Word problems can feel like puzzles written in a foreign language, but they’re actually one of the most practical skills you’ll use on the ACT Math section—and in real life. The key to conquering them isn’t memorizing formulas; it’s learning to translate everyday language into the precise language of algebra. Once you master this translation skill, word problems transform from intimidating obstacles into straightforward point-earning opportunities. Let’s break down exactly how to make that translation happen, step by step.

This topic appears in 5-10 questions on the ACT Math section. Understanding it thoroughly can add 2-4 points to your composite score. Let’s break it down with proven strategies that work!

⚡ Quick Answer: The 5-Step Translation Method

Example: “Sarah has three times as many books as Tom. If Sarah has 24 books, how many books does Tom have?”

What we’re solving for: The number of books Tom has

For our example: Let $$x$$ = the number of books Tom has

Continuing our example:
“Sarah has three times as many books as Tom” translates to:
Sarah’s books = $$3 \times$$ Tom’s books

We know Sarah has 24 books, and Tom has $$x$$ books, so:
$$24 = 3x$$

Solution:
$$24 = 3x$$
$$\frac{24}{3} = x$$
$$x = 8$$

Verification: If Tom has 8 books, then Sarah has $$3 \times 8 = 24$$ books. ✓ This matches the problem!

Problem Statement

Jessica is 4 years older than her brother Mike. The sum of their ages is 28. How old is Mike?

Solution Process

Step 1: Identify what we’re solving for
We need to find Mike’s age.

Step 2: Assign variables
Let $$x$$ = Mike’s age
Then Jessica’s age = $$x + 4$$ (since she’s 4 years older)

Step 3: Translate to equation
“The sum of their ages is 28” means:
Mike’s age + Jessica’s age = 28
$$x + (x + 4) = 28$$

Step 4: Solve
$$x + x + 4 = 28$$
$$2x + 4 = 28$$
$$2x = 24$$
$$x = 12$$

Step 5: Verify
Mike is 12 years old, Jessica is $$12 + 4 = 16$$ years old
Sum: $$12 + 16 = 28$$ ✓

🎨 Visual Solution Breakdown

Mike's Age:        [====x====]
Jessica's Age:     [====x====][+4]
                   
Combined:          [====x====] + [====x====][+4] = 28
                   
Simplified:        [====2x====][+4] = 28
                   
Remove +4:         [====2x====] = 24
                   
Divide by 2:       [====x====] = 12

Result: Mike = 12, Jessica = 16

Problem Statement

A movie ticket costs $12, and a popcorn costs $6. If Alex spent $54 total and bought 3 popcorns, how many movie tickets did he buy?

Solution Process

Step 1: Identify what we’re solving for
Number of movie tickets Alex bought.

Step 2: Assign variables
Let $$x$$ = number of movie tickets

Step 3: Translate to equation
Cost of tickets + Cost of popcorns = Total spent
$$12x + 6(3) = 54$$

Step 4: Solve
$$12x + 18 = 54$$
$$12x = 36$$
$$x = 3$$

Step 5: Verify
3 tickets at $12 each: $$3 \times 12 = 36$$
3 popcorns at $6 each: $$3 \times 6 = 18$$
Total: $$36 + 18 = 54$$ ✓

Problem Statement

The sum of three consecutive even integers is 78. What is the smallest of these integers?

Solution Process

Step 1: Identify what we’re solving for
The smallest of three consecutive even integers.

Step 2: Assign variables
Let $$x$$ = the smallest even integer
Then $$x + 2$$ = the second even integer
And $$x + 4$$ = the third even integer
(We add 2 each time because consecutive even integers differ by 2)

Step 3: Translate to equation
“The sum of three consecutive even integers is 78”:
$$x + (x + 2) + (x + 4) = 78$$

Step 4: Solve
$$x + x + 2 + x + 4 = 78$$
$$3x + 6 = 78$$
$$3x = 72$$
$$x = 24$$

Step 5: Verify
The three integers are: 24, 26, 28
Sum: $$24 + 26 + 28 = 78$$ ✓
All are even ✓
They are consecutive ✓

❌ Mistake #1: Mixing Up “Less Than” Order

Wrong: “5 less than x” → $$5 – x$$
Correct: “5 less than x” → $$x – 5$$

Why it matters: The phrase “less than” reverses the order. Think of it as “x with 5 taken away.”

❌ Mistake #2: Forgetting to Define All Variables

In problems with multiple unknowns, students often define only one variable and forget to express the others in terms of it.

Example: “John has twice as many apples as Mary”
Don’t just write $$x$$ for John’s apples. Also write: Mary has $$\frac{x}{2}$$ apples (or let $$x$$ be Mary’s and John has $$2x$$).

❌ Mistake #3: Not Verifying Your Answer

You might solve the equation correctly but get the wrong answer to the actual question asked. Always plug your solution back into the original problem to check.

Example: If the problem asks for “the larger number” and you solved for $$x$$ (the smaller number), make sure to calculate and report the larger number, not $$x$$.

❌ Mistake #4: Confusing “Of” with Addition

Wrong: “Half of x” → $$\frac{1}{2} + x$$
Correct: “Half of x” → $$\frac{1}{2} \times x$$ or $$\frac{x}{2}$$

Remember: The word “of” in math almost always means multiplication, especially with fractions and percentages.

🎥 Video Explanation

Watch this detailed video explanation to understand the concept better with visual demonstrations and step-by-step guidance.

Practice Question 1 (Basic)

A number decreased by 7 equals 15. What is the number?

Practice Question 2 (Intermediate)

The length of a rectangle is 3 times its width. If the perimeter is 48 inches, what is the width of the rectangle?

Practice Question 3 (Intermediate)

Maria has $5 more than twice the amount of money that Carlos has. If Maria has $37, how much money does Carlos have?

Practice Question 4 (Advanced)

In a class, there are 8 more girls than boys. If the total number of students is 32, how many boys are in the class?

Practice Question 5 (Advanced)

A store sells notebooks for $3 each and pens for $2 each. If a student bought a total of 15 items and spent $38, how many notebooks did the student buy?

📚 Related ACT Math Resources

• Explore more comprehensive ACT preparation materials
• Practice with our full-length ACT practice tests
• Visit the official ACT website for test dates and registration

🎓 You’ve Got This!

Translating word problems into algebraic equations is a skill that improves with practice. Every problem you solve makes the next one easier. Keep practicing, stay confident, and watch your ACT Math score soar! Remember: the ACT isn’t testing whether you’re “good at math”—it’s testing whether you can recognize patterns and apply strategies. You’ve learned those strategies today. Now go use them!

Addition Keywords: sum, more than, increased by, total, plus, combined, added to

Subtraction Keywords: difference, less than, decreased by, minus, reduced by, fewer than

Multiplication Keywords: product, times, multiplied by, of, twice, double, triple

Division Keywords: quotient, divided by, per, ratio, out of, split

Equality Keywords: is, are, will be, gives, equals, results in, yields

Problem: “Three times a number decreased by 4 equals 11. What is the number?”

Solution Process:

1. Identify the unknown: Let $$x$$ represent the unknown number

2. Translate each phrase:

• “Three times a number” → $$3x$$
• “Decreased by 4” → $$3x – 4$$
• “Equals 11” → $$= 11$$

3. Write the equation: $$3x – 4 = 11$$

4. Solve the equation:

$$3x – 4 = 11$$

$$3x = 15$$ (add 4 to both sides)

$$x = 5$$ (divide both sides by 3)

Answer: The number is 5

Important Note: “Four less than x” means “x minus 4,” NOT “4 minus x.” Test this with real numbers: if someone earns four dollars less per hour than you, and you earn $$p$$ dollars per hour, they earn $$p – 4$$, not $$4 – p$$.

Problem: “Twenty gallons of crude oil were poured into two containers of different sizes. Express the amount poured into the smaller container in terms of the amount $$g$$ poured into the larger container.”

Reasoning:

• Total amount: 20 gallons
• Amount in larger container: $$g$$ gallons
• Amount in smaller container: what’s left over

Solution: The amount left equals the total minus what’s been used:

$$20 – g$$ gallons

Problem 1: A number decreased by 4 equals 10. Find the number.

Solution:

Let $$x$$ = the unknown number

Equation: $$x – 4 = 10$$

Solve: $$x = 14$$

Answer: 14

Problem 2: The product of a number and 5 equals 35. Find the number.

Solution:

Let $$n$$ = the unknown number

Equation: $$5n = 35$$

Solve: $$n = 7$$

Answer: 7

Problem 3: The length of a rectangle is twice its width. If the perimeter is 36 units, find the dimensions.

Solution:

Let $$w$$ = width, then length = $$2w$$

Perimeter formula: $$P = 2l + 2w$$

Equation: $$2(2w) + 2w = 36$$

Simplify: $$6w = 36$$, so $$w = 6$$

Answer: Width = 6 units, Length = 12 units

Problem 4: A father is three times as old as his son. If the sum of their ages is 48 years, find their ages.

Solution:

Let $$s$$ = son’s age, then father’s age = $$3s$$

Equation: $$s + 3s = 48$$

Simplify: $$4s = 48$$, so $$s = 12$$

Answer: Son = 12 years, Father = 36 years

Problem 5: Two numbers differ by 8 and their sum is 48. Find the numbers.

Solution:

Let $$x$$ = smaller number, then larger number = $$x + 8$$

Equation: $$x + (x + 8) = 48$$

Simplify: $$2x + 8 = 48$$, so $$2x = 40$$, thus $$x = 20$$

Answer: The numbers are 20 and 28

Problem 6: The sum of a number and twice another number is 22. If the second number is 3 less than the first number, find the numbers.

Solution:

Let $$x$$ = first number, then second number = $$x – 3$$

Equation: $$x + 2(x – 3) = 22$$

Simplify: $$x + 2x – 6 = 22$$, so $$3x = 28$$, thus $$x = \frac{28}{3}$$ or approximately 9.33

Second number: $$\frac{28}{3} – 3 = \frac{19}{3}$$ or approximately 6.33

Answer: First number = $$\frac{28}{3}$$, Second number = $$\frac{19}{3}$$

Problem 7: A shop sells pencils at $2 each and erasers at $3 each. If a student buys a total of 10 items and spends $24, how many pencils and erasers did the student buy?

Solution:

Let $$p$$ = number of pencils, then erasers = $$10 – p$$

Equation: $$2p + 3(10 – p) = 24$$

Simplify: $$2p + 30 – 3p = 24$$, so $$-p = -6$$, thus $$p = 6$$

Erasers: $$10 – 6 = 4$$

Answer: 6 pencils and 4 erasers

Problem 8: The difference between a number and 7 equals twice the number decreased by 5. Find the number.

Solution:

Let $$x$$ = the unknown number

Equation: $$x – 7 = 2x – 5$$

Solve: $$-7 + 5 = 2x – x$$, so $$-2 = x$$

Answer: -2

Problem 9: The sum of three consecutive integers is 51. Find the integers.

Solution:

Let $$n$$ = first integer, then $$n + 1$$ and $$n + 2$$ are the next two

Equation: $$n + (n + 1) + (n + 2) = 51$$

Simplify: $$3n + 3 = 51$$, so $$3n = 48$$, thus $$n = 16$$

Answer: The integers are 16, 17, and 18

Problem 10: A car rental company charges a flat fee of $30 plus $0.20 per mile driven. If a customer paid $50 for a rental, how many miles did they drive?

Solution:

Let $$m$$ = number of miles driven

Equation: $$30 + 0.20m = 50$$

Solve: $$0.20m = 20$$, so $$m = 100$$

Answer: 100 miles

• Age Problems: Determining people’s ages at different times
• Geometry Problems: Finding dimensions using perimeter, area, and volume formulas
• Coin Problems: Calculating quantities of different coin denominations
• Distance Problems: Using the formula $$d = rt$$ (distance = rate × time)
• Investment Problems: Applying interest formulas $$I = Prt$$
• Mixture Problems: Combining substances with different concentrations or prices
• Number Problems: Finding unknown numbers based on relationships
• Percent Problems: Calculating discounts, increases, and percentages
• Work Problems: Determining completion times when multiple people work together

1. Don’t treat keywords as absolute rules—use them as helpful guides while applying logical thinking
2. Test your translations with real numbers to verify they make sense
3. Write down what your variable represents before setting up equations
4. Pay special attention to order in subtraction and division problems
5. Check your final answer by substituting it back into the original problem
6. Practice explaining your work to others—if you can teach it, you’ve mastered it
7. Draw diagrams when appropriate to visualize the problem
8. Break complex problems into smaller steps rather than attempting everything at once

Final Reminder: The journey to mastering algebraic word problems requires patience and persistence. Keep practicing, stay curious, and always verify your answers. Your problem-solving abilities will improve dramatically with each problem you tackle!