--- title: "CCAT Word Problems: Setups, Shortcuts and Practice" url: https://cognitivecatprep.com/ccat-word-problems updated: 2026-10-06 description: "CCAT word problems with worked, step-by-step answers: work rates, speed, averages, percents, ratios and algebra. Try 8 practice questions free." --- > Cognitive Cat Prep is an independent CCAT practice site. It is not affiliated with Criteria Corp or any employer. CCAT question types · Math · Word problems # CCAT word problems: translate, set up, solve By [Avi Flombaum](https://avi.nyc) · Updated October 6, 2026 Word problems are the slowest math questions on the Criteria Cognitive Aptitude Test (CCAT), the hiring test, because you have to turn a paragraph into an equation before you can do any arithmetic. Six setups cover most of them. Learn those, learn when to estimate, and learn when to skip. ## What counts as a word problem on the CCAT A word problem is any math question where the numbers come wrapped in a story. Two pipes fill a tank. A courier drives out and back. A price goes up, then down. The arithmetic is usually easy. The work is in the translation. In our bank these come from six families: algebra and word equations, work rates, speed and unit rates, averages, percentages, and ratios. They share one skill: turning a sentence into an equation fast. With about 18 seconds per question on a 50-question, 15-minute test, you cannot afford to reread the paragraph three times. ## Translate first: words to an equation Name the unknown once, in writing, before you touch a number. "Let s be the son's age now." Then translate one sentence at a time. | The words say | You write | | --- | --- | | is, was, will be | = | | more than, older than, added to | + | | less than, younger than, fewer | − | | twice, three times, of | × | | per, for every, out of | ÷ or a ratio | Here is the whole method on one problem. Maya is three times as old as her son. In 12 years, she will be twice his age. How old is Maya now? Let s be the son's age now. Maya is 3s. In 12 years, Maya is 3s + 12 and the son is s + 12. "She will be twice his age" becomes 3s + 12 = 2(s + 12). Distribute: 3s + 12 = 2s + 24. Subtract 2s from both sides: s + 12 = 24. Subtract 12: s = 12. Maya is 36. Check it: in 12 years she is 48 and he is 24, and 48 is twice 24. The common miss is adding 12 years to only one person. Write both future ages before you write the equation. ## Six setups to know cold ### Work rates: add rates, not times If A finishes a job in a hours, A does 1/a of it per hour. Working together, the rates add: 1/a + 1/b = 1/t. A takes 6 hours and B takes 3: 1/6 + 1/3 = 1/2, so together they take 2 hours. For two workers there is a shortcut: t = ab ÷ (a + b) = 18 ÷ 9 = 2. When you know the team's time and one worker's time, subtract: the other worker's rate is 1/t − 1/a. ### Distance = rate × time Write the units. Convert minutes to hours before you multiply: 40 minutes is 2/3 of an hour. Two travellers heading toward each other close the gap at the sum of their speeds. For a round trip, average speed is total distance over total time, never the average of the two speeds. Drive 60 miles at 30 mph and back at 60 mph: 2 hours out, 1 hour back, 120 miles in 3 hours is 40 mph, not 45. ### Averages: turn the mean back into a total Sum = mean × count. You averaged 82 on four tests and want 84 across five. You need 5 × 84 = 420 points and have 4 × 82 = 328, so the fifth test needs 92. Faster: the new test must cover the target, plus 2 points for each of the four earlier tests: 84 + 8 = 92. ### Percent change: use multipliers Up 25% is × 1.25. Down 20% is × 0.8. Chain them by multiplying: up 20% then down 20% is 1.2 × 0.8 = 0.96, a 4% drop, not zero. To find the original after an increase, divide: a price that rose 25% to $150 was 150 ÷ 1.25 = $120. ### Ratios: count parts A 3:5 split of 64 has 8 parts, so each part is 8, and the shares are 24 and 40. When one side changes and the ratio shifts, write each side as a multiple of k and solve for k. ### Consecutive integers: start from the middle Five consecutive integers sum to 85. The middle one is 85 ÷ 5 = 17, so they run 15 to 19 and the largest is 19. No equation needed. ## Estimate, then eliminate Most word problems have a range the answer must fall in. Find it before you calculate, and you can often cross out three choices. - **Work together:** the team is faster than its fastest member. If one worker takes 3 hours alone, the combined time is under 3 hours. - **Round-trip speed:** the answer sits between the two speeds and below their plain average. If the plain average is a choice, it is probably the trap. - **Original price after an increase:** smaller than the final price. After a decrease: larger. - **Mixtures:** a blend's percentage lands between the two ingredients, closer to the one you used more of. - **Averages:** adding a number above the mean raises it; below, lowers it. Estimation is not guessing. It is the check you would do at the end, moved to the start, where it saves time. ## What the step-by-step solutions show All 180 algebra questions in our bank, plus a few ratio and overlapping-group questions (184 questions in all), come with a full worked solution, not just a one-line answer. It names the unknown, shows the sentence-by-sentence translation, then solves with one operation per line, each labelled with the rule it uses: "Distribute: a(b + c) = ab + ac", "Subtract the same amount from both sides." A program checks the arithmetic in every step with exact math. You can see them in the practice questions below. In the app, they open automatically when you miss a question, and the AI coach can walk through any step you want explained. ## When to skip a word problem Skipping is a skill. A wrong answer and a blank cost you the same, so the question is only whether 30 more seconds here earns more than 30 seconds somewhere else. Skip, or guess and move on, when: - You have read it twice and still cannot name the unknown. - The setup needs three or more equations. - The numbers are ugly (fractions of fractions, four-digit division) and estimation does not narrow the choices. - It is a work problem with someone joining or leaving partway and you are already behind the clock. Before you skip, eliminate what you can and guess from what's left. ## Eight practice questions with worked answers These come from our bank, at medium and hard difficulty, one or two from each setup above. Pick an answer, then open the explanation. Questions 1, 2, 7 and 8 include the full step-by-step solution. Question 1 Algebra & word equations Harder Lea is 32 years older than Jo. In 8 years, Lea will be twice Jo’s age. How old is Lea now? - A. 61 - B. 55 - C. 58 - D. 57 - E. 56 **Show the answer and the fast route** **Answer: E, 56** Let Jo’s age be j. Then Lea is j + 32. In 8 years: j + 40 = 2(j + 8), so j = 24. Lea is 56. The fast route Name the unknown once; translate the relationship before doing arithmetic. Solve it step by step 1. #### 1 Name the unknown Let j = Jo's age now 2. #### 2 Translate the words into math - “Lea is 32 years older than Jo” `Lea = j + 32` - “In 8 years, Lea will be twice Jo's age” `(j + 32) + 8 = 2(j + 8)` 3. #### 3 Solve the equation, one move at a time Whatever you do to one side, do to the other to keep the two sides equal. 1. Start The equation j + 40 = 2(j + 8) 2. 1 Rule Distribute: a(b + c) = ab + ac Multiply 2 by each term inside the parentheses j + 40 = 2j + 16 3. 2 Rule Subtract the same amount from both sides Subtract j from both sides 40 = j + 16 4. 3 Rule Subtract the same amount from both sides Subtract 16 from both sides 24 = j 5. 4 Rule Flip the sides Write j on the left j = 24 4. #### 4 Answer Jo is 24, so Lea is 24 + 32 = 56. **Why it's here.** An age problem. Open the answer to see the full step-by-step solution. Question 2 Algebra & word equations Medium Three consecutive integers have a sum of 99. What is the largest integer? - A. 35 - B. 33 - C. 39 - D. 36 - E. 34 **Show the answer and the fast route** **Answer: E, 34** The middle integer is the total divided by 3: 99 ÷ 3 = 33. The largest is one more, 34. The fast route Name the unknown once; translate the relationship before doing arithmetic. Solve it step by step 1. #### 1 Name the unknown Let m = the middle integer 2. #### 2 Translate the words into math - “three consecutive integers” `m - 1, m, m + 1` - “have a sum of 99” `(m - 1) + m + (m + 1) = 99` 3. #### 3 Solve the equation, one move at a time Whatever you do to one side, do to the other to keep the two sides equal. 1. Start The equation (m - 1) + m + (m + 1) = 99 2. 1 Rule Combine like terms Add the three m terms and the numbers: -1 and +1 cancel 3m = 99 3. 2 Rule Divide both sides by the same nonzero number Divide both sides by 3 m = 33 4. #### 4 Answer The integers are 32, 33, and 34, so the largest is 34. **Why it's here.** Consecutive integers: the middle-first shortcut skips the equation. Question 3 Work rates Medium Pat and Sam together can paint a fence in 12 hours. Sam alone takes 20 hours. How many hours would Pat take to paint the fence alone? - A. 8 - B. 32 - C. 24 - D. 30 - E. 16 **Show the answer and the fast route** **Answer: D, 30** Pat's rate is 1/12 − 1/20 = 5/60 − 3/60 = 2/60 = 1/30 per hour, so 30 hours. The fast route Subtract rates, then flip. **Why it's here.** Subtract rates, then flip. Two wrong choices come from subtracting and adding the times. Question 4 Speed & unit rates Harder A courier drives 120 miles to a client at 40 miles per hour and returns the same 120 miles at 60 miles per hour. What is the average speed for the round trip, in miles per hour? - A. 50 - B. 48 - C. 45 - D. 52 - E. 46 **Show the answer and the fast route** **Answer: B, 48** Out: 120 ÷ 40 = 3 hours. Back: 120 ÷ 60 = 2 hours. Average = 240 ÷ 5 = 48 miles per hour. The fast route Equal distances: never average the speeds, divide total distance by total time. **Why it's here.** The round-trip trap: one choice is the plain average of the two speeds. Question 5 Averages Medium Jordan has averaged 84 on his first four tests. What score does he need on the fifth test to raise his average to 86? - A. 90 - B. 92 - C. 88 - D. 94 - E. 96 **Show the answer and the fast route** **Answer: D, 94** Needed total: 5 × 86 = 430. Current total: 4 × 84 = 336. 430 - 336 = 94. The fast route He needs 86 plus 2 points for each of the 4 earlier tests: 86 + 8 = 94. **Why it's here.** Turn the mean back into a total, or use the points-per-earlier-test shortcut. Question 6 Percentages Harder An item's price rose 25%, and then the new price fell 20%, ending at $360. What was the original price? - A. $360 - B. $345 - C. $375 - D. $400 - E. $384 **Show the answer and the fast route** **Answer: A, $360** The combined multiplier is 1.25 × 0.8 = 1, so the final price equals the original: 360. The fast route Multiply the factors before working backward: 1.25 × 0.8 = 1. **Why it's here.** Multipliers: up 25% then down 20% cancels exactly. Question 7 Ratios & proportions Harder A team has boys and girls in the ratio 3:2. After 10 more girls join and no one leaves, the numbers of boys and girls are equal. How many boys are on the team? - A. 20 - B. 30 - C. 15 - D. 25 - E. 10 **Show the answer and the fast route** **Answer: B, 30** Boys = 3k and girls = 2k. Then 2k + 10 = 3k, so k = 10 and boys = 30. The fast route The 10 new girls fill a 1-part gap, so one part is 10. Solve it step by step 1. #### 1 Name the unknown Let k = the size of one ratio part 2. #### 2 Translate the words into math - “boys and girls in the ratio 3:2” `boys = 3k, girls = 2k` - “After 10 more girls join, the numbers of boys and girls are equal” `2k + 10 = 3k` 3. #### 3 Solve the equation, one move at a time Whatever you do to one side, do to the other to keep the two sides equal. 1. Start The equation 2k + 10 = 3k 2. 1 Rule Subtract the same amount from both sides Subtract 2k from both sides 10 = k 3. 2 Rule Flip the sides Write k on the left k = 10 4. #### 4 Answer Each part is 10, so there are 3 × 10 = 30 boys. **Why it's here.** Ratio parts with a change: the new girls fill a one-part gap. Question 8 Ratios & proportions Harder How many liters of a 50% acid solution must be added to 20 liters of a 20% acid solution to produce a 30% acid solution? - A. 20 - B. 10 - C. 5 - D. 15 - E. 8 **Show the answer and the fast route** **Answer: B, 10** Acid balance: 0.20 × 20 + 0.50x = 0.30(20 + x), so 4 + 0.5x = 6 + 0.3x, 0.2x = 2, x = 10 liters. The fast route 30% is 10 points from 20% and 20 points from 50%, so the amounts are in ratio 2:1: 20 liters to 10 liters. Solve it step by step 1. #### 1 Name the unknown Let x = the liters of 50% acid solution added 2. #### 2 Translate the words into math - “20 liters of a 20% acid solution” `acid = 0.2 × 20` - “liters of a 50% acid solution must be added” `acid = 0.5x` - “to produce a 30% acid solution” `0.2 × 20 + 0.5x = 0.3(20 + x)` 3. #### 3 Solve the equation, one move at a time Whatever you do to one side, do to the other to keep the two sides equal. 1. Start The equation 0.2 × 20 + 0.5x = 0.3(20 + x) 2. 1 Rule Distribute: a(b + c) = ab + ac Multiply 0.3 by each term inside the parentheses 0.2 × 20 + 0.5x = 6 + 0.3x 3. 2 Rule Combine like terms Multiply 0.2 × 20 to get 4 4 + 0.5x = 6 + 0.3x 4. 3 Rule Subtract the same amount from both sides Subtract 0.3x from both sides 4 + 0.2x = 6 5. 4 Rule Subtract the same amount from both sides Subtract 4 from both sides 0.2x = 2 6. 5 Rule Divide both sides by the same nonzero number Divide both sides by 0.2 x = 10 4. #### 4 Answer You must add 10 liters of the 50% solution. **Why it's here.** A mixture: the equation is in the steps, and a faster route is in the shortcut. ## Where word problems fit in the CCAT In our simulator, each 50-question test has 18 math questions drawn from 12 math families, and six of those families are word-problem families. That is our simulator's mix, built to match the format; Criteria does not publish a per-type breakdown. The rest of the math slots go to mental arithmetic, fractions, probability, tables and charts, and number sequences. Number sequences are the opposite kind of math question: no story, just a pattern. If those slow you down too, the [number series guide](https://cognitivecatprep.com/ccat-number-series-questions) has the checklist. For the full list of math types, see the [CCAT math questions](https://cognitivecatprep.com/ccat-math-questions) page. ## How to practice word problems Practice them by setup, not at random. Do ten work-rate questions in a row until 1/a + 1/b is automatic, then ten round-trip and meeting problems, then averages. In the app, a focus drill on one family times every answer and, on questions that have worked steps, opens them when you miss. The coach tracks which setups cost you the most time. When the setups feel automatic, take a [full CCAT practice test](https://cognitivecatprep.com/ccat-practice-test) under the 15-minute clock, where word problems compete with 32 non-math questions for your time. More question types are on the [CCAT practice questions hub](https://cognitivecatprep.com/ccat-practice-questions). ## Questions, answered **What kinds of word problems are on the CCAT?** Expect short story problems on rates and travel, work done together, averages, percent change, ratios, and simple algebra like ages or consecutive numbers. Our practice bank covers each of these as its own question type. **Can I use a calculator for CCAT practice here?** Our simulator has no calculator, so you practice by hand and by estimating. Check your employer's test instructions for the rules on your real test. **How do I solve work rate problems fast?** Add rates, not times. Each worker's rate is 1 divided by their time, the rates add, and the combined time is 1 divided by the total. For two workers, multiply the times and divide by their sum. **How do I find the average speed of a round trip?** Divide the total distance by the total time. Do not average the two speeds; the slower leg takes longer, so the true average is lower than the plain average. **Should I skip hard word problems on the CCAT?** Often, yes. A wrong answer and a blank count the same, and a long setup can cost you two easier questions. Eliminate what you can, guess, and move on. **How should I practice word problems?** Identify the unknown and turn the story into an equation before calculating. After answering, compare your setup with the worked solution and look for a shorter route. ## Keep going - [Question types: Syllogisms](https://cognitivecatprep.com/ccat-syllogisms) - [Question types: Math questions](https://cognitivecatprep.com/ccat-math-questions) - [Question types: Verbal questions](https://cognitivecatprep.com/ccat-verbal-questions) - [Question types: Logic questions](https://cognitivecatprep.com/ccat-logic-questions)