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.
Flashcards
Solve: The sum of two consecutive integers is 47. Find the integers.
Answer
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 ✓
Solve: One number is 15 more than twice another. Their sum is 78. Find both numbers.
Answer
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 ✓
Solve: The sum of three consecutive odd integers is 105. Find the integers.
Answer
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 ✓
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.
Answer
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 ✓
Solve: Maria is 8 years older than Jose. In 5 years, Maria will be twice as old as Jose. Find their current ages.
Answer
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) ✓
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.
Answer
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) ✓
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?
Answer
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
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?
Answer
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
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?
Answer
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
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?
Answer
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 ✓
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?
Answer
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).
Solve: How many liters of 20% acid solution must be mixed with 5 liters of 60% acid solution to get a 40% acid solution?
Answer
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% ✓
Solve: A chemist has 3 liters of 25% salt solution. How much pure water should be added to make it 15% salt solution?
Answer
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% ✓
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?
Answer
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) ✓
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?
Answer
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 ✓
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?
Answer
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 ✓
When do you use the formula: Distance = Rate × Time?
Answer
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.
When do you use the work formula: 1/a + 1/b = 1/t?
Answer
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.
What's the key strategy for translating word problems into equations?
Answer
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).
Common mistake: Why is 'x less than 10' written as 10 - x, not x - 10?
Answer
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.
Diagrams
Word Problems Classification and Approach
A comprehensive mind map showing the six major categories of word problems and their subcategories, helping students identify problem types and choose appropriate solution strategies.
Word Problem Solving Strategy (3 R's + ESP Method)
A step-by-step flowchart illustrating the systematic approach to solving any word problem using the 3 R's and ESP method, with decision points for verification.
Formula Selection Guide
A decision tree to help students quickly identify which formula or approach to use based on the type of word problem they encounter.
UPCAT Word Problem Distribution
Distribution of different word problem types commonly found in UPCAT and similar college entrance exams, helping students prioritize their study focus.