Mathematics • Year 9 • Unit 1 • Lesson 13
Quotients and Powers in the Real World
Use the quotient and power-of-a-product rules in real contexts: scaling a recipe, sharing screen pixels, packaging cubes, lab dilutions and tile patterns. Then communicate one of the most common student slips clearly.
1. Word problems
Each problem uses the quotient rule, the power-of-a-product rule, or both. Show your working — final answers alone earn half marks.
1.1 — Scaling down a recipe. A cake recipe uses $12 x^5$ grams of flour and serves $4 x^2$ people.
(a) Write the flour PER PERSON as a single simplified term.
(b) State which rule from Lesson 13 you used and why. 3 marks
1.2 — Storage cubes. A packing company makes a cubic box with side length $2 x^3$ cm.
(a) Write the volume of one box as a single simplified term using power-of-a-product.
(b) Evaluate the volume when $x = 2$. 3 marks
1.3 — Phone screen pixels per icon. A phone screen has $18 p^7 q^2$ pixels in total, arranged into $6 p^3 q$ identical app icons.
(a) Write the pixels per icon as a single simplified term.
(b) If $p = 2$ and $q = 5$, how many pixels does ONE icon have? 3 marks
1.4 — Lab dilutions ("scale-cubed"). A scientist scales a culture in a $3 x^2 y$-litre flask up to a cube of that size for storage (so the storage volume is $(3 x^2 y)^3$ litres).
(a) Write the storage volume as a single simplified term.
(b) Explain why the coefficient $3$ also gets the outside power, not just the variables. 3 marks
1.5 — Tile patterns. A tile shop produces patterns that contain $(3 x^2)^3$ small tiles in the design plus $2x$ extra border tiles. The total tile count is then $(3 x^2)^3 \times 2 x$ once the design is repeated.
(a) Write the total number of tiles as a single simplified term using power-of-a-product then the product rule.
(b) If they then divide that total by $6 x^4$ (one box's worth of stock), write the result simplified — using the quotient rule. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate writes "$(2 x^3)^4 = 2 x^{12}$". They're sure they're right because they "did the index times index part correctly". In your own words, explain (i) what they got right, (ii) what they got wrong, (iii) the correct answer with working, and (iv) one practical check they could have done in 5 seconds to catch the slip. Use the phrase "every factor inside the bracket" somewhere in your reply.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Scaling down a recipe
(a) $\dfrac{12 x^5}{4 x^2} = \dfrac{12}{4} \cdot x^{5-2} = \mathbf{3 x^3}$ grams per person.
(b) Quotient rule for the variable ($x^{m-n}$); ordinary division for the coefficients ($12 \div 4$).
1.2 — Storage cubes
(a) Volume $= (2 x^3)^3 = 2^3 \cdot x^{3 \times 3} = \mathbf{8 x^9}$ cm$^3$.
(b) At $x = 2$: $V = 8 \times 2^9 = 8 \times 512 = \mathbf{4096}$ cm$^3$.
1.3 — Phone screen pixels per icon
(a) Pixels per icon $= \dfrac{18 p^7 q^2}{6 p^3 q} = \dfrac{18}{6} \cdot p^{7-3} \cdot q^{2-1} = \mathbf{3 p^4 q}$.
(b) At $p = 2, q = 5$: $3 \times 2^4 \times 5 = 3 \times 16 \times 5 = \mathbf{240}$ pixels per icon.
1.4 — Lab dilution storage volume
(a) $(3 x^2 y)^3 = 3^3 \cdot (x^2)^3 \cdot y^3 = \mathbf{27 x^6 y^3}$ L.
(b) Inside the brackets there are THREE factors being multiplied: $3$, $x^2$, and $y$. The outside power applies to the WHOLE multiplication, so every factor — including the coefficient $3$ — gets the outer index. That's why $3$ becomes $3^3 = 27$, not just stays as $3$.
1.5 — Tile patterns
(a) $(3 x^2)^3 = 3^3 \cdot x^{2 \times 3} = 27 x^6$. Then $27 x^6 \times 2x = (27 \times 2) \cdot x^{6+1} = \mathbf{54 x^7}$ tiles.
(b) $\dfrac{54 x^7}{6 x^4} = \dfrac{54}{6} \cdot x^{7-4} = \mathbf{9 x^3}$ box-units.
2.1 — Explain your thinking (sample response)
My classmate did the index-times-index part correctly: $(x^3)^4 = x^{3 \times 4} = x^{12}$. That part is right. The mistake is that they left the coefficient as $2$ instead of raising it to the outside power as well. In a power-of-a-product, every factor inside the bracket gets the outer index — and the $2$ is inside the bracket. The correct working is: $2^4 = 16$, and $x^{3 \times 4} = x^{12}$, so $(2 x^3)^4 = \mathbf{16 x^{12}}$, NOT $2 x^{12}$. A 5-second check: try $x = 1$. Then $(2 \times 1^3)^4 = 2^4 = 16$, but their answer gives $2 \times 1^{12} = 2$. The two clearly disagree, so something is wrong — and that something is the missing $2^4$.
Marking: 1 mark for noting what was right; 1 for naming the missed step; 1 for the correct answer with working; 1 for a sensible 5-second check.