Word Problems — Speed/Distance/Age, Discount & Interest - Slides
This comprehensive chapter covers essential word problem types commonly encountered in Philippine examinations including UPCAT, CSE, and other standardized tests. Students will master problem-solving techniques for speed-distance-time relationships, age problems, discount calculations, and simple interest computations. These skills are fundamental for numerical ability sections and practical life applications.
Slides
Chapter Overview: Word Problems in Numerical Ability
Notes
Start with overview to show how all topics connect
Examples
- A jeepney travels from Manila to Baguio - speed problem
- Maria is twice as old as her brother - age problem
- Department store offers 20% discount - discount problem
- Bank loan with 5% annual interest - simple interest problem
Word problems test your ability to translate real-world situations into mathematical equations. These problem types frequently appear in entrance exams and civil service tests, requiring both mathematical skills and logical reasoning.
Main points
- Four main categories of word problems in Philippine exams
- Speed/Distance/Time relationships and applications
- Age problem solving strategies and techniques
- Discount calculations in business mathematics
- Simple interest formulas and real-world applications
Diagram
Mind map showing the four main categories of word problems and their key components
Speed, Distance, and Time Fundamentals
Notes
Essential foundation formulas that students must memorize
Examples
- A car travels 90 km in 1.5 hours. Speed = 90 ÷ 1.5 = 60 km/h
- At 50 km/h for 3 hours. Distance = 50 × 3 = 150 km
- 300 km at 75 km/h. Time = 300 ÷ 75 = 4 hours
The relationship between speed, distance, and time forms the foundation for solving motion problems. These three quantities are interconnected, and knowing any two allows you to find the third using basic formulas.
Main points
- Speed = Distance ÷ Time
- Distance = Speed × Time
- Time = Distance ÷ Speed
- Units: km/h, m/s, mph for speed; km, m, miles for distance; hours, minutes, seconds for time
- Always check unit consistency in problems
Diagram
Decision flowchart for choosing the correct speed-distance-time formula
Unit Conversion in Speed Problems
Notes
Critical skill for avoiding errors in speed problems
Examples
- 72 km/h to m/s: 72 × 5/18 = 20 m/s
- 25 m/s to km/h: 25 × 18/5 = 90 km/h
- 2.5 hours = 2.5 × 60 = 150 minutes
- Convert everything to same units before solving
Unit conversion is crucial in speed problems as different units are often mixed. Converting to consistent units prevents calculation errors and ensures accurate answers.
Main points
- km/h to m/s: multiply by 5/18 or divide by 3.6
- m/s to km/h: multiply by 18/5 or multiply by 3.6
- Time conversions: 1 hour = 60 minutes = 3600 seconds
- Distance conversions: 1 km = 1000 m = 100,000 cm
- Always convert to consistent units before calculating
Diagram
Unit conversion relationships for speed, time, and distance
Average Speed Problems
Notes
Emphasize this is NOT arithmetic mean of speeds
Examples
- Trip 1: 120 km in 2 hours (60 km/h)
- Trip 2: 180 km in 3 hours (60 km/h)
- Total: 300 km in 5 hours
- Average speed = 300 ÷ 5 = 60 km/h (not 65 km/h)
Average speed problems require careful calculation of total distance and total time. Students often make the mistake of averaging the speeds directly, which gives incorrect results.
Main points
- Average Speed = Total Distance ÷ Total Time
- NOT the arithmetic mean of individual speeds
- Calculate total distance by adding all segments
- Calculate total time by adding all time intervals
- Common in round-trip and multi-segment journey problems
Diagram
Step-by-step process for calculating average speed in multi-segment journeys
Solving Speed-Distance-Time Word Problems
Notes
Systematic approach prevents errors
Examples
- A bus travels 240 km from Manila to Baguio in 4 hours. Find speed.
- Given: D = 240 km, T = 4 hours, Find: S = ?
- Formula: S = D ÷ T = 240 ÷ 4 = 60 km/h
A systematic approach ensures accuracy in speed problems. Following these steps helps organize information and prevents common calculation errors.
Main points
- Step 1: Identify what is given and what to find
- Step 2: Check and convert units if necessary
- Step 3: Choose appropriate formula (S=D/T, D=S×T, T=D/S)
- Step 4: Substitute values and calculate
- Step 5: Verify answer makes sense and has correct units
Diagram
Problem-solving flowchart for speed-distance-time word problems
Age Problems: Basic Concepts
Notes
Emphasize that age differences never change
Examples
- Maria is currently 20, Juan is 15. Maria is 5 years older.
- In 10 years: Maria will be 30, Juan will be 25. Still 5 years difference.
- 5 years ago: Maria was 15, Juan was 10. Still 5 years difference.
Age problems test logical thinking about how ages change over time while maintaining constant relationships. Understanding that age differences never change is crucial for solving these problems.
Main points
- Age problems involve relationships between ages at different times
- Present age, past age, and future age relationships
- Key principle: Age differences remain constant over time
- If A is 5 years older than B now, A will always be 5 years older than B
- Use variables to represent unknown ages
Diagram
Timeline showing how age difference remains constant over time
Setting Up Age Problem Equations
Notes
Show systematic approach to variable assignment
Examples
- Let x = Ana's present age
- Ana is 3 times as old as Ben: Ben's age = x/3
- In 5 years: Ana = x+5, Ben = x/3+5
- 5 years ago: Ana = x-5, Ben = x/3-5
Age problems require careful translation of word statements into algebraic equations. Using variables and expressing relationships mathematically is key to finding solutions.
Main points
- Let x = present age of one person
- Express other ages in terms of x
- Past ages: subtract years from present age
- Future ages: add years to present age
- Set up equation based on given relationship
Diagram
Process for setting up and solving age problem equations
Age Problem Examples and Solutions
Notes
Complete worked example with verification
Examples
- Son's present age: x = 10 years
- Father's present age: 4x = 40 years
- Check: In 20 years, son = 30, father = 60
- 60 = 2 × 30 ✓ (Father is twice as old as son)
This example demonstrates the complete process of solving age problems, from variable assignment through equation setup to final solution verification.
Main points
- Example: Father is 4 times as old as son. In 20 years, father will be twice as old as son.
- Let x = son's present age, then father's present age = 4x
- In 20 years: son = x + 20, father = 4x + 20
- Equation: 4x + 20 = 2(x + 20)
- Solution: 4x + 20 = 2x + 40, so 2x = 20, x = 10
Diagram
Step-by-step solution of age problem example
Discount Problems: Business Mathematics Basics
Notes
Foundation concepts for discount calculations
Examples
- Shirt has ₱500 list price with 20% discount
- Discount amount = ₱500 × 0.20 = ₱100
- Selling price = ₱500 - ₱100 = ₱400
- Customer pays ₱400 for the shirt
Discount problems are common in business mathematics and consumer situations. Understanding the relationship between list price, discount, and selling price is essential for solving these problems.
Main points
- List Price (Marked Price): Original price tag on item
- Selling Price: Actual price paid by customer
- Discount: Amount reduced from list price
- Discount Rate: Percentage reduction from list price
- Profit/Loss: Based on cost price vs selling price
Diagram
Calculation flow for finding selling price after discount
Discount Formulas and Calculations
Notes
Different formulas for different problem types
Examples
- Find selling price: ₱1500 with 20% discount = ₱1500 × 0.8 = ₱1200
- Find discount rate: ₱150 list, ₱120 selling = (30÷150) × 100 = 20%
- Find list price: ₱800 selling with 20% discount = ₱800 ÷ 0.8 = ₱1000
These formulas provide different approaches to solve discount problems depending on what information is given and what needs to be found.
Main points
- Discount Amount = List Price × Discount Rate
- Selling Price = List Price - Discount Amount
- Alternative: Selling Price = List Price × (1 - Discount Rate)
- Discount Rate % = (Discount Amount ÷ List Price) × 100
- List Price = Selling Price ÷ (1 - Discount Rate)
Diagram
Decision tree for choosing appropriate discount formula
Types of Discounts
Notes
Important to distinguish between discount types
Examples
- Trade: Wholesaler gives 15% to retailer
- Quantity: Buy 100+ items, get 10% off
- Promotional: New product launch, 25% off
- Multiple: 20% + 10% = 28% total discount, not 30%
Different types of discounts serve various business purposes. Understanding these types helps in solving real-world discount problems and business situations.
Main points
- Trade Discount: Given by distributor to retailer (not to end customer)
- Quantity Discount: For buying large quantities or bulk purchases
- Promotional Discount: For new product promotion or stock clearance
- Seasonal Discount: During specific seasons or holidays
- Multiple discounts: Applied successively, not additively
Diagram
Mind map showing different types of discounts and their characteristics
Simple Interest: Basic Concepts
Notes
Emphasize interest only on principal, not on interest
Examples
- Borrow ₱10,000 at 8% annual simple interest for 3 years
- Principal = ₱10,000, Rate = 8% = 0.08, Time = 3 years
- Interest earned only on the original ₱10,000
Simple interest is the most basic form of interest calculation, used in many loans and investments. Unlike compound interest, it doesn't earn interest on previously earned interest.
Main points
- Simple Interest: Interest calculated only on principal amount
- Principal (P): Original amount borrowed or invested
- Rate (R): Annual interest rate (as decimal: 5% = 0.05)
- Time (T): Period in years
- Interest (I): Amount earned or paid for use of money
Diagram
Components of simple interest calculation
Simple Interest Formulas and Applications
Notes
Students need to identify what's given and what to find
Examples
- Find interest: I = ₱5000 × 0.06 × 2 = ₱600
- Find principal: P = ₱600 ÷ (0.06 × 2) = ₱5000
- Find rate: R = ₱600 ÷ (₱5000 × 2) = 0.06 = 6%
- Maturity value: F = ₱5000 + ₱600 = ₱5600
These formulas allow calculation of any unknown component when the other three are given. The maturity value represents the total amount to be received or paid.
Main points
- Interest Formula: I = P × R × T
- Principal Formula: P = I ÷ (R × T)
- Rate Formula: R = I ÷ (P × T)
- Time Formula: T = I ÷ (P × R)
- Maturity Value: F = P + I = P(1 + RT)
Diagram
Decision tree for choosing appropriate simple interest formula
Time Conversion in Interest Problems
Notes
Critical step often overlooked by students
Examples
- 9 months = 9/12 = 0.75 years
- 18 months = 18/12 = 1.5 years
- 90 days = 90/365 ≈ 0.247 years
- 2 years 6 months = 2.5 years
Time conversion is critical in interest problems because the standard formula assumes annual rates and time in years. Incorrect time conversion leads to wrong answers.
Main points
- Interest rate is typically annual, so time must be in years
- Months to years: divide by 12 (6 months = 6/12 = 0.5 years)
- Days to years: divide by 365 (73 days = 73/365 = 0.2 years)
- Quarters to years: divide by 4 (3 quarters = 3/4 = 0.75 years)
- Always convert time to years before using I = PRT
Diagram
Time conversion process for simple interest calculations
Problem-Solving Strategies and Tips
Notes
Emphasize systematic approach for exam success
Examples
- Speed problem: Look for distance and time clues
- Age problem: Watch for time references (past, present, future)
- Discount: Identify list price vs selling price
- Interest: Check if rate is annual and convert time accordingly
Successful problem-solving requires systematic approach and careful attention to details. These strategies help avoid common errors and improve accuracy in examinations.
Main points
- Read problems carefully and identify given information
- Draw diagrams or tables to organize information when helpful
- Convert units consistently before calculations
- Check if answer is reasonable and has correct units
- Practice identifying problem type quickly in exams
Diagram
Universal problem-solving strategy for all word problem types
In summary
Mastering word problems in speed/distance/time, age, discount, and simple interest requires understanding fundamental formulas, systematic problem-solving approaches, and consistent practice. These problem types are essential for success in Philippine entrance examinations and civil service tests. Remember to always identify what is given and what to find, convert units appropriately, and verify that your answers are reasonable. Regular practice with varied problem types will build confidence and speed in examinations.