Strategies for Word Problems

1) Familiarize.

Read through the problem more than once. Don’t immediately worry about how you’re going to solve the problem. Begin by clearly establishing what the problem is talking about. What is the question asking you? How is that related to what you know? What formulas relate the things you know to the things you don’t know?

2) Translate.

Write what you know in mathematical terms. Assign variables to the quantities you’ll be working with. Write equations that represent the relationships between them.

3) Solve.

Find values that satisfy the equations you’ve written. You will almost always use the same techniques to solve the problem as the exercises that preceded it. If you’ve just learned how to work out problems with two variables, the word problems you get will probably involve two variables.

4) Check.

Make sure your answer works. Does it make sense? Does it sound reasonable? Try putting your answer into the equations you’ve come up with. Are they true?

5) Answer.

Write your answer as a complete sentence. Make sure it answers the original question.

Example Word Problems

Example 1

The area of a rectangle is 18 square centimeters. If the length is 9 centimeters, what is the width?

Familiarize. Read over the problem until you’re comfortable with it. This problem asks about the area of a rectangle, which should remind you that a rectangle’s area is related to its width and length by the formula

A = L×W

This tells us that the area of a rectangle is the product of its length and width. If we know any two of these quantities, we can find the remaining one.

Translate. Take the information the problem gives you, and put it into a form that you can work with. Substitute the given area (18) for A, and the given length (9) for L.

18 = 9×W

Solve. Now that you’ve set up the problem, solve it for the given quantity. In this case, we want to isolate W. To do that, we do the opposite operation. W has been multiplied by nine, so the opposite operation is division by nine.

18 ÷ 9 = (9×W) ÷ 9

2 = W

Check. We’ve found that W is equal to 2. Does this make sense in the original problem? Try putting it into the original context. If a rectangle has a length of 9 centimeters, and a width of 2 centimeters, does it have an area of 18 square centimeters?

Answer. State your answer clearly:

The width of the rectangle is two centimeters.

Example 2

Bob’s Bunnies sells fluffy bunnies for $3.00 a piece. Decrepit, fluff-less bunnies are sold for $2.00. In one day Bob sells 11 bunnies. If Bob made $27.00 that day, how many of each type of bunny did he sell?

1.) As stated above, read through the problem several times. Don’t try to solve it all at once. As soon as you stop freaking out, move on to step 2.

2.) To begin this step, we need to define our variables. In this case, since the problem asks “how many of each type of bunny” your variables are going to be the number of fluffy bunnies (I’ll call this f) and the number of decrepit, fluff-less bunnies (I’ll call this d).

From there we would try to find at least two true equations, since we have two variables. The first equation is easy, and it is derived from the sentence, “in one day Bob sells 11 bunnies.” We know then that the amount of fluffy bunnies, f, and the amount of decrepit bunnies, d, must add up to 11. Therefore;

f + d = 11

The next equation takes more work. We have to think about the cost of each type of bunny. The cost of the fluffy bunnies is going to be the number of fluffy bunnies times the cost per fluffy bunny, or f*3. The cost of the decrepit bunnies similarly is going to be the number of decrepit bunnies times the cost per decrepit bunny, or d*2. So, the total cost of all the bunnies sold was f*3 + d*2. We also know that the total cost was $27.00. So since the cost is both f*3 + d*2 and $27.00 then f*3 + d*2=27.

Our two true equations then are f + d = 11 and f*3 + d*2=27.

3.) To solve this system of equations, we use substitution. We start with the simpler equation

f + d =11.

f + d =11

-d -d

f = 11 – d

Next, substitute 11 – d for f in the other equation.

f*3 + d*2=27

(11 – d)*3 + d*2=27

33 – 3 *d + 2*d = 27

33 – d = 27

-33 = -33

33 – d – 33 = -6

-d = -6

d = 6

We get a solution of d = 6. d, remember is the number of decrepit bunnies sold. So we have half of our answer. The other half will come from substituting 6 in for d in the first, easy equation.

f + d =11

f + 6 =11

-6 = -6

f = 5

We get the last piece of the puzzle; f = 5. This says that the number of fluffy bunnies sold was 5. But before we close our book and turn in the homework, we’d better do step 4; check our answer.

4.) Checking the answer is easy, even with a system of two equations like this. We simply need to show that using the two values that we got for our variables, d and f, makes both equations true.

f + d =11

5 + 6 = 11

11 = 11

The first equation is true. This is a good sign

f*3 + d*2=27

5*3 + 6*2 = 27

15+12 = 27

27 = 27

Both equations are true. We’re almost done!

5.) Finally, we simply write out the answer in words stating what we found to be true.

5 fluffy bunnies were sold and 6 decrepit fluff-less bunnies were sold.