Mathematics • Year 9 • Unit 1 • Lesson 18
Scientific Notation — Mixed Challenge
Pull together every technique from Lesson 18: multiply / divide in scientific notation, re-normalise into $[1, 10)$, compare orders of magnitude, round to significant figures, and use a power on a scientific-notation number. Plus a "find the mistake" and an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of techniques from Lesson 18. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 Calculate $(6 \times 10^4) \times (5 \times 10^7)$, then re-normalise. Give the answer in standard form.
1.2 Calculate $\dfrac{4.8 \times 10^{-3}}{1.6 \times 10^{2}}$, giving the answer in standard form.
1.3 Calculate $(2 \times 10^5)^3$, giving the answer in standard form. (Hint: use power-of-a-product on the bracket — cube both the $2$ and the $10^5$.)
1.4 Round $3.05471 \times 10^7$ to 3 significant figures, then to 2 significant figures, then to 1 significant figure.
1.5 Light travels at $3.0 \times 10^8$ m/s. The Moon is about $3.8 \times 10^8$ m from Earth. How long does light take from the Moon to Earth, to 2 significant figures?
1.6 Order these four numbers from smallest to largest:
$3 \times 10^{-5}$, $8 \times 10^{-4}$, $1.2 \times 10^{-4}$, $9 \times 10^{-6}$. Briefly justify your reasoning.
2. Find the mistake
Another student has tried to calculate $(4 \times 10^5) \times (3 \times 10^6)$. 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 — calculate $(4 \times 10^5) \times (3 \times 10^6)$:
Line 1: Coefficients: $4 \times 3 = 12$.
Line 2: Indices: $10^{5 + 6} = 10^{11}$.
Line 3: Combine: $12 \times 10^{11}$.
Line 4: So $(4 \times 10^5) \times (3 \times 10^6) = 12 \times 10^{11}$ in scientific notation.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected final line and final answer.
Stuck? Check whether the coefficient $12$ is in $[1, 10)$ — if not, the form is wrong even though the value is correct.3. Open-ended challenge — atom-counting
This question is open ended — there's no single right answer, but every estimate must be in scientific notation and justified. 4 marks
3.1 — A single grain of rice. A single grain of rice is about $0.03$ g and is mostly water and starch. A single water molecule has mass about $3 \times 10^{-23}$ g.
(a) Convert $0.03$ g to scientific notation.
(b) Assuming (for a rough estimate) that the grain is entirely water, calculate the number of water molecules in one grain of rice, to 2 significant figures, in scientific notation.
(c) State the order of magnitude of your answer.
(d) Briefly comment: how does your number compare to the world's population (about $8 \times 10^9$)?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $(6 \times 10^4) \times (5 \times 10^7)$
Coefficients: $6 \times 5 = 30$. Indices: $10^{4 + 7} = 10^{11}$. Current: $30 \times 10^{11}$. Re-normalise: $\mathbf{3 \times 10^{12}}$ (decimal one left, index up by 1).
1.2 — $\dfrac{4.8 \times 10^{-3}}{1.6 \times 10^{2}}$
$\dfrac{4.8}{1.6} = 3$. Indices: $10^{-3 - 2} = 10^{-5}$. Answer: $\mathbf{3 \times 10^{-5}}$.
1.3 — $(2 \times 10^5)^3$
Power of a product: $2^3 \times (10^5)^3 = 8 \times 10^{5 \times 3} = \mathbf{8 \times 10^{15}}$. Coefficient already in $[1, 10)$.
1.4 — Round $3.05471 \times 10^7$
To 3 s.f.: keep $3$, $0$, $5$; next digit is $4 < 5$, round down $\to \mathbf{3.05 \times 10^7}$.
To 2 s.f.: keep $3$, $0$; next digit is $5$, round up $\to \mathbf{3.1 \times 10^7}$.
To 1 s.f.: keep $3$; next digit is $0 < 5$, round down $\to \mathbf{3 \times 10^7}$.
1.5 — Light from the Moon
$t = \dfrac{3.8 \times 10^8}{3.0 \times 10^8} = \dfrac{3.8}{3.0} \times 10^{8 - 8} \approx 1.267 \times 10^0 \approx \mathbf{1.3}$ s (2 s.f.).
Sanity check: moon-light reaches us in just over a second — that's why a "live" link to the Moon would have a noticeable delay.
1.6 — Order from smallest to largest
Compare indices first (more negative $\to$ smaller). Indices: $-6, -5, -4, -4$. So $9 \times 10^{-6}$ is smallest. Next, $3 \times 10^{-5}$. The two with index $-4$ tie on power, so compare coefficients: $1.2 < 8$, so $1.2 \times 10^{-4}$ before $8 \times 10^{-4}$.
Smallest to largest: $\mathbf{9 \times 10^{-6} \;<\; 3 \times 10^{-5} \;<\; 1.2 \times 10^{-4} \;<\; 8 \times 10^{-4}}$.
Rule: bigger (more positive) index wins; only check coefficients when indices tie.
2 — Find the mistake
(a) The mistake is on Line 4 (or equivalently the conclusion in Line 3 — the value $12 \times 10^{11}$ is correct, but it is not in scientific notation).
(b) The coefficient $12$ is outside the allowed range $[1, 10)$, so $12 \times 10^{11}$ is not in standard scientific-notation form. The student forgot to re-normalise — slide the decimal one place left and bump the index up by 1.
(c) Corrected final line: $12 \times 10^{11} = \mathbf{1.2 \times 10^{12}}$.
The value was already right ($12 \times 10^{11} = 1.2 \times 10^{12} = 1{,}200{,}000{,}000{,}000$) — only the form needed fixing.
3 — Open-ended (sample solution)
(a) $0.03 = \mathbf{3 \times 10^{-2}}$ g.
(b) Number of molecules $= \dfrac{3 \times 10^{-2}}{3 \times 10^{-23}} = \dfrac{3}{3} \times 10^{-2 - (-23)} = 1 \times 10^{21} = \mathbf{1 \times 10^{21}}$ molecules (2 s.f.).
(c) Order of magnitude $\approx \mathbf{10^{21}}$.
(d) Population is $\approx 10^{10}$ orders of magnitude. The grain has $10^{21}$ molecules vs $10^{10}$ people on Earth, so the grain holds about $10^{21 - 10} = 10^{11}$ times more molecules than there are people on Earth — every person would get about a hundred billion molecules. A single grain of rice contains vastly more molecules than the planet has humans.
Marking: 1 mark for the conversion in (a); 1 for the calculation in (b); 1 for the order of magnitude in (c); 1 for a reasonable comparison in (d) referencing order-of-magnitude difference.