Standard form and calculations at different scales
Write very large and very small quantities compactly, then calculate without losing their scale.
Section Paper 2 · Extended · Non-calculator
Core and Extended. Core standard-form calculations are assessed on the calculator paper. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use decimal place value.
- Use positive and negative integer powers of ten.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which is standard form for 0.00072?
What you will learn
- Convert numbers to and from standard form.
- Multiply, divide, add and subtract standard-form values.
- Interpret calculator scientific notation.
One leading non-zero digit
Standard form writes a positive magnitude as A × 10ⁿ, with 1 ≤ A < 10 and integer n. For 4 700 000, A is 4.7 and n is 6. For 0.00047, A is still 4.7, but n is −4. The power restores the original decimal scale; moving a decimal point is a method, not the definition.
A negative number can carry a minus sign outside this form, such as −4.7 × 10⁶. Zero has no representation with 1 ≤ A < 10. In 47 × 10⁵ the value is correct but the coefficient is not normalised, so rewrite it as 4.7 × 10⁶ when standard form is requested.
Multiply or divide the two parts
For multiplication, multiply the coefficients and add the powers of ten. Thus (3 × 10⁴)(4 × 10⁻²) = 12 × 10² = 1.2 × 10³. Renormalise only after calculating: replacing 12 by 1.2 needs one extra power of ten to preserve the value.
For division, divide coefficients and subtract exponents. For example, (6 × 10⁻³)/(2 × 10⁴) = 3 × 10⁻⁷. Keep parentheses around the complete denominator when entering a quotient. A scale estimate can check whether the answer should be very small or very large.
Addition needs a common power
You cannot add exponents to add quantities. Express both terms with the same power first: 3.2 × 10⁵ + 4.5 × 10⁴ = 3.2 × 10⁵ + 0.45 × 10⁵ = 3.65 × 10⁵. This is the same idea as giving fractions a common denominator before adding their numerators.
Subtraction may require renormalising. If 6.1 × 10⁻⁴ − 5.7 × 10⁻⁴ = 0.4 × 10⁻⁴, rewrite the result as 4 × 10⁻⁵. Retain units throughout: subtracting a mass from a length is not meaningful even if their powers of ten match.
Read calculator notation and precision
A display such as 2.4E−6 means 2.4 × 10⁻⁶; the E is an exponent separator, not a new unit. Scientific calculators have different labels for the exponent key. Enter the coefficient, exponent key and signed exponent; do not also multiply by ten unless the chosen input method requires it.
Standard form does not determine how many significant figures to give. If the question requests three significant figures, 1.234 × 10⁷ becomes 1.23 × 10⁷. Keep the unrounded value for any later step, and write the final answer with the multiplication sign and power rather than copying an ambiguous display string.
Pause and explain
How should 2 × 10³ + 3 × 10² be calculated?
Worked example
Calculate (2.4 × 10⁻³)(5 × 10⁶), giving standard form.
Show the worked solution
- Multiply coefficients: 2.4 × 5 = 12.
- Add exponents: −3 + 6 = 3, giving 12 × 10³.
- Normalise the coefficient: 12 × 10³ = 1.2 × 10⁴.
Answer 1.2 × 10⁴
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Divide 7.2 × 10⁻⁴ by 3 × 10².
Give me a hint
Divide coefficients and subtract the exponents in their original order.
Compare my reasoning
- 7.2 ÷ 3 = 2.4.
- The exponent is −4 − 2 = −6.
- The normalised result is 2.4 × 10⁻⁶, equal to 0.0000024.
2.4 × 10⁻⁶
Look for these in your work
- Subtracted the divisor's exponent.
- Checked that the quotient is smaller than the dividend.
Calculate 6.4 × 10⁵ − 8 × 10⁴ in standard form.
Give me a hint
Write the second term using 10⁵.
Compare my reasoning
- 8 × 10⁴ = 0.8 × 10⁵.
- Subtract coefficients: 6.4 − 0.8 = 5.6.
- The result is 5.6 × 10⁵, or 560 000.
5.6 × 10⁵
Look for these in your work
- Used a shared power before subtraction.
- Kept the coefficient in the standard-form range.
A file is 3.6 × 10⁷ bytes. A link transfers 1.2 × 10⁶ bytes each second at a constant rate. Find the transfer time, stating one assumption.
Give me a hint
Time is the amount divided by the amount per second.
Compare my reasoning
- Divide: (3.6/1.2) × 10⁷⁻⁶ = 3 × 10¹.
- This equals 30 seconds; byte units cancel.
- The model assumes the stated rate remains constant, with no additional overhead or interruptions.
30 seconds, assuming a constant effective transfer rate.
Look for these in your work
- Used a quotient with compatible units.
- Stated a limitation of the constant-rate model.
Common mistakes
- Giving a coefficient outside 1 ≤ A < 10.
- Adding exponents for addition or subtraction.
- Entering a denominator without grouping its coefficient and power.
What can you explain now?
Write 0.0000056 in standard form.
Compare with the explanation
5.6 × 10⁻⁶
Multiplying 5.6 by one millionth restores the six-place decimal shift.
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Make it stick
Convert 0.000083 and 9 100 000 to standard form, then explain why adding them requires matching powers but multiplying them does not.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.