Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Number
Practise Standard form. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Writing very large and very small numbers in standard form \(A \times 10^n\), converting to and from ordinary numbers, ordering them and calculating with them, with and without a calculator. Expect 1 to 3 mark questions on both papers, often set in science or computing contexts.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Write large numbers in standard formPlace the point after the first digit and count the moves: 45,000 = \(4.5 \times 10^4\).
4
Write small numbers in standard formA number between 0 and 1 has a negative power: 0.00072 = \(7.2 \times 10^{-4}\).
4
Change standard form to ordinary numbers\(3.06 \times 10^5\) = 306,000 and \(3.06 \times 10^{-3}\) = 0.00306.
4
Order numbers in standard formCompare the powers of 10 first, then the values of \(A\).
4
Use a calculator with standard formUse the \(\times 10^x\) key, and write the answer in proper standard form, not calculator notation.
5
Multiply and divide in standard formDeal with the numbers and the powers of 10 separately, then adjust: \(12 \times 10^9 = 1.2 \times 10^{10}\).
5
Add and subtract in standard formChange to ordinary numbers first: \(3.2 \times 10^4 + 5 \times 10^3\) = 32,000 + 5000 = \(3.7 \times 10^4\).
Notes
What standard form is
A number in standard form is written \(A \times 10^n\), where \(1 \le A \lt 10\) and \(n\) is an integer.
Large numbers have a positive power: 6,300,000 = \(6.3 \times 10^6\).
Numbers between 0 and 1 have a negative power: 0.00058 = \(5.8 \times 10^{-4}\).
\(n\) tells you how many places the digits move. To write the ordinary number, move the decimal point right for a positive \(n\), left for a negative \(n\).
\(45 \times 10^3\) is not in standard form, because 45 is not between 1 and 10. It is \(4.5 \times 10^4\).
Ordering
Compare the powers of 10 first: the bigger power gives the bigger number. \(2 \times 10^{-3}\) is bigger than \(9 \times 10^{-5}\), because −3 is greater than −5.
If the powers are the same, compare the values of \(A\).
Multiplying and dividing
Multiply or divide the \(A\) parts, and use the laws of indices on the powers of 10.
Fixing \(A\): if it is 10 or more, divide it by 10 and add 1 to the power. If it is less than 1, multiply it by 10 and subtract 1 from the power.
Adding, subtracting and calculators
To add or subtract, change both to ordinary numbers (or to the same power of 10): \(6.4 \times 10^5 - 8 \times 10^4\) = 640,000 − 80,000 = 560,000 = \(5.6 \times 10^5\).
On a calculator, use the \(\times 10^x\) (or EXP) key: \(3.2 \times 10^{-5}\) is typed 3.2, \(\times 10^x\), (−), 5.
Some calculators show answers like 1.8E−02. Always write this as \(1.8 \times 10^{-2}\).
Cheatsheet
Standard form: \(A \times 10^n\), with \(1 \le A \lt 10\) and \(n\) an integer
Large number → positive \(n\); number between 0 and 1 → negative \(n\)
The mass of the Earth is \(5.97 \times 10^{24}\) kg. The mass of the Moon is \(7.35 \times 10^{22}\) kg.
(a) Work out how many times heavier the Earth is than the Moon. Give your answer correct to 3 significant figures.[2]
(b) Hana estimates the answer to part (a). She writes \(6 \div 7 \approx 0.86\) and \(10^{24} \div 10^{22} = 10^{\frac{24}{22}} \approx 10\), so the Earth is about 8.6 times heavier. Explain the mistake in Hana's working.[1]
Show the answer and mark scheme
(a)Answer: 81.2
M1 for \((5.97 \times 10^{24}) \div (7.35 \times 10^{22})\)
A1 for 81.2
Worked solution: \(\frac{5.97 \times 10^{24}}{7.35 \times 10^{22}} = 0.81224\ldots \times 10^2 = 81.224\ldots\), so 81.2 times.
(b)Answer: To divide powers of 10 you subtract the indices: \(10^{24} \div 10^{22} = 10^{2} = 100\), not \(10^{\frac{24}{22}}\).
C1 for explaining that the indices should be subtracted, \(10^{24} \div 10^{22} = 10^2\) (= 100)
Worked solution: \(10^{24} \div 10^{22} = 10^{24 - 22} = 10^2 = 100\), so the estimate should be about \(0.86 \times 100 = 86\).