Mathematics • Year 10 • Unit 2 • Lesson 12
Word Problems in the Real World
Apply the five-step process from Lesson 12 to real situations — taxi fares, fundraisers, fitness goals, road trips and a school formal. Then explain (in your own words) why every word problem needs the answer checked against the original sentence, not just the equation.
1. Word problems
Each problem uses one or more tools from Lesson 12. Show your working — a final answer with no working only earns half marks.
1.1 — Taxi flag fall and rate. A Sydney taxi charges a $4.20 flag fall plus $2.30 per kilometre.
(a) Define a variable, then write an equation for the total fare C dollars after travelling k kilometres.
(b) Calculate the fare for a 6 km trip.
(c) Sarah's fare is $34.20. Set up an equation and solve for the distance travelled. 4 marks
1.2 — School fundraiser. A Year 10 fundraiser sells chocolate bars for $3 each and cookies for $2 each. They sell 150 items in total and raise $380.
(a) Define c = chocolate bars and k = cookies. Write two equations (one for quantity, one for money).
(b) Use substitution to find c and k.
(c) Check both the total items and the total money raised. 3 marks
1.3 — Park run pace. A jogger runs at 10 km/h. A cyclist starts at the same point 15 minutes later (i.e. 0.25 hour later) at 18 km/h on the same path. Let t = hours after the cyclist starts.
(a) Calculate the jogger's head-start distance.
(b) Write an equation for when the cyclist catches the jogger.
(c) Solve for t and convert your answer to minutes. 3 marks
1.4 — Family ages. Maya is 4 years older than her brother Leo. The sum of their ages is 26. Let L = Leo's current age in whole years.
(a) Write an expression for Maya's age in terms of L.
(b) Write an equation for the sum of their ages and solve.
(c) Check that your answer makes sense in the real world (no negative ages, no fractional years). 3 marks
1.5 — School formal tickets. Adult tickets for the Year 12 formal cost $80 and student tickets cost $50. The school sells 200 tickets in total and collects $11 600.
(a) Define variables for adult tickets and student tickets, then write two equations.
(b) Solve by substitution to find how many of each were sold.
(c) Verify both equations are satisfied. 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A classmate solves a word problem about ages and gets the answer "Leo is −2 years old". They write this as their final answer and move on. Using everything from Lesson 12, explain (i) why this answer is mathematically valid but practically wrong, (ii) what step in the five-step process they skipped, and (iii) the one habit they should adopt to never miss this kind of error again. Use the words "check", "real world" and "sentence" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Taxi flag fall and rate
(a) Let k = distance in km, C = total fare. C = 4.20 + 2.30k.
(b) C = 4.20 + 2.30(6) = 4.20 + 13.80 = $18.00.
(c) 4.20 + 2.30k = 34.20 → 2.30k = 30 → k ≈ 13.04 km (or exactly 30/2.3 km). Check: 4.20 + 2.30(13.04) ≈ 34.20 ✓.
1.2 — School fundraiser, 150 items, $380
(a) c + k = 150 and 3c + 2k = 380.
(b) k = 150 − c. Substitute: 3c + 2(150 − c) = 380 → 3c + 300 − 2c = 380 → c = 80. Then k = 150 − 80 = 70. So 80 chocolate bars and 70 cookies.
(c) Check items: 80 + 70 = 150 ✓. Check money: 3(80) + 2(70) = 240 + 140 = $380 ✓.
1.3 — Park run pace
(a) Jogger head start = 10 × 0.25 = 2.5 km.
(b) 2.5 + 10t = 18t.
(c) 2.5 = 8t → t = 0.3125 hours = 18.75 minutes. Check: jogger distance at t = 0.3125 is 2.5 + 10(0.3125) = 5.625 km; cyclist = 18(0.3125) = 5.625 km ✓.
1.4 — Family ages
(a) Maya's age = L + 4.
(b) L + (L + 4) = 26 → 2L + 4 = 26 → 2L = 22 → L = 11. So Leo is 11, Maya is 15.
(c) Both ages are whole positive numbers — sensible in real-world terms (no negative ages, no fractional years). Check sum: 11 + 15 = 26 ✓.
1.5 — School formal tickets, 200 sold, $11 600
(a) Let a = adult tickets, s = student tickets. a + s = 200 and 80a + 50s = 11600.
(b) s = 200 − a. Substitute: 80a + 50(200 − a) = 11600 → 80a + 10000 − 50a = 11600 → 30a = 1600 → a ≈ 53.33. Since tickets must be whole, this set of numbers does not split evenly — likely 53 adults and 147 students would collect $80(53) + $50(147) = $4240 + $7350 = $11 590 (close, but $10 short). For the algebra mark, give the exact algebra answer a = 160/3, s = 440/3. In a real exam, students would flag this and check whether the data is consistent.
(c) The check exposes the mismatch — this is exactly why Step 5 (Check) matters.
2.1 — Explain your thinking (sample response)
(i) "Leo is −2 years old" is mathematically valid — the algebra produced a real number that satisfies the equation — but it is not a possible real world age. A person cannot have a negative age, so the answer fails the practical context of the problem. (ii) The classmate skipped Step 5 of the five-step process: Check. Specifically, they did not substitute the answer back into the original problem sentence to ask "Does this make sense?". (iii) The habit to adopt is to always re-read the original problem and ask, in plain English, "Does my answer make sense for what the question is asking?" — a negative age, a fractional number of people, or a distance that exceeds the total road length all signal that you set the equation up wrong (often a sign error or a swapped variable) and need to revisit Steps 2 or 3. Always check against the words of the problem, not only the equation.
Marking: 1 for distinguishing mathematically valid vs practically wrong, 1 for naming Step 5 / Check as the missing step, 1 for proposing the "re-read and ask Does this make sense" habit, 1 for using all three required words coherently.