Algebra Word Problems

This section covers:

Note that Using Systems to Solve Algebra Word Problems can be found here in the Systems of Linear Equations section.

Now that you can do these difficult algebra problems, you can trick your friends by doing some fancy word problems; these are a lot of fun.  The problems here only involve one variable; later we’ll work on some that involve more than one.

English to Math Translation

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:

Words to Math Table

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.

Note that most of these word problems can also be solved with Algebraic Linear Systems, here in the Systems of Linear Equations section.

Now let’s do some problems that use some of the translations above.  We’ll get to more difficult algebra word problems later.  Trick your friends with these problems!

Unit Rate Problem:

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


To get the unit rate, we want the amount for one pound of apples; this is when “x” (apples) is 1.  So we can set up a ratio:  \(\frac{5}{{3.75}}=\frac{1}{x};\,\,\,x=\$.75\).

“Find the Numbers” Word Problems:

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


Ok, so 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!)  But I knew the sum of the two numbers had to be 18, so do you see how you’d take 10 and subtract it from 18 to get the other number?   See how much easier it is to think of real numbers, instead of variables when you’re coming up with the expressions?

We don’t need to worry about “n” being the smaller number (instead of “18 – n“); the problem will just work out this way!

So let’s translate the English into math:

Smaller Larger Number Word Problem

Another 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?


Ok, so we always have to define a variable, and we can look at what they are asking.  The problem is asking for a number, so let’s make that n. Now let’s try to 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.  So 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 need to subtract 3.   Also, “33 less than 133” is 100, so for the “33 less than”, we need to subtract 33 at the end:

Another Numbers Word Problem

Percent Word Problem:

60 is 20% of what number?  (We saw similar problems in the Percents, Ratios, and Proportions section!)


The translation is pretty straight forward; note that we had to turn 20% into a decimal. (Remember: we need to get rid of the % – we’re afraid of it – so we move the decimal 2 places away from it).

Percent Word Problem

Percent Increase Word Problem:

The price of a pair of (cheap) shoes has increased by 15%.  The original price of the shoes was $20.  What is the new price?


Percent Increase Word Problem

Ratio/Proportion Word Problems

Relating Two Things Together: a Rate

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.


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 that we’ll learn about in the Coordinate System and Graphing Lines including Inequalities section. Here’s the math:

Ratio Rate Problem

Ratio/Proportion Problem:

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


This is a ratio problem; we learned about ratios in the Percents, Ratios, and Proportions section.  A ratio is a comparison of two numbers; a ratio of 5 to 2 (also written 5 : 2 or ) means you have 5 boys for every 2 girls in your class.  So 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 see how we can 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 here earlier.  Let x = the number of boys in the class.

Proportion Word Problem

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

Here’s a ratio problem that’s pretty tricky; we have to do it in a lot of steps: Problem: 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: Let’s work backwards on this problem, and first work with solution Z, since we know there are 1260 ounces of it.   It’s good to start with the parts of the problems with numbers first!

Difficult Ratio Problem

Weighted Average Word 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 in the class for the semester?


Ok, we always have to define a variable, and we can look at what they are asking.

So let x = what you need to make on the final.  So 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.

Let’s use the equation for an average:

Weighted Average Word Problem

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 (say, an 80) counting 40% of your grade, and test 3 (say, a 78) counting 40% of your grade, you will 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).  So when using decimals, your denominator should be 1:

Consecutive Integer Word 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?


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.

Consecutive Number Word Problem

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

Age Word 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?


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

So let M = the age of sister Molly now.  Then 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), and 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):

Age Word Problem

Money (Coins) Word Problem:

Suppose Briley has 10 coins in quarters and dimes and has a total of $1.45. How many of each coin does she have?


So we always have to define a variable, and we can look at what they are asking.

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).

Since she has a total of $1.45, and each quarter is worth $.25 and each dime is worth $.10, then 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).

Money Word Problem

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

Mixture Word 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 10 pounds of candy (both kinds) that sells for $80?


Let’s first define a variable, and use another table like we did before.  Let J = the number of pounds of jelly candy that is used in the mixture.   Then 10 – J equals the number of pounds of the chocolate candy.

Mixture Word Problem

Percent Mixture Word 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?


Let’s first define a variable, and use another table like we did before.  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).

Let’s put this in a chart again – it’s not too bad:

Another Mixture Word Problem

This one was a little more difficult since we had to multiply across for the Total row, too, since we wanted a 30% solution of the total.  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.  Also, remember that if the problem calls for a pure solution or concentrate, use 100%.

Rate/Distance Word Problem:

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?


Remember always that distance = rate x time.  Let’s 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.

It’s always good to draw pictures for these types of problems:

Distance Problem

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

Profit Word 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


Let’s make a table to store the information.  Let  x = the number of programs that Hannah bought.  Let’s 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 is 100 – 20, so the number sold would be x – 20. Profit Word Problem

Converting Repeating Decimal to Fraction Word Problem:

Convert  (.4252525…) to a fraction.


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 we’ll 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 we subtract the two equations that we just created, and solve for x:

Repeating Decimal to Fraction Problem

Inequality Word Problem:

4/5 of a number is less than 2 less than the same number.  Solve the inequality and graph the results.


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).

Again, for an inequality, we’ll set up and solve inequalities like we do regular equations.  So let “x” be the number, and translate the problem word-for-word: .  Now solve:

Inequality Word Problem

Another Inequality Word 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 what’s 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?


First we define a variable h, which will be 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 “”.

So 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”.  This number must be at least 12.  So let’s set up the equation and solve:

Inequality Word Problem Two

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!

Learn these rules, and practice, practice, practice!

On to Coordinate Systems and Graphing Lines including Inequalities – you are ready!

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