Mathematics • Year 7 • Unit 1 • Lesson 14
Ratios & Rates — Mixed Challenge
Pull everything from Lesson 14 together: simplify ratios with mixed units, divide an amount in a three-part ratio, find a unit rate, compare best value, and spot a common rate mistake. Finish with an open-ended planning puzzle.
1. Mixed problems — choose the right idea
Each question uses a different part of Lesson 14. Decide which method applies before you start. Show working. 2 marks each
1.1 A motorcycle covers 144 km in 1 hour 30 minutes. Find its speed in km/h.
1.2 Simplify the ratio 250 g : 1 kg.
1.3 A printer ink cartridge prints 4500 pages and costs $90. (a) What is the price per page (in cents)? (b) How many pages can you print per dollar?
1.4 Share $200 in the ratio 1:3:6. List all three shares.
1.5 A water tank fills at a rate of 12 L per minute. How long (in minutes) will it take to fill a 540 L tank?
1.6 Compare three packs of pasta: 250 g for $1.80, 500 g for $3.20, 1 kg for $6.20. Find the price per kg for each pack and decide which is best value.
2. Find the mistake
Another Year 7 student tried to find the speed of a car that travelled 180 km in 2 hours 15 minutes. Their working is below. Exactly one line contains the mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — speed for 180 km in 2 h 15 min:
Line 1: Speed = distance ÷ time.
Line 2: Convert time to a decimal: 2 h 15 min = 2.15 h.
Line 3: Calculate: 180 ÷ 2.15 ≈ 83.7 km/h.
Line 4: Final answer: speed ≈ 83.7 km/h.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? 15 minutes is not 0.15 of an hour. 60 minutes makes 1 hour, so 15 min = 15/60 = 0.25 hours. Be careful — time units are 60-based, not 100-based.3. Open-ended challenge — design a class trip budget
This question has more than one correct answer. Show one that works and explain. 4 marks
3.1 Your Year 7 class (30 students plus 2 teachers) is going on an excursion. The bus charges $2.50 per kilometre, the venue charges $8 per student (free for teachers), and there is a flat fee of $50 for booking.
(i) Pick a venue distance between 20 km and 80 km. Calculate the total cost of the trip with your distance.
(ii) Then calculate the cost per student if the cost is shared evenly among the 30 students.
Bonus: What is the largest distance the class can choose so that the cost per student is no more than $20? Show your reasoning.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Motorcycle speed
1 h 30 min = 1.5 h. Speed = 144 ÷ 1.5. Shift dots: 1440 ÷ 15 = 96 km/h.
1.2 — Simplify 250 g : 1 kg
Convert to the same unit: 1 kg = 1000 g. Ratio: 250 : 1000. HCF(250, 1000) = 250. 250 ÷ 250 = 1, 1000 ÷ 250 = 4. Answer: 1:4.
1.3 — Ink cartridge
(a) $90 ÷ 4500 pages = $0.02/page = 2 cents per page.
(b) 4500 pages ÷ $90 = 50 pages per dollar.
1.4 — $200 in ratio 1:3:6
Total parts = 1 + 3 + 6 = 10. One part = $200 ÷ 10 = $20.
Shares: 1 × $20 = $20; 3 × $20 = $60; 6 × $20 = $120. Check: $20 + $60 + $120 = $200 ✓.
1.5 — Water tank
Time = volume ÷ rate = 540 ÷ 12 = 45 minutes.
1.6 — Pasta packs
250 g = 0.25 kg → $1.80 ÷ 0.25 = $7.20/kg.
500 g = 0.5 kg → $3.20 ÷ 0.5 = $6.40/kg.
1 kg → $6.20 ÷ 1 = $6.20/kg.
Best value: 1 kg pack at $6.20/kg. Worst: 250 g pack at $7.20/kg.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) 15 minutes is not 0.15 of an hour — time isn't decimal-based, it's based on 60. 15 minutes = 15/60 = 0.25 hours. So 2 h 15 min = 2.25 h, not 2.15 h.
(c) Corrected working:
Line 1 (kept): Speed = distance ÷ time.
Line 2 (fixed): 2 h 15 min = 2 + 15/60 = 2 + 0.25 = 2.25 h.
Line 3 (fixed): 180 ÷ 2.25 = 18000 ÷ 225 = 80 km/h.
Line 4: speed = 80 km/h.
Sanity check: 180 km in 2¼ hours ≈ 90 km in 1 hour 7 min, so 80 km/h is in the right ballpark.
3 — Class trip puzzle (sample solution)
Sample distance: 60 km.
Bus: 2.50 × 60 = $150. Venue: 8 × 30 = $240. Flat fee: $50. Total: $150 + $240 + $50 = $440.
Per student: $440 ÷ 30 = $14.67 (to the nearest cent).
Bonus: Cost per student ≤ $20 means total ≤ 30 × 20 = $600. So 2.50d + 290 ≤ 600 → 2.50d ≤ 310 → d ≤ 124 km. Since the rules cap distance at 80 km, every distance in the allowed range (20-80 km) keeps the per-student cost under $20. The maximum allowed distance is 80 km, giving per-student cost = (2.50 × 80 + 290) ÷ 30 = ($200 + $290) ÷ 30 = $490 ÷ 30 ≈ $16.33 per student.
Marking: 2 marks for a valid distance with correct total cost and per-student calculation; 1 mark for setting up the per-student formula; 1 bonus mark for the inequality work in part (ii) — accept either the absolute limit of 124 km or the bounded answer of 80 km.