Word Problems

Word Problems

About 35% of your total questions on the Math Section will be made up of word problems. Though the actual math topics can vary, it is important to develop a consistent process for answering them.

Almost any word problem can be broken down into four steps:

  1. Identify what value the question is asking for. We’ll call this the desired value
  2. Identify unknown values and label them with variables
  3. Identify relationships and translate them into equations
  4. Use the equations to solve for the desired value

In essence, you need to turn a word problem into a system of equations and use those equations to solve for the desired value.

Example: A candy company sells premium chocolate at $5 per pound and regular chocolates at $4 per pound. If Barrett buys a 7-pound box of chocolates that costs him $31, how many pounds of premium chocolates are in the box?

Step 1. Identify the desired value.
The question asks for the number of pounds of premium chocolate that Barrett bought.

Step 2. Identify unknown values and label them with variables.
In this question, there are two basic unknown values: the number of pounds of premium chocolate and the number of pounds of regular chocolate.

Step 3. Identify relationships and translate them into equations.
One fairly straightforward relationship concerns the total number of pounds of chocolate. Barrett bought a 7-pound box of chocolate. If the box contains only regular and premium chocolate, then you can write the following equation:

r + p = 7 ................................... Equation 1

The other relationship concerns the total cost of the box. The total cost of the box is equal to the cost of the premium chocolates plus the cost of the regular chocolates. We are given the following:

Total Cost of Box = $31
Cost of Premiums = (5 $/pound) x (p pounds) = 5p
Cost of Regulars = (4 $/pound) x (r pounds) = 4r
31 = 5p + 4r ................................... Equation 2

Step 4. Use the equations to solve for the desired value.
These are the two equations you have to work with:
r + p = 7
31 = 5p + 4r

Solving simultaneously, we get 3 = p.

The ACT has many ways of making various stages of a word problem more difficult, which is why it is so important to have a good process. Train yourself to use these four steps to stay on track and continually work toward a solution.

In order to translate your ACT word problems into actionable math equations you can solve, you’ll need to understand and know how to utilize some key math terms. Whenever you see these words, you can translate them into the proper mathematical action.

Here is a chart with ALL THE KEY TERMS and symbols you should know for word problems.

Solved Example 1
A food truck sells salad for $6.50 each and drinks for $2.00 each. The food truck’s revenue from selling a total of 209 salads and drinks in one day was $836.50. How many salads were sold that day?
A. 77
B. 93
C. 99
D. 105

Solution: We'll say that the number of salads sold = S, and the number of drinks sold = D. The two equations we can create from the information in the question is

S + D = 209
6.50S + 2D = 836.5

Now, just subtract S from both sides to get what D equals:

D = 209 − S ................................Equation 1
6.50S + 2(209 − S) = 836.5 ................................Equation 2

6.50S + 418 − 2S = 836.5
6.50S − 2S = 418.5
4.5S = 418.5
S = 93

The correct answer is B.

Replace Variables with Numbers

Which of the following QUESTIONS is easier to answer?

Most likely, you found the question on the right easier. In general, numbers are easier to work with than variables. Notice that the questions are actually identical, except for the fact that x has been replaced by 40, and y has been replaced by 3. Problems that have variables in the answer choices can almost always be answered by replacing variables with numbers.

There are three basic steps to remember when replacing a variable with numbers:

  1. Identify unknowns and replace them with numbers. In the above example, you replaced X with 40 and replaced Y with 3.
  2. Use these numbers to calculate the answer to the problem. A car traveling 40 miles per hour for 3 hours travels 120 miles. Remember the number you calculate in this step. You should refer to this as the target.
  3. Plug the same numbers into the answer choices. The correct answer will be equal to the target number. Notice that the answers to the problem on the right are in the same form as the answers to the problem on the left. All xs have been replaced with 40s and all Ys have been replaced with 3s.

The correct answer is C because it equals 120, which is the same number we calculated in Step 2. This technique can also be used with more complicated word problems.

Solved Example 2 
A park ranger travels from his base to a campsite via truck at r miles per hour. Upon arriving, he collects a snowmobile and uses it to return to base. If the campsite is d miles from the park ranger's base and the entire trip took t hours to complete, what was his speed on the snowmobile, in terms of t, d, and r?
tr - d
B. td - r
C. \(\frac{dr}{rt-d}\)
D. \(\frac{td-r}{d}\)

Solution: The key to replacing variables with numbers is to pick numbers that make calculation easy. You should pick the rate and the distance so that the time travelled is an integer. You also want to pick numbers that are small enough that they will be easy to plug in to the answer choices.

Say that r = 10 and d = 20. That way it takes 2 hours for the ranger to reach the camp. To keep things easy, you can say that the whole trip took 4 hours. That would mean that the ranger traveled 10 miles per hour on the snowmobile as well.

You’ve picked numbers for your variables and calculated a target: r = 10 d = 20 t= 4
Target = 10

Now you need to figure out which answer choice matches the target:
A. tr - d = (4)(10) - (20) = 20
B. td - r = (4)(20) - (10)
C. \(\frac{dr}{rt-d}=\frac{(20)(10)}{(10)(4)-(20)}=\frac{200}{20}=10\)
D. \(\frac{td-r}{d}=\frac{(4)(20)-(10)}{(20)}=\frac{70}{20}=35\)

The correct answer is C.


The standard way to solve word problems is to translate the words into equations and solve for the desired value. Some word problems, however, contain equations that make the desired value difficult to solve for. Instead, you can test answer choices. This technique is called BACKSOLVING.

Solved Example 3 
Preeti has money in two separate bank accounts. Account X earns 8% interest annually and account Y earns 15% interest annually. Preeti earned a total of $53 in interest last year. If the total amount of money in the accounts at the beginning of last year was $400, and there were no other deposits or withdrawals, how much money was in account Y?
B. $200
C. $250
D. $300

Solution: Let x be the total amount of money in account X and let y be the total amount of money in account Y. The question provides you with enough information to create two equations:

x + y = 400 0.08x + 0.15y = 53

You need to solve for y, and with backsolving, you will start by assuming the value of y. If the correct answer were B, y would be 200. You can use the first equation to find the value of x. Then, you will plug x and y into the second equation. If it is true, you know you have the correct answer.

If y = 200, x = 200 (because x + y = 400). Plug x = 200 and y = 200 into the second equation:
0.08(200) + 0.15(200) = 53
16 + 30 = 53
46 ≠ 53

Answer choice B is incorrect. Backsolving typically works best when you test answer choice B or C first, as it will allow you to eliminate two answer choices (if B is wrong). If, in contrast, you tested answer choice A first and it was wrong, you would only be able to eliminate A.

Now that you know the answer is either C or D, you need to do the math only one more time. If the answer choice you choose to test works, you know that’s the answer. If it doesn’t work, you know the other answer choice must be correct.

Looking at the answer choices, it actually looks like D would be easier to test. The numbers 300 and 100 will be easier to use than 250 and 150. If you assume that y = 300, then x must equal 100. Plug these values into the second equation:
0.08(100) + 0.15(300) = 53
8 + 45 = 53

The equation is true, so answer choice D is correct.

Backsolving can be a time saving technique because, at most, you will need to test two answer choices. Test answer choice B first to really save time. Although back solving does not allow you to avoid algebra entirely, it can be useful when it seems that solving for the desired value will involve messy equations.

Use Charts to Organize Variables

When a word problem involves several quantities and multiple relationships, a chart or table can be used to organize the information effectively.

Example: A circus earned $150,000 in ticket revenue by selling 1,800 V.I.P. and Standard tickets. They sold 25% more Standard tickets than V.I.P. tickets. If the revenue from Standard tickets represents one-third of the total ticket revenue, what is the price of a V.I.P. ticket?

In this problem, there are several unknowns: the number of Standard and V.I.P tickets, the cost per ticket for each type of ticket, revenue generated, etc. Trying to create a variable for each of these values could be quite tedious and confusing.

One way to deal with all these unknowns is to create a table.

The relationship: (Price) x (Quantity) = (Revenue) is relevant for both Standard and V.I.P. tickets.

Notice that you’ve used this strategy before; charts are useful on rate problems and overlapping sets problems. You have a similar advantage here: the chart allows you to keep track of a number of different relationships all at the same time.

Now you can enter information into the chart. You know that the total revenue is $150,000, and the total number of tickets sold is 1,800. You also know that the circus sold 25% more Standard tickets than V.I.P. tickets. If you let * equal the number of V.I.P. tickets sold, then the number of Standard tickets sold is 1.25x:

The revenue from Standard tickets is one-third of the total revenue, which is $150,000. So the revenue from Standard tickets is $50,000 and the revenue from V.I.P. tickets is $100,000.

Additionally, you now have an equation for the number of tickets: 1.25x + x = 1,800
x = 800

800 V.I.P. tickets were sold, and 1,000 Standard tickets were sold:

Now, at last, you can solve for the price of a V.I.P. ticket. Price per ticket times the number of tickets equals the revenue:
p x 800 = 100,000
p = 125

The price of a V.I.P. ticket is $125.

A chart was useful for this problem because of the number of relationships present in the argument. Instead of writing down a lot of equations, you could organize the information in a chart.


In general optimization problems, you are asked to maximize or minimize some quantity, given constraints on other quantities. These quantities are all related through an equation.

Example: The guests at a football banquet consumed a total of 401 pounds of food. If no individual guest consumed more than 2.5 pounds of food, what is the minimum number of guests that could have attended the banquet?

You can visualize the underlying equation in the following table:

Notice that finding the minimum value of the number of guests involves using the maximum pounds of food per guest because the two quantities multiply to a constant. This sort of inversion is typical.

Begin by considering the extreme case in which each guest eats as much food as possible, or 2.5 pounds apiece. The corresponding number of guests at the banquet works out to 401/2.5 = 160.4 people.

However, you obviously cannot have a fractional number of guests at the banquet. Thus, the answer must be rounded. To determine whether to round up or down, consider the explicit constraint: the amount of food per guest is a maximum of 2.5 pounds per guest. Therefore, the minimum number of guests is 160.4 (if guests could be fractional), and you must round up to make the number of guests an integer: 161.

Note the careful reasoning required! Although the phrase “minimum number of guests” may tempt you to round down, you will get an incorrect answer if you do so.


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