Edexcel International GCSE Mathematics A 4MA1: Degree of accuracy and standard form – Revision Notes

For full explanations and worked examples, use the Degree of accuracy and standard form study guide. These notes condense sections 1.8 and 1.9 of Topic 1 in the Pearson Edexcel International GCSE Mathematics A (4MA1) specification, Issue 2 (November 2017), for the January and June series examined on it. Rounding, identifying bounds, estimation and calculating with standard form are on both tiers. Solving problems with bounds and solving problems involving standard form are Higher tier only and are labelled. A calculator may be used on every 4MA1 paper.

Course hub: Edexcel IGCSE Mathematics. Tick off each statement on the printable checklist, then try the practice questions or a free diagnostic. The wider topic is covered in the Numbers and the number system revision notes.

Key words

Term	Meaning
Degree of accuracy	How a value was rounded: to a power of 10, a number of d.p., or a number of s.f.
Decimal places (d.p.)	Digits after the decimal point
Significant figures (s.f.)	Digits counted from the first non-zero digit
Lower bound	Smallest value that rounds to the given value
Upper bound	The value at the top of the range; anything just below it rounds to the given value
Estimate	Answer worked out after rounding every number to 1 s.f.
Standard form	a × 10ⁿ with 1 ≤ a < 10 and n an integer

1.8A and 1.8B Rounding – both tiers

Method in steps

1. Find the last digit you are keeping (the place value, the d.p., or the s.f.).
2. Look at the next digit only.
3. 5 or more: round the kept digit up. 4 or less: leave it.
4. Replace dropped digits before the decimal point with zeros; drop digits after it.

Which zeros count as significant figures?

Zeros	Significant?	Example
Leading zeros in a decimal	No	0.0004983: first s.f. is 4
Zeros between other digits	Yes	305 has 3 s.f.
Trailing zeros you write after rounding	Yes, they show accuracy	12.00 to 2 d.p.

Small reminders

• 3 s.f. of 0.080 96 is 0.0810. Keep the final 0: it is the third significant figure.
• 2 s.f. of 61 540 is 62 000, not 62.
• 1 d.p. of 2.96 is 3.0, not 3.

1.8C Identifying bounds – both tiers

Method in steps

1. Name the unit of rounding (nearest 5? 0.1? 100?).
2. Halve it.
3. Lower bound = value − half unit. Upper bound = value + half unit.

Given	Unit	Bounds
40 cm to nearest 5 cm	5	37.5 cm and 42.5 cm
0.3 kg to 1 d.p.	0.1	0.25 kg and 0.35 kg
120 000 to 2 s.f.	10 000	115 000 and 125 000
7.06 to 3 s.f.	0.01	7.055 and 7.065

As an inequality: 37.5 ≤ x < 42.5. “Nearest 5” catches people out: the half unit is 2.5.

1.8D Estimation – both tiers

Method in steps

1. Round every number to 1 s.f. Write the rounded calculation down.
2. Work it out without a calculator.
3. If the question asks, say whether it is an overestimate or underestimate by checking which way each number was rounded.

Reminder: dividing by 0.2 is multiplying by 5; dividing by 0.05 is multiplying by 20.

(0.587 x 3.94) / 0.213  ->  (0.6 x 4) / 0.2 = 2.4 / 0.2 = 12

1.8A Solving problems with bounds – Higher tier only

Combining bounds

Want	Use
Biggest sum or product	both upper bounds
Smallest sum or product	both lower bounds
Biggest difference a − b	upper a, lower b
Smallest difference a − b	lower a, upper b
Biggest quotient a ÷ b	upper a, lower b
Smallest quotient a ÷ b	lower a, upper b

Worked reminder (difference). P = 17.6 to 1 d.p. and Q = 9.25 to 2 d.p.

P: 17.55 to 17.65        Q: 9.245 to 9.255
upper bound of P - Q = 17.65 - 9.245 = 8.405
lower bound of P - Q = 17.55 - 9.255 = 8.295

Suitable degree of accuracy. Work out both bounds. Round each to 1 s.f., then 2 s.f., then 3 s.f. The answer is the most accurate value on which both bounds agree. Above, to 2 s.f. the bounds give 8.4 and 8.3, so they disagree. To 1 s.f. both give 8. So P − Q = 8 to 1 s.f.

“Must be sure” decisions. If a quantity must definitely be enough (fence, fabric, paint), use the upper bound of what is needed. If something must definitely fit, compare the upper bound of the object with the lower bound of the space.

1.9A Standard form – both tiers

Converting

Ordinary number	Standard form	Note
92 000	9.2 × 10⁴	big number, positive power
0.000 605	6.05 × 10⁻⁴	small number, negative power
7	7 × 10⁰	10⁰ = 1

Calculating

Operation	Rule	Then
×	multiply the a parts, add the powers	adjust so 1 ≤ a < 10
÷	divide the a parts, subtract the powers	adjust so 1 ≤ a < 10
+ or −	convert to ordinary numbers or match the powers	convert back
squaring	square a, double the power	adjust

Adjusting. If a ≥ 10, divide a by 10 and add 1 to the power. If a < 1, multiply a by 10 and subtract 1 from the power.

25 x 10^-6  = 2.5 x 10^-5
0.8 x 10^9  = 8 x 10^8

Calculator display. 4.7E-03 or 4.7 × 10⁻⁰³ means 4.7 × 10⁻³. Write it in proper notation in your answer.

1.9A Solving problems involving standard form – Higher tier only

• “How many times bigger?” Divide the larger by the smaller: (6 × 10⁸) ÷ (1.5 × 10⁵) = 4 × 10³.
• Square roots. Make the power even first: √(3.6 × 10⁷) = √(36 × 10⁶) = 6 × 10³.
• Expressions in n. Treat 10ⁿ with index laws, then adjust a: (2 × 10ⁿ)³ = 8 × 10³ⁿ; (6 × 10ⁿ)² = 36 × 10²ⁿ = 3.6 × 10²ⁿ⁺¹.
• Units. 1 km = 10³ m, 1 m = 10³ mm. Convert first, then calculate.

Must-know distinctions

• d.p. vs s.f. 0.0467 to 2 d.p. is 0.05; to 2 s.f. it is 0.047.
• Identify bounds (both tiers) vs solve problems with bounds (Higher). Foundation stops at writing the two bounds.
• Estimate vs exact. An estimate uses 1 s.f. values; a calculator value earns no estimate marks.
• 13 × 10⁴ vs 1.3 × 10⁵. Equal value, but only the second is standard form.

Quick self-test

1. Round 7649 to the nearest 100.
2. Write 0.000 498 3 to 2 significant figures.
3. Write 12.0451 to 2 decimal places.
4. A length is 9.0 m to 1 d.p. Write down its bounds.
5. A crowd is recorded as 600 to the nearest 50. Write down the bounds.
6. Estimate (48.7 × 21.3) ÷ 9.8.
7. Write 0.002 81 in standard form.
8. Work out (5 × 10⁻³)², giving your answer in standard form.
9. Work out 2.4 × 10⁵ + 6 × 10⁴ in standard form.
10. (Higher) a = 12 and b = 5, both to the nearest whole number. Find the upper bound of a ÷ b to 3 s.f.
11. (Higher) Work out √(4.9 × 10⁷) as an ordinary number.
12. (Higher) a = 30 to the nearest 10 and b = 12 to the nearest whole number. Find the lower bound of a − b.

Answers

1. 7600
2. 0.000 50 (the trailing 0 is the second s.f.)
3. 12.05
4. 8.95 m and 9.05 m
5. 575 and 625
6. (50 × 20) ÷ 10 = 100 (calculator: 105.8)
7. 2.81 × 10⁻³
8. 25 × 10⁻⁶ = 2.5 × 10⁻⁵
9. 240 000 + 60 000 = 300 000 = 3 × 10⁵
10. 12.5 ÷ 4.5 = 2.78
11. √(49 × 10⁶) = 7 × 10³ = 7000
12. 25 − 12.5 = 12.5

Where marks are usually lost

• Writing 0.0005 for 0.000 498 3 to 2 s.f. The answer needs the trailing zero: 0.000 50.
• Rounding 61 540 to 2 s.f. as 62 and losing the place value.
• Using 0.1 rather than 0.05 as the half unit for 1 d.p. values.
• Taking the half unit of “nearest 5” as 0.5 instead of 2.5.
• Estimating with 2 s.f. values, or not writing the rounded values down.
• (Higher) Using upper − upper for the biggest difference.
• (Higher) Giving a “suitable degree of accuracy” without showing both bounds rounded.
• Leaving answers like 25 × 10⁻⁶ or 0.8 × 10⁹, which are not in standard form.
• Adding 2.4 and 6 when the powers differ (2.4 × 10⁵ + 6 × 10⁴ is not 8.4 × 10⁵).
• Copying calculator notation such as 4.7E-03 as the final answer.

Official syllabus

Pearson Edexcel International GCSE in Mathematics (Specification A) (4MA1), Specification, Issue 2, November 2017, first examination June 2018. Published by Pearson Education Limited.