Mathematics • Year 9 • Unit 1 • Lesson 14
Mixed Index Laws in the Real World
Use the four-step plan — power $\to$ product $\to$ quotient $\to$ zero — on real situations: scaling cubes, pixels per icon, stage lighting, fuel rationing and recipe maths. Then explain why "the answer is just a number" sometimes happens.
1. Word problems
Each problem mixes at least two index laws. Show every step. Watch for $x^0 = 1$ — sometimes the answer is just a number.
1.1 — Scaling factor reduction. A digital scale-up factor is $(2 x^2)^3$, applied to a product of $5 x$ items. The result is then shared evenly among $10 x^7$ users.
(a) Write the amount each user receives as a single simplified value.
(b) Comment on why the answer ends up as a plain number rather than an expression in $x$. 3 marks
1.2 — Pixels per app icon. A phone's display can render $(3 m^2)^2$ pixel units for each row, multiplied by $2 m$ rows worth of icons. The total is then split across $m^3$ icons.
(a) Write the pixels per icon as a single simplified term.
(b) Evaluate the answer when $m = 3$. 3 marks
1.3 — Stage lighting cubes. A theatre uses $(2 x y)^3$ small LED bulbs in its main array, then adds $5 x$ extra spotlights. The total is checked against a stock allocation of $20 x^4 y^3$ bulbs.
(a) Write the ratio $\dfrac{(2xy)^3 \cdot 5 x}{20 x^4 y^3}$ as a single simplified value.
(b) Explain in one sentence what your answer tells you about the bulb stock. 3 marks
1.4 — Fuel ration per truck. A logistics company delivers $8 p^5 q^2 \times p$ litres of fuel in total. Each route uses $(2 p^3 q)^2$ litres.
(a) Write the number of routes that can be run on this fuel as a single simplified value.
(b) Why does the variable $p$ disappear from your answer? 3 marks
1.5 — Recipe scale-up. A baker has $(2 x^2 y)^3$ grams of flour and a recipe that uses $6 x y^2$ grams per cake. The kitchen produces enough cakes to feed $(3 x^3 y)^2$ guests.
(a) Write $\dfrac{(2 x^2 y)^3 \cdot 6 x y^2}{(3 x^3 y)^2}$ as a single simplified term — the "grams of flour per guest".
(b) Evaluate your answer when $x = 1$ and $y = 2$. 3 marks
2. Explain your thinking
This question is about communication. Use full sentences. 4 marks
2.1 A classmate simplifies $\dfrac{5 y^4 \cdot 2 y^3}{10 y^7}$ and writes their final answer as $\mathbf{0}$, because "everything cancels". In your own words, explain (i) what the correct answer is, (ii) where their reasoning went wrong (be specific about $y^0$ vs $0$), (iii) the four-step plan from Lesson 14 they should have followed, and (iv) why the answer is a number with no $y$ in it.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Scaling factor reduction
(a) Setup: $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7} = \dfrac{8 x^6 \cdot 5 x}{10 x^7} = \dfrac{40 x^7}{10 x^7} = 4 x^0 = \mathbf{4}$.
(b) The numerator and denominator end up with the SAME power of $x$ (both $x^7$), so the variables cancel ($x^{7-7} = x^0 = 1$). What's left is just the ratio of coefficients $\dfrac{40}{10} = 4$.
1.2 — Pixels per app icon
(a) Setup: $\dfrac{(3 m^2)^2 \cdot 2 m}{m^3} = \dfrac{9 m^4 \cdot 2 m}{m^3} = \dfrac{18 m^5}{m^3} = \mathbf{18 m^2}$ pixels per icon.
(b) At $m = 3$: $18 \times 9 = \mathbf{162}$ pixels per icon.
1.3 — Stage lighting cubes
(a) $(2 xy)^3 = 8 x^3 y^3$. Numerator $= 8 x^3 y^3 \cdot 5 x = 40 x^4 y^3$. Quotient $= \dfrac{40 x^4 y^3}{20 x^4 y^3} = 2 \cdot x^0 \cdot y^0 = \mathbf{2}$.
(b) The total bulbs needed are exactly $2 \times$ the stock allocation — twice as many bulbs as one allocation provides, so they need a second box of stock.
1.4 — Fuel ration per truck
(a) Denominator $(2 p^3 q)^2 = 4 p^6 q^2$. Numerator $8 p^5 q^2 \cdot p = 8 p^6 q^2$. Routes $= \dfrac{8 p^6 q^2}{4 p^6 q^2} = 2 \cdot p^0 \cdot q^0 = \mathbf{2}$ routes.
(b) The $p^6$ in the numerator and denominator are EQUAL after simplifying, so $p^{6-6} = p^0 = 1$. Same for $q$. With both variables collapsing to $1$, only the ratio $\dfrac{8}{4} = 2$ remains.
1.5 — Recipe scale-up
(a) Powers: $(2 x^2 y)^3 = 8 x^6 y^3$; $(3 x^3 y)^2 = 9 x^6 y^2$.
Numerator: $8 x^6 y^3 \cdot 6 x y^2 = 48 x^7 y^5$.
Quotient: $\dfrac{48 x^7 y^5}{9 x^6 y^2} = \dfrac{48}{9} \cdot x^{7-6} \cdot y^{5-2} = \mathbf{\dfrac{16}{3} x y^3}$ grams per guest.
(b) At $x = 1, y = 2$: $\dfrac{16}{3} \times 1 \times 8 = \dfrac{128}{3} \approx \mathbf{42.7}$ grams per guest.
2.1 — Explain your thinking (sample response)
The correct answer is $\mathbf{1}$, not $\mathbf{0}$. My classmate confused $y^0$ with $0$ — but the zero index rule says $y^0 = 1$ for any non-zero $y$, not zero. The four-step plan they should have followed is power $\to$ product $\to$ quotient $\to$ zero. With no brackets there's no power step; the product step gives numerator $5 y^4 \cdot 2 y^3 = 10 y^7$; the quotient step gives $\dfrac{10 y^7}{10 y^7} = 1 \cdot y^{7-7} = 1 \cdot y^0$; the zero step then collapses $y^0$ to $1$, giving a final answer of $1$. The reason there's no $y$ in the answer is that the numerator and denominator had the SAME power of $y$ ($y^7$), so the $y$s cancel completely — leaving just the coefficient ratio $\dfrac{10}{10} = 1$.
Marking: 1 mark for the correct answer (1); 1 for naming the $y^0 \ne 0$ confusion; 1 for the four-step plan; 1 for explaining the $y$ disappearing.