Word Problems Application — Study Notes
Overview
Word problems are the bridge between classroom mathematics and real-world scenarios. In SOF IMO, these problems test whether you can read a situation, translate it into mathematical language, solve it using the correct formulas or methods, and interpret the answer back in context. They typically appear in both the Everyday Mathematics section and the Achievers Section.
Expect problems drawn from percentage, profit-loss, time-speed-distance, ratios, linear equations, geometry (areas, perimeters), mensuration (volumes, surface areas), statistics, and number theory. The key skill is **problem comprehension**: identifying what is given, what is asked, and which mathematical operation or formula applies. Most students lose marks not because they can't do the math, but because they misread the question or set up the equation incorrectly.
Mastering word problems requires practice in recognizing common patterns (age problems, mixture problems, train problems), translating words into algebraic expressions, and always checking if your numerical answer makes practical sense. A negative age or a speed of 500 km/h for a bicycle should immediately signal an error.
Key Concepts
- **Read twice, solve once**: Read the entire problem carefully before writing anything. Identify the unknown quantity you need to find and label it clearly (often as *x* or another variable).
- **Unit consistency**: All quantities in a calculation must use the same units. Convert hours to minutes, kilometers to meters, or rupees to paise before applying formulas.
- **Translate phrases into math**: "5 more than a number" → x + 5; "twice the sum of two numbers" → 2(x + y); "A is 20% less than B" → A = 0.8B.
- **Draw diagrams where possible**: For geometry, distance-time, and mixture problems, a simple sketch or number line clarifies relationships and prevents errors.
- **Check reasonableness**: After computing, ask: Does this answer fit the story? If a man's age comes out as 150 years or a discount exceeds 100%, revisit your setup.
- **Work backwards for verification**: Substitute your answer back into the original conditions to confirm it satisfies all the given constraints.
Formulas / Key Facts
1. **Percentage**: Part = (Percentage/100) × Whole; Increase = Original × (1 + r/100); Decrease = Original × (1 − r/100).
2. **Profit and Loss**: Profit = Selling Price − Cost Price; Profit% = (Profit/CP) × 100; Loss% = (Loss/CP) × 100.
3. **Discount**: Discount = Marked Price − Selling Price; Discount% = (Discount/MP) × 100; SP = MP × (1 − Discount%/100).
4. **Simple Interest**: SI = (P × R × T)/100; Amount = Principal + SI.
5. **Compound Interest**: Amount = P(1 + R/100)^T; CI = Amount − Principal.
6. **Speed, Distance, Time**: Distance = Speed × Time; Speed = Distance/Time; Time = Distance/Speed. For relative speed: same direction subtract, opposite direction add.
7. **Time and Work**: Work = Rate × Time; If A does work in *a* days and B in *b* days, together they finish in 1/(1/a + 1/b) days.
8. **Average**: Average = Sum of observations / Number of observations.
9. **Ratio and Proportion**: If a:b = c:d, then ad = bc (cross-multiplication).
10. **Ages**: If current ages are *x* and *y*, then *n* years ago they were x−n and y−n; *n* years hence they will be x+n and y+n.
Worked Examples
**Example 1 — Profit and Loss** *A shopkeeper buys 50 kg of rice at ₹40 per kg. He sells 30 kg at ₹45 per kg and the remaining at ₹38 per kg. Find his overall profit or loss percentage.*
**Solution**: Total Cost Price = 50 × 40 = ₹2000. Selling Price of 30 kg = 30 × 45 = ₹1350. Selling Price of remaining 20 kg = 20 × 38 = ₹760. Total Selling Price = 1350 + 760 = ₹2110. Profit = 2110 − 2000 = ₹110. Profit% = (110/2000) × 100 = 5.5%.
**Example 2 — Time, Speed, Distance** *A train 150 m long passes a pole in 15 seconds. What is its speed in km/h?*
**Solution**: Distance covered = length of train = 150 m (a pole has negligible length). Time = 15 s. Speed = 150/15 = 10 m/s. Convert to km/h: 10 m/s = 10 × (18/5) = 36 km/h.
**Example 3 — Age Problem** *The sum of the ages of a father and son is 60 years. Six years ago, the father was five times as old as the son. Find their present ages.*
**Solution**: Let present age of son = *x* years, father = (60 − *x*) years. Six years ago: son was *x* − 6, father was 60 − *x* − 6 = 54 − *x*. Given: 54 − *x* = 5(*x* − 6) 54 − *x* = 5*x* − 30 54 + 30 = 6*x* *x* = 84/6 = 14. Son's age = 14 years, Father's age = 60 − 14 = 46 years. Verify: Six years ago son was 8, father was 40; 40 = 5×8 ✓.
Common Mistakes
- **Ignoring units**: Adding 2 hours and 30 minutes directly as 2+30=32 instead of converting to the same unit (2.5 hours or 150 minutes). **Fix**: Always convert to a single unit before calculation.
- **Misinterpreting "percentage of"**: Saying "A is 25% more than B" and writing B = 1.25A instead of A = 1.25B. **Fix**: Identify the base quantity carefully; "more than B" means B is the reference, so A = B + 0.25B.
- **Confusing relative speed**: In overtaking or meeting problems, forgetting to add or subtract speeds correctly. **Fix**: Same direction → subtract speeds; opposite direction → add speeds.
- **Setting up wrong variable**: In age or number problems, defining the variable for the wrong quantity or time frame. **Fix**: Clearly write "Let x = present age of son" before forming equations.
- **Skipping unit check at the end**: Computing speed as 10 m/s but the question asks for km/h, and forgetting to convert. **Fix**: Underline the unit requested in the question and convert your final answer.
Quick Reference
- Translate the problem into an equation by identifying knowns, unknowns, and relationships.
- Always use consistent units throughout the calculation.
- Draw a diagram or table for complex word problems (speeds, mixtures, geometry).
- Verify your answer by plugging it back into the original conditions.
- Common IMO word-problem topics: profit-loss, time-work, speed-distance, ages, ratios, mensuration applications.
- Practice recognizing standard problem types to speed up setup and avoid errors.