CEUET Mathematics — Word Problems — Number, Age, Work, Motion, Mixture, InvestmentFlash Cards
Flashcards specifically for Word Problems — Number, Age, Work, Motion, Mixture, Investment in the CEUET 2026. Every card has been tuned to match Centro Escolar University's preferred question style. Practise them during your commute, at lunch, or before bed — they are designed for short bursts of high-retention review.
Exam context
On the CEUET 2026, the Mathematics subtest carries a "Core" weight in Centro Escolar University's pattern. Word Problems — Number, Age, Work, Motion, Mixture, Investment lands at position 4th out of 9 in the standard review order. Target score is Competitive overall score, and roughly a meaningful share of items come from Mathematics on a typical CEUET paper.
Word Problems — Number, Age, Work, Motion, Mixture, Investment - Flashcards
Master the art of solving word problems by learning systematic approaches to translate real-world scenarios into mathematical equations. This comprehensive set covers the six major types of word problems commonly found in UPCAT and other college entrance exams: number problems, age problems, work problems, motion problems, mixture problems, and investment problems. Each card focuses on problem-solving techniques with step-by-step solutions to build your confidence and analytical skills.
Cards
Solve: The sum of two consecutive integers is 47. Find the integers.
Step 1: Let x = first integer, then x + 1 = second integer. Step 2: Set up equation: x + (x + 1) = 47. Step 3: Simplify: 2x + 1 = 47. Step 4: Solve: 2x = 46, so x = 23. Step 5: The integers are 23 and 24. Check: 23 + 24 = 47 ✓
Tags
- consecutive_integers
- basic_algebra
- moderate
Topic
Number Problems
Card Id
FC1
Difficulty
moderate
Image Prompt
Solve: One number is 15 more than twice another. Their sum is 78. Find both numbers.
Step 1: Let x = smaller number, then 2x + 15 = larger number. Step 2: Set up equation: x + (2x + 15) = 78. Step 3: Simplify: 3x + 15 = 78. Step 4: Solve: 3x = 63, so x = 21. Step 5: Smaller number = 21, larger number = 2(21) + 15 = 57. Check: 21 + 57 = 78 ✓
Tags
- two_numbers
- algebraic_translation
- moderate
Topic
Number Problems
Card Id
FC2
Difficulty
moderate
Image Prompt
Solve: The sum of three consecutive odd integers is 105. Find the integers.
Step 1: Let x = first odd integer, then x + 2 = second, x + 4 = third. Step 2: Set up equation: x + (x + 2) + (x + 4) = 105. Step 3: Simplify: 3x + 6 = 105. Step 4: Solve: 3x = 99, so x = 33. Step 5: The integers are 33, 35, and 37. Check: 33 + 35 + 37 = 105 ✓
Tags
- consecutive_odd
- three_numbers
- moderate
Topic
Number Problems
Card Id
FC3
Difficulty
moderate
Image Prompt
Solve: A two-digit number has digits whose sum is 12. If the digits are reversed, the new number is 36 less than the original. Find the number.
Step 1: Let x = tens digit, y = ones digit. Original number = 10x + y, reversed = 10y + x. Step 2: From sum of digits: x + y = 12, so y = 12 - x. Step 3: From reversal condition: 10x + y = (10y + x) + 36. Step 4: Substitute: 10x + (12 - x) = 10(12 - x) + x + 36. Step 5: Simplify: 9x + 12 = 120 - 9x + 36, so 18x = 144, x = 8. Step 6: y = 12 - 8 = 4. Answer: 84. Check: 8 + 4 = 12 ✓, 84 - 48 = 36 ✓
Tags
- digit_problems
- two_variables
- challenging
Topic
Number Problems
Card Id
FC4
Difficulty
challenging
Image Prompt
Solve: Maria is 8 years older than Jose. In 5 years, Maria will be twice as old as Jose. Find their current ages.
Step 1: Let x = Jose's current age, then x + 8 = Maria's current age. Step 2: In 5 years: Jose = x + 5, Maria = (x + 8) + 5 = x + 13. Step 3: Set up equation: x + 13 = 2(x + 5). Step 4: Expand: x + 13 = 2x + 10. Step 5: Solve: 13 - 10 = 2x - x, so x = 3. Step 6: Jose = 3 years old, Maria = 11 years old. Check: In 5 years, Jose = 8, Maria = 16, and 16 = 2(8) ✓
Tags
- current_future_ages
- basic_algebra
- moderate
Topic
Age Problems
Card Id
FC5
Difficulty
moderate
Image Prompt
Solve: A father is 32 years older than his son. Ten years ago, the father was three times as old as his son. Find their current ages.
Step 1: Let x = son's current age, then x + 32 = father's current age. Step 2: Ten years ago: son = x - 10, father = (x + 32) - 10 = x + 22. Step 3: Set up equation: x + 22 = 3(x - 10). Step 4: Expand: x + 22 = 3x - 30. Step 5: Solve: 22 + 30 = 3x - x, so 52 = 2x, x = 26. Step 6: Son = 26 years old, Father = 58 years old. Check: 10 years ago, son = 16, father = 48, and 48 = 3(16) ✓
Tags
- past_ages
- father_son
- moderate
Topic
Age Problems
Card Id
FC6
Difficulty
moderate
Image Prompt
Solve: Ana can finish a job in 6 hours. Ben can finish the same job in 4 hours. How long will it take them working together?
Step 1: Ana's rate = 1/6 job per hour, Ben's rate = 1/4 job per hour. Step 2: Combined rate = 1/6 + 1/4 = 2/12 + 3/12 = 5/12 job per hour. Step 3: Time = 1 job ÷ (5/12 job per hour) = 12/5 hours. Step 4: Convert: 12/5 = 2.4 hours = 2 hours 24 minutes. Answer: 2 hours 24 minutes or 2.4 hours
Tags
- combined_work
- fractions
- moderate
Topic
Work Problems
Card Id
FC7
Difficulty
moderate
Image Prompt
Solve: Machine A can produce 200 items in 5 hours. Machine B can produce 200 items in 8 hours. Working together, how long to produce 200 items?
Step 1: Machine A's rate = 200/5 = 40 items per hour. Step 2: Machine B's rate = 200/8 = 25 items per hour. Step 3: Combined rate = 40 + 25 = 65 items per hour. Step 4: Time = 200 items ÷ 65 items per hour = 200/65 = 40/13 hours. Step 5: Convert: 40/13 ≈ 3.08 hours = 3 hours 5 minutes. Answer: 40/13 hours or approximately 3 hours 5 minutes
Tags
- machine_production
- rates
- moderate
Topic
Work Problems
Card Id
FC8
Difficulty
moderate
Image Prompt
Solve: Two cars start from towns 240 km apart and drive toward each other. Car A travels at 60 kph, Car B at 80 kph. When will they meet?
Step 1: Let t = time until they meet (in hours). Step 2: Distance covered by Car A = 60t km. Step 3: Distance covered by Car B = 80t km. Step 4: Total distance = 60t + 80t = 240. Step 5: Solve: 140t = 240, so t = 240/140 = 12/7 hours. Step 6: Convert: 12/7 ≈ 1.71 hours = 1 hour 43 minutes. Answer: 12/7 hours or 1 hour 43 minutes
Tags
- meeting_problems
- opposite_directions
- moderate
Topic
Motion Problems
Card Id
FC9
Difficulty
moderate
Image Prompt
Solve: A bus travels 180 km in 3 hours. On the return trip, traffic reduces its speed by 15 kph. How long is the return trip?
Step 1: Original speed = 180 km ÷ 3 hours = 60 kph. Step 2: Return speed = 60 - 15 = 45 kph. Step 3: Return time = 180 km ÷ 45 kph = 4 hours. Answer: 4 hours. Check: Distance = speed × time = 45 × 4 = 180 km ✓
Tags
- speed_change
- return_trip
- easy
Topic
Motion Problems
Card Id
FC10
Difficulty
easy
Image Prompt
Solve: A runner starts 30 minutes after a cyclist. The cyclist travels at 20 kph, the runner at 12 kph. When will the runner catch up if they travel the same route?
Step 1: Let t = time runner travels (hours). Cyclist travels for t + 0.5 hours. Step 2: When they meet, distances are equal: 12t = 20(t + 0.5). Step 3: Expand: 12t = 20t + 10. Step 4: Solve: 12t - 20t = 10, so -8t = 10, t = -10/8 = -1.25. Step 5: Since negative time is impossible, the runner will NEVER catch up to the cyclist (cyclist is faster and has a head start).
Tags
- catch_up
- head_start
- challenging
Topic
Motion Problems
Card Id
FC11
Difficulty
challenging
Image Prompt
Solve: How many liters of 20% acid solution must be mixed with 5 liters of 60% acid solution to get a 40% acid solution?
Step 1: Let x = liters of 20% solution needed. Step 2: Set up equation for pure acid: 0.20x + 0.60(5) = 0.40(x + 5). Step 3: Simplify: 0.20x + 3 = 0.40x + 2. Step 4: Solve: 3 - 2 = 0.40x - 0.20x, so 1 = 0.20x, x = 5. Step 5: Need 5 liters of 20% solution. Check: 0.20(5) + 0.60(5) = 1 + 3 = 4 liters pure acid. Total volume = 10 liters. Concentration = 4/10 = 40% ✓
Tags
- acid_solutions
- percentages
- moderate
Topic
Mixture Problems
Card Id
FC12
Difficulty
moderate
Image Prompt
Solve: A chemist has 3 liters of 25% salt solution. How much pure water should be added to make it 15% salt solution?
Step 1: Let x = liters of water to add. Step 2: Pure salt remains constant: 0.25(3) = 0.75 liters. Step 3: New volume = 3 + x liters. Step 4: Set up equation: 0.75 = 0.15(3 + x). Step 5: Solve: 0.75 = 0.45 + 0.15x, so 0.30 = 0.15x, x = 2. Step 6: Add 2 liters of water. Check: 0.75/(3 + 2) = 0.75/5 = 0.15 = 15% ✓
Tags
- dilution
- pure_water
- moderate
Topic
Mixture Problems
Card Id
FC13
Difficulty
moderate
Image Prompt
Solve: Coffee worth ₱120 per kg is mixed with coffee worth ₱180 per kg. How many kg of each to make 50 kg of blend worth ₱150 per kg?
Step 1: Let x = kg of ₱120 coffee, then 50 - x = kg of ₱180 coffee. Step 2: Set up value equation: 120x + 180(50 - x) = 150(50). Step 3: Expand: 120x + 9000 - 180x = 7500. Step 4: Simplify: -60x = 7500 - 9000 = -1500. Step 5: Solve: x = 25. Step 6: Use 25 kg of ₱120 coffee and 25 kg of ₱180 coffee. Check: 120(25) + 180(25) = 3000 + 4500 = 7500 = 150(50) ✓
Tags
- coffee_blend
- cost_per_unit
- moderate
Topic
Mixture Problems
Card Id
FC14
Difficulty
moderate
Image Prompt
Solve: ₱12,000 is invested, part at 6% annual interest and part at 8% annual interest. If the total annual interest is ₱840, how much is invested at each rate?
Step 1: Let x = amount at 6%, then 12000 - x = amount at 8%. Step 2: Set up interest equation: 0.06x + 0.08(12000 - x) = 840. Step 3: Expand: 0.06x + 960 - 0.08x = 840. Step 4: Simplify: -0.02x = 840 - 960 = -120. Step 5: Solve: x = 6000. Step 6: ₱6,000 at 6% and ₱6,000 at 8%. Check: 0.06(6000) + 0.08(6000) = 360 + 480 = 840 ✓
Tags
- simple_interest
- split_investment
- moderate
Topic
Investment Problems
Card Id
FC15
Difficulty
moderate
Image Prompt
Solve: ₱8,000 is invested at 5% annual simple interest. How much additional money must be invested at 9% to earn ₱700 total annual interest?
Step 1: Interest from first investment = 0.05(8000) = ₱400. Step 2: Additional interest needed = 700 - 400 = ₱300. Step 3: Let x = additional amount at 9%. Step 4: Set up equation: 0.09x = 300. Step 5: Solve: x = 300/0.09 = ₱3,333.33. Step 6: Must invest ₱3,333.33 more at 9%. Check: 400 + 0.09(3333.33) = 400 + 300 = 700 ✓
Tags
- additional_investment
- target_interest
- moderate
Topic
Investment Problems
Card Id
FC16
Difficulty
moderate
Image Prompt
When do you use the formula: Distance = Rate × Time?
Use for motion problems involving: 1) Objects traveling at constant speed, 2) Meeting problems (objects approaching each other), 3) Overtaking problems (faster object catching slower one), 4) Round-trip problems. Example: Car travels 60 kph for 3 hours → Distance = 60 × 3 = 180 km. Always identify what you're solving for: D, R, or T.
Tags
- formula_application
- distance_formula
- basic
Topic
Motion Problems
Card Id
FC17
Difficulty
basic
Image Prompt
When do you use the work formula: 1/a + 1/b = 1/t?
Use when two people/machines work together on the same job. Where: a = time for person A alone, b = time for person B alone, t = time working together. Example: Ana finishes job in 4 hours, Ben in 6 hours. Together: 1/4 + 1/6 = 1/t → 3/12 + 2/12 = 5/12 = 1/t → t = 12/5 = 2.4 hours.
Tags
- formula_application
- work_formula
- basic
Topic
Work Problems
Card Id
FC18
Difficulty
basic
Image Prompt
What's the key strategy for translating word problems into equations?
Use the 3 R's + ESP method: 1) READ thoroughly, 2) REPRESENT unknowns with variables, 3) RELATE variables using given information, 4) EQUATE using problem facts, 5) SOLVE the equation, 6) PROVE by checking. Key phrases: 'is/was/will be' = equals sign, 'more than' = addition (but switch order), 'less than' = subtraction (switch order).
Tags
- problem_solving
- translation
- strategy
Topic
General Strategy
Card Id
FC19
Difficulty
basic
Image Prompt
Common mistake: Why is 'x less than 10' written as 10 - x, not x - 10?
The phrase 'x less than 10' means '10 decreased by x' or '10 minus x', so it's 10 - x. Similarly, '5 more than x' means x + 5. The number mentioned FIRST in the phrase comes SECOND in the mathematical expression when using 'less than' or 'subtracted from'. Always read carefully: '4 less than a number' → x - 4, but 'a number less than 4' → 4 - x.
Tags
- common_mistakes
- phrase_translation
- basic
Topic
Translation Skills
Card Id
FC20
Difficulty
basic
Image Prompt
Tag Distribution
Easy
1
Basic
4
Moderate
12
Challenging
3
Problem Solving
15
Formula Application
3
Algebraic Translation
8
Topic Distribution
Age Problems
2
Work Problems
2
Motion Problems
3
Number Problems
4
General Strategy
2
Mixture Problems
3
Translation Skills
2
Investment Problems
2
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