Understanding and Calculating with Standard Form Revision Notes | GCSE AQA Higher Maths

1. Why Standard Form Exists

The distance from Earth to the Sun is approximately 150,000,000,000 metres. The diameter of a hydrogen atom is roughly 0.0000000001 metres. Writing either of those out in full is slow, error-prone, and hard to compare at a glance.

Standard form — also called scientific notation — solves this by expressing any number as a value between 1 and 10 multiplied by a power of ten. The Sun's distance becomes $1.5 \times 10^{11}$ m; the hydrogen atom's diameter becomes $1 \times 10^{-10}$ m. Both are compact, unambiguous, and instantly comparable.

The two conditions on $A$ and $n$ are non-negotiable on the mark scheme. $A$ must be at least 1 but strictly less than 10 — so exactly one non-zero digit sits before the decimal point. $n$ must be a whole number, positive or negative.

2. Converting to and from Standard Form

The key move is locating where the decimal point needs to sit to give a value of $A$ between 1 and 10, then counting how many places it has shifted. Each place the decimal moves right corresponds to multiplying by 10; each place left corresponds to dividing by 10.

Worked example: Writing a Large Number in Standard Form

Write $4\,730\,000$ in standard form.

1. $4\,730\,000 = 4730000.$ — The decimal point is currently at the far right of the number.
2. $4.730000$ — Move the decimal point 6 places to the left so the number reads 4.73. This gives $A = 4.73$, which satisfies $1 \leq 4.73 < 10$. [M1]
3. $n = +6$ — The decimal moved 6 places to the left, so we compensate by multiplying by $10^6$. Moving left means a positive power.
4. $4.73 \times 10^6$ — Check: $4.73 \times 10^6 = 4\,730\,000$. Correct. [A1]

Answer: $4.73 \times 10^6$

Worked example: Writing a Small Number in Standard Form

Write $0.000052$ in standard form.

1. $0.000052$ — The decimal point is currently after the leading zero.
2. $5.2$ — Move the decimal point 5 places to the right to reach 5.2. This gives $A = 5.2$, satisfying $1 \leq 5.2 < 10$. [M1]
3. $n = -5$ — The decimal moved 5 places to the right, so we compensate by multiplying by $10^{-5}$. Moving right means a negative power.
4. $5.2 \times 10^{-5}$ — Check: $5.2 \times 10^{-5} = 0.000052$. Correct. [A1]

Answer: $5.2 \times 10^{-5}$

Converting back to an ordinary number reverses the process. A positive power of ten means shift the decimal right (the number gets larger); a negative power means shift left (the number gets smaller).

3. Multiplying and Dividing in Standard Form

Now that converting is secure, calculations with standard form follow naturally. The key insight: treat the coefficients and the powers of ten separately, then recombine.

For multiplication, multiply the $A$ values together and add the indices. For division, divide the $A$ values and subtract the indices. This works because of the index law $10^a \times 10^b = 10^{a+b}$.

Worked example: Multiplying Two Numbers in Standard Form

Calculate $(3 \times 10^4) \times (2.5 \times 10^3)$. Give your answer in standard form.

1. $3 \times 2.5 = 7.5$ — Deal with the $A$ values first. $3 \times 2.5 = 7.5$. [M1]
2. $10^4 \times 10^3 = 10^{4+3} = 10^7$ — Use the index law: add the exponents. $4 + 3 = 7$.
3. $7.5 \times 10^7$ — $7.5$ satisfies $1 \leq 7.5 < 10$, so no adjustment is needed. [A1]

Answer: $7.5 \times 10^7$

Worked example: Dividing Two Numbers in Standard Form

Calculate $\frac{9.6 \times 10^8}{3 \times 10^5}$. Give your answer in standard form.

1. $9.6 \div 3 = 3.2$ — Deal with the $A$ values first. $9.6 \div 3 = 3.2$. [M1]
2. $10^8 \div 10^5 = 10^{8-5} = 10^3$ — Use the index law: subtract the exponents. $8 - 5 = 3$.
3. $3.2 \times 10^3$ — $3.2$ satisfies $1 \leq 3.2 < 10$, so no adjustment is needed. [A1]

Answer: $3.2 \times 10^3$

There is one catch worth flagging: after multiplying the coefficients, the result might fall outside $[1, 10)$. For example, $6 \times 8 = 48$, which is not between 1 and 10. You must rewrite $48$ as $4.8 \times 10^1$ and then adjust the overall power accordingly.

Worked example: Multiplying When the Coefficient Needs Adjusting

Calculate $(6 \times 10^5) \times (8 \times 10^4)$. Give your answer in standard form.

1. $6 \times 8 = 48$ — $48$ is not in the range $[1, 10)$, so an adjustment will be needed.
2. $10^5 \times 10^4 = 10^{5+4} = 10^9$ — Index law: add the exponents.
3. $48 \times 10^9$ — This is not yet in standard form because $48 \geq 10$.
4. $48 = 4.8 \times 10^1$ — Express the out-of-range coefficient as its own standard form value. [M1]
5. $4.8 \times 10^1 \times 10^9 = 4.8 \times 10^{10}$ — $1 + 9 = 10$. Now $4.8$ is in range. [A1]

Answer: $4.8 \times 10^{10}$

4. Adding and Subtracting in Standard Form

Adding and subtracting in standard form requires a different approach from multiplying and dividing. You cannot add the powers — you must first make the powers of ten identical, then add or subtract the coefficients.

Think of it like adding fractions: you need a common denominator before you can combine the numerators. Here, the 'common denominator' is the same power of ten.

Worked example: Adding Two Numbers in Standard Form

Calculate $(4.2 \times 10^6) + (3 \times 10^5)$. Give your answer in standard form.

1. Use $10^6$ as the common power. — Pick the larger power to keep the arithmetic clean. Either power works, but $10^6$ avoids decimals in the next step.
2. $3 \times 10^5 = 0.3 \times 10^6$ — $3 \times 10^5 = 3 \times 10^{-1} \times 10^6 = 0.3 \times 10^6$. [M1]
3. $(4.2 + 0.3) \times 10^6 = 4.5 \times 10^6$ — Now both terms share $10^6$, so the coefficients can be added directly.
4. $4.5 \times 10^6$ — $4.5$ is in $[1, 10)$. No adjustment needed. [A1]

Answer: $4.5 \times 10^6$

Worked example: Subtracting Two Numbers in Standard Form

Calculate $(7.5 \times 10^8) - (2.5 \times 10^7)$. Give your answer in standard form.

1. Use $10^8$ as the common power. — Rewrite the second number to match the larger power.
2. $2.5 \times 10^7 = 0.25 \times 10^8$ — $2.5 \times 10^7 = 2.5 \times 10^{-1} \times 10^8 = 0.25 \times 10^8$. [M1]
3. $(7.5 - 0.25) \times 10^8 = 7.25 \times 10^8$ — Coefficients can now be subtracted directly.
4. $7.25 \times 10^8$ — $7.25$ is in $[1, 10)$. No adjustment needed. [A1]

Answer: $7.25 \times 10^8$

5. Deliberate Practice

The methods for all four operations are now in place. The following problems cover the full range — converting, multiplying, dividing, adding, and subtracting. Work through each one before checking the solution.

6. Exam Technique and Common Mistakes

Standard form questions appear on both questions without a calculator and calculator-allowed exam questions. In questions without a calculator, you must show all arithmetic working — the examiner cannot award method marks if the working is absent. On calculator-allowed exam questions, you can use the EXP or $\times 10^x$ key, but always write the answer in the form $A \times 10^n$ rather than reading off a decimal display.

When a question says 'give your answer in standard form', the mark scheme will penalise an answer that is not fully adjusted — for example, $20 \times 10^9$ instead of $2 \times 10^{10}$. Always check that your coefficient $A$ satisfies $1 \leq A < 10$ before writing your final answer.

Every spec requirement for standard form — writing numbers in the form $A \times 10^n$ where $1 \leq A < 10$ and $n$ is an integer, and calculating using all four operations — has now been covered with worked methods and practice. The mark-winning habit is the same throughout: check your coefficient is in range, show your index arithmetic explicitly, and state units when the context requires them.