Word Problems — Number, Age, Work, Motion, Mixture, Investment - Study notes
Word problems are mathematical problems expressed in everyday language that require translation into mathematical equations. Mastering word problems is essential for UPCAT success as they test both your mathematical skills and logical thinking. This chapter covers six major types: number problems, age problems, work problems, motion problems, mixture problems, and investment problems. Each type follows the 3 R's and ESP method: Read, Represent, Relate, Equate, Solve, and Prove.
The 3 R's and ESP Method for Word Problems
Every word problem can be solved systematically using this proven method:
READ - Read the problem thoroughly and understand what is being asked
REPRESENT - Use variables (usually x, y) to represent unknown quantities
RELATE - Establish relationships between variables and given information
EQUATE - Set up equations based on the relationships
SOLVE - Solve the equation(s) algebraically
PROVE - Check your answer by substituting back into the original problem
This systematic approach ensures you don't miss important details and helps organize your solution clearly.
Examples
- Problem: 'Four more than twice a number equals 14. Find the number.' Step 1 (Read): We need to find an unknown number Step 2 (Represent): Let x = the number Step 3 (Relate): 'Four more than twice the number' means 2x + 4 Step 4 (Equate): 2x + 4 = 14 Step 5 (Solve): 2x = 10, so x = 5 Step 6 (Prove): 'Four more than twice 5' = 4 + 2(5) = 4 + 10 = 14 ✓
Key points
- Always read the problem at least twice before setting up equations
- Choose clear, meaningful variables to represent unknowns
- Identify what equals what in the problem statement
- Always verify your answer makes sense in the original context
Translating Word Phrases to Mathematical Expressions
Before solving word problems, you must master translating English phrases into mathematical expressions. Here are the most common translations:
Addition: 'plus', 'more than', 'increased by', 'sum of', 'added to'
Subtraction: 'minus', 'less than', 'decreased by', 'difference of', 'subtracted from'
Multiplication: 'times', 'product of', 'of', 'twice' (×2), 'thrice' (×3)
Division: 'divided by', 'quotient of', 'per', 'ratio of'
Equals: 'is', 'equals', 'results in', 'gives', 'will be'
Warning: Pay attention to order! 'x less than y' means y - x, not x - y. 'x subtracted from y' means y - x.
Examples
- '5 more than x' → x + 5
- '5 less than x' → x - 5 (order switched!)
- '5 subtracted from x' → x - 5 (order switched!)
- 'The product of 3 and x' → 3x
- 'x divided by 5' → x/5
- 'Twice x decreased by 7' → 2x - 7
Key points
- Order matters in subtraction and division phrases
- The word 'of' usually means multiplication
- Consecutive integers differ by 1; consecutive even/odd integers differ by 2
- Practice translating phrases before attempting full problems
Number Problems
Number problems involve finding unknown numbers based on given relationships. Common types include consecutive integers, digit problems, and simple number relationships.
Consecutive Integers: Use x, x+1, x+2, ... for consecutive integers. For consecutive odd/even integers, use x, x+2, x+4, ...
Worked Example: Find three consecutive integers whose sum is 51.
Step 1: Let x = first integer, x+1 = second integer, x+2 = third integer
Step 2: Set up equation: x + (x+1) + (x+2) = 51
Step 3: Solve: 3x + 3 = 51, so 3x = 48, therefore x = 16
Step 4: The integers are 16, 17, 18
Step 5: Check: 16 + 17 + 18 = 51 ✓
Examples
- Find two consecutive odd integers whose sum is 76. Let x = first odd integer, x+2 = second Equation: x + (x+2) = 76 Solve: 2x + 2 = 76, so x = 37 Answer: 37 and 39
- The sum of digits of a two-digit number is 12. If digits are reversed, the new number is 18 more than original. Find the number. Let original number = 10x + y where x and y are digits Equations: x + y = 12 and 10y + x = 10x + y + 18 Solving: x = 5, y = 7, so number is 57
Key points
- Consecutive integers: x, x+1, x+2, ...
- Consecutive odd/even: x, x+2, x+4, ...
- For digit problems, a two-digit number ab = 10a + b
- Always check that your answer satisfies the original conditions
Age Problems
Age problems involve relationships between people's ages at different times. The key insight is that everyone ages at the same rate.
Strategy: Create a table showing ages at different times (past, present, future). If someone is x years old now, they were (x-n) years old n years ago, and will be (x+n) years old in n years.
Worked Example: Maria is 8 years older than Jose now. In 5 years, Maria will be twice as old as Jose. Find their present ages.
Step 1: Let J = Jose's current age, then Maria's age = J + 8
Step 2: Create age table:
| Now | In 5 years |
|---|
| Jose | J | J + 5 |
|---|---|---|
| Maria | J+8 | J + 8 + 5 |
Step 3: In 5 years, Maria will be twice Jose's age: J + 8 + 5 = 2(J + 5)
Step 4: Solve: J + 13 = 2J + 10, so J = 3
Step 5: Jose is 3, Maria is 11
Step 6: Check: In 5 years, Jose will be 8, Maria will be 16 = 2(8) ✓
Examples
- A father is 30 years older than his son. In 10 years, he will be twice as old as his son. Find their current ages. Let s = son's age now In 10 years: s + 30 + 10 = 2(s + 10) Solve: s + 40 = 2s + 20, so s = 20 Son is 20, father is 50
Key points
- Use a table to organize ages at different times
- Everyone ages at the same rate
- If current age is x, then past age is (x - years ago)
- Future age is (x + years from now)
Work Problems
Work problems involve rates of completing jobs. The key formula is: Work = Rate × Time, where total work = 1 job.
Key Principle: If person A can complete a job in n hours, their work rate is 1/n jobs per hour. When working together, rates are added.
Formula for combined work: If A works alone in time a, B works alone in time b, then together they work in time t where: 1/a + 1/b = 1/t
Worked Example: Ana can clean a house in 6 hours. Beth can clean the same house in 4 hours. How long will it take them working together?
Step 1: Ana's rate = 1/6 house per hour
Step 2: Beth's rate = 1/4 house per hour
Step 3: Combined rate = 1/6 + 1/4 = 2/12 + 3/12 = 5/12 house per hour
Step 4: Time together = 1 ÷ (5/12) = 12/5 = 2.4 hours
Step 5: Check: In 2.4 hours, Ana completes 2.4/6 = 0.4 of the job, Beth completes 2.4/4 = 0.6 of the job. Total: 0.4 + 0.6 = 1.0 ✓
Examples
- Machine A can produce 100 items in 5 hours. Machine B can produce 100 items in 3 hours. Working together, how long to produce 100 items? Rate A = 1/5 jobs per hour, Rate B = 1/3 jobs per hour Combined rate = 1/5 + 1/3 = 8/15 jobs per hour Time = 1 ÷ (8/15) = 15/8 = 1.875 hours
Key points
- Work rate = 1/(time to complete alone)
- Combined rates add: Rate₁ + Rate₂ = Combined Rate
- Use fractions to represent partial work completed
- Time together = 1 ÷ (combined rate)
Motion Problems
Motion problems use the fundamental relationship: Distance = Rate × Time (D = RT)
Common scenarios:
- Meeting problems: Objects moving toward each other - add distances
- Overtaking problems: One object catching another - distances are equal
- Round trip problems: Different speeds in different directions
Worked Example: Two cars start from cities 240 km apart and drive toward each other. Car A travels at 60 kph, Car B at 40 kph. When do they meet?
Step 1: Let t = time until they meet
Step 2: Distance covered by A = 60t km
Step 3: Distance covered by B = 40t km
Step 4: Total distance = 60t + 40t = 240
Step 5: Solve: 100t = 240, so t = 2.4 hours
Step 6: Check: A travels 60(2.4) = 144 km, B travels 40(2.4) = 96 km. Total: 144 + 96 = 240 km ✓
Examples
- A train leaves Manila at 60 kph. Two hours later, an express train follows at 90 kph. When does the express train catch up? Let t = time express train travels Regular train travels for (t + 2) hours Distances equal: 60(t + 2) = 90t Solve: 60t + 120 = 90t, so 30t = 120, t = 4 hours Express train catches up after 4 hours
Key points
- Always identify what the distances represent in the problem
- Meeting problems: add distances
- Overtaking problems: set distances equal
- Keep units consistent (kph and hours, or mph and hours)
Mixture Problems
Mixture problems involve combining solutions with different concentrations. Key concept: Amount of pure substance = Concentration × Total volume
Setup strategy:
- Create a table with columns: % concentration, Volume, Amount of pure substance
- Use the fact that pure substances add up in the final mixture
Worked Example: How much water must be added to 40 liters of 25% salt solution to make it 10% salt solution?
Step 1: Let x = liters of water to add
Step 2: Set up table:
| % Salt | Volume | Pure Salt |
|---|
| Original | 0.25 | 40 | 10 |
|---|---|---|---|
| Water | 0.00 | x | 0 |
| Final | 0.10 | 40 + x | 0.10(40+x) |
Step 3: Pure salt remains constant: 10 = 0.10(40 + x)
Step 4: Solve: 10 = 4 + 0.1x, so 6 = 0.1x, therefore x = 60
Step 5: Add 60 liters of water
Step 6: Check: Final solution has 10 liters salt in 100 liters total = 10% ✓
Examples
- Mix 30 L of 20% alcohol solution with 70 L of 60% alcohol solution. Find final concentration. Pure alcohol: 30(0.20) + 70(0.60) = 6 + 42 = 48 L Total volume: 30 + 70 = 100 L Final concentration: 48/100 = 48%
Key points
- Pure substance amount = % concentration × total volume
- When mixing, pure substances add up
- Water or pure solvent has 0% concentration of solute
- Pure substance has 100% concentration
Investment Problems
Investment problems use the formula: Interest = Principal × Rate × Time (I = PRT)
Key concepts:
- Principal: Amount invested
- Rate: Interest rate (as decimal)
- Time: Duration (usually in years)
- Simple Interest: Interest calculated only on principal
- Compound Interest: Interest calculated on principal plus accumulated interest
Worked Example: Rosa invests ₱15,000 total in two accounts. One earns 4% annually, the other 6% annually. If total annual interest is ₱800, how much is invested at each rate?
Step 1: Let x = amount at 4%, then (15000 - x) = amount at 6%
Step 2: Set up equation using total interest:
0.04x + 0.06(15000 - x) = 800
Step 3: Solve: 0.04x + 900 - 0.06x = 800
-0.02x = -100
x = 5000
Step 4: ₱5,000 at 4%, ₱10,000 at 6%
Step 5: Check: 5000(0.04) + 10000(0.06) = 200 + 600 = 800 ✓
Examples
- ₱8,000 invested at 5% simple interest for 3 years. Find total interest. I = PRT = 8000 × 0.05 × 3 = ₱1,200 Final amount = ₱8,000 + ₱1,200 = ₱9,200
Key points
- Convert percentage rates to decimals (5% = 0.05)
- Annual interest means t = 1 in the I = PRT formula
- Total investment = sum of individual investments
- Total interest = sum of individual interests
Diagrams
Word Problems Overview
Comprehensive overview showing the six types of word problems, the systematic solution method, and key skills needed for success
Word Problem Solution Strategy
Decision tree showing how to identify problem types and apply appropriate solution strategies with verification steps
Common Translation Patterns
Flowchart showing how to translate common word phrases into mathematical expressions with emphasis on order-sensitive operations
In summary
Word problems are systematic mathematical challenges that can be mastered using the 3 R's and ESP method. Success requires: (1) Careful reading and identification of given information and unknowns, (2) Proper translation of word phrases to mathematical expressions, (3) Setting up correct equations based on problem relationships, and (4) Systematic solving and verification. The six main types covered—number, age, work, motion, mixture, and investment problems—each have specific formulas and approaches, but all follow the same logical problem-solving framework. Practice with various problem types and always verify answers by checking against original conditions.