Standard Form · Study Notes · Edexcel IGCSE 4MA1 Higher

Compact notation for very large and very small numbers. Edexcel expects fluent conversion + arithmetic in both Paper 1H and Paper 2H.

At a glance

Standard form: $a \times 10^{n}$ with $1 \le a < 10$ and $n$ integer.
Coefficient ALWAYS in $[1, 10)$ .
Convert TO: count decimal places shifted; positive shift LEFT = positive $n$ .
Convert FROM: opposite direction.
Mult/div: combine coefficients separately, combine powers separately.
Add/sub: convert to SAME exponent first.
Adjust if coefficient out of range.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.11 — Convert between ordinary numbers and standard form.
1.11 — Calculate with numbers in standard form (multiply, divide, add, subtract).
1.11 — Order numbers in standard form.

Converting to and from standard form

▼

Move the decimal until exactly one non-zero digit is to its left. Count the moves.

Form. A number in standard form is written:

$a \times 10^{n}$

with $1 \le a < 10$ and $n$ an integer.

Converting TO standard form.

Move the decimal until ONE non-zero digit is to its left.
Count moves. Right $\to$ negative $n$ . Left $\to$ positive $n$ .

Examples:

Number	$a$	Moves	$n$	Standard form
$345{,}000$	$3.45$	$5$ left	$+5$	$3.45 \times 10^{5}$
$0.000247$	$2.47$	$4$ right	$-4$	$2.47 \times 10^{-4}$
$7$	$7$	$0$	$0$	$7 \times 10^{0}$
$0.5$	$5$	$1$ right	$-1$	$5 \times 10^{-1}$

Moving left gives a positive power; moving right gives a negative power.

Converting FROM standard form.

Look at $n$ .
Positive $n$ : move decimal $n$ places RIGHT (large number).
Negative $n$ : move decimal $|n|$ places LEFT (small number).

Examples:

Standard form	$n$	Direction	Number
$7.2 \times 10^{4}$	$+4$	right	$72{,}000$
$5.6 \times 10^{-3}$	$-3$	left	$0.0056$
$1 \times 10^{0}$	$0$	none	$1$

Worked qualitative. Why is $1 \le a < 10$ (strictly less than $10$ )?

If $a = 10$ , then $a \times 10^{n} = 10 \times 10^{n} = 10^{n+1}$ — should have been written with the higher exponent.
The bounds force a UNIQUE representation.
Every number has exactly one standard-form expression.

Edexcel tip. Always show the moves: 'I moved the decimal $5$ places left, so the exponent is $+5$ .' Mark schemes credit the working.

$a \in [1, 10)$ , $n$ integer.
TO: move decimal, count moves.
FROM: opposite direction.
Positive $n$ = large number.
Negative $n$ = small number.

Arithmetic in standard form

▼

Mult/div: split coefficients and powers. Add/sub: same exponent first.

Multiplication and division use index laws.

$\boxed{(a \times 10^{m}) \times (b \times 10^{n}) = (a \times b) \times 10^{m+n}}$

$\boxed{(a \times 10^{m}) \div (b \times 10^{n}) = (a/b) \times 10^{m-n}}$

Example. $(3 \times 10^{6}) \times (4 \times 10^{-2})$ .

Coefficients: $3 \times 4 = 12$ .
Powers: $10^{6} \times 10^{-2} = 10^{4}$ .
Result: $12 \times 10^{4}$ .
ADJUST: $12$ is too big. Divide by $10$ , multiply $n$ by another $10$ : $1.2 \times 10^{5}$ .

Example. $(8 \times 10^{6}) \div (2 \times 10^{4})$ .

Coefficients: $8/2 = 4$ .
Powers: $10^{6}/10^{4} = 10^{2}$ .
Result: $4 \times 10^{2}$ .

Addition and subtraction.

You CANNOT just add/subtract coefficients if the powers differ. Convert to the SAME power first.

Example. $(4.5 \times 10^{6}) + (2.7 \times 10^{5})$ .

Convert smaller to larger power: $2.7 \times 10^{5} = 0.27 \times 10^{6}$ .

Now: $(4.5 + 0.27) \times 10^{6} = 4.77 \times 10^{6}$ .

Adjust at the end.

ALWAYS check: is the coefficient in $[1, 10)$ ?

Too big ( $\ge 10$ ): divide by $10$ , increase $n$ . $12 \times 10^{4} \to 1.2 \times 10^{5}$ .
Too small ( $< 1$ ): multiply by $10$ , decrease $n$ . $0.5 \times 10^{4} \to 5 \times 10^{3}$ .

Worked qualitative. Why use standard form if calculator does the work?

Numbers like $6{,}023{,}000{,}000{,}000{,}000{,}000{,}000{,}000$ (Avogadro's number) are unwieldy.
Standard form: $6.023 \times 10^{23}$ . Compact, readable, comparable.
Used universally in physics, chemistry, biology, engineering.

Edexcel tip. Always do the adjustment step. Mark schemes deduct for answers like $12 \times 10^{4}$ even if the value is correct.

Mult/div: split into coefficients and powers.
Add/sub: convert to same exponent first.
ALWAYS adjust if coefficient $\ge 10$ or $< 1$ .
Common in physics, chemistry, biology.

Ordering numbers in standard form

▼

Compare exponents first. Same exponents: compare coefficients.

When comparing or ordering numbers in standard form:

Compare the EXPONENTS first.
Same exponent: compare the COEFFICIENTS.
Negative exponents: a SMALLER (more negative) exponent means a SMALLER number.

Example. Order ascending: $3.4 \times 10^{4}$ , $2.1 \times 10^{5}$ , $9.7 \times 10^{3}$ , $5 \times 10^{4}$ .

Exponents: $4, 5, 3, 4$ .
Smallest exponent first: $9.7 \times 10^{3}$ (smallest).
Then exponent $4$ : compare $3.4$ vs $5$ . So $3.4 \times 10^{4} < 5 \times 10^{4}$ .
Then largest exponent: $2.1 \times 10^{5}$ .

Order: $9.7 \times 10^{3}, 3.4 \times 10^{4}, 5 \times 10^{4}, 2.1 \times 10^{5}$ .

Bigger exponent wins; when exponents tie, the larger coefficient wins.

Negative exponents.

For small numbers: $5 \times 10^{-3}$ vs $2 \times 10^{-4}$ .

Compare exponents: $-3 > -4$ .
So $5 \times 10^{-3} > 2 \times 10^{-4}$ .
The number with the LESS NEGATIVE exponent is bigger.

Worked qualitative. Compare $3.7 \times 10^{-5}$ and $1.2 \times 10^{-4}$ .

Exponents: $-5 < -4$ .
$3.7 \times 10^{-5} = 0.000037$ .
$1.2 \times 10^{-4} = 0.00012$ .
$0.000037 < 0.00012$ , confirming the exponent comparison.

Edexcel tip. Convert to ordinary numbers if exponents differ widely or you're unsure. Slower but bulletproof.

Exponents first.
Same exponent: coefficients.
Less negative exponent = bigger.
Convert to ordinary if uncertain.

Quick recap

$a \times 10^{n}$ , $1 \le a < 10$ .
TO standard form: move decimal, count.
Mult/div: split into coefficients and powers.
Add/sub: same exponent first.
ALWAYS check coefficient in $[1, 10)$ .
Order: exponents first, coefficients second.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Standard form: $a \times 10^{n}$ , $1 \le a < 10$ .
Mult: $(a \times 10^{m}) \times (b \times 10^{n}) = ab \times 10^{m+n}$ .
Div: $(a \times 10^{m}) / (b \times 10^{n}) = (a/b) \times 10^{m-n}$ .
Add/sub: same exponent FIRST.
Adjust: $12 \times 10^{4} = 1.2 \times 10^{5}$ .

How it’s examined

Standard form appears every Paper 1H AND 2H (3-5 marks). Typical: convert to/from, multiply/divide in standard form, problem with very large/small physical quantities. Examiner reports flag (1) coefficient out of range, (2) wrong addition without converting exponents, and (3) decimal-shift direction errors.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on standard form, written exactly the way a tutor would explain them at the board.

1Convert to standard form
Foundation• standard form
▼
Question
Express in standard form: (a) $345{,}000$ (b) $0.000247$
Step-by-step solution
1. Step 1
  (a) $345{,}000$ . Move decimal so just one non-zero digit is left of it: $3.45000$ . Moved $5$ places LEFT, so multiply by $10^{5}$ .
  $345{,}000 = 3.45 \times 10^{5}$
2. Step 2
  (b) $0.000247$ . Move decimal RIGHT to get $2.47$ . Moved $4$ places, so multiply by $10^{-4}$ .
  $0.000247 = 2.47 \times 10^{-4}$
Answer
(a) $3.45 \times 10^{5}$ (b) $2.47 \times 10^{-4}$
Examiner tip
Standard form: $a \times 10^{n}$ where $1 \le a < 10$ and $n$ is an integer. The coefficient $a$ MUST be at least $1$ but less than $10$ .
2Convert from standard form
Foundation• standard form
▼
Question
Write as ordinary numbers: (a) $7.2 \times 10^{4}$ (b) $5.6 \times 10^{-3}$
Step-by-step solution
1. Step 1
  (a) Multiplied by $10^{4}$ → move decimal $4$ places RIGHT.
  $7.2 \times 10^{4} = 72{,}000$
2. Step 2
  (b) Multiplied by $10^{-3}$ → move decimal $3$ places LEFT.
  $5.6 \times 10^{-3} = 0.0056$
Answer
(a) $72{,}000$ (b) $0.0056$
3Multiplying in standard form
Higher• Adapted from 4MA1/2H Jan 2024 Q8• standard form, multiplication
▼
Question
Calculate $(3 \times 10^{6}) \times (4 \times 10^{-2})$ . Give your answer in standard form.
Step-by-step solution
1. Step 1
  Multiply coefficients separately and powers of $10$ separately.
  $(3 \times 4) \times (10^{6} \times 10^{-2}) = 12 \times 10^{4}$
2. Step 2
  But $12$ isn't between $1$ and $10$ . Adjust: $12 = 1.2 \times 10$ .
  $12 \times 10^{4} = 1.2 \times 10 \times 10^{4} = 1.2 \times 10^{5}$
Answer
$1.2 \times 10^{5}$
Examiner tip
After computing, ALWAYS check the coefficient is in $[1, 10)$ . Adjust the power of $10$ if not.
4Adding in standard form (different powers)
Higher• standard form, addition
▼
Question
Calculate $(4.5 \times 10^{6}) + (2.7 \times 10^{5})$ . Give your answer in standard form.
Step-by-step solution
1. Step 1
  Convert both to the SAME power of $10$ (use the larger):
  $2.7 \times 10^{5} = 0.27 \times 10^{6}$
2. Step 2
  Add coefficients: $4.5 + 0.27 = 4.77$ .
  $(4.5 + 0.27) \times 10^{6} = 4.77 \times 10^{6}$
Answer
$4.77 \times 10^{6}$
Examiner tip
Addition/subtraction in standard form: convert to the SAME exponent first. Multiplication/division: keep separate exponents and combine via index laws.
5Order of magnitude estimate
Higher• standard form, estimation
▼
Question
Estimate $(6.7 \times 10^{8}) \div (3.2 \times 10^{4})$ in standard form.
Step-by-step solution
1. Step 1
  Divide coefficients: $6.7 / 3.2 \approx 2.09$ .
2. Step 2
  Divide powers: $10^{8} / 10^{4} = 10^{4}$ .
3. Step 3
  Combine: $\approx 2.09 \times 10^{4}$ .
Answer
$\approx 2.1 \times 10^{4}$
Examiner tip
On Paper 1H (non-calculator), use approximations. $6.7/3.2 \approx 7/3 \approx 2.3$ . On Paper 2H, calculate exactly.
6A number given in words [1 mark]
Getting started• Paper 1H/2H style — 1 mark• standard form, converting
▼
Question
Write sixty-four million in standard form. [1]
Step-by-step solution
1. Step 1
  Sixty-four million is $64$ followed by six zeros.
  $64\,000\,000$
2. Step 2
  B1 The point moves $7$ places left to give $6.4$ , so the index is $7$ .
  $64\,000\,000 = 6.4 \times 10^{7}$
Answer
$6.4 \times 10^{7}$
Examiner tip
Examiner tip: Write the ordinary number out first. One million has six zeros, so sixty-four million is $64$ followed by six zeros.

Common mistake: Writing $64 \times 10^{6}$ . It has the right value, but $64$ is not between $1$ and $10$ , so it scores B0.
7Ordering small numbers in mixed forms [2 marks]
Getting started• Paper 1H/2H style — 2 marks• standard form, ordering
▼
Question
Write these numbers in order of size. Start with the smallest number.

$5.3 \times 10^{-4}$ $\qquad 0.000\,52$ $\qquad 51 \times 10^{-5}$ $\qquad 5.1 \times 10^{-3}$ $\qquad 0.0005$ [2]
Step-by-step solution
1. Step 1
  B1 Write every number in the same form, here as ordinary numbers. Note that $51 \times 10^{-5}$ is not in standard form: it equals $5.1 \times 10^{-4}$ .
  $0.000\,53, \quad 0.000\,52, \quad 0.000\,51, \quad 0.0051, \quad 0.0005$
2. Step 2
  B1 Compare the ordinary numbers and list them smallest first, in their original form.
  $0.0005,\ \ 51 \times 10^{-5},\ \ 0.000\,52,\ \ 5.3 \times 10^{-4},\ \ 5.1 \times 10^{-3}$
Answer
$0.0005$ , $51 \times 10^{-5}$ , $0.000\,52$ , $5.3 \times 10^{-4}$ , $5.1 \times 10^{-3}$
Examiner tip
Examiner tip: Give the answer using the numbers exactly as printed. B1 is awarded for converting all five to one form correctly, or for four in the correct order.

Common mistake: Treating $51 \times 10^{-5}$ as the largest because $51$ is the biggest coefficient, or putting $5.1 \times 10^{-3}$ next to $5.3 \times 10^{-4}$ because the coefficients look close. The index decides the size before the coefficient does.
8Reading and rounding a calculator display [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• standard form, calculator, rounding
▼
Question
(a) Work out $\dfrac{4.7 \times 10^{5}}{(3.6 \times 10^{-2})^{2}}$

Write down all the figures on your calculator display. [2]

(b) Give your answer to part (a) in standard form correct to 2 significant figures. [1]
Step-by-step solution
1. Step 1
  M1 (a) Work out the denominator first and write it down.
  $(3.6 \times 10^{-2})^{2} = 12.96 \times 10^{-4} = 1.296 \times 10^{-3}$
2. Step 2
  A1 Divide. The display shows the value in standard form; write it as an ordinary number or copy it exactly.
  $\dfrac{4.7 \times 10^{5}}{1.296 \times 10^{-3}} = 362\,654\,320.98\ldots$
3. Step 3
  B1 (b) Round to 2 significant figures: the third figure is $2$ , so round down.
  $362\,654\,320.98\ldots \approx 3.6 \times 10^{8}$
Answer
(a) $362\,654\,320.98\ldots$ (a display of $362\,654\,321$ or $3.626\,543\,21 \times 10^{8}$ is accepted) (b) $3.6 \times 10^{8}$
Examiner tip
Examiner tip: A display such as 3.62654321 × 10⁸ must be written with the power of 10. Copying only "3.62654321" answers a different number and scores A0.

Common mistake: Keying the denominator without brackets, so the calculator squares only $10^{-2}$ . That gives $\dfrac{4.7 \times 10^{5}}{3.6} \times 10^{4}$ , about $1.3 \times 10^{9}$ . Always bracket a squared term.
9Packing a year of production [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• standard form, percentages, context
▼
Question
A factory makes $2.4 \times 10^{8}$ screws in one year.

6% of the screws are faulty and are thrown away.

The rest are packed in boxes of $1.5 \times 10^{3}$ screws.

Work out the number of boxes that are filled in the year.

Give your answer in standard form. [4]
Step-by-step solution
1. Step 1
  M1 The screws kept are $100\% - 6\% = 94\%$ of the total.
  $0.94 \times 2.4 \times 10^{8}$
2. Step 2
  A1 Evaluate.
  $= 2.256 \times 10^{8}\ \text{screws}$
3. Step 3
  M1 Divide by the number of screws in one box.
  $\dfrac{2.256 \times 10^{8}}{1.5 \times 10^{3}} = 1.504 \times 10^{5}$
4. Step 4
  A1 The division is exact, so every box is full.
  $150\,400 = 1.504 \times 10^{5}$
Answer
$1.504 \times 10^{5}$ boxes
Examiner tip
Examiner tip: Using the multiplier $0.94$ does the percentage in one step. Subtracting $0.06 \times 2.4 \times 10^{8} = 1.44 \times 10^{7}$ from the total is equally valid.

Common mistake: Working with the 6% that are thrown away instead of the 94% that are kept: $(0.06 \times 2.4 \times 10^{8}) \div (1.5 \times 10^{3})$ gives $9.6 \times 10^{3}$ boxes, which answers a different question. Another is dividing the full $2.4 \times 10^{8}$ and forgetting the faulty screws, giving $1.6 \times 10^{5}$ .
10Oxygen and carbon dioxide in the atmosphere [5 marks]
Stretch• Paper 1H/2H style — 5 marks• standard form, percentages, how many times
▼
Question
The mass of the Earth's atmosphere is $5.1 \times 10^{18}$ kg.

23% of this mass is oxygen.

(a) Work out the mass of oxygen in the atmosphere. Give your answer in standard form. [2]

The mass of carbon dioxide in the atmosphere is $3.2 \times 10^{15}$ kg.

(b) How many times greater is the mass of oxygen than the mass of carbon dioxide? Give your answer correct to 3 significant figures. [3]
Step-by-step solution
1. Step 1
  M1 (a) Find 23% using the multiplier $0.23$ .
  $0.23 \times 5.1 \times 10^{18}$
2. Step 2
  A1 Evaluate. $0.23 \times 5.1 = 1.173$ , which is already between $1$ and $10$ , so the index stays $18$ .
  $= 1.173 \times 10^{18}\ \text{kg}$
3. Step 3
  M1 (b) Divide the mass of oxygen by the mass of carbon dioxide, using the unrounded value from (a).
  $\dfrac{1.173 \times 10^{18}}{3.2 \times 10^{15}}$
4. Step 4
  A1 Evaluate in full.
  $= 366.5625$
5. Step 5
  B1 Round to 3 significant figures.
  $366.5625 \approx 367$
Answer
(a) $1.173 \times 10^{18}$ kg (b) $367$
Examiner tip
Examiner tip: The answer to (a) is exact, so carry all four figures into (b). An answer to a "how many times" question can be an ordinary number when it is not large.

Common mistake: Rounding (a) to $1.17 \times 10^{18}$ and using that in (b), which gives $365.625$ and the answer $366$ . This is premature rounding and loses the A1.

Question 1
Paper 1H/2H style — 1 mark1 mark
Here are four numbers.

$0.7 \times 10^{4}$ $\qquad 7 \times 10^{0.5}$ $\qquad 7.0 \times 10^{-4}$ $\qquad 70 \times 10^{3}$

Write down the number that is written in standard form. [1]
Model answer
$7.0 \times 10^{-4}$
Why this scores
Mark scheme:
- B1 for $7.0 \times 10^{-4}$ cao
Why this scores full marks: Each option is tested against both conditions. $0.7$ and $70$ fail $1 \leqslant a < 10$ , and $10^{0.5}$ fails because the index must be an integer. A negative index is allowed, so $7.0 \times 10^{-4}$ is the only one that meets both.
Question 2
Paper 1H/2H style — 2 marks2 marks
Work out $\dfrac{7.2 \times 10^{5} + 4.8 \times 10^{4}}{1.6 \times 10^{-2}}$

Give your answer in standard form. [2]
Model answer
$7.2 \times 10^{5} + 4.8 \times 10^{4} = 720\,000 + 48\,000 = 768\,000$

$\frac{768\,000}{1.6 \times 10^{-2}} = 48\,000\,000$

$4.8 \times 10^{7}$
Why this scores
Mark scheme:
- M1 for $768\,000$ oe, e.g. $7.68 \times 10^{5}$
- A1 for $4.8 \times 10^{7}$ cao
Why this scores full marks: The numerator is worked out first, as its own line, so the M1 is visible. Dividing by a number less than 1 makes the answer bigger, and the candidate's answer is bigger than the numerator, which is a quick sense check.
Question 3
Paper 1H/2H style — 2 marks2 marks
$A = 3.2 \times 10^{6}$ and $B = 8 \times 10^{-2}$

(a) Work out $AB$ . Give your answer in standard form. [1]

(b) Work out $\dfrac{A}{B}$ . Give your answer in standard form. [1]
Model answer
(a)

$AB = 25.6 \times 10^{4} = 2.56 \times 10^{5}$

(b)

$\frac{A}{B} = 0.4 \times 10^{8} = 4 \times 10^{7}$
Why this scores
Mark scheme:
- B1 for $2.56 \times 10^{5}$ cao
- B1 for $4 \times 10^{7}$ cao
Why this scores full marks: Both answers are adjusted into standard form. $25.6 \times 10^{4}$ and $0.4 \times 10^{8}$ have the right value but would each score B0, because the coefficient is outside $1 \leqslant a < 10$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
The speed of light is $3 \times 10^{8}$ m/s.

The speed of sound in air is $3.4 \times 10^{2}$ m/s.

How many times faster is the speed of light than the speed of sound?

Give your answer in standard form correct to 3 significant figures. [3]
Model answer
$\frac{3 \times 10^{8}}{3.4 \times 10^{2}} = 882\,352.94\ldots$

$8.82 \times 10^{5}$
Why this scores
Mark scheme:
- M1 for $\dfrac{3 \times 10^{8}}{3.4 \times 10^{2}}$ oe
- A1 for awrt $882\,000$
- B1 ft for their answer in standard form to 3 significant figures
Why this scores full marks: The larger speed is divided by the smaller, so the answer is greater than 1. The full display is written before rounding, and the final line is in the form and accuracy the question asks for.
Question 5
Paper 1H/2H style — 3 marks3 marks
An empty lorry has a mass of $1.2 \times 10^{4}$ kg.

The lorry carries $4.5 \times 10^{3}$ bricks. Each brick has a mass of $2.6$ kg.

Work out the total mass of the lorry and the bricks.

Give your answer in standard form. [3]
Model answer
Mass of bricks:

$4.5 \times 10^{3} \times 2.6 = 11\,700 = 1.17 \times 10^{4}\ \text{kg}$

Total:

$1.2 \times 10^{4} + 1.17 \times 10^{4} = 2.37 \times 10^{4}\ \text{kg}$
Why this scores
Mark scheme:
- M1 for $4.5 \times 10^{3} \times 2.6$ oe ( $11\,700$ )
- M1 for $1.2 \times 10^{4} +$ their mass of bricks
- A1 for $2.37 \times 10^{4}$ cao
Why this scores full marks: The two masses are both written with $10^{4}$ , so the coefficients add directly. Candidates who add the indices as well, writing $2.37 \times 10^{8}$ , lose the A1.
Question 6
Paper 1H/2H style — 4 marks4 marks
Three space probes are at these distances from the Earth.

$4.2 \times 10^{5}$ km $\qquad 3.9 \times 10^{6}$ km $\qquad 8.1 \times 10^{4}$ km

Work out the mean distance of the three probes from the Earth.

Give your answer in standard form. [4]
Model answer
$420\,000 + 3\,900\,000 + 81\,000 = 4\,401\,000$

$\text{Mean} = \frac{4\,401\,000}{3} = 1\,467\,000$

$1.467 \times 10^{6}$ km
Why this scores
Mark scheme:
- M1 for writing all three distances in a common form, e.g. as ordinary numbers or all with $10^{4}$
- M1 for a sum of $4\,401\,000$ oe
- M1 for their sum $\div 3$
- A1 for $1.467 \times 10^{6}$ cao
Why this scores full marks: The candidate converts before adding, so the sum is right. Adding the coefficients $4.2 + 3.9 + 8.1$ is the classic error and scores at most the M1 for dividing by $3$ . The mean is exact, so no rounding is needed.
Question 7
Paper 1H/2H style — 4 marks4 marks
The mass of a virus is $6 \times 10^{-16}$ g.

The mass of a bacterium is $1.5 \times 10^{-12}$ g.

(a) Write the ratio mass of virus : mass of bacterium in the form $1 : n$ [2]

(b) Work out how many of these viruses have the same total mass as $5$ of these bacteria. Give your answer in standard form. [2]
Model answer
(a)

$n = \frac{1.5 \times 10^{-12}}{6 \times 10^{-16}} = 0.25 \times 10^{4} = 2500$

$1 : 2500$

(b)

$5 \times 2500 = 12\,500 = 1.25 \times 10^{4}$
Why this scores
Mark scheme:
- M1 for $\dfrac{1.5 \times 10^{-12}}{6 \times 10^{-16}}$ oe
- A1 for $1 : 2500$ oe
- M1 for $5 \times$ their $n$ , or $\dfrac{5 \times 1.5 \times 10^{-12}}{6 \times 10^{-16}}$
- A1 for $1.25 \times 10^{4}$ cao
Why this scores full marks: Dividing both sides of the ratio by the mass of the virus makes the first part 1. The index subtraction $-12 - (-16) = 4$ is written out, which is where most sign slips happen.
Question 8
Paper 1H/2H style — 5 marks5 marks
$\dfrac{(6 \times 10^{n})^{2}}{4 \times 10^{3}} = 9 \times 10^{11}$

(a) Find the value of $n$ . Show your working clearly. [3]

(b) Work out $\dfrac{1}{6 \times 10^{n}}$ for this value of $n$ . Give your answer in standard form correct to 3 significant figures. [2]
Model answer
(a)

$(6 \times 10^{n})^{2} = 36 \times 10^{2n}$

$\frac{36 \times 10^{2n}}{4 \times 10^{3}} = 9 \times 10^{2n-3}$

$2n - 3 = 11 \quad\Rightarrow\quad n = 7$

(b)

$\frac{1}{6 \times 10^{7}} = 1.666\ldots \times 10^{-8}$

$1.67 \times 10^{-8}$
Why this scores
Mark scheme:
- M1 for $36 \times 10^{2n}$
- M1 for $9 \times 10^{2n-3}$ oe, or $2n - 3 = 11$
- A1 for $n = 7$ cao
- M1 for $\dfrac{1}{6 \times 10^{7}}$ or $1.666\ldots \times 10^{-8}$ ft their $n$
- A1 for $1.67 \times 10^{-8}$ cao
Why this scores full marks: The coefficient $36 \div 4 = 9$ already matches, so the equation is formed from the indices alone. In (b) the reciprocal of a large number is small, so the index is negative; $0.1\dot{6} \times 10^{-7}$ would not be standard form.
Question 9
Paper 1H/2H style — 5 marks5 marks
Country A has a population of $6.72 \times 10^{7}$ and an area of $2.4 \times 10^{5}$ km².

Country B has a population of $1.28 \times 10^{8}$ and an area of $3.2 \times 10^{5}$ km².

Population density $= \dfrac{\text{population}}{\text{area}}$

Which country has the greater population density, and by how many people per km²? Show your working clearly. [5]
Model answer
Country A:

$\frac{6.72 \times 10^{7}}{2.4 \times 10^{5}} = 2.8 \times 10^{2} = 280\ \text{people per km}^{2}$

Country B:

$\frac{1.28 \times 10^{8}}{3.2 \times 10^{5}} = 0.4 \times 10^{3} = 400\ \text{people per km}^{2}$

$400 - 280 = 120$

Country B has the greater population density, by 120 people per km².
Why this scores
Mark scheme:
- M1 for $\dfrac{6.72 \times 10^{7}}{2.4 \times 10^{5}}$
- A1 for $280$
- M1 for $\dfrac{1.28 \times 10^{8}}{3.2 \times 10^{5}}$
- A1 for $400$
- B1 for Country B by $120$ people per km², dep on both M marks
Why this scores full marks: Both densities are worked out and labelled before comparing, and the conclusion names the country and gives the difference with units. Country B has the larger population but also the larger area, so the comparison cannot be made by looking at populations alone.
Question 10
Paper 1H/2H style — 5 marks5 marks
One light-year is $9.46 \times 10^{12}$ km.

A star is $4.02 \times 10^{13}$ km from the Earth.

(a) Work out the distance of the star from the Earth in light-years. Give your answer correct to 3 significant figures. [2]

A space probe travels at a constant speed of $6.2 \times 10^{4}$ km/h. Take one year to be $365$ days.

(b) Work out the number of years the probe takes to travel to the star. Give your answer in standard form correct to 2 significant figures. [3]
Model answer
(a)

$\frac{4.02 \times 10^{13}}{9.46 \times 10^{12}} = 4.2494\ldots$

$4.25$ light-years

(b)

$\text{Time} = \frac{4.02 \times 10^{13}}{6.2 \times 10^{4}} = 648\,387\,096.8\ldots\ \text{hours}$

$\frac{648\,387\,096.8\ldots}{365 \times 24} = 74\,016.79\ldots\ \text{years}$

$7.4 \times 10^{4}$ years
Why this scores
Mark scheme:
- M1 for $\dfrac{4.02 \times 10^{13}}{9.46 \times 10^{12}}$
- A1 for $4.25$ cao
- M1 for $\dfrac{4.02 \times 10^{13}}{6.2 \times 10^{4}}$ oe (awrt $6.48 \times 10^{8}$ )
- M1 for their time in hours $\div (365 \times 24)$
- A1 for $7.4 \times 10^{4}$ cao
Why this scores full marks: Time $=$ distance $\div$ speed is used with distance in km, matching the speed in km/h. The hours are then changed to years with both factors, $365$ and $24$ . The full value is kept in the calculator until the last line, which is rounded to the 2 significant figures asked for.

4MA1 Standard Form

At a glance

What you’ll learn

Converting to and from standard form

Arithmetic in standard form

Ordering numbers in standard form

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Convert to standard form

2Convert from standard form

3Multiplying in standard form

4Adding in standard form (different powers)

5Order of magnitude estimate

6A number given in words [1 mark]

7Ordering small numbers in mixed forms [2 marks]

8Reading and rounding a calculator display [3 marks]

9Packing a year of production [4 marks]

10Oxygen and carbon dioxide in the atmosphere [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10