answer. Check our work: If we take the product of 6 and –7 (–42) and reduce it by 3, we get –45. Is this number 33 less than twice the opposite of 6? Twice the opposite of 6 is –12, and 33 less than –12 is (-12-33=-45). We got it!   √

Simple Interest Rate Problem:

Problem: How much money would need to be invested at 2% annual simple interest for 10 years to earn $1200? Use the formula (text{Interest}=text{Principle }times ,text{Rate},times ,text{Time}).

Solution: Using the formula (I=Prt), solve for (displaystyle P:,P=frac{I}{{rt}}=frac{{1200}}{{left( {.02} right)10}};,,,P=$6000). Note that we had to turn (2%) into a decimal: we need to get rid of the % (pretend we’re afraid of it), so we move the decimal 2 places away from it.

Note: Compound Interest problems can be found here in the Exponential Functions section.

Percent Word Problem:

(We saw similar problems in the Percentages, Ratios, and Proportions section!)

Percent Problem/Math	Notes
60 is 20% of what number? (displaystyle begin{align}60&=.20times n\frac{{60}}{{.2}}&=frac{{.2n}}{{.2}}\300&=,n\n&=300end{align})	Translate word-for-word, and remember again that of = times, and (20%=.20=.2). Check our work: (20%,,,text{of},,,300=.2times 300=60)  √

Percent Increase Word Problem:

Percent Increase Problem/Math	Notes
The price of a pair of shoes has increased by 15%. The original price of the shoes was $20. What is the new price?    (displaystyle begin{array}{l}x=$20+left( {15%,times 20} right)\x=$20+left( {.15times 20} right)\x=$20+$3=$23\x=$23,end{array})     or       (displaystyle begin{array}{l}x=$20times left( {1+15%} right)\x=$20times left( {1+.15} right)\x=$20times left( {1.15} right)\x=$23,end{array})	15% of the original amount (=15%times 20), since of = times. Turn the 15% into a decimal and add the product to the original amount. The second way to do it is to multiply the original amount ($20) by (1.15,,(100%+15%)), which adds (displaystyle 15%) to the original amount before multiplying.

Ratio/Proportion Word Problems

Relating Two Things Together: a Rate

Problem: It takes 2 minutes to print out 3 color photos on Erin’s printer. Write an equation relating the number of color photos (p) to the number of minutes (m).

Solution: This problem seems easy, but you have to think about what the problem is asking. When we are asked to relate something to something else, typically we use the last thing (the “to the” part) as the (y), or the dependent variable. I like to set up these types of problems as proportions, but what we’re looking for is actually a rate of minutes to photos, or how many minutes to print 1 photo. Remember that rate is “how many (y)” to “one (x)”, or in our case, how many “(m)” to one “(p)”. We will see later that this is like a Slope from the Coordinate System and Graphing Lines including Inequalities section. Here’s the math:

Ratio Problem/Math	Notes
It takes 2 minutes to print out 3 color photos on Erin’s printer. Write an equation relating the number of color photos (p) to the number of minutes (m). (displaystyle begin{align}frac{{text{2 minutes}}}{{text{3 color photos}}}&=frac{{text{how many minutes}}}{{text{1 color photo}}}\frac{text{2}}{text{3}}&=frac{m}{{1p}}\3m&=2p\m&=frac{2}{3}pend{align})	To get the rate of minutes to photos, we can set up a proportion with the minutes on the top and the photos on the bottom, and then cross-multiply. The equation relating the number of color photos (p) to the number of minutes (m) is (displaystyle m=frac{2}{3}p).     √

Ratio/Proportion Problem:

Problem: The ratio of boys to girls in your new class is 5 : 2. The sum of the kids in the class is 28. How many boys are in the class?   

Solution: This is a rati0 problem; we learned about ratios in the Percentages, Ratios, and Proportions section. A ratio is a comparison of two numbers; a ratio of 5 to 2 (also written 5:2 or (displaystyle frac{5}{2})) means you have 5 boys for every 2 girls in your class. For example, if you had only 7 in your class, you’d have 5 boys and 2 girls. But what if you had 14? You’d have 10 boys and 4 girls, since 10 is 5 times 2, and 4 is 2 times 2. Let’s set this up in an equation, though, so we can do the algebra!

There are actually a couple of different ways to do this type of problem. Probably the most common is to set up a proportion like we did in the Percents, Ratios, and Proportions section. Let (x=) the number of boys in the class.

Ratio Problem/Math

Notes

The ratio of boys to girls in your new class is 5:2. The sum of the kids in the class is 28. How many boys are in the class? (require{cancel} displaystyle begin{align}frac{{text{boys}}}{{text{total in class}}}&=frac{x}{{28}}=frac{5}{7}\\7x&=5times 28=140\frac{{cancel{7}x}}{{cancel{7}}}&=frac{{140}}{7}\x&=20text{ boys}end{align})

We know that the sum of the numbers in the ratio is 7 (boys and girls: 5 + 2), and the sum of the kids in the class is 28. We also know the “boys” part of the ratio is 5. We can set up a proportion with the same things on top or on bottom; our ratios will have “boys” on top and “total in class” on bottom. In other words, we need to see how many boys out of 28 will keep a ratio of 5 boys to 7 total in the class. Cross-multiply to get (x=20). There are 20 boys and 8 girls (28 – 20) in the new class. Check our answer: the ratio of 20 to 8 is the same as the ratio of 5 to 2. And the number of boys and girls add up to 28!       √

There’s another common way to handle these types of problems, but this way can be a little trickier since the variable in the equation is not what the problem is asking for; we will make the variable a “multiplier” for the ratio. The advantage to this way is we don’t have to use fractions.

Ratio Word Problem/Math

Notes

The ratio of boys to girls in your new class is 5:2. The sum of the kids in the class is 28. How many boys are in the class? (begin{align}5x+2x&=28\7x=28\frac{{7x}}{7}&=frac{{28}}{7}\x&=4\\5times 4&=20,,,,text{boys}\2times 4&=8,,,,text{girls}end{align})

We can multiply both numbers by the same thing to keep the ratio the same – try this with some numbers to see this. 5 times a number, and 2 times that same number must equal 28. Let (x) be the multiplier – not the number of boys or girls. We get 4 as the multiplier, but we’re looking for the number of boys in the class (5 times 4 = 20) and the number of girls in the class (2 times 4 = 8). There are 20 boys and 8 girls. Check our answer: the ratio of 20 to 8 is the same as the ratio of 5 to 2 (each is divided by 4 – the multiplier!) And the number of boys and girls add up to 28!       √

Here’s a ratio problem that’s pretty tricky; we have to do it in a lot of steps:

Ratio Problem/Math	Notes
One ounce of solution X contains ingredients a and b in a ratio of 2:3. One ounce of solution Y contains ingredients a and b in a ratio of 1:2. If solution Z is made by mixing solutions X and Y in a ratio of 3:11, then 1260 ounces of solution Z contains how many ounces of ingredient a? Solution Z: (begin{array}{c}3u+11u=1260;,,,,,u=90\3times 90=270,,,text{oz}text{. solution X}\11times 90=990,,,text{oz}text{. solution Y}end{array})	Work backwards on this problem, and first work with solution Z, since there are 1260 ounces of it. It’s good to start with the parts of the problems with numbers first! Since the ratio of X and Y is 3:11 in solution Z, we can find the ratio multiplier ((u)), and find how much of solutions X and Y are in Z. There are 270 ounces of X and 990 of Y in solution Z.
Solution X: (begin{array}{c}2v+3v=270;,,,,v=54\2times 54=108,,,text{oz}text{. ingredient a}\3times 54=162,,,,text{oz}text{. ingredient b}end{array}) Solution Y: (begin{array}{c}1w+2w=990;,,,,w=330\1times 330=330,,,text{oz}text{. ingredient a}\,2times 330=660,,,text{oz}text{. ingredient b}end{array})	From above, there are 270 ounces of solution X in solution Z. We can find out how much of ingredients a and b are in solution X by using a ratio multiplier again ((v)), since one ounce of solution X contains ingredients a and b in a ratio of 2:3. There are 108 ounces of ingredient a in solution X. Do the same for solution Y (using ratio multiplier (w)), which contains ingredients a and b in a ratio of 1:2. There are 330 ounces of ingredient a in solution Y.
(108+330=438)	The problem asks for the amount of ingredient a in solution Z, so add the amounts of ingredient a in Solutions X and Y to get 438. 1260 ounces of solution Z contains 438 ounces of ingredient a.     √

Weighted Average Word Problem:

Weighted average problems have to do with taking averages, but assigning different weights to the elements (for example, counting them more than once).   

Problem: You’ve taken four tests in your Algebra II class and made an 89, 92, 78, and 83. The final is worth two test grades. What do you need to make on the final to make an A (a 90) in the class for the semester?

Solution: Define a variable, and look at what is being asked. Let (x=) what you need to make on the final. Now we have 6 test grades that will count towards our semester grade: 4 regular tests and 2 test grades that will be what you get on the final (since it counts twice, we need to add it 2 times). This is called a weighted average, since we “weighted” the final test grade twice. Use the equation for an average:

Weighted Average Problem/Math

Notes

You’ve taken four tests in your Algebra II class and made an 89, 92, 78, and 83. The final is worth two test grades. What do you need to make on the final to make an A (a 90) in the class for the semester? (require{cancel} displaystyle begin{align}frac{{89+92+78+83+x+x}}{6}&=90\frac{{89+92+78+83+2x}}{{cancel{6}}}times frac{{cancel{6}}}{1}&=90times frac{6}{1}\342+2x&=540\2x&=198\frac{{2x}}{2}&=frac{{198}}{2}\x&=99end{align})

The average or mean equation is just adding up all the values, and then dividing by the number of values that we just added up. For this example, divide by 6, since we have 4 tests given, and the final is worth 2 test grades. You have to make a 99 on the final to make an A in the class! Yikes! Good luck – you can do it! Let’s see if it works: (displaystyle frac{{86+92+78+83+99+99}}{6}=frac{{540}}{6},=90,,,,,surd )

HINT:  For any problem with weighted averages, you can multiply each value by the weight in the numerator, and then divide by the sum of all the weights that you’ve used. For example, if you had test 1 (say, an 89) counting 20% of your grade, test 2 (80) counting 40% of your grade, and test 3 (78) counting 40% of your grade, take the weighted average as in the formula below. Don’t forget to turn percentages into decimals and make sure that all the percentages that you use (the “weights”) add up to 100 (all the decimals you use as weights should add up to 1). When using decimals, your denominator should be 1:

(displaystyle frac{{left( {89times .2} right),+,left( {80times .4} right),+,left( {78times .4} right)}}{{.2+.4+.4}}=frac{{17.8+32+31.2}}{1}=81)

Consecutive Integer Word Problem:

Problem: The sum of the least and greatest of 3 consecutive integers (numbers in a row) is 60. What are the values of the 3 integers?

Solution: You’ll see these “consecutive integer” problems a lot in algebra. When you see these, you always have to assign “(n)” to the first number, “(n+1)” to the second, “(n+2)” to the third, and so on. This makes sense, since consecutive means “in a row” and we’re always adding 1 to get to the next number.

Note: If the problem asks for even or odd consecutive numbers, use “(n)”, “(n+2)”, “(n+4)”, and so on – for both even and odd numbers! It will work; trust me!

Consecutive Integer Problem/Math	Notes
The sum of the least and greatest of three consecutive integers (numbers in a row) is 60. What are the values of the three integers? (displaystyle begin{align}n+n+2&=60\2n+2&=60\2n&=58\frac{{2n}}{2}&=frac{{58}}{2}end{align}) (displaystyle n=29,,,,,,n+1=30,,,,,,n+2=31)	Assign “(n)” to the first number, “(n+1)” to the second, and “(n+2)” to the third, since they are consecutive numbers. Translate the English into math. The least of the three consecutive numbers is “(n)“, and the greatest is “(n+2)”. Add the least number and the greatest to get 60. The three consecutive numbers are 29, 30, and 31. Check our answer:  (29+31=60,,,, surd )

Age Word Problem:

Problem: Your little sister Molly is one third the age of your mom. In 12 years, Molly will be half the age of your mom. How old is Molly and your mom now?   

Solution: Doesn’t this one sound complicated? It’s a great one to try on your friends! It’s not that bad though – first define a variable by looking at what the problem is asking.

Age Word Problem/Math

Notes

Your little sister Molly is one third the age of your mom. In 12 years, Molly will be half the age of your mom. How old is Molly and your mom now? (begin{align}M+12&=frac{1}{2}left( {3M+12} right)\2times left( {M+12} right)&=3M+12\2M+24&=3M+12\\24-12&=3M-2M\12&=Mend{align}) (M=12,,,,,,,,,3M=36)

Let (M=) the age of sister Molly now. From this, we know that your mom is (3M) (make it into an easier problem – if Molly is 10, your mom is 30). Turn English into math (second sentence). Remember that we have to add 12 years to both ages ((M+12) for Molly and (3M+12) for your mom), since we’re talking about 12 years from now (unfortunately, moms have to age, too). Multiply both sides by 2 to get rid of the fraction, and then “push” the 2 through the parentheses. Get the variables to one side, and the constants to the other. Molly is 12, and your mother is 36. Let’s see if it works: In 12 years, Molly will be 24, and her mom will be 48. Aha! Molly will be half of her mom’s age in 12 years.    (surd )

Money (Coins) Word Problem:

Note that problems like this with two variables to solve for are more easily solved with a System of Equations, which we will cover in the next section.

Money Word Problem/Math

Notes

Suppose Briley has 10 coins in quarters and dimes and has a total of $1.45. How many of each coin does she have? (displaystyle begin{align}.25Q+.10(10-Q)&=1.45\.25Q+1-.1Q&=1.45\.15Q+1&=1.45\.15Q&=.45\frac{{.15Q}}{{.15}}&=frac{{.45}}{{.15}}\Q&=3\D&=10-3=7end{align})

Let (Q=) the number of quarters that Briley has. Then we know that she has (10-Q) dimes (turn into easier problem – if she has 2 quarters, she has 10 minus 2, or 8 dimes). Each quarter is worth $.25 and each dime is worth $.10; thus, the number of quarters times .25 plus the number of dimes times .10 must equal her total, or $1.45. (Again, turn into easier problem: if you have 4 quarters, you have .25 times 4 = $1.00 total). Briley has 3 quarters, and 7 dimes. Let’s see if it works: 3 quarters is $.75 and 7 dimes is $.70. If we add $.75 and $.70, we get $1.45.    (surd )

We could have also done this problem (and many problems like these) with a table:

	Amount	Price	Total
Quarters	Q	.25	.25 Q	Multiply across
Dimes	10 – Q	.10	.10 (10 – Q)	Multiply across
Total	10		1.45	Do Nothing Here
	Add Down	Do Nothing Here	Add Down: (.25Q+.10left( {10-Q} right)=1.45);  then solve to get (Q=3,,,10-Q=7).

Mixture Word Problem:

Mixture Word Problems are where different quantities are mixed together, and you are asked to find certain quantities. (They are also solved with multiple variables with Systems of Equations). Here is an example using one variable:

Problem: One kind of candy (jelly) sells for $5 a pound and another (chocolate) for $10 a pound. How many pounds of each should be used to make a mixture of 12 pounds of candy (both kinds) that sells for $80?   
Solution: First define a variable, and it helps to use a table. Let (J=) the number of pounds of jelly candy that is used in the mixture. Then (12-J) equals the number of pounds of the chocolate candy.

	Amount	Price	Total
Jelly Candy	J	5	5J	Multiply across
Chocolate Candy	12 – J	10	10 (12 – J)	Multiply across
Total	12		80	Do Nothing Here
	Add Down	Do Nothing Here	Add Down: (5J+10left( {12-J} right)=80); then solve to get (J=8).

Here’s whole problem:

Mixture Word Problem	Notes
One kind of candy (jelly) sells for $5 a pound and another (chocolate) for $10 a pound. How many pounds of each should be used to make a mixture of 12 pounds of candy (both kinds) that sells for $80?	From the table above: (begin{align}5J+10(12-J)&=80\5J+120-10J&=80\-5J&=-40\J&=8end{align}) We would need 8 pounds of the jelly candy and (10-8=2) pounds of the chocolate candy.

Percent Mixture Word Problem:

Problem: A 20% concentrate is to be mixed with a mixture having a concentration of 60% to obtain 80 liters of a mixture with a concentration of 30%. How much of the 20% concentrate and the 60% concentrate will be needed?   

Solution: First define a variable, and it helps to use a table. Let (T=) the number of liters we need from the 20% concentrate, and then (80-T) will be the number of liters from the 60% concentrate. (Put in real numbers to check this).

Put this in a chart again – it’s not too bad. This one is a little more difficult since we have to multiply across for the Total row, too, since we want a 30% solution of the total.

	Amount	% Concentrate	Total
20% Concentrate	T	.20	.2T	Multiply across
60% Concentrate	80 – T	.60	.60 (80 – T)	Multiply across
Total (What we want)	80	.30	24	Multiply across
	Add Down	Do Nothing Here	Add Down: (.2T+.6left( {80-T} right)=24);  then solve to get (t=60).

Here’s the whole problem:

Percent Mixture Word Problem	Notes
A 20% concentrate is to be mixed with a mixture having a concentration of 60% to obtain 80 liters of a mixture with a concentration of 30%. How much of the 20% concentrate and the 60% concentrate will be needed?	From the table above: (begin{align}.2T+.6left( {80-T} right)&=24\.2T+48-.6T&=24\-.4T&=-24\T&=60end{align}) We would need 60 liters of the 20% solutions and (80-60=20) liters of the 60% solution.

Remember that if the problem calls for a pure solution or concentrate, use 100% (if the percentage is that solution) or 0% (if the percentage is another solution). Here’s an example:

Problem: How many ounces of pure water must be added to 100 ounces of a 30% saline solution to make it a 10% saline solution?

Solution:

	Amount	% Saline	Total
Pure Water	W	0% = 0	0W = 0	Multiply across
30% Saline (Salt) Solution	100	.30	30	Multiply across
10% Saline Solution (What we want)	W + 100	.10	.1(W + 100)	Multiply across
	Add Down	Do Nothing Here	Add Down: (0+30=.1left( {W+100} right);,,W=200).

Don’t worry if you don’t totally get these; as you do more, they’ll get easier. We’ll do more of these when we get to the Systems of Linear Equations and Word Problems topics.

Rate/Distance Word Problem:

Distance problems have to do with how far, how fast and how long objects have travelled. Here is an example:

Distance Word Problem/Math

Notes

A train and a car start at the same place. The train is going 40 miles per hour and a car is going in the opposite direction at 60 miles per hour. How long will it be until they are 100 miles apart? (text{Distance},,=,,text{Rate},,times ,,text{Time}) Total Distance:  (100=60t+40t) Solve: (begin{array}{l}100=60t+40t\100=100t;,,,,t=1end{array})

Always draw pictures! Remember always that (text{Distance}=text{Rate},times ,text{Time}). The rates of the train and car are 40 and 60, respectively. Usually a rate is “something per something”. Let (t) equal the how long (in hours) it will be until the train and the car are 100 miles apart. We must figure the distance of the train and car separately, and then we can add distances together to get 100. Again, you can always add distances; look at them separately first, and then you can put them together to equal the total distance (100). The math was pretty easy on this one! In one hour, the train and the car will be 100 miles apart.

Note that there’s an example of a System Equations Distance Problem here, and a Parametric Distance Problem here in the Parametric Equations section.

Profit Word Problem

Profit word problems have to do with the amount of money made for products or services in business. Here is an example:

Problem: Hannah paid $1.50 each for programs to her play. She sold all but 20 of them for $3 each and made a profit of $15 total. How many programs did Hannah buy? How many did she sell? Hint:  Profit = Selling Price – Purchase Price

Solution: It’s easiest to a table to store the information. Let (x=) the number of programs that Hannah bought. Put in real numbers to see how we’d get the number that she sold: if she bought 100 programs and sold all but 20 of them, she would have sold 80 of them. We can see that (80=100-20), so the number sold would be (x-20).

	Number of Programs	Price	Total
Sold	x – 20	3.00	3(x – 20)	Multiply across
Bought	x	1.50	1.5x	Multiply across
Profit			15	Total Profit Given
	Do Nothing Here	Do Nothing Here	Subtract Down: To get profit, subtract Total Bought from Total Sold, and set to Profit (15): (3left( {x-20} right)-1.5x=15); solve to get (x=50).

Here’s the whole problem:

Profit Word Problem	Notes
Hannah paid $1.50 each for programs to her play. She sold all but 20 of them for $3 each and made a profit of $15 total. How many programs did Hannah buy? How many did she sell?	From the table above: (begin{align}3left( {x-20} right)-1.5x&=15\3x-60-1.5x&=15\1.5x&=75\x&=50end{align}) Hannah bought 50 programs to make a total profit of $15. Since she sold 20 less than she bought, she sold 50 – 20 = 30 programs.

Converting Repeating Decimal to Fraction Word Problem:

Problem: Convert (.4overline{{25}},,,(.4252525…)) to a fraction.

Solution: Converting repeating decimal to fraction problems can be easily solved with a little trick; we have to set it up as a subtraction, so the repeating part of the decimal is gone. To do this, let (x=) the repeating fraction, and then figure out ways to multiply (x) by 10, 100, and so on (multiples of 10) so we can subtract two numbers and eliminate the repeating part.
The rule of thumb is to multiply the repeating decimal by a multiple of 10 so we get the repeating digit(s) just to the left of the decimal point, and then multiply the repeating digit again by a multiple of 10 so we get repeating digit(s) just to the right of the decimal. Then subtract the two equations that we just created, and solve for (x):

Repeating Decimal Problem/Math

Notes

Convert (.4overline{{25}},,,(.4252525…)) to a fraction. (begin{array}{l},,,,,,,,,,,x=4.2525252…\text{(original repeating decimal)}\\,,,,,,,1000x=425.252525…\,,,,,,,,underline{{,,,,10x=,,,,,,4.252525…}}\,,,,,,,,,990x=421.0\,,,,,,,,,,,,,,,,,x=frac{{421}}{{990}}end{array})

First, let (x=.4overline{{25}}). Since (.4overline{{25}}) has repeating digits 25, we first have to figure out how to multiply (.4overline{{25}}) to get the repeating digits just to the left of the decimal place. We can do this by multiplying (.4overline{{25}}) by 1000; we get (1000x=4underline{{25}}.252525…) Also, if we multiply (x) by 10, we get the repeating part (25) just to the right of the decimal point; we get (10x=4.underline{{25}}2525…). Now line up and subtract the two equations on the left and solve for (x); we get (displaystyle x=frac{{421}}{{990}}). Pretty cool! Let’s see if it works: Put (displaystyle frac{{421}}{{990}}) in your graphing calculator, and then hit Enter; you should something like .4252525253.  √

Inequality Word Problems

Note that inequalities are very common in real-world situation, since we commonly hear expressions like “is less than” ((<)), “is more than” ((>)),“is no more than” ((le )), “is at least” ((ge )), and “is at most” ((le )). There are several inequality word problems the Solving Inequalities sections; here’s an example:

Problem: Erica must tutor at least 12 hour per week in order to be eligible for her work-study program at her university. She must also study 10 more hours than the time she’s tutoring, so she can keep up her grades in the program. What is the minimum number of hours Erica must study in order to be eligible for her work-study program?

Solution: First, define a variable (h), which is the number of hours that Erica must study (look at what the problem is asking). We know from above that “at least” can be translated to “(ge)”.

If Erica works, let’s say, 30 hours in her work study program, she’d have to study 40 hours (it’s easier to put in real numbers). So, the amount of time she works in her work study program would be “(h-10)”, and this number must be at least 12. Let’s set up the equation and solve:

Inequality Word Problem/Math

Notes

Erica must tutor at least 12 hour per week in order to be eligible for her work-study program at her university. She must also study 10 more hours than the time she’s tutoring, so she can keep up her grades in the program. What is the minimum number of hours Erica must study in order to be eligible for her work-study program? (begin{array}{l}h-10ge 12\underline{{,,,,,,+10ge ,,,+10}}\h,,,,,,,,,,,,ge 22end{array})

The algebra is simple, and we don’t have to worry about changing the sign, since we’re not multiplying or dividing by a negative number. Erica would have to tutor at least 22 hours. Notice that 22 hours works, since the problems asked for “at least”. To check your answer, try numbers right around the answer, like 21 hours (which wouldn’t be enough), and 22 hours (which would work!)

This inequality word problem is a little more advanced:

Problem: (displaystyle frac{4}{5}) of a number is less than 2 less than the same number. Solve the inequality and graph the results.

Solution: This is a little tricky since we have two different meanings of the words “less than”. The words “is less than” means we should use “(<)” in the problem; it’s an inequality. The words “2 less than the same number” means “(x-2)” (try it with “real” numbers).

Inequality Word Problem/Math

Notes

(displaystyle frac{4}{5}) of a number is less than 2 less than the same number. Solve the inequality and graph the results. (displaystyle begin{array}{l},,,,,,,,,,,,,,frac{4}{5}x,,,,,,,,<,,,,,,,x-2\,,,,,,,,,,,,,,,,underline{{,,,,,,-x<-x,,,,}}\,,,,,,,,,,frac{4}{5}x-x,,<,,,,,,,,-2\,,,,,,,,,,,,,,,,-frac{1}{5}x,,<,,-2\left( {-frac{1}{5}x} right)left( {-5} right),>left( {-2} right)left( {-5} right)\,,,,,,,,,,,,,,,,,,,,,,,,,,x>10,,,,,text{(watch sign!)}end{array})

Set up and solve inequalities like we do regular equations. Let “(x)” be the number, and translate the problem word-for-word: (displaystyle frac{4}{5}x<x-2). Get all the “(x)”s to one side and all the “numbers” to the other sign. (We could have also multiplied both sides by 5 earlier to get rid of the fraction.) Remember to change the sign when we multiply both sides by –5, since we’re multiplying by a negative number. Once we get the answer, we can also graph the solution. Try numbers close to 10, like 9 and 11, to make sure it works.

Rational Inequality Word Problem

Technically, this next problem contains a rational function, but it’s relatively easy to solve.

Inequality Word Problem

Math/Notes

A school group wants to rent part of a bowling alley to have a party. The bowling alley costs $500 to rent, plus an additional charge of $5 per student to bowl. The group doesn’t want any student to pay more than $15 total to attend this party. What is an inequality that could represent this situation? How many students would need to attend so each student would pay at most $15?

Since there is a one-time cost in addition to a per-person cost, the cost per person will depend on the number of students attending the party: the more students, the lower the cost. For (x) students attending, each would have to pay (displaystyle frac{{500}}{x}) for the bowling alley rent; try it with real numbers! In addition, each student needs to pay their $5 to bowl. Therefore, each student will need to pay (displaystyle frac{{500}}{x}+5), and since we don’t want any student to pay more than $15, the inequality that represents this situation is (displaystyle frac{{500}}{x}+5le 15). To see how many students would have to attend to keep the cost at $15 per person, solve for (x). In this example, we can multiply both sides by (x) without worrying about reversing the inequality sign, since (x) can only be positive: (displaystyle frac{{500}}{x}+5le 15;,,,frac{{500}}{x}le 10;,,,500le 10x;,,,xge 50). At least 50 students would have to attend so that each student pays no more than $15.

Integer Function Problem (a little bit more advanced…)

Problem: The fee for hiring a tour guide to explore Italy is $1000. One guide can only take 10 tourists and additional tour guides may be hired if needed. What is the cost of hiring tour guides, as a function of the number of tourists who go on the tour? If there are 72 tourists, what is the cost of hiring guides?

Solution: Let’s think about this by using some real numbers. From 1–10 tourists the fee is (1times 1000=$1000), for 11–20 tourists, the fee is (2times 2000=$2000), and so on. Do you see how if we divide the number of tourists by 10, and go up to the next integer, we’ll get the number of tour guides we need? This is because any fraction of a set of ten tourists requires another tour guide.

To get the function we need, we can use the Least Integer Function, or Ceiling Function, which gives the least integer greater than or equal to a number (think of this as rounding up to the closest integer). The integer function is designated by (y=leftlceil x rightrceil ). (We saw a graph of a similar function, the Greatest Integer Function, in the  Parent Functions and Transformations section.)

Thus, the cost of hiring tour guides is (displaystyle 1000times leftlceil {frac{x}{{10}}} rightrceil ). For 72 tourists, the cost is (displaystyle 1000times leftlceil {frac{{72}}{{10}}} rightrceil =1000times leftlceil {7.2} rightrceil =1000times 8=$8000). Makes sense!

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Now, all these types of problems can get much more difficult (and we will see later how to use two variables with some of them), but it’s important to take baby steps with them. Don’t worry if they seem difficult at this time, but as long as you get the general idea of how we’re doing the translations, you’re in great shape! And don’t forget:

• When assigning variables (letters), look at what the problem is asking. You’ll typically find what the variables should be there.
• If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!