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FEUCAT MathematicsWord Problems — Number, Age, Work, Motion, Mixture, InvestmentFlash Cards

Flashcards for Word Problems — Number, Age, Work, Motion, Mixture, Investment — the active-recall tool for FEUCAT Mathematics aspirants. Each card tests a key concept, formula, or definition from the FEUCAT 2026 syllabus. Use them daily in the final month before exam day.

Exam context

The Far Eastern University College Admission Test is conducted by Far Eastern University and is scheduled for Q3–Q4 2026. The Mathematics subtest is marked as "Core section" in the official pattern, and Word Problems — Number, Age, Work, Motion, Mixture, Investment appears in position 4th of 9 in the FEUCAT Mathematics review rotation. Passing mark: Competitive overall score. Recent FEUCAT 2026 papers have drawn roughly a meaningful share of questions from this subject.

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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