FEUCAT Mathematics — Word Problems — Number, Age, Work, Motion, Mixture, InvestmentStudy Notes

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.

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.

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 ✓

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) ✓

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 ✓

Motion problems use the fundamental relationship: Distance = Rate × Time (D = RT) **Common scenarios**: 1. **Meeting problems**: Objects moving toward each other - add distances 2. **Overtaking problems**: One object catching another - distances are equal 3. **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 ✓

Mixture problems involve combining solutions with different concentrations. Key concept: Amount of pure substance = Concentration × Total volume **Setup strategy**: 1. Create a table with columns: % concentration, Volume, Amount of pure substance 2. 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% ✓

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 ✓