Mathematics • Year 10 • Unit 2 • Lesson 11
Inequalities in the Real World
Apply linear inequalities to real situations — theme-park height limits, budget caps, casual-job pay, mobile data plans and a school excursion. Then explain (in your own words) why dividing by a negative flips the sign.
1. Word problems
Each problem uses one or more tools from Lesson 11. Show your working — a final answer with no working only earns half marks.
1.1 — Theme-park ride. A roller-coaster requires riders to be at least 120 cm and no taller than 195 cm. Let h be a rider's height in cm.
(a) Write a compound inequality for the allowed heights.
(b) A 115 cm child wants to ride. Show using the inequality why they cannot.
(c) A 120 cm child arrives. Can they ride? Justify using the correct inequality symbol. 3 marks
1.2 — Budget cap on lunches. A canteen lunch costs $8.50. Mia has $50 in her lunch account and refuses to let the balance drop below $0. Let n be the number of lunches she can buy this week.
(a) Write an inequality for the total spent in terms of n.
(b) Solve the inequality for n.
(c) Because n must be a whole number, state the largest value of n she can choose. 3 marks
1.3 — Casual job hours. Jordan is paid a $25 weekly bonus plus $22 per hour. To save for a $300 phone, the weekly pay must be at least $300. Let h be the number of hours worked in a week.
(a) Write an inequality for "weekly pay is at least $300".
(b) Solve to find the minimum number of hours h.
(c) Because hours are usually rounded to the next quarter, state the smallest whole quarter-hour value Jordan can work. 3 marks
1.4 — Mobile data plan. A plan charges $20 per month plus $4 per extra GB. Sam wants the monthly bill to stay strictly less than $60. Let g be extra GB used.
(a) Write an inequality for the monthly cost staying under $60.
(b) Solve the inequality for g.
(c) Sketch the solution on a number line (open/closed circle, direction of shading) and write it in interval notation. 3 marks
1.5 — School excursion target. A Year 10 excursion needs at least 30 paying students to run, but the bus seats no more than 48. Let s be the number of paying students.
(a) Write a compound inequality describing the valid range of s.
(b) Write your answer in interval notation.
(c) If 29 students sign up, the excursion is cancelled. Explain using your inequality why the cancellation is mathematically correct. 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A classmate solves −4x > 20 and writes "x > −5" as the answer. Using everything from Lesson 11, explain (i) which step they did wrong, (ii) what the correct answer is, and (iii) the one habit a student can adopt to never make this mistake again. Use the words "FLIP", "negative" and "reflect" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Theme-park ride heights
(a) 120 ≤ h ≤ 195.
(b) 115 does not satisfy h ≥ 120 (115 < 120), so the child is below the minimum height.
(c) Yes — h = 120 satisfies h ≥ 120 because the symbol is ≥ (greater than or equal to). On a number line, 120 would be a closed circle.
1.2 — Budget cap on lunches
(a) 8.5n ≤ 50.
(b) n ≤ 50 ÷ 8.5 = 5.88… (no flip — dividing by positive 8.5).
(c) Largest whole-number value is n = 5 lunches. (n = 6 would cost $51, breaking the cap.)
1.3 — Casual job hours
(a) 25 + 22h ≥ 300.
(b) 22h ≥ 275 → h ≥ 12.5.
(c) Smallest quarter-hour value satisfying h ≥ 12.5 is h = 12.5 hours exactly (12.5 is already a quarter-hour value). At 12.5 h: pay = 25 + 22(12.5) = $300 ✓.
1.4 — Mobile data plan
(a) 20 + 4g < 60 (and g ≥ 0).
(b) 4g < 40 → g < 10.
(c) Combined with g ≥ 0: 0 ≤ g < 10. Number line: closed circle at 0, open circle at 10, shade between. Interval: [0, 10).
1.5 — School excursion target
(a) 30 ≤ s ≤ 48.
(b) Interval: [30, 48] (both endpoints included).
(c) 29 does not satisfy s ≥ 30 (29 < 30), so 29 falls outside [30, 48] and the minimum-numbers condition is broken — the excursion is cancelled.
2.1 — Explain your thinking (sample response)
(i) The mistake is at the divide step. To get x by itself from −4x > 20, the classmate divided both sides by −4 but forgot to FLIP the inequality sign. Because the divisor is negative, every value gets reflected across zero, which reverses the order of inequalities. (ii) The correct answer is x < −5 — the > reverses to < when we divide by the negative. (iii) The habit to adopt is to write the word "FLIP" next to the step whenever you see a negative multiplier or divisor; then immediately reverse the inequality before continuing. A second check is to substitute a value: x = −10 should satisfy the original inequality, and −4(−10) = 40 > 20 is true, but x = 0 should fail, and −4(0) = 0 > 20 is false — both consistent with x < −5.
Marking: 1 for identifying the missing FLIP, 1 for the correct answer x < −5, 1 for explaining why dividing by a negative reverses the sign, 1 for naming the FLIP habit (or test-a-value habit) using the three required words.