Civil Service Exam (Subprofessional) Numerical Ability — Word Problems — Speed/Distance/Age, Discount & InterestStudy Notes
Study notes for Word Problems — Speed/Distance/Age, Discount & Interest that match the Civil Service Exam (Subprofessional) 2026 syllabus. Built to mirror how Civil Service Commission (CSC) structures Civil Service Exam (Subprofessional) Numerical Ability questions, these notes walk through each concept with examples, formulas, and practice questions designed for time-pressured exam conditions.
Exam context
The Career Service Examination — Subprofessional Level is conducted by Civil Service Commission (CSC) and is scheduled for Bi-annual — March and August 2026. The Numerical Ability subtest is marked as "~25% weightage" in the official pattern, and Word Problems — Speed/Distance/Age, Discount & Interest appears in position 6th of 9 in the Civil Service Exam (Subprofessional) Numerical Ability review rotation. Passing mark: 80%. Recent Civil Service Exam (Subprofessional) 2026 papers have drawn roughly 17 questions from this subject.
Word Problems — Speed/Distance/Age, Discount & Interest - Study notes
Word problems are one of the most challenging yet important parts of numerical ability tests in Philippine entrance examinations like UPCAT, CSE, and others. This chapter covers four essential types of word problems: Speed/Distance/Time relationships, Age problems, Discount calculations, and Interest computations. Mastering these concepts requires understanding the underlying formulas, identifying key information in word problems, and applying logical problem-solving strategies. These topics are frequently tested because they combine mathematical skills with real-world applications that students will encounter in business, finance, and everyday life.
Summary
Word problems in Speed/Distance/Age, Discount, and Interest are essential skills for Philippine entrance examinations. Speed problems use the relationship between distance, speed, and time, with special attention to average speed calculations. Age problems require setting up equations based on relationships between ages in different time periods. Discount problems involve calculating price reductions and percentages in business contexts. Simple interest grows linearly on the principal amount, while compound interest grows exponentially on the accumulated balance. Success in these problems requires: systematic problem-solving approaches, careful identification of given information and requirements, proper use of formulas and unit conversions, organized presentation of work, and verification of answers. Regular practice with various problem types builds pattern recognition and solution speed, essential for exam success.
Sections
Speed, distance, and time problems are fundamental in physics and everyday applications. The basic relationship is expressed through three interconnected formulas: Speed = Distance ÷ Time, Distance = Speed × Time, and Time = Distance ÷ Speed. These problems often involve unit conversions (km/hr to m/s, etc.) and require careful attention to given information. For conversion: to change km/hr to m/s, multiply by 5/18; to change m/s to km/hr, multiply by 18/5. Average speed problems require special attention - average speed equals total distance divided by total time, NOT the average of individual speeds. When solving these problems, create a systematic approach: identify what's given, what's missing, choose the appropriate formula, and solve step by step.
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Speed, Distance, and Time Problems
Examples
- A car travels 210 km in 3 hours. Find the speed: Speed = 210 km ÷ 3 hrs = 70 km/hr
- Convert 70 km/hr to m/s: 70 × (5/18) = 19.44 m/s
- Average speed problem: Drive 180 miles at 60 mph, return at 90 mph. Time1 = 180÷60 = 3 hrs, Time2 = 180÷90 = 2 hrs. Average speed = 360 miles ÷ 5 hrs = 72 mph
Key Points
- Basic formula triangle: Speed = Distance ÷ Time
- Always check units and convert when necessary
- Average speed = Total distance ÷ Total time
- Never average individual speeds to get average speed
- Organize information in tables for complex problems
Age problems involve relationships between people's ages in the past, present, or future. The key to solving age problems is setting up equations based on the given relationships and using variables to represent unknown ages. Important relationships include: if present age is x, then age n years ago is (x-n), age n years hence is (x+n), and if someone is n times as old, multiply by n. Create a systematic table showing past, present, and future ages for all people involved. Set up equations based on the relationships described in the problem, then solve for the unknown variables. Always verify your answer by checking if it satisfies all given conditions.
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Age Word Problems
Examples
- 4 years ago, Karylle was 3 times as old as Kylle. In 3 years, Karylle will be 2 times as old as Kylle. Let K = Karylle's current age, y = Kylle's current age. Set up: K-4 = 3(y-4) and K+3 = 2(y+3). Solve to get y = 11 years
- Father is 3 times as old as son. In 12 years, father will be twice as old as son. If son's current age is x, then father's age is 3x. Set up: 3x + 12 = 2(x + 12). Solve: x = 12 years (son), father = 36 years
Key Points
- Use variables to represent unknown current ages
- Present age ± n years = past/future age
- Set up equations based on given relationships
- Create tables to organize age information
- Always verify answers against all given conditions
Discount problems involve calculating reduced prices and discount percentages in business transactions. Key concepts include List Price (original marked price), Selling Price (actual selling price), and Discount (amount reduced). The basic formulas are: Discount = List Price - Selling Price, Discount % = (Discount ÷ List Price) × 100, and Selling Price = List Price - Discount. Alternative method: Selling Price = List Price × (100% - Discount%). There are different types of discounts: Trade discount (distributor to retailer), Quantity discount (bulk purchases), and Promotional discount (marketing purposes). When solving discount problems, identify the given values, determine what needs to be calculated, apply the appropriate formula, and always express percentages correctly.
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Discount Problems
Examples
- List price ₱1500, 20% discount. Discount = ₱1500 × 0.20 = ₱300. Selling price = ₱1500 - ₱300 = ₱1200
- Alternative method: Selling price = ₱1500 × (100% - 20%) = ₱1500 × 80% = ₱1200
- Find discount rate: List price ₱150, discount ₱30. Discount rate = (₱30 ÷ ₱150) × 100 = 20%
Key Points
- Discount = List Price - Selling Price
- Discount % = (Discount ÷ List Price) × 100
- Selling Price = List Price × (100% - Discount%)
- Different types: Trade, Quantity, Promotional discounts
- Always convert percentages to decimals for calculations
Simple interest is calculated only on the principal amount throughout the entire period. The fundamental formula is I = PRT, where I = Interest, P = Principal, R = Rate (as decimal), T = Time (in years). Related formulas include: P = I ÷ (RT), R = I ÷ (PT), T = I ÷ (PR), and Maturity Value F = P + I or F = P(1 + RT). Key terms include Principal (original amount), Interest (amount paid for use of money), Rate (annual percentage), Time (period in years), and Maturity Value (final amount). When solving, convert percentages to decimals, convert months to years by dividing by 12, identify what's being asked, substitute into appropriate formula, and solve systematically.
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Simple Interest Problems
Examples
- ₱60,000 borrowed for 9 months at 10% annual rate. I = ₱60,000 × 0.10 × (9÷12) = ₱60,000 × 0.10 × 0.75 = ₱4,500
- Find principal: Interest ₱11,500, rate 8%, time 2 years. P = ₱11,500 ÷ (0.08 × 2) = ₱71,875
- Maturity value: ₱2M at 0.50% for 5 years. F = ₱2M(1 + 0.005 × 5) = ₱2M × 1.025 = ₱2.05M
Key Points
- Simple Interest formula: I = PRT
- Convert rate percentages to decimals
- Convert months to years (divide by 12)
- Maturity Value = Principal + Interest
- Interest is calculated only on principal amount
Compound interest is calculated on both the principal and previously earned interest, causing exponential growth. The key formula is F = P(1 + r)^t for future value, P = F ÷ (1 + r)^t for present value, and Compound Interest = F - P. Unlike simple interest, compound interest grows faster because interest earns interest. When solving compound interest problems, identify the principal, rate (as decimal), and time period, substitute into the appropriate formula, use calculator for exponential calculations, and find the required value (future value, present value, or compound interest). The difference between simple and compound interest becomes more significant over longer time periods and higher interest rates.
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Compound Interest Problems
Examples
- ₱15,000 compounded annually at 3% for 3 years. F = ₱15,000(1 + 0.03)^3 = ₱15,000(1.092727) = ₱16,390.91
- Compound interest = ₱16,390.91 - ₱15,000 = ₱1,390.91
- Present value: ₱40,000 due in 5 years at 9%. P = ₱40,000 ÷ (1.09)^5 = ₱40,000 ÷ 1.5386 = ₱25,997.26
Key Points
- Formula: F = P(1 + r)^t for future value
- Interest is calculated on principal plus accumulated interest
- Compound interest = Future value - Principal
- Growth is exponential, not linear like simple interest
- Use calculator for exponential calculations
Successful word problem solving requires a systematic approach. Start by reading the problem carefully and identifying what information is given and what needs to be found. Look for keywords that indicate the type of problem and the operations needed. Set up equations or use appropriate formulas based on the problem type. For complex problems, create tables or diagrams to organize information. Always check units and convert when necessary. Verify your answer by substituting back into the original problem conditions. Practice identifying problem types quickly by recognizing common patterns and keywords. Time management is crucial in exams - if stuck on a problem, move on and return later.
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Problem-Solving Strategies
Examples
- Speed problems: Look for words like 'travels', 'speed', 'distance', 'time', 'average speed'
- Age problems: Look for phrases like 'years ago', 'years from now', 'times as old'
- Discount problems: Look for 'list price', 'selling price', 'discount', 'marked down'
- Interest problems: Look for 'borrowed', 'invested', 'interest rate', 'annual', 'compound'
Key Points
- Read problems carefully and identify given information
- Look for keywords that indicate problem type
- Create tables or diagrams for complex problems
- Always verify answers by checking original conditions
- Practice pattern recognition for quick problem identification
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