Word Problems — Number, Age, Work, Motion, Mixture, Investment - Detailed explanation
Word problems are mathematical scenarios presented in written form that require you to translate real-world situations into mathematical equations and solve them. These problems are fundamental in Philippine college entrance exams like UPCAT, ACET, and USTET because they test both your mathematical skills and your ability to understand and analyze practical situations. Success in word problems requires mastering the 3 R's and ESP approach: Read thoroughly, Represent variables, Relate quantities, Equate using problem facts, Solve the equation, and Prove your answer. This systematic approach will help you tackle any word problem with confidence.
Translating Word Problems into Mathematical Expressions
Examples
Translate: '5 more than twice a number is 17'
Solution
Let x = the number. '5 more than twice x' means 2x + 5. 'is 17' means equals 17. Therefore: 2x + 5 = 17
Start with the variable (x), identify 'twice' as 2x, then '5 more than' adds 5 to get 2x + 5. The word 'is' creates the equation 2x + 5 = 17.
Translate: '3 less than a number is twice the number decreased by 8'
Solution
Let x = the number. '3 less than x' means x - 3. 'twice the number decreased by 8' means 2x - 8. Therefore: x - 3 = 2x - 8
For 'less than', we subtract 3 from x (not 3 - x). 'Decreased by' means subtract 8 from 2x. Connect both sides with equals sign.
The foundation of solving word problems is accurately translating English phrases into mathematical expressions. Every mathematical operation has specific keywords and phrases that signal when to use addition, subtraction, multiplication, or division. Understanding these translation patterns is crucial because even a small misinterpretation can lead to completely wrong answers.
Applications
- Setting up equations for number problems
- Converting real-world scenarios into solvable mathematical forms
- Understanding problem requirements before attempting solutions
Common misconceptions
- Confusing 'less than' with simple subtraction (x less than 5 is 5-x, not x-5)
- Mixing up the order in 'subtracted from' expressions
- Forgetting to define variables clearly before translating
Related concepts
- Algebraic expressions
- Linear equations
- Variable representation
Common exam questions
Example
The sum of a number and 12 is equal to 4 times the number minus 3
Approach
Identify the variable, translate each phrase step by step, then form the equation
Key points to remember
- Addition keywords: plus, more than, increased by, sum of, when added, total
- Subtraction keywords: minus, less than, subtracted from, decreased by, difference of
- Multiplication keywords: times, product of, of, twice, thrice, half
- Division keywords: divided by, quotient of, per
- Equality keywords: is, was, will be, equal, results to
- For 'less than' and 'subtracted from', always switch the order of numbers
Number Problems and Consecutive Integers
Examples
Find three consecutive integers whose sum is 54.
Solution
Let x = first integer, x+1 = second integer, x+2 = third integer. x + (x+1) + (x+2) = 54. 3x + 3 = 54. 3x = 51. x = 17. The integers are 17, 18, 19.
We represent consecutive integers as x, x+1, x+2 because each increases by 1. Adding them gives 3x+3=54, solving gives x=17.
Find three consecutive even integers whose sum is 48.
Solution
Let x = first even integer, x+2 = second, x+4 = third. x + (x+2) + (x+4) = 48. 3x + 6 = 48. 3x = 42. x = 14. The integers are 14, 16, 18.
Consecutive even integers differ by 2, so we use x, x+2, x+4. Since x=14 is even, our answer is correct.
One number is 7 more than another. Their sum is 23. Find both numbers.
Solution
Let x = smaller number, x+7 = larger number. x + (x+7) = 23. 2x + 7 = 23. 2x = 16. x = 8. The numbers are 8 and 15.
We define one number in terms of the other using their relationship, then use the sum condition to solve.
Number problems involve finding unknown numbers based on relationships described in words. Consecutive integer problems are a special type where numbers follow each other in sequence. For regular consecutive integers (like 7, 8, 9), each number is 1 more than the previous. For consecutive odd or even integers (like 3, 5, 7 or 4, 6, 8), each number is 2 more than the previous.
Applications
- Finding unknown quantities with known relationships
- Solving problems involving sequences and patterns
- Real-world applications like house numbers, page numbers, seat numbers
Common misconceptions
- Using x, x+1, x+2 for consecutive odd/even integers (should be x, x+2, x+4)
- Forgetting to check if the first number has the required property (odd/even)
- Incorrectly setting up relationships between numbers
Related concepts
- Linear equations
- Systems of equations
- Number sequences and patterns
Common exam questions
Example
The sum of four consecutive odd integers is 56
Approach
Represent integers as x, x+1, x+2, etc., set up sum equation, solve for x
Example
One number is 12 less than twice another, their sum is 39
Approach
Express one number in terms of the other, use given conditions to form equations
Key points to remember
- For consecutive integers: n, n+1, n+2, n+3, etc.
- For consecutive odd/even integers: n, n+2, n+4, n+6, etc.
- Always check if your first number matches the required type (odd/even)
- The sum of n consecutive integers can be found using the middle value
- Use variables to represent the first number in the sequence
Age Problems
Examples
Maria is 4 years older than Juan today. Three years ago, Maria was twice as old as Juan. Find their present ages.
Solution
Let x = Juan's present age, x+4 = Maria's present age. Three years ago: Juan was x-3, Maria was (x+4)-3 = x+1. Maria was twice Juan's age: x+1 = 2(x-3). x+1 = 2x-6. 7 = x. Juan is 7, Maria is 11.
We create a time table: Present (Juan: x, Maria: x+4), Past (Juan: x-3, Maria: x+1). The relationship 'Maria was twice Juan's age' gives us the equation.
A father is 28 years older than his son. In 6 years, the father will be 3 times as old as his son. Find their present ages.
Solution
Let x = son's present age, x+28 = father's present age. In 6 years: son will be x+6, father will be (x+28)+6 = x+34. Father will be 3 times son's age: x+34 = 3(x+6). x+34 = 3x+18. 16 = 2x. x = 8. Son is 8, father is 36.
We set up future ages by adding 6 to present ages, then use the future relationship to create our equation.
Age problems involve relationships between people's ages at different time periods. The key insight is that everyone ages at the same rate - if 3 years pass, everyone becomes 3 years older. Creating a table showing present ages and ages at different times helps organize the information and set up equations correctly.
Applications
- Family age relationships and planning
- Historical timeline problems
- Comparative age analysis in various contexts
Common misconceptions
- Forgetting that everyone ages at the same rate
- Incorrectly calculating past or future ages
- Setting up relationships in the wrong time period
Related concepts
- Linear equations
- Time and chronology
- Proportional relationships
Common exam questions
Example
Five years ago, Anna was three times as old as Ben. Now Anna is twice as old as Ben.
Approach
Set up present ages, subtract years for past ages, use given past relationship
Example
In 4 years, the sum of their ages will be 50
Approach
Set up present ages, add years for future ages, use given future relationship
Key points to remember
- Everyone ages at the same rate over time
- Use a table to organize present ages and ages at different time periods
- Assign variables to ages in the time frame asked about in the question
- Past ages = present age - number of years
- Future ages = present age + number of years
- Relationships between ages remain proportionally consistent over time
Work Problems
Examples
Carlos can paint a house in 8 hours. David can paint the same house in 6 hours. How long will it take them working together?
Solution
Carlos's rate: 1/8 house per hour. David's rate: 1/6 house per hour. Combined rate: 1/8 + 1/6 = 3/24 + 4/24 = 7/24 house per hour. Time = Work/Rate = 1 ÷ (7/24) = 24/7 ≈ 3.43 hours.
We find each person's rate by taking 1/time, add the rates for combined rate, then divide 1 complete job by combined rate to get time.
Machine A can produce 200 items in 4 hours. Machine B can produce the same 200 items in 5 hours. If they work together for 2 hours, how many items will they produce?
Solution
Machine A rate: 200/4 = 50 items/hour. Machine B rate: 200/5 = 40 items/hour. Combined rate: 50 + 40 = 90 items/hour. In 2 hours: 90 × 2 = 180 items.
We calculate rates in items per hour, add them for combined rate, then multiply by time to get total items produced.
Work problems involve calculating how long it takes people or machines to complete jobs working alone or together. The fundamental principle is Work = Rate × Time. If someone can complete a job in n hours, their rate is 1/n of the job per hour. When people work together, their combined rate equals the sum of their individual rates.
Applications
- Construction and manufacturing scheduling
- Team productivity calculations
- Resource allocation and planning
Common misconceptions
- Adding times instead of rates when working together
- Forgetting to convert rates to common units
- Incorrectly calculating individual rates from given information
Related concepts
- Rates and ratios
- Fractions and decimals
- Time and efficiency
Common exam questions
Example
Person A completes job in 6 hours, Person B in 4 hours, find time working together
Approach
Find individual rates, add for combined rate, use Work = Rate × Time
Example
A works for 3 hours alone, then A and B work together for 2 hours to finish
Approach
Calculate work done by each person/machine, ensure total equals 1
Key points to remember
- Work = Rate × Time, so Rate = Work/Time and Time = Work/Rate
- If a job takes n hours alone, the rate is 1/n job per hour
- Combined rate = sum of individual rates when working together
- Use fractions to represent portions of work completed
- Total work completed by all workers = 1 (complete job)
Motion Problems
Examples
Two cars start 240 km apart and drive toward each other. Car A drives at 60 kph, Car B at 40 kph. When will they meet?
Solution
Let t = time until they meet. Car A travels 60t km, Car B travels 40t km. Total distance: 60t + 40t = 240. 100t = 240. t = 2.4 hours.
Since they drive toward each other, their combined distances equal the initial separation. We add their distances and set equal to 240 km.
A train leaves Manila at 8 AM traveling at 80 kph. Another train leaves at 10 AM traveling at 100 kph in the same direction. When will the second train overtake the first?
Solution
Let t = hours after 10 AM. First train travels for (t+2) hours at 80 kph: 80(t+2) km. Second train travels for t hours at 100 kph: 100t km. When they meet: 80(t+2) = 100t. 80t + 160 = 100t. 160 = 20t. t = 8 hours after 10 AM = 6 PM.
The first train has a 2-hour head start, so it travels for t+2 hours. When overtaking occurs, both trains have traveled the same distance from Manila.
Motion problems use the fundamental relationship Distance = Speed × Time (D = ST). The key is understanding different motion scenarios: objects moving toward each other (distances add), objects moving in the same direction (distance equals difference), and overtaking problems (distances become equal). Using a table to organize distance, speed, and time for each object helps solve these problems systematically.
Applications
- Transportation planning and scheduling
- Sports and racing calculations
- Navigation and travel time estimation
Common misconceptions
- Forgetting to account for head starts in time calculations
- Using wrong relationship for objects moving in same vs opposite directions
- Mixing up when to add versus equate distances
Related concepts
- Speed, distance, and time relationships
- Linear equations
- Rate problems
Common exam questions
Example
Two cyclists 90 km apart ride toward each other at 15 kph and 12 kph
Approach
Add distances when moving toward each other, set sum equal to initial separation
Example
Faster car leaves 1 hour later but catches up to slower car
Approach
Set distances equal when faster object catches slower one, account for head starts
Key points to remember
- Distance = Speed × Time (D = ST), so Speed = D/T and Time = D/S
- When objects move toward each other, total distance = sum of individual distances
- When one object overtakes another, their distances traveled are equal
- When objects move apart, total separation = sum of distances
- Always use consistent units (km and hours, or meters and seconds)
Mixture Problems
Examples
How much pure alcohol must be added to 400 mL of 20% alcohol solution to make it 35% alcohol?
Solution
Let x = mL of pure alcohol to add. Original solution: 400 mL at 20% = 0.20 × 400 = 80 mL alcohol. Pure alcohol: x mL at 100% = x mL alcohol. Final: (400+x) mL at 35% = 0.35(400+x) mL alcohol. 80 + x = 0.35(400+x). 80 + x = 140 + 0.35x. 0.65x = 60. x = 92.31 mL.
We track alcohol content: original alcohol + added alcohol = final alcohol content. The equation balances solute amounts before and after mixing.
A chemist has 300 mL of 15% acid solution. How much 45% acid solution must be added to make a 30% solution?
Solution
Let x = mL of 45% solution to add. Original: 300 × 0.15 = 45 mL pure acid. Added: x × 0.45 = 0.45x mL pure acid. Final: (300+x) × 0.30 = 0.30(300+x) mL pure acid. 45 + 0.45x = 0.30(300+x). 45 + 0.45x = 90 + 0.30x. 0.15x = 45. x = 300 mL.
We set up the equation by equating the total acid before and after mixing. The 45% solution contributes 0.45x mL of pure acid.
Mixture problems involve combining solutions with different concentrations to achieve a desired final concentration. The key principle is that the amount of pure substance (solute) in the final mixture equals the sum of pure substances from all original solutions. We track both the total volume and the volume of pure solute in each solution.
Applications
- Chemistry laboratory preparations
- Manufacturing and quality control
- Food and beverage production
- Pharmaceutical preparations
Common misconceptions
- Adding percentages directly instead of calculating actual solute amounts
- Forgetting to convert percentages to decimals in calculations
- Incorrectly setting up the final concentration equation
Related concepts
- Percentages and decimals
- Ratios and proportions
- Systems of linear equations
Common exam questions
Example
How much water to add to 25% salt solution to make it 15%?
Approach
Add pure solvent (0% concentration) to reduce overall concentration
Example
Mix 10% and 40% solutions to get 25% concentration
Approach
Add higher concentration solution or pure solute to increase overall concentration
Key points to remember
- Amount of solute = Concentration × Total volume
- Final solute amount = sum of all original solute amounts
- Concentration is usually given as a percentage (divide by 100 for calculations)
- Pure substance = 100% concentration, pure solvent = 0% concentration
- Use a table to organize concentration, volume, and solute amount for each solution
Investment Problems
Examples
Ana invested ₱15,000, part at 4% annual interest and the rest at 6% annual interest. If her total annual interest is ₱780, how much did she invest at each rate?
Solution
Let x = amount at 4%, then 15000-x = amount at 6%. Interest from 4%: 0.04x. Interest from 6%: 0.06(15000-x). Total interest: 0.04x + 0.06(15000-x) = 780. 0.04x + 900 - 0.06x = 780. -0.02x = -120. x = 6000. She invested ₱6,000 at 4% and ₱9,000 at 6%.
We express one investment in terms of the other, calculate interest from each using I=PRT, then sum the interests to equal the total given interest.
₱8,000 is invested at 5% simple annual interest. How much interest is earned and what is the total value after 18 months?
Solution
P = ₱8,000, R = 5% = 0.05, T = 18 months = 1.5 years. Interest = P × R × T = 8000 × 0.05 × 1.5 = ₱600. Total value = Principal + Interest = ₱8,000 + ₱600 = ₱8,600.
We convert 18 months to 1.5 years to match the annual rate, then apply the simple interest formula directly.
Investment problems involve calculating interest earned on money invested at different rates for different time periods. The fundamental formula is Interest = Principal × Rate × Time. For problems involving multiple investments, we set up equations based on the total amount invested and the relationship between the interest earned from different investments.
Applications
- Personal financial planning
- Bank investment strategies
- Business capital allocation
- Retirement and savings planning
Common misconceptions
- Forgetting to convert time units to match the interest rate period
- Using percentages directly instead of converting to decimals
- Confusing simple interest with compound interest formulas
Related concepts
- Percentages and decimals
- Systems of linear equations
- Financial mathematics
Common exam questions
Example
₱20,000 invested at two different rates totaling ₱1,200 annual interest
Approach
Set up equations for total principal and total interest, solve the system
Example
Find interest on ₱5,000 at 6% annual rate for 8 months
Approach
Apply I = PRT directly, ensuring time units match rate units
Key points to remember
- Simple Interest: I = P × R × T (Interest = Principal × Rate × Time)
- Total amount after interest = Principal + Interest
- Rate is usually given as annual percentage (divide by 100 for calculations)
- Time must match the rate period (annual rate needs time in years)
- For multiple investments: sum of principals = total invested, sum of interests = total interest
Practice problems
The sum of three consecutive odd integers is 75. Find the three integers.
Solution
Let x = first odd integer, x+2 = second, x+4 = third. x + (x+2) + (x+4) = 75. 3x + 6 = 75. 3x = 69. x = 23. The integers are 23, 25, 27.
Consecutive odd integers differ by 2, so we use x, x+2, x+4. We verify: 23+25+27 = 75 ✓, and all three are odd consecutive integers.
Rosa is twice as old as her brother today. In 5 years, Rosa will be 1.5 times as old as her brother. Find their current ages.
Solution
Let x = brother's current age, 2x = Rosa's current age. In 5 years: brother is x+5, Rosa is 2x+5. Rosa will be 1.5 times brother's age: 2x+5 = 1.5(x+5). 2x+5 = 1.5x+7.5. 0.5x = 2.5. x = 5. Brother is 5, Rosa is 10.
We set up the future relationship and solve. Check: In 5 years, brother will be 10, Rosa will be 15. 15 = 1.5 × 10 ✓
Machine X can complete a job in 12 hours. Machine Y can complete the same job in 8 hours. If they work together for 3 hours, what fraction of the job is completed?
Solution
Machine X rate: 1/12 job per hour. Machine Y rate: 1/8 job per hour. Combined rate: 1/12 + 1/8 = 2/24 + 3/24 = 5/24 job per hour. In 3 hours: (5/24) × 3 = 15/24 = 5/8 of the job completed.
We find individual rates, add them for combined rate, then multiply by time to get the fraction of work completed.
Two cars start from the same point and travel in opposite directions. Car A travels at 65 kph and Car B at 55 kph. After how many hours will they be 360 km apart?
Solution
Let t = time in hours. Car A travels 65t km, Car B travels 55t km. Since they go in opposite directions, total distance = 65t + 55t = 120t. 120t = 360. t = 3 hours.
When traveling in opposite directions, we add their distances to get total separation. After 3 hours: A travels 195 km, B travels 165 km, total 360 km apart.
How much pure water should be added to 200 mL of 30% salt solution to reduce it to 18% concentration?
Solution
Let x = mL of pure water to add. Original salt: 200 × 0.30 = 60 mL. Added water has 0 mL salt. Final: (200+x) mL at 18% = 0.18(200+x) mL salt. 60 = 0.18(200+x). 60 = 36 + 0.18x. 24 = 0.18x. x = 133.33 mL.
Pure water adds volume but no salt, so salt amount stays 60 mL while total volume increases, reducing concentration.
₱12,000 is invested in two accounts. One earns 3% annual interest, the other 5%. If the total annual interest is ₱480, how much is invested at each rate?
Solution
Let x = amount at 3%, then 12000-x = amount at 5%. 0.03x + 0.05(12000-x) = 480. 0.03x + 600 - 0.05x = 480. -0.02x = -120. x = 6000. ₱6,000 at 3%, ₱6,000 at 5%.
We set up the interest equation: interest from first account plus interest from second account equals total interest. Check: 6000×0.03 + 6000×0.05 = 180 + 300 = 480 ✓
Exam preparation tips
- Always start by clearly defining your variables and what they represent
- Create tables or organize information systematically for complex problems
- Check your answer by substituting back into the original problem conditions
- Practice translating common phrases into mathematical expressions
- Pay attention to units and ensure they are consistent throughout your calculations
- Draw diagrams for motion problems to visualize the situation
- For mixture problems, always track the amount of pure substance, not just percentages
- In age problems, remember that everyone ages at the same rate
- For work problems, focus on rates (work per unit time) rather than total times
- Read the problem multiple times to ensure you understand what is being asked
- Practice different types of word problems to recognize patterns and approaches
- Time yourself on practice problems to improve speed for exam conditions
Diagrams
Word Problem Types and Solution Strategies
A comprehensive mind map showing all six major types of word problems and their key subtypes, helping students organize their approach to different problem categories.
The 3 R's and ESP Problem-Solving Process
A step-by-step flowchart illustrating the systematic 3 R's and ESP approach to solving any word problem, with a feedback loop for verification.
Motion Problem Decision Tree
A decision tree helping students determine which approach to use for different motion problem scenarios based on the direction and relative speeds of moving objects.
Mixture Problem Solution Strategy
A systematic flowchart for approaching mixture problems, emphasizing the importance of tracking pure substance amounts and using a structured table format.
Common Word Problem Types in UPCAT
A pie chart showing the typical distribution of different word problem types in UPCAT and similar college entrance exams, helping students prioritize their study focus.
In summary
Mastering word problems requires systematic approach, consistent practice, and strong foundational skills in translating between English and mathematical expressions. The 3 R's and ESP method (Read, Represent, Relate, Equate, Solve, Prove) provides a reliable framework for tackling any word problem type. Remember that success comes from understanding the underlying mathematical relationships rather than memorizing specific problem types. Regular practice with various problem categories will build your confidence and speed, essential for success in college entrance examinations like UPCAT, ACET, and USTET. Focus on accuracy first, then work on improving your solving speed through consistent practice and pattern recognition.