Mathematics • Year 8 • Unit 1 • Lesson 13
Unitary Method — Mixed Challenge
Pull together everything from Lesson 13: dividing to find 1, multiplying to scale, distance-speed-time, and spotting when the method DOESN'T apply. Six mixed problems, one “find the mistake”, and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different combination of ideas from Lesson 13. Decide which move applies before you start writing. Show your working. 3 marks each
1.1 5 pens cost $\$8$. Use the unitary method to find the cost of 12 pens.
1.2 At 90 km/h, find the time taken to travel 270 km.
1.3 A factory makes 240 widgets in 6 hours. (a) Find the production rate in widgets per hour. (b) How many widgets are made in a 40-hour week?
1.4 4 kg of dog food lasts a puppy 10 days. At the same rate, (a) how long would 7 kg last? (b) How much food is needed for 30 days?
1.5 $\$45$ pays for 3 hours of tutoring. (a) Find the per-hour cost. (b) Find the cost of 5.5 hours at the same rate.
1.6 A train travels at 110 km/h. (a) How far does it go in 90 minutes? (b) How long does it take to cover 275 km? Show both calculations using a unit rate.
2. Find the mistake
Another student is trying to find the cost of 7 kg of tomatoes when 4 kg cost $\$24$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — cost of 7 kg of tomatoes:
Line 1: 4 kg cost $\$24$, so per kg it's $24 \div 4 = \$6$/kg.
Line 2: I want 7 kg. Difference from 4 is $7 - 4 = 3$ kg extra.
Line 3: Extra cost = $3 \times \$6 = \$18$.
Line 4: So 7 kg costs $\$24 + \$18 = \$32$.
(a) Is the final answer ($\$32$) actually wrong, or has the student just taken a long route?
(b) The student's answer is wrong. Find the mistake: which line first goes wrong, and what's the value error? (Hint: double-check Line 3.)
(c) Re-do the working using the simpler unitary method (per kg × 7 kg) and give the corrected final answer.
Stuck? Per kg is $\$6$. For 7 kg it's $7 \times \$6 = \$42$. So the student's $\$32$ is wrong — check whether the extra-cost arithmetic in Line 3 makes sense.3. Open-ended challenge — design a unitary method problem
This question has more than one valid answer. 4 marks
3.1 Design a real-world word problem where:
- the unitary method is the natural way to solve it,
- the “per 1” rate is a clean whole-dollar (or whole-km, or whole-hour) amount, and
- the answer is between $\$20$ and $\$80$.
Write:
(i) Your word problem (a couple of sentences).
(ii) The full unitary-method working in two clear steps (find 1, then scale).
(iii) The final answer with units.
(iv) A one-sentence sanity check (“is this in the right ballpark? Why?”).
Bonus: Modify your own problem so the unitary method DOES NOT work directly (e.g., a bulk discount applies after a certain quantity). Explain in one line why the unitary method fails there.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 12 pens
Per pen: $\$8 \div 5 = \$1.60$. 12 pens: $12 \times \$1.60 = \textbf{\$19.20}$.
1.2 — Time for 270 km
$270 \div 90 = \textbf{3 hours}$.
1.3 — Factory production
(a) $240 \div 6 = \textbf{40 widgets/hour}$.
(b) 40-hour week: $40 \times 40 = \textbf{1600 widgets}$.
1.4 — Dog food
Per kg: 10 ÷ 4 = 2.5 days/kg.
(a) 7 kg: $7 \times 2.5 = \textbf{17.5 days}$.
(b) For 30 days: $30 \div 2.5 = \textbf{12 kg}$ of food.
1.5 — Tutoring
(a) $\$45 \div 3 = \textbf{\$15/h}$.
(b) $5.5 \times \$15 = \textbf{\$82.50}$.
1.6 — Train
(a) 90 min = 1.5 h, distance = $110 \times 1.5 = \textbf{165 km}$.
(b) Time = $275 \div 110 = \textbf{2.5 hours}$.
2 — Find the mistake
(a) The final answer $\$32$ is wrong — the correct cost of 7 kg is $\$42$.
(b) Line 3 is where it first goes wrong: 3 extra kg at $\$6$/kg is indeed $\$18$, BUT the issue is in Line 4 / Line 2 arithmetic when adding: $\$24 + \$18 = \$42$, not $\$32$. The student added $24 + 18$ as $32$ (probably did $24 + 8 = 32$, dropping the “1” from the tens column). The arithmetic is the slip.
(c) Using the unitary method directly: per kg = $\$6$; 7 kg = $7 \times \$6 = \textbf{\$42}$. Sanity check: 7 kg should cost much more than 4 kg ($\$24$) but less than 8 kg ($\$48$). $\$42$ fits. ✓
3 — Open-ended challenge (sample solution)
(i) Word problem: A casual gardener earns $\$60$ for 4 hours of work. At the same hourly rate, what does she earn for 5 hours of work?
(ii) Working: Step 1 (find 1): per hour = $\$60 \div 4 = \$15/h$. Step 2 (scale): 5 hours = $5 \times \$15 = \$75$.
(iii) Answer: $\$75$ (between $\$20$ and $\$80$ — ✓).
(iv) Sanity check: 5 hours should pay a bit more than 4 hours ($\$60$) but less than 6 hours ($\$90$). $\$75$ fits.
Bonus modification: “A casual gardener earns $\$15$/h for the first 4 hours, then $\$22.50$/h (time-and-a-half) for every extra hour. What does she earn for 5 hours?” The unitary method (just multiply per-hour by 5) fails here because the pay rate isn't constant — once you cross 4 hours, each extra hour costs MORE. You have to handle the first 4 hours at one rate and the 5th hour at the higher rate separately ($60 + 22.50 = \$82.50$).
Marking: 1 mark for a realistic word problem; 1 mark for clean two-step working; 1 mark for an in-range correct answer with sanity check; 1 bonus mark for a valid non-proportional twist with an explanation.