Mathematics · AQA 8300 §N6
Standard Form
- Convert between standard form and ordinary numbers
- Order numbers written in standard form
- Add, subtract, multiply and divide in standard form
- Apply standard form to scientific and real-world problems
- Estimate and check the reasonableness of standard form answers
🔑 Core Concepts
What is Standard Form?
Standard form (also called scientific notation) is a compact way of writing very large or very small numbers. Scientists, engineers, and astronomers use it constantly — the distance to Proxima Centauri ($4 \times 10^{16}$ m) or the diameter of a hydrogen atom ($1.06 \times 10^{-10}$ m) would be unwieldy to write in full. Standard form makes such numbers easy to compare, calculate with, and communicate.
DEFINITION: Standard Form
A number is in standard form when it is written as $A \times 10^n$, where:
- $1 \leq A < 10$ — the coefficient $A$ is at least 1 but strictly less than 10
- $n$ is an integer — the power $n$ is a whole number (positive, negative, or zero)
IMPORTANT: The Boundary Values
$A = 1$ is allowed; $A = 10$ is NOT allowed. So $1.0 \times 10^3$ is valid, but $10 \times 10^2$ is not — even though both equal 1000. Always ensure $A \in [1, 10)$ (closed on 1, open on 10).
The Standard Form Rule
$$A \times 10^n \quad \text{where } 1 \leq A < 10 \text{ and } n \in \mathbb{Z}$$
$A$ = coefficient (between 1 and 10)
$10^n$ = the power of ten
$n$ = integer power (positive, negative, or zero)
Converting Ordinary Numbers to Standard Form
The key insight is that multiplying by $10^n$ shifts the decimal point $n$ places. So to find $n$, we count how many places the decimal must move to transform the ordinary number into something with $1 \leq A < 10$.
Write the number
→
Move decimal to give $1 \leq A < 10$
→
Count places moved = $|n|$
→
Left move → $n > 0$
Right move → $n < 0$
Right move → $n < 0$
→
Write $A \times 10^n$
EXAM TIP: Which Direction — Which Sign?
Large numbers (greater than 10): decimal moves left → $n$ is positive.$\quad$ Example: $47{,}500 \rightarrow 4.75 \times 10^4$ (moved 4 left, $n = +4$)
Small numbers (between 0 and 1): decimal moves right → $n$ is negative.
$\quad$ Example: $0.000308 \rightarrow 3.08 \times 10^{-4}$ (moved 4 right, $n = -4$)
Memory aid: "Large = positive, Small = negative."
COMMON MISTAKE: Positive Power for Small Numbers
A common error is writing $0.00045 = 4.5 \times 10^4$ (positive power). But $4.5 \times 10^4 = 45{,}000$ — far too large. The correct answer is $4.5 \times 10^{-4}$. Check: if the original number is less than 1, the power must be negative.
Converting Standard Form to Ordinary Numbers
Reverse the process: move the decimal point in $A$ by $|n|$ places. Positive $n$ means move right (number grows); negative $n$ means move left (number shrinks).
WORKED EXAMPLE: Standard Form to Ordinary
$3.7 \times 10^5$: move decimal 5 places right $\rightarrow 370{,}000$$6.02 \times 10^{-3}$: move decimal 3 places left $\rightarrow 0.00602$
$1.0 \times 10^0$: no movement ($10^0 = 1$) $\rightarrow 1.0$
$8.5 \times 10^{-1}$: move 1 place left $\rightarrow 0.85$
Ordering Numbers in Standard Form
To compare $A_1 \times 10^{n_1}$ and $A_2 \times 10^{n_2}$: first compare the powers. The larger power gives the larger number (when $A$ values are in $[1,10)$). Only compare $A$ values directly when the powers are equal.
DEFINITION: Ordering Rule
If $n_1 > n_2$: then $A_1 \times 10^{n_1} > A_2 \times 10^{n_2}$ regardless of $A$ values (since both $A \geq 1$).If $n_1 = n_2$: compare $A_1$ and $A_2$ directly — larger $A$ gives larger number.
EXAM TIP: Ordering with Negative Powers
Be careful: $10^{-2} > 10^{-5}$ because $-2 > -5$ on the number line. So $3.1 \times 10^{-2} > 9.9 \times 10^{-5}$, even though 9.9 > 3.1. Always compare the powers first.
Multiplying Numbers in Standard Form
Multiplication exploits the index law $10^a \times 10^b = 10^{a+b}$. We handle the coefficient and the power of 10 separately, then combine. This works because multiplication is commutative and associative:
$(A_1 \times 10^a) \times (A_2 \times 10^b) = (A_1 \times A_2) \times (10^a \times 10^b) = (A_1 A_2) \times 10^{a+b}$
Multiplication Rule
$$(A_1 \times 10^a) \times (A_2 \times 10^b) = (A_1 \times A_2) \times 10^{a+b}$$
Multiply the $A$ values together
Add the powers of 10
Re-express if result's $A \notin [1,10)$
COMMON MISTAKE: Product Outside Range — Must Re-express
If $A_1 \times A_2 \geq 10$, the answer is NOT in standard form yet. For example: $(5 \times 10^3) \times (4 \times 10^2) = 20 \times 10^5$. Since $20 \geq 10$, re-express: $20 \times 10^5 = 2 \times 10^1 \times 10^5 = 2 \times 10^6$. Always add an extra step to check.
Dividing Numbers in Standard Form
Division uses $10^a \div 10^b = 10^{a-b}$. Divide the $A$ values, subtract the powers. If the quotient of $A$ values falls below 1, re-express by adjusting both $A$ and the power.
Division Rule
$$(A_1 \times 10^a) \div (A_2 \times 10^b) = \left(\frac{A_1}{A_2}\right) \times 10^{a-b}$$
Divide the $A$ values
Subtract the powers of 10
Re-express if result's $A \notin [1,10)$
EXAM TIP: Subtracting a Negative Power
Be very careful with signs: $10^5 \div 10^{-2} = 10^{5-(-2)} = 10^{5+2} = 10^7$. The minus signs cancel! Write it out fully to avoid errors.
EXAM TIP: Re-expressing After Division
If division gives $A < 1$: multiply $A$ by 10 and reduce the power by 1.Example: $0.5 \times 10^7$ — here $0.5 < 1$, so write $0.5 = 5 \times 10^{-1}$, giving $5 \times 10^{-1} \times 10^7 = 5 \times 10^6$.
Adding and Subtracting in Standard Form
Unlike multiplication, you cannot simply combine $A$ values unless both numbers have the same power of 10. Adding $3.2 \times 10^5$ and $4.1 \times 10^4$ is like adding 32 tens-of-thousands and 41 thousands — you must express them in the same "unit" first.
DEFINITION: Addition/Subtraction Method
To compute $A_1 \times 10^a \pm A_2 \times 10^b$ (where $a \neq b$):Step 1. Convert both to the same power of 10 (use the larger power).
Step 2. Add or subtract the $A$ values.
Step 3. Re-express in standard form if the result's $A \notin [1,10)$.
WORKED EXAMPLE: Addition with Different Powers
$3.2 \times 10^5 + 4.1 \times 10^4$$= 32 \times 10^4 + 4.1 \times 10^4$ (convert $3.2 \times 10^5$ to power $10^4$)
$= (32 + 4.1) \times 10^4$ (add $A$ values — powers are now the same)
$= 36.1 \times 10^4$ (but $36.1 \geq 10$, so re-express)
$= 3.61 \times 10^5$
COMMON MISTAKE: Adding A Values with Different Powers
$4.2 \times 10^6 + 3.1 \times 10^5 \neq 7.3 \times 10^6$. The correct answer is $4.51 \times 10^6$ (since $3.1 \times 10^5 = 0.31 \times 10^6$). Think of it like adding 4.2 metres and 31 centimetres — you must convert to the same unit first.
MEMORY TRICK: Operations Summary
✕ and ÷: Operate on both parts separately — multiply/divide $A$, add/subtract powers.+ and −: Align powers first, then operate on $A$ — like adding fractions needing a common denominator.
Always: Re-express if $A$ ends up outside $[1,10)$.
Grade 9: Re-expressing Results
At Grade 9, examiners expect you to always present answers in correct standard form. The re-expressing step is often worth a dedicated method mark. Know the rules:
| Situation | Result Type | Correction | Example |
|---|---|---|---|
| $A \geq 10$ | Too large | $A \div 10$, power $+1$ | $25 \times 10^4 = 2.5 \times 10^5$ |
| $A \geq 100$ | Much too large | $A \div 100$, power $+2$ | $350 \times 10^3 = 3.5 \times 10^5$ |
| $A < 1$ | Too small | $A \times 10$, power $-1$ | $0.4 \times 10^7 = 4 \times 10^6$ |
| $A < 0.1$ | Much too small | $A \times 100$, power $-2$ | $0.035 \times 10^5 = 3.5 \times 10^3$ |
EXAM TIP: Estimating in Standard Form
For multi-step problems, estimate at each stage to check reasonableness. For $(3.1 \times 10^8) \div (4.9 \times 10^2)$, estimate as $3 \div 5 = 0.6$, re-express to give $\approx 6 \times 10^{-1}$, power adjustment: $10^{8-2} = 10^6$, rough answer $\approx 6 \times 10^5$. If your calculated answer is wildly different, you have made an error.
🗺️ Visual Notes
Standard Form
$A \times 10^n$
$A \times 10^n$
Definition
- $A \times 10^n$ format
- $1 \leq A < 10$ always
- $n$ is an integer
- $n$ can be +, −, or 0
Converting
- Large number → $n > 0$
- Small number → $n < 0$
- Count decimal places moved
- Check $1 \leq A < 10$ at end
✕ and ÷
- Multiply: $A_1 \times A_2$; add powers
- Divide: $A_1 \div A_2$; subtract powers
- Use index laws throughout
- Re-express if $A \notin [1,10)$
+ and −
- Align powers first
- Convert to same $10^n$
- Then add/subtract $A$ values
- Re-express result if needed
Scientific Use
- Distances in astronomy
- Atomic & nuclear sizes
- Speed of light: $3 \times 10^8$ m/s
- Population & financial data
Grade 9 Skills
- Multi-step calculations
- Mix standard & ordinary
- Estimate then verify
- Scientific context problems
Standard Form Reference: Real-World Examples
| Ordinary Number | Standard Form | Power $n$ | Sign | Real-World Context |
|---|---|---|---|---|
| 300,000,000 | $3 \times 10^8$ | +8 | Positive | Speed of light (m/s) |
| 150,000,000,000 | $1.5 \times 10^{11}$ | +11 | Positive | Earth–Sun distance (m) |
| 8,100,000,000 | $8.1 \times 10^{9}$ | +9 | Positive | World population (approx.) |
| 6,400,000 | $6.4 \times 10^{6}$ | +6 | Positive | Radius of Earth (m) |
| 0.001 | $1 \times 10^{-3}$ | −3 | Negative | 1 millimetre in metres |
| 0.000001 | $1 \times 10^{-6}$ | −6 | Negative | 1 micrometre |
| 0.00000000094 | $9.4 \times 10^{-10}$ | −10 | Negative | Atomic radius of hydrogen (m) |
| 0.00000000000000000016 | $1.6 \times 10^{-19}$ | −19 | Negative | Charge of an electron (C) |
Choosing the Right Method: Decision Flow
Numbers in standard form
→
Operation?
→
✕ or ÷
Operate on $A$, then on powers
Operate on $A$, then on powers
→
+ or −
Align powers first
Align powers first
→
Is $1 \leq A < 10$?
→
Yes → Done
No → Re-express
No → Re-express
Powers of 10 at a Glance
| Power | Value | Prefix Name | Symbol |
|---|---|---|---|
| $10^{12}$ | 1,000,000,000,000 | Tera | T |
| $10^{9}$ | 1,000,000,000 | Giga | G |
| $10^{6}$ | 1,000,000 | Mega | M |
| $10^{3}$ | 1,000 | Kilo | k |
| $10^{0}$ | 1 | — | — |
| $10^{-3}$ | 0.001 | Milli | m |
| $10^{-6}$ | 0.000001 | Micro | μ |
| $10^{-9}$ | 0.000000001 | Nano | n |
| $10^{-12}$ | 0.000000000001 | Pico | p |
✏️ Worked Examples
Grade 4–5 · Basic Conversion
Write (a) 47,500 and (b) 0.0000308 in standard form.
1
Part (a): Find the coefficient $A$
Start at 47,500. Move the decimal point left until the number is between 1 and 10.$47500. \rightarrow 4750.0 \rightarrow 475.00 \rightarrow 47.500 \rightarrow 4.7500$
Four moves left. So $A = 4.75$.
2
Part (a): Determine the power $n$
Moved left 4 places → $n = +4$ (left = positive, large number). Check: $4.75 \times 10^4 = 47500$. ✓
3
Part (b): Find the coefficient $A$
Start at $0.0000308$. Move the decimal point right until the number is between 1 and 10.$0.0000308 \rightarrow 0.000308 \rightarrow 0.00308 \rightarrow 0.0308 \rightarrow 0.308 \rightarrow 3.08$
Five moves right. So $A = 3.08$.
4
Part (b): Determine the power $n$
Moved right 5 places → $n = -5$ (right = negative, small number). Check: $3.08 \times 10^{-5} = 0.0000308$. ✓
(a) $47{,}500 = 4.75 \times 10^{4}$ (b) $0.0000308 = 3.08 \times 10^{-5}$
Grade 6–7 · Multiplication with Re-expressing
Calculate $(3.6 \times 10^7) \times (4.5 \times 10^{-3})$. Give your answer in standard form.
1
Multiply the $A$ values
$3.6 \times 4.5$$= 3.6 \times 4 + 3.6 \times 0.5 = 14.4 + 1.8 = 16.2$
2
Add the powers using the index law
$10^7 \times 10^{-3} = 10^{7+(-3)} = 10^{7-3} = 10^4$
3
Combine and check
Current result: $16.2 \times 10^4$Check: is $16.2 \in [1,10)$? No — $16.2 \geq 10$. Must re-express.
4
Re-express in standard form
$16.2 = 1.62 \times 10^1$So $16.2 \times 10^4 = 1.62 \times 10^1 \times 10^4 = 1.62 \times 10^{1+4} = 1.62 \times 10^5$
Check: $1.62 \in [1,10)$ ✓, $5$ is an integer ✓.
$(3.6 \times 10^7) \times (4.5 \times 10^{-3}) = 1.62 \times 10^5$
Grade 9 · Multi-step Scientific Context
The mass of the Sun is $1.99 \times 10^{30}$ kg. The mass of the Earth is $5.97 \times 10^{24}$ kg.
(a) How many times greater is the mass of the Sun than the mass of the Earth? Give your answer in standard form to 3 significant figures. [3 marks]
(b) The mass of the Moon is approximately $\frac{1}{81}$ of the mass of the Earth. Find the combined mass of the Earth and Moon in standard form to 3 significant figures. [4 marks]
(a) How many times greater is the mass of the Sun than the mass of the Earth? Give your answer in standard form to 3 significant figures. [3 marks]
(b) The mass of the Moon is approximately $\frac{1}{81}$ of the mass of the Earth. Find the combined mass of the Earth and Moon in standard form to 3 significant figures. [4 marks]
1
Part (a): Set up the calculation
"How many times greater" means we divide: $\dfrac{\text{mass of Sun}}{\text{mass of Earth}} = \dfrac{1.99 \times 10^{30}}{5.97 \times 10^{24}}$
2
Part (a): Divide the $A$ values
$\dfrac{1.99}{5.97} = 0.33333\ldots$Note: $0.333\ldots < 1$, so re-expressing will be needed.
3
Part (a): Subtract the powers
$10^{30} \div 10^{24} = 10^{30-24} = 10^6$Combined: $0.33333\ldots \times 10^6$
4
Part (a): Re-express and round
$0.33333\ldots < 1$, so: $0.33333 \times 10^6 = 3.3333 \times 10^{-1} \times 10^6 = 3.3333 \times 10^5$Rounded to 3 s.f.: $3.33 \times 10^5$
5
Part (b): Find the mass of the Moon
Mass of Moon $= \dfrac{5.97 \times 10^{24}}{81}$$= \dfrac{5.97}{81} \times 10^{24} = 0.073703\ldots \times 10^{24}$
Re-express: $0.073703 \times 10^{24} = 7.3703 \times 10^{-2} \times 10^{24} = 7.3703 \times 10^{22}$ kg
6
Part (b): Add Earth mass + Moon mass (align powers)
Earth: $5.97 \times 10^{24}$ kgMoon: $7.3703 \times 10^{22} = 0.073703 \times 10^{24}$ kg (convert to power $10^{24}$)
Sum: $(5.97 + 0.073703) \times 10^{24} = 6.043703 \times 10^{24}$ kg
7
Part (b): Round and verify standard form
$6.043703 \times 10^{24} \approx 6.04 \times 10^{24}$ kg (3 s.f.)Check: $6.04 \in [1,10)$ ✓, $n = 24$ is an integer ✓.
Reasonableness check: Moon is much lighter than Earth, so combined mass should be only slightly more than Earth's $5.97 \times 10^{24}$. Our $6.04 \times 10^{24}$ is about 1.2% more — consistent. ✓
(a) The Sun is $3.33 \times 10^5$ times heavier than the Earth (3 s.f.).
(b) Combined mass of Earth and Moon $= 6.04 \times 10^{24}$ kg (3 s.f.)
(b) Combined mass of Earth and Moon $= 6.04 \times 10^{24}$ kg (3 s.f.)
❓ Exam Questions
Q1
1 mark
Write $0.000072$ in standard form.
Answer: $7.2 \times 10^{-5}$
Mark scheme:
B1 for $7.2 \times 10^{-5}$ (accept exact equivalent). The decimal moves 5 places right to give $A = 7.2$; since the original number is less than 1, $n$ is negative: $n = -5$.
Mark scheme:
B1 for $7.2 \times 10^{-5}$ (accept exact equivalent). The decimal moves 5 places right to give $A = 7.2$; since the original number is less than 1, $n$ is negative: $n = -5$.
Q2
2 marks
Write these numbers in order of size, starting with the smallest:
$\quad 5.2 \times 10^{3}, \quad 8.9 \times 10^{-2}, \quad 3.1 \times 10^{4}, \quad 7.6 \times 10^{-3}$
Answer (smallest to largest):
$7.6 \times 10^{-3}, \quad 8.9 \times 10^{-2}, \quad 5.2 \times 10^{3}, \quad 3.1 \times 10^{4}$
Mark scheme:
M1 for correct method — comparing powers first, then $A$ values when powers match.
A1 for fully correct order.
Verification (ordinary form): $0.0076$; $0.089$; $5200$; $31{,}000$.
$7.6 \times 10^{-3}, \quad 8.9 \times 10^{-2}, \quad 5.2 \times 10^{3}, \quad 3.1 \times 10^{4}$
Mark scheme:
M1 for correct method — comparing powers first, then $A$ values when powers match.
A1 for fully correct order.
Verification (ordinary form): $0.0076$; $0.089$; $5200$; $31{,}000$.
Q3
3 marks
Calculate $(8.4 \times 10^{5}) \div (2.4 \times 10^{-2})$. Give your answer in standard form.
Answer: $3.5 \times 10^{7}$
Working:
Divide $A$ values: $8.4 \div 2.4 = 3.5$ (M1)
Subtract powers: $10^5 \div 10^{-2} = 10^{5-(-2)} = 10^{5+2} = 10^7$ (M1)
Combine: $3.5 \times 10^7$ — check: $3.5 \in [1,10)$ ✓ (A1)
Mark scheme: M1 for dividing $A$ values correctly; M1 for correctly handling the negative power in subtraction ($5-(-2)=7$); A1 for correct final answer in standard form.
Working:
Divide $A$ values: $8.4 \div 2.4 = 3.5$ (M1)
Subtract powers: $10^5 \div 10^{-2} = 10^{5-(-2)} = 10^{5+2} = 10^7$ (M1)
Combine: $3.5 \times 10^7$ — check: $3.5 \in [1,10)$ ✓ (A1)
Mark scheme: M1 for dividing $A$ values correctly; M1 for correctly handling the negative power in subtraction ($5-(-2)=7$); A1 for correct final answer in standard form.
Q4
4 marks
Calculate $6.3 \times 10^{8} - 4.7 \times 10^{7}$. Give your answer in standard form.
Answer: $5.83 \times 10^{8}$
Working:
Align powers — convert $6.3 \times 10^8$ to power $10^7$: $6.3 \times 10^8 = 63 \times 10^7$ (M1)
Subtract $A$ values: $63 \times 10^7 - 4.7 \times 10^7 = (63 - 4.7) \times 10^7 = 58.3 \times 10^7$ (M1)
Re-express — $58.3 \geq 10$: $58.3 \times 10^7 = 5.83 \times 10^8$ (M1)
Final answer: $5.83 \times 10^8$ (A1)
Mark scheme: M1 aligning powers; M1 correct subtraction of aligned values; M1 re-expressing in standard form; A1 correct answer.
Working:
Align powers — convert $6.3 \times 10^8$ to power $10^7$: $6.3 \times 10^8 = 63 \times 10^7$ (M1)
Subtract $A$ values: $63 \times 10^7 - 4.7 \times 10^7 = (63 - 4.7) \times 10^7 = 58.3 \times 10^7$ (M1)
Re-express — $58.3 \geq 10$: $58.3 \times 10^7 = 5.83 \times 10^8$ (M1)
Final answer: $5.83 \times 10^8$ (A1)
Mark scheme: M1 aligning powers; M1 correct subtraction of aligned values; M1 re-expressing in standard form; A1 correct answer.
Q5
6 marks
Light travels at $3 \times 10^{8}$ metres per second. The distance from Earth to Proxima Centauri is $4.013 \times 10^{16}$ metres.
(a) Calculate the time in seconds for light to travel from Proxima Centauri to Earth. Give your answer in standard form to 2 significant figures. [3 marks]
(b) One year is approximately $3.15 \times 10^{7}$ seconds. Using your answer to part (a), express this journey time in years in standard form to 2 significant figures. [3 marks]
Part (a):
Method: time $= \dfrac{\text{distance}}{\text{speed}}$ (M1 for selecting correct formula)
$t = \dfrac{4.013 \times 10^{16}}{3 \times 10^8}$ (M1 for correct substitution)
$= \dfrac{4.013}{3} \times 10^{16-8} = 1.337\overline{6} \times 10^8$
Rounded to 2 s.f.: $1.3 \times 10^8$ seconds (A1)
Part (b):
Time in years $= \dfrac{1.3 \times 10^8}{3.15 \times 10^7}$ (M1 for using (a))
$= \dfrac{1.3}{3.15} \times 10^{8-7} = 0.41\overline{26} \times 10^1$ (M1 for correct calculation)
$= 4.1\overline{26}$ — re-express: $4.1 \times 10^0$ or simply $4.1$ years (A1)
Context check: Proxima Centauri is 4.24 light-years away. Our rounded answer of 4.1 light-years is consistent — the small difference arises from rounding the speed of light and in part (a). A Grade 9 response should note this.
Method: time $= \dfrac{\text{distance}}{\text{speed}}$ (M1 for selecting correct formula)
$t = \dfrac{4.013 \times 10^{16}}{3 \times 10^8}$ (M1 for correct substitution)
$= \dfrac{4.013}{3} \times 10^{16-8} = 1.337\overline{6} \times 10^8$
Rounded to 2 s.f.: $1.3 \times 10^8$ seconds (A1)
Part (b):
Time in years $= \dfrac{1.3 \times 10^8}{3.15 \times 10^7}$ (M1 for using (a))
$= \dfrac{1.3}{3.15} \times 10^{8-7} = 0.41\overline{26} \times 10^1$ (M1 for correct calculation)
$= 4.1\overline{26}$ — re-express: $4.1 \times 10^0$ or simply $4.1$ years (A1)
Context check: Proxima Centauri is 4.24 light-years away. Our rounded answer of 4.1 light-years is consistent — the small difference arises from rounding the speed of light and in part (a). A Grade 9 response should note this.
Q6
3 marks
Given that $p = 3.2 \times 10^4$ and $q = 8 \times 10^{-3}$, calculate $p + q^2$. Give your answer in standard form to 3 significant figures.
Answer: $3.20 \times 10^4$ (3 s.f.) — see working for why.
Working:
Step 1 — Calculate $q^2$: $(8 \times 10^{-3})^2 = 8^2 \times (10^{-3})^2 = 64 \times 10^{-6} = 6.4 \times 10^{-5}$ (M1)
Step 2 — Add $p + q^2$: need to add $3.2 \times 10^4$ and $6.4 \times 10^{-5}$
$6.4 \times 10^{-5} = 0.0000000064 \times 10^4$ (convert to power $10^4$) (M1)
Sum $= (3.2 + 0.0000000064) \times 10^4 \approx 3.2 \times 10^4$ (A1 for recognising $q^2$ is negligible at 3 s.f.)
Key insight: $q^2 = 6.4 \times 10^{-5}$ is negligibly small compared to $p = 3.2 \times 10^4$. To 3 s.f., the answer is $3.20 \times 10^4$. This tests whether students understand the scale of their calculations.
Working:
Step 1 — Calculate $q^2$: $(8 \times 10^{-3})^2 = 8^2 \times (10^{-3})^2 = 64 \times 10^{-6} = 6.4 \times 10^{-5}$ (M1)
Step 2 — Add $p + q^2$: need to add $3.2 \times 10^4$ and $6.4 \times 10^{-5}$
$6.4 \times 10^{-5} = 0.0000000064 \times 10^4$ (convert to power $10^4$) (M1)
Sum $= (3.2 + 0.0000000064) \times 10^4 \approx 3.2 \times 10^4$ (A1 for recognising $q^2$ is negligible at 3 s.f.)
Key insight: $q^2 = 6.4 \times 10^{-5}$ is negligibly small compared to $p = 3.2 \times 10^4$. To 3 s.f., the answer is $3.20 \times 10^4$. This tests whether students understand the scale of their calculations.
⭐ Grade 9 Model Answers
Full Annotated Answer: Q5 (6 marks)
This question examines multi-step standard form in a scientific context — a classic Grade 9 style. The model answer below demonstrates examiner-approved method, presentation, and checking.
PART (a) — Model Answer with Examiner Commentary
Identify the formula:$\text{time} = \dfrac{\text{distance}}{\text{speed}}$ ← M1: Selecting the correct formula (this mark is earned even if subsequent arithmetic is wrong)
Substitute values:
$t = \dfrac{4.013 \times 10^{16}}{3 \times 10^8}$ ← M1: Correct substitution in standard form
Divide systematically:
Divide $A$ values: $\dfrac{4.013}{3} = 1.337\overline{6}$
Subtract powers: $10^{16} \div 10^{8} = 10^{16-8} = 10^8$
Result: $1.337\overline{6} \times 10^8$
Round and check standard form:
$1.337\ldots$ rounded to 2 s.f. is $1.3$. Is $1.3 \in [1,10)$? Yes ✓
Final answer: $1.3 \times 10^8$ seconds ← A1: Correct, rounded to 2 s.f., in standard form
PART (b) — Model Answer with Examiner Commentary
Set up using part (a):Time in years $= \dfrac{1.3 \times 10^8 \text{ s}}{3.15 \times 10^7 \text{ s year}^{-1}}$ ← M1: Using answer from (a) as numerator (follow-through mark)
Divide:
$A$ values: $\dfrac{1.3}{3.15} = 0.41\overline{26}$ (Note: less than 1 — re-expression needed)
Powers: $10^{8-7} = 10^1$
Current result: $0.4126\ldots \times 10^1$ ← M1: Correct division and power subtraction
Re-express:
$0.4126 \times 10^1 = 4.126$ — this is an ordinary number. Write as $4.1 \times 10^0$ (2 s.f.).
Alternatively, $4.1$ years is an acceptable form for a final answer in context.
Final answer: $4.1$ years ← A1: Correct to 2 s.f., accept $4.1 \times 10^0$ or $4.1$
Reasonableness check (Grade 9 habit): Proxima Centauri is known to be ~4.24 light-years away. Our answer of 4.1 years is close — the difference is due to rounding the speed of light to $3 \times 10^8$ (actual: $2.998 \times 10^8$) and the 2 s.f. rounding in (a). This demonstrates the answer is physically plausible. ✓
EXAM TIP: Understanding the Mark Types
B mark (behaviour): Awarded independently — no method needed. Usually for simple conversions or stating values.M mark (method): Awarded for showing the correct process, even if the arithmetic is wrong. Always show your working.
A mark (accuracy): Awarded for the correct final answer. Often only available if preceding M marks are earned.
ft mark (follow-through): Awarded for correct method applied to a wrong previous answer — you can still earn it.
In a 6-mark question, partial credit at every step can save a significant portion of marks even if you make an error.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Standard form | $A \times 10^n$, where $1 \leq A < 10$ and $n \in \mathbb{Z}$ |
| Coefficient | The value of $A$; always in the range $[1, 10)$ |
| Index / power | The integer $n$ indicating the power of 10 |
| Positive power | Number $\geq 10$ (large number) |
| Negative power | Number between 0 and 1 (small number) |
| Zero power | $10^0 = 1$; number between 1 and 10 |
| Re-expressing | Adjusting $A$ and $n$ so $A \in [1,10)$ after a calculation |
Essential Formulae
Definition: $A \times 10^n$, $\; 1 \leq A < 10$, $\; n \in \mathbb{Z}$
Multiply: $(A_1 \times 10^a)(A_2 \times 10^b) = A_1 A_2 \times 10^{a+b}$
Divide: $\dfrac{A_1 \times 10^a}{A_2 \times 10^b} = \dfrac{A_1}{A_2} \times 10^{a-b}$
Add/Subtract: Align powers, then $\pm A$ values
If $A \geq 10$: $A \times 10^n = \frac{A}{10} \times 10^{n+1}$
If $A < 1$: $A \times 10^n = (10A) \times 10^{n-1}$
Index laws used: $10^a \times 10^b = 10^{a+b}$; $\; 10^a \div 10^b = 10^{a-b}$
Memory Hooks
- "Large = Positive": Numbers $> 1$ have $n > 0$
- "Small = Negative": Numbers $< 1$ have $n < 0$
- "Multiply = Add powers": $\times \Rightarrow$ powers ADD
- "Divide = Subtract powers": $\div \Rightarrow$ powers SUBTRACT
- "Add/sub = Align first": $+ / - \Rightarrow$ same power needed
- "ARSE check": $A$ in Range [1,10), $n$ is an integer, Standard form expressed, Extra check for re-expression
- "Minus a minus = plus": $a - (-b) = a + b$ for power subtraction
Exam Tips
- Always write units in scientific context questions
- Show the re-expressing step explicitly — it earns its own method mark
- $10^5 \div 10^{-2} = 10^7$ (subtracting a negative adds)
- For add/subtract: convert to the larger power first
- Read the question: "2 s.f." and "3 s.f." lose marks if ignored
- Always check: does your answer make sense in context?
- $n$ must be an integer — $10^{2.5}$ is NOT standard form
- $A$ can equal 1 but cannot equal 10
- In multi-step problems, keep full precision until the final answer
🔄 Flashcards
Click each card to reveal the answer. Use these for quick self-testing.
✗ Common Mistakes
MISTAKE 1: Coefficient Not in Range [1, 10)
What students do: Leave answers as $34 \times 10^3$ or $0.5 \times 10^8$.Why marks are lost: Neither is in standard form — the answer mark requires the result to be correctly expressed. $34 \times 10^3 = 3.4 \times 10^4$ and $0.5 \times 10^8 = 5 \times 10^7$.
How to avoid it: After every calculation, perform the ARSE check: is $A$ in Range [1,10)? If not, fix it before writing your final answer. Make this your automatic last step.
MISTAKE 2: Wrong Sign on the Power
What students do: Write $0.00045 = 4.5 \times 10^4$ (positive power for a small number).Why marks are lost: $4.5 \times 10^4 = 45{,}000$, which is completely wrong. All marks for that conversion are lost.
How to avoid it: Memorise the rule: any number less than 1 must have a negative power. Before writing your answer, ask: "Is my original number less than 1? Then my power must be negative."
MISTAKE 3: Adding A Values Instead of Multiplying Them
What students do: $(2 \times 10^3) \times (3 \times 10^4) = 5 \times 10^7$ (added 2+3 instead of $2 \times 3$).Why marks are lost: The correct answer is $6 \times 10^7$. Adding gives the wrong $A$ value and loses the accuracy mark.
How to avoid it: The rule is: for multiplication, you MULTIPLY the $A$ values AND ADD the powers. Two separate operations on two separate parts. Write them out explicitly.
MISTAKE 4: Adding A Values Without Aligning Powers
What students do: $4.2 \times 10^6 + 3.1 \times 10^5 = 7.3 \times 10^6$ (added 4.2 + 3.1 directly).Why marks are lost: $3.1 \times 10^5 = 0.31 \times 10^6$, so the correct answer is $4.51 \times 10^6$. The answer $7.3 \times 10^6$ is about 60% too large.
How to avoid it: For addition and subtraction, always align powers first — write both numbers to the same power of 10 before combining $A$ values. Think of it like adding distances: you cannot add 4.2 km and 3.1 m without converting to the same unit.
MISTAKE 5: Sign Error When Subtracting a Negative Power
What students do: $10^5 \div 10^{-2} = 10^{5-2} = 10^3$ instead of $10^{5-(-2)} = 10^7$.Why marks are lost: $10^3 = 1000$ but the correct answer is $10^7 = 10{,}000{,}000$ — a factor of $10{,}000$ out. This is one of the most heavily penalised errors in standard form questions.
How to avoid it: Always write the subtraction explicitly: $a - (-b) = a + b$. With a negative divisor's power, you are effectively adding, not subtracting.
MISTAKE 6: Rounding A Before Completing the Calculation
What students do: In a multi-step problem, round an intermediate $A$ value (e.g. to 2 s.f.) partway through, then use the rounded value in the next step.Why marks are lost: Early rounding introduces compounding errors, causing the final answer to differ from the correct answer — even if the method is perfect.
How to avoid it: Keep full calculator precision throughout all steps. Only round to the required number of significant figures at the very last step. Write "= ... (unrounded)" as an intermediate step to remind yourself.
✅ Final Checklist
Click each item when you feel confident. Your progress is saved automatically.
- I can state the two conditions for standard form: $1 \leq A < 10$ and $n \in \mathbb{Z}$
- I can convert large numbers (e.g. 380,000) into standard form without errors
- I can convert small numbers (e.g. 0.0000047) into standard form without errors
- I can convert from standard form back to ordinary numbers
- I can order a list of numbers given in standard form, including with negative powers
- I can multiply two numbers in standard form and always re-express the result correctly
- I can divide two numbers in standard form, including when subtracting a negative power
- I can add two numbers in standard form by first converting to the same power of 10
- I can subtract two numbers in standard form by first aligning the powers
- I can identify when a calculated result is not in standard form and re-express it correctly
- I can solve multi-step problems involving standard form in scientific contexts
- I can estimate the magnitude of a standard form answer to check reasonableness
- I know common scientific values in standard form (speed of light, Earth–Sun distance, etc.)
- I keep full precision during multi-step calculations and only round at the final step
- I can confidently answer Grade 9, multi-mark standard form questions and explain each step