Civil Service Exam (Subprofessional) Numerical Ability — Word Problems — Speed/Distance/Age, Discount & InterestDetailed Explanation
Detailed explanations for Civil Service Exam (Subprofessional) Numerical Ability — Word Problems — Speed/Distance/Age, Discount & Interest. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Civil Service Commission (CSC) frames Word Problems — Speed/Distance/Age, Discount & Interest questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Civil Service Commission (CSC) runs the Career Service Examination — Subprofessional Level on Bi-annual — March and August 2026. Its Numerical Ability section sits under a "~25% weightage" weighting, and Word Problems — Speed/Distance/Age, Discount & Interest is the 6th chapter in the 9-chapter Civil Service Exam (Subprofessional) Numerical Ability rotation. The Civil Service Exam (Subprofessional) passing mark is 80%, and the most recent 2026 paper drew about 17 questions from Numerical Ability.
Word Problems — Speed/Distance/Age, Discount & Interest - Detailed explanation
Word problems are mathematical scenarios presented in narrative form that require you to extract key information, identify relationships between variables, and apply appropriate formulas to find solutions. This chapter focuses on four fundamental types of word problems commonly found in Philippine entrance exams: Speed/Distance/Time problems, Age problems, Discount problems, and Interest calculations. These problem types are essential for success in major Philippine entrance examinations including UPCAT, CSE, LET, NLE, NMAT, ACET, and USTET. Mastering these concepts requires understanding the underlying mathematical relationships, recognizing problem patterns, and applying systematic problem-solving approaches.
Concepts
Speed, Distance, and Time Problems
Speed, distance, and time problems involve the relationship between how fast an object moves (speed), how far it travels (distance), and how long it takes (time). The fundamental relationship is expressed through three interconnected formulas: Distance = Speed × Time, Speed = Distance ÷ Time, and Time = Distance ÷ Speed. These problems often involve unit conversions, average speed calculations, and scenarios with multiple segments of travel. Understanding these relationships is crucial for solving transportation problems, physics applications, and real-world scenarios involving motion.
Examples
This is a direct application of the speed formula. Simply divide the total distance by the total time to get the average speed.
Scenario
A jeepney travels 120 km in 3 hours. What is its average speed?
Solution
Speed = Distance ÷ Time = 120 km ÷ 3 hours = 40 km/h
For average speed in round trips, calculate the total distance and total time separately. Never average the individual speeds directly.
Scenario
Maria drives to Baguio at 60 km/h and returns via the same route at 40 km/h. If the distance to Baguio is 180 km, what is her average speed for the entire trip?
Solution
Going: Time = 180 km ÷ 60 km/h = 3 hours. Returning: Time = 180 km ÷ 40 km/h = 4.5 hours. Total distance = 360 km, Total time = 7.5 hours. Average speed = 360 km ÷ 7.5 hours = 48 km/h
Applications
- Transportation planning and logistics
- Physics problems involving motion
- Sports timing and performance analysis
- Engineering calculations for vehicles and machinery
- Navigation and GPS systems
Misconceptions
- Thinking average speed is the arithmetic mean of individual speeds
- Forgetting to convert units before calculation
- Confusing distance traveled with displacement in physics contexts
- Not accounting for direction changes in relative motion problems
Related Concepts
- Unit conversion and dimensional analysis
- Linear equations and proportions
- Rate problems in general mathematics
- Physics concepts of velocity and acceleration
Common Exam Questions
Example
A bus travels the first 100 km at 50 km/h and the next 150 km at 75 km/h. Find the average speed.
Approach
Calculate total distance and total time separately, then divide
Question Type
Average Speed Calculation
Example
Two cars start from the same point. One travels at 60 km/h, the other at 80 km/h. When will they be 40 km apart?
Approach
Set up equations based on when distances become equal
Question Type
Meeting/Overtaking Problems
Example
Two trains approaching each other at 70 km/h and 50 km/h. How long to pass if combined length is 300 m?
Approach
Add speeds when moving in opposite directions, subtract when moving in same direction
Question Type
Relative Speed Problems
Key Points To Remember
- The basic formula triangle: D = S × T, S = D ÷ T, T = D ÷ S
- Average speed = Total distance ÷ Total time (not the average of individual speeds)
- Always check units and convert when necessary (km/h to m/s: multiply by 5/18)
- For round trips or multiple segments, calculate each segment separately then find totals
- Speed is always positive; use absolute values when dealing with direction changes
Age Problems
Age problems involve relationships between the ages of two or more people at different points in time - past, present, or future. These problems typically provide information about age ratios, age differences, or relative age comparisons and ask you to determine current ages or ages at specific times. The key to solving age problems is setting up variables for current ages and expressing past or future ages in terms of these variables. Age problems test your ability to work with linear equations, understand time relationships, and manipulate algebraic expressions systematically.
Examples
Set up variables for current ages, express future ages in terms of these variables, then create an equation based on the future relationship.
Scenario
Father is currently 3 times as old as his son. In 12 years, he will be twice as old as his son. Find their current ages.
Solution
Let son's current age = x, father's current age = 3x. In 12 years: Father = 3x + 12, Son = x + 12. Equation: 3x + 12 = 2(x + 12). Solving: 3x + 12 = 2x + 24, x = 12. Son is 12, father is 36.
Use two variables when dealing with two people, set up a system of equations based on the given conditions, and solve simultaneously.
Scenario
The sum of ages of Ana and Ben is 45. Five years ago, Ana was twice as old as Ben. Find their current ages.
Solution
Let Ana's current age = x, Ben's current age = y. x + y = 45. Five years ago: Ana = x - 5, Ben = y - 5. Equation: x - 5 = 2(y - 5). Solving: x - 5 = 2y - 10, x = 2y - 5. Substituting: (2y - 5) + y = 45, 3y = 50, y = 16.67, x = 28.33
Applications
- Family relationship problems in mathematics
- Logic puzzles and brain teasers
- Real-world planning scenarios involving time
- Algebraic word problem foundations
- Critical thinking and problem-solving skills
Misconceptions
- Confusing age differences with age ratios
- Forgetting that age differences remain constant over time
- Setting up incorrect time relationships (adding instead of subtracting for past ages)
- Not checking solutions against all given conditions
Related Concepts
- Linear equations and systems of equations
- Ratio and proportion problems
- Algebraic expression manipulation
- Time-based mathematical modeling
Common Exam Questions
Example
Ages of A and B are in ratio 3:4 now, and will be 4:5 in 6 years. Find current ages.
Approach
Set up ratios for different time periods and create equations
Question Type
Ratio-based Age Problems
Example
Sum of ages of three siblings is 48. Eldest is twice the youngest, middle child is 4 years older than youngest.
Approach
Use multiple variables and create system of equations
Question Type
Multiple Person Age Problems
Example
10 years ago, A was half of B's age. 10 years from now, A will be 3/4 of B's age. Find current ages.
Approach
Express ages at different times in terms of current age variables
Question Type
Past-Present-Future Age Problems
Key Points To Remember
- Let variables represent current ages, then express other ages in terms of these variables
- If current age is x, then age n years ago is (x - n) and age n years hence is (x + n)
- Age differences remain constant over time - this is often the key insight
- Set up equations based on the given relationships and solve systematically
- Always verify your answer by checking it against the original problem conditions
Discount Problems
Discount problems involve calculating reduced prices, percentage savings, and the relationship between original price, discount amount, and final selling price. These problems are essential for business mathematics and consumer applications. Key concepts include understanding different types of discounts (trade, quantity, promotional), calculating discount percentages, and determining selling prices after discounts. The fundamental relationships are: Discount = List Price - Selling Price, Discount Rate = (Discount ÷ List Price) × 100%, and Selling Price = List Price - Discount = List Price × (1 - Discount Rate).
Examples
Calculate the discount amount first, then subtract from original price. Alternatively, multiply original price by (1 - discount rate).
Scenario
A shirt originally costs ₱1,500 and is offered at 20% discount. What is the selling price?
Solution
Discount = ₱1,500 × 0.20 = ₱300. Selling Price = ₱1,500 - ₱300 = ₱1,200. Alternative: Selling Price = ₱1,500 × (1 - 0.20) = ₱1,500 × 0.80 = ₱1,200
Add the discount amount to the selling price to get the original price, then calculate the discount rate as a percentage of the original price.
Scenario
A laptop is sold for ₱45,000 after a ₱15,000 discount. What was the original price and discount rate?
Solution
Original Price = ₱45,000 + ₱15,000 = ₱60,000. Discount Rate = (₱15,000 ÷ ₱60,000) × 100% = 25%
Applications
- Retail and consumer mathematics
- Business profit and loss calculations
- Sales and marketing analysis
- Personal finance and budgeting
- Economic analysis and pricing strategies
Misconceptions
- Adding multiple discount percentages instead of applying them successively
- Confusing discount amount with discount rate
- Calculating discount based on selling price instead of original price
- Forgetting to convert percentages to decimals in calculations
Related Concepts
- Percentage calculations and applications
- Business mathematics and profit/loss
- Consumer mathematics and economics
- Proportional reasoning and scaling
Common Exam Questions
Example
Find selling price of ₱2,000 item with 15% discount
Approach
Apply discount rate to list price to find selling price
Question Type
Single Discount Calculation
Example
Item sold for ₱850 after 15% discount. Find original price.
Approach
Work backwards from selling price to find original price or discount rate
Question Type
Reverse Discount Problems
Example
Item with 20% discount followed by additional 10% discount on reduced price
Approach
Apply discounts successively, not additively
Question Type
Multiple Discount Problems
Key Points To Remember
- List Price is the original marked price; Selling Price is the final price after discount
- Discount Rate = (Discount Amount ÷ List Price) × 100%
- Selling Price = List Price × (100% - Discount Rate) = List Price × (1 - Discount Rate)
- Multiple discounts are applied successively, not additively
- Always convert percentages to decimals before calculation
Simple Interest Problems
Simple interest is calculated only on the principal amount throughout the entire period. The formula I = PRT gives the interest earned, where P is principal, R is the annual rate (as decimal), and T is time in years. The total amount after time T is A = P + I = P(1 + RT). Simple interest problems involve finding any of these variables when the others are known. These calculations are fundamental to personal finance, banking, and investment decisions. Understanding simple interest helps in comparing loan options, calculating returns on investments, and making informed financial decisions.
Examples
Apply the simple interest formula directly, then add interest to principal for total amount.
Scenario
₱50,000 is invested at 6% annual simple interest for 3 years. Find the interest earned and total amount.
Solution
I = PRT = ₱50,000 × 0.06 × 3 = ₱9,000. Total Amount = ₱50,000 + ₱9,000 = ₱59,000
Rearrange the simple interest formula to solve for principal when interest, rate, and time are known.
Scenario
What principal amount will earn ₱2,400 interest in 2 years at 8% annual rate?
Solution
I = PRT, so P = I/(RT) = ₱2,400/(0.08 × 2) = ₱2,400/0.16 = ₱15,000
Applications
- Personal savings and investment planning
- Loan calculations and banking
- Business finance and cash flow management
- Government bonds and treasury bills
- Educational financial literacy
Misconceptions
- Confusing simple interest with compound interest calculations
- Forgetting to convert percentage rates to decimal form
- Using months or days directly without converting to years
- Calculating interest on total amount instead of principal only
Related Concepts
- Compound interest and exponential growth
- Percentage calculations and applications
- Linear growth and arithmetic progressions
- Financial mathematics and economics
Common Exam Questions
Example
Find interest on ₱25,000 at 7% for 4 years
Approach
Apply I = PRT formula directly
Question Type
Direct Simple Interest Calculation
Example
At what rate will ₱10,000 earn ₱1,500 interest in 3 years?
Approach
Rearrange formula to solve for unknown variable
Question Type
Finding Principal, Rate, or Time
Example
Find interest on ₱8,000 at 9% for 8 months
Approach
Convert months/days to years before applying formula
Question Type
Time Conversion Problems
Key Points To Remember
- Formula: I = PRT where I=Interest, P=Principal, R=Rate(decimal), T=Time(years)
- Total Amount = Principal + Interest = P(1 + RT)
- Convert percentage rates to decimals (5% = 0.05)
- Convert time to years if given in months or days
- Interest is calculated only on the original principal amount
Compound Interest Problems
Compound interest is calculated on both the principal amount and previously earned interest. The formula A = P(1 + r)^t gives the final amount, where P is principal, r is the rate per period, and t is the number of periods. The compound interest earned is I = A - P. Unlike simple interest, compound interest grows exponentially because interest earns interest. This concept is crucial for understanding long-term investments, loans with compound interest, and the time value of money. Compound interest demonstrates the power of exponential growth in financial applications.
Examples
Apply the compound interest formula, calculate the power term, then subtract principal to find interest earned.
Scenario
₱100,000 is invested at 5% annual compound interest for 3 years. Find the final amount and interest earned.
Solution
A = P(1 + r)^t = ₱100,000(1 + 0.05)^3 = ₱100,000(1.05)^3 = ₱100,000(1.157625) = ₱115,762.50. Interest = ₱115,762.50 - ₱100,000 = ₱15,762.50
Rearrange the compound interest formula to solve for principal, which gives the present value formula.
Scenario
What principal will grow to ₱200,000 in 4 years at 8% annual compound interest?
Solution
A = P(1 + r)^t, so P = A/(1 + r)^t = ₱200,000/(1.08)^4 = ₱200,000/1.3605 = ₱147,058
Applications
- Long-term investment planning and retirement funds
- Mortgage and loan calculations with compound interest
- Business valuation and future value analysis
- Inflation calculations and purchasing power
- Educational savings plans and financial planning
Misconceptions
- Confusing compound interest formula with simple interest
- Forgetting to raise (1 + r) to the power t
- Mixing up present value and future value concepts
- Not understanding the exponential nature of compound growth
Related Concepts
- Exponential functions and logarithms
- Present value and future value analysis
- Geometric sequences and series
- Financial mathematics and time value of money
Common Exam Questions
Example
Find final amount of ₱75,000 at 6% compounded annually for 5 years
Approach
Apply A = P(1 + r)^t formula directly
Question Type
Future Value Calculation
Example
What amount invested today at 7% will grow to ₱500,000 in 10 years?
Approach
Rearrange formula to P = A/(1 + r)^t
Question Type
Present Value Problems
Example
Compare compound and simple interest on ₱50,000 at 8% for 6 years
Approach
Calculate both and find the difference
Question Type
Compound vs Simple Interest Comparison
Key Points To Remember
- Formula: A = P(1 + r)^t for final amount, I = A - P for compound interest
- Interest is calculated on principal plus all previously earned interest
- Growth is exponential, not linear like simple interest
- More frequent compounding (monthly vs. annually) increases total return
- Small differences in rate or time have large effects due to exponential nature
Practice Problems
Calculate time for each segment separately, then use total distance divided by total time for average speed.
Problem
A tricycle travels the first 15 km of its route at 25 km/h and the remaining 10 km at 20 km/h. What is the average speed for the entire journey?
Solution
Time for first part: 15 km ÷ 25 km/h = 0.6 hours. Time for second part: 10 km ÷ 20 km/h = 0.5 hours. Total distance = 25 km, Total time = 1.1 hours. Average speed = 25 km ÷ 1.1 hours = 22.73 km/h
Set up variables for current ages, use the constant age difference, and create an equation based on the future relationship.
Problem
A mother is currently 28 years older than her daughter. In 8 years, she will be twice as old as her daughter. Find their current ages.
Solution
Let daughter's current age = x, mother's current age = x + 28. In 8 years: daughter = x + 8, mother = x + 28 + 8 = x + 36. Equation: x + 36 = 2(x + 8). Solving: x + 36 = 2x + 16, x = 20. Daughter is 20, mother is 48.
Apply discounts successively, not additively. Each discount is calculated on the current price, not the original price.
Problem
A smartphone originally priced at ₱25,000 is first given a 15% discount, then an additional 10% discount on the reduced price. What is the final selling price?
Solution
After first discount: ₱25,000 × (1 - 0.15) = ₱25,000 × 0.85 = ₱21,250. After second discount: ₱21,250 × (1 - 0.10) = ₱21,250 × 0.90 = ₱19,125
Find the interest amount by subtracting principal from total repaid, then use the simple interest formula to solve for time.
Problem
₱80,000 is borrowed at 9% annual simple interest. If ₱93,600 is repaid after some time, how long was the money borrowed?
Solution
Interest paid = ₱93,600 - ₱80,000 = ₱13,600. Using I = PRT: ₱13,600 = ₱80,000 × 0.09 × T. T = ₱13,600 ÷ (₱80,000 × 0.09) = ₱13,600 ÷ ₱7,200 = 1.89 years ≈ 1 year 11 months
Calculate both simple and compound interest using their respective formulas, then compare the final amounts to see the advantage of compounding.
Problem
Compare the final amounts of ₱60,000 invested for 4 years at 7% annual rate under simple interest versus compound interest.
Solution
Simple Interest: I = ₱60,000 × 0.07 × 4 = ₱16,800. Amount = ₱60,000 + ₱16,800 = ₱76,800. Compound Interest: A = ₱60,000(1.07)^4 = ₱60,000 × 1.3108 = ₱78,648. Difference = ₱78,648 - ₱76,800 = ₱1,848
Exam Preparation Tips
- Master the fundamental formulas for each problem type and practice rearranging them to solve for different variables
- Always identify what type of problem you're dealing with by looking for key words: 'speed', 'distance', 'time', 'age', 'years ago/hence', 'discount', 'interest', etc.
- For speed problems, pay careful attention to units and convert when necessary (km/h to m/s, etc.)
- In age problems, remember that age differences remain constant over time - this is often the key insight needed
- For discount problems, be careful with multiple discounts - apply them successively, not additively
- In interest problems, distinguish between simple and compound interest, and always convert percentages to decimals
- Set up organized tables or charts for complex problems to keep track of given information and relationships
- Always check your answers by substituting back into the original problem to verify they make sense
- Practice identifying problem patterns - many exam questions follow standard templates with different numbers
- Time management is crucial - spend more time on setup and less time on calculation by being systematic
In summary
Mastering word problems in Speed/Distance/Age, Discount, and Interest requires understanding the underlying mathematical relationships, recognizing problem patterns, and applying systematic problem-solving strategies. These four types of problems form the foundation of quantitative reasoning in major Philippine entrance examinations. Success depends on: (1) Accurately identifying problem types and extracting key information, (2) Selecting and applying appropriate formulas correctly, (3) Performing careful calculations with proper unit conversions, (4) Verifying answers for reasonableness and accuracy. Regular practice with varied problem scenarios will build confidence and speed in exam situations. Remember that word problems test not just mathematical computation, but also reading comprehension, logical reasoning, and real-world application skills. Focus on understanding the concepts rather than memorizing solutions, as this will enable you to tackle novel problem variations confidently in actual examinations.
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