Degree of Accuracy · Study Notes · Edexcel IGCSE 4MA1 Higher

Rounding, significant figures, and bounds. The technique behind A* answers when measurements have been rounded — used in any 'find the largest/smallest possible' question.

At a glance

d.p. = digits after decimal point.
s.f. = digits from the first non-zero, including trailing zeros after a decimal.
Round half UP for exact $5$ .
Bounds: $\pm$ half the rounding unit.
Multiplication: max × max for upper bound.
Subtraction: max(a) - min(b) for upper bound.
Division: max(a) / min(b) for upper bound.

What you’ll learn

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

1.10 — Round to a given number of decimal places or significant figures.
1.10 — Identify upper and lower bounds for measurements.
1.10 — Use bounds in calculations (multiplication, division, subtraction).

Rounding to d.p. and s.f.

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d.p. counts after the decimal. s.f. counts from the first non-zero digit.

Decimal places (d.p.). Count digits AFTER the decimal point.

$3.14159$ to $2$ d.p.: look at 3rd d.p. ( $1$ ) → round down → $3.14$ .
$3.14159$ to $3$ d.p.: look at 4th d.p. ( $5$ ) → round UP → $3.142$ .

Significant figures (s.f.). Start counting at the FIRST non-zero digit.

$0.0067$ has $2$ s.f. ( $6, 7$ ).
$0.0670$ has $3$ s.f. ( $6, 7, 0$ — trailing zero after decimal IS significant).
$5670$ has $3$ s.f. (trailing zeros in integers are AMBIGUOUS — usually $4$ or $3$ ).
$5670.0$ has $5$ s.f. (decimal forces all trailing zeros to count).

Rules:

Leading zeros (before first non-zero) are NEVER significant.
Zeros between non-zero digits ARE significant. ( $304$ has $3$ s.f.)
Trailing zeros after a decimal ARE significant. ( $30.0$ has $3$ s.f.)
Trailing zeros in integers are ambiguous — treat per context.

Round half up. When the digit being rounded is exactly $5$ , round UP.

$0.65$ to $1$ d.p. → $0.7$ .
$1.85$ to $1$ d.p. → $1.9$ .

Worked qualitative. Why is $0.0067$ said to have $2$ s.f., not $5$ ?

Significant figures convey ACTUAL PRECISION, not how many digits are written.
The leading zeros just mark the place value (telling us this is a small number) — they don't add precision.
Two non-zero digits = two significant figures.

Edexcel tip. Always state both numbers when rounding: 'Look at the next digit ( $X$ ); since $X \ge 5$ / $X < 5$ , round up/down'.

d.p. counts after decimal.
s.f. starts from first non-zero.
Leading zeros: not significant.
Trailing zeros after decimal: significant.
Half UP for exact $5$ .

Upper and lower bounds — single value

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When a value is rounded, the true value lies in a half-interval around it.

The rule: A value rounded to a unit $u$ lies within $\pm \dfrac{u}{2}$ of the stated value.

Stated as	Rounding unit $u$	Half-unit	Bounds
$7.4$ ( $1$ d.p.)	$0.1$	$0.05$	$[7.35, 7.45)$
$7.40$ ( $2$ d.p.)	$0.01$	$0.005$	$[7.395, 7.405)$
$250$ (nearest $10$ )	$10$	$5$	$[245, 255)$
$250$ (nearest $50$ )	$50$	$25$	$[225, 275)$
$5.6$ ( $2$ s.f.)	$0.1$	$0.05$	$[5.55, 5.65)$

Notation: square bracket $[$ for inclusive (lower bound), parenthesis $)$ for exclusive (upper bound). Conventionally:

LOWER bound is INCLUDED.
UPPER bound is EXCLUDED (because $7.45$ would round UP to $7.5$ ).

Worked example. A measurement is given as $52$ kg to the nearest kg. Find bounds.

$u = 1$ kg. Half-unit $= 0.5$ kg.

Lower bound = $51.5$ kg. Upper bound = $52.5$ kg.

The true value sits within half a rounding unit either side: lower bound included, upper bound excluded.

Worked qualitative. Why is the upper bound 'excluded'?

$52.5$ would round UP to $53$ (because of round half up).
So if the original value was exactly $52.5$ , it would be reported as $53$ , not $52$ .
The actual range of values reported as $52$ is $[51.5, 52.5)$ .

Edexcel tip. State the rounding unit and halve it explicitly. 'Rounded to $1$ d.p., so half-unit is $0.05$ .' This earns the M1 mark.

Bounds = stated value $\pm u/2$ .
Lower bound INCLUDED (round-up convention).
Upper bound EXCLUDED.
Always state $u$ explicitly.

Bounds in calculations

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Different operations need different combinations of upper and lower bounds.

Why this matters. When computing $a \times b$ where both are rounded, the answer has its own range. To find the range, combine the bounds correctly.

Addition. Both bounds add naturally:

$UB(a + b) = UB(a) + UB(b)$ .
$LB(a + b) = LB(a) + LB(b)$ .

Subtraction. REVERSE one bound:

$UB(a - b) = UB(a) - LB(b)$ .
$LB(a - b) = LB(a) - UB(b)$ .

(Think: biggest gap when first is largest, second is smallest.)

Multiplication. Both same direction (for positive values):

$UB(a \times b) = UB(a) \times UB(b)$ .
$LB(a \times b) = LB(a) \times LB(b)$ .

Division. Mixed (for positive values):

$UB(a \div b) = UB(a) \div LB(b)$ .
$LB(a \div b) = LB(a) \div UB(b)$ .

Memory aid.

Want big? Make top big, bottom small.
Want small? Make top small, bottom big.

Addition and multiplication keep the same direction; subtraction and division flip the second bound.

Worked example. $a = 8.2$ ( $1$ d.p.), $b = 4.5$ ( $1$ d.p.). Find UB and LB of $a/b$ .

Bounds: $a \in [8.15, 8.25]$ , $b \in [4.45, 4.55]$ .

$UB(a/b) = 8.25 / 4.45 = 1.853\ldots$
$LB(a/b) = 8.15 / 4.55 = 1.791\ldots$

So $a/b \in [1.791, 1.854]$ approximately.

Worked qualitative. Why does division use 'big over small' for the upper bound?

Bigger numerator → bigger answer.
Smaller denominator → bigger answer.
For maximum, want both effects: max numerator, min denominator.

Edexcel tip. Set up bounds explicitly: $a \in [\ldots, \ldots]$ , $b \in [\ldots, \ldots]$ . Then choose the right combination. Mark schemes credit each bound calculation step.

Addition: same direction.
Subtraction: reverse one.
Multiplication: same direction.
Division: max top, min bottom.

Quick recap

Rounding: d.p. or s.f. — different conventions.
Half UP for exact $5$ .
Bounds: $\pm u/2$ .
Lower included, upper excluded.
Addition/multiplication: same direction.
Subtraction: reverse one. Division: top big, bottom small for UB.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

d.p. — decimal places, count after dot.
s.f. — significant figures, from first non-zero.
Bounds: $\pm$ half the rounding unit.
$UB(a/b) = UB(a) / LB(b)$ — top big, bottom small.
$UB(a-b) = UB(a) - LB(b)$ — biggest gap.

How it’s examined

Degree of accuracy appears Paper 1H (rounding, 2-3 marks) AND Paper 2H (bounds in calculations, 4-6 marks). Examiner reports flag (1) confusing leading zeros vs significant trailing zeros, (2) using $u$ (not $u/2$ ) for bounds, and (3) wrong bound combinations in division and subtraction.

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 May/Jun 2024. Last reviewed 2026-05-07.

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

1Round to a given accuracy
Foundation• rounding
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Question
Round $0.06743$ to (a) $1$ d.p., (b) $2$ s.f., (c) $3$ s.f.
Step-by-step solution
1. Step 1
  (a) $1$ d.p.: look at the 2nd decimal place ( $6$ ) — round up. $0.06743 \to 0.1$ .
2. Step 2
  (b) $2$ s.f.: leading zeros don't count. The first two significant figures are $6$ and $7$ . Look at the next ( $4$ ) — round down. $0.06743 \to 0.067$ .
3. Step 3
  (c) $3$ s.f.: include $4$ . Next digit is $3$ — round down. $0.06743 \to 0.0674$ .
Answer
(a) $0.1$ (b) $0.067$ (c) $0.0674$
Examiner tip
Significant figures count from the FIRST non-zero digit. Leading zeros are not significant. Trailing zeros after a decimal ARE significant.
2Find upper and lower bounds
Higher• bounds
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Question
A length is measured as $7.4$ cm to $1$ d.p. Find the upper and lower bounds.
Step-by-step solution
1. Step 1
  Half of the smallest unit ( $0.1$ ) is $0.05$ .
2. Step 2
  Lower bound: $7.4 - 0.05 = 7.35$ cm.
3. Step 3
  Upper bound: $7.4 + 0.05 = 7.45$ cm.
Answer
Lower bound $7.35$ cm; upper bound $7.45$ cm.
Examiner tip
ALWAYS halve the rounding unit. For $1$ d.p. ( $0.1$ ): bounds $\pm 0.05$ . For $2$ s.f.: bounds $\pm$ half of last sig fig place value.
3Bounds with multiplication
Higher• Adapted from 4MA1/2H May/Jun 2024 Q19• bounds, multiplication
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Question
$a = 8.2$ to $1$ d.p., $b = 4.5$ to $1$ d.p. Find the upper bound of $a \times b$ .
Step-by-step solution
1. Step 1
  Bounds: $a \in [8.15, 8.25]$ , $b \in [4.45, 4.55]$ .
2. Step 2
  For multiplication, max product = max × max:
  $(\text{UB of}\ a)\times(\text{UB of}\ b) = 8.25 \times 4.55$
3. Step 3
  Compute: $8.25 \times 4.55 = 37.5375$ .
Answer
Upper bound = $37.5375$
Examiner tip
For multiplication: max × max for upper bound; min × min for lower bound. For division: max ÷ min for upper bound; min ÷ max for lower bound.
4Bounds with subtraction
Higher• bounds, subtraction
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Question
$a = 12.6$ to $1$ d.p., $b = 5.3$ to $1$ d.p. Find the lower bound of $a - b$ .
Step-by-step solution
1. Step 1
  Bounds: $a \in [12.55, 12.65]$ , $b \in [5.25, 5.35]$ .
2. Step 2
  For SUBTRACTION, MIN of $a - b$ uses MIN of $a$ and MAX of $b$ :
  $\min(a - b) = \min(a) - \max(b) = 12.55 - 5.35 = 7.2$
Answer
Lower bound = $7.2$
Examiner tip
Subtraction is the trickiest case. Lower bound: smallest $a$ minus largest $b$ . Upper bound: largest $a$ minus smallest $b$ . Many students get this WRONG by 'min minus min'.
5Truncate vs round
Foundation• truncate
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Question
Truncate $\pi = 3.14159\ldots$ to $3$ decimal places. Compare with rounding to $3$ d.p.
Step-by-step solution
1. Step 1
  TRUNCATING: just chop off after $3$ d.p. → $3.141$ .
2. Step 2
  ROUNDING: look at next digit ( $5$ ) → round up → $3.142$ .
Answer
Truncate: $3.141$ . Round: $3.142$ .
Examiner tip
Edexcel rarely asks for truncation explicitly, but it's used in some money calculations. Always read the question carefully — 'round' and 'truncate' give different answers.
6Rounding a large number to 2 significant figures [1 mark]
Getting started• Paper 1H/2H style — 1 mark• rounding, significant figures
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Question
Write $2\,049\,713$ correct to $2$ significant figures. [1]
Step-by-step solution
1. Step 1
  The first two significant figures are $2$ and $0$ . The zero between non-zero digits counts. The next digit is $4$ , so round down.
2. Step 2
  B1 Fill the remaining places with zeros so the size is unchanged.
  $2\,049\,713 \to 2\,000\,000$
Answer
$2\,000\,000$
Examiner tip
Examiner tip: A zero between non-zero digits is significant, so $2\,0$ are the first two significant figures here.

Common mistake: Skipping the zero and rounding at the $4$ , giving $2\,050\,000$ (that is $3$ s.f.), or writing $2.0$ (the size of the number is lost).
7Error interval for a population to 2 significant figures [2 marks]
Getting started• Paper 1H/2H style — 2 marks• error interval, bounds
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Question
The population, $P$ , of a country is $5\,400\,000$ , correct to $2$ significant figures.

Write down the error interval for $P$ . [2]
Step-by-step solution
1. Step 1
  To $2$ s.f. this number is rounded to the nearest $100\,000$ , so the bounds are $\pm 50\,000$ .
2. Step 2
  B1 Lower bound:
  $5\,400\,000 - 50\,000 = 5\,350\,000$
3. Step 3
  B1 Upper bound, written with the correct inequality signs:
  $5\,350\,000 \leqslant P < 5\,450\,000$
Answer
$5\,350\,000 \leqslant P < 5\,450\,000$
Examiner tip
Examiner tip: Find the place value of the last significant figure first (here the hundred-thousands), then halve it.

Common mistake: Using $\pm 5$ or $\pm 0.5$ because " $2$ s.f." was read as a number of decimal places, giving $5\,399\,995 \leqslant P < 5\,400\,005$ .
8Estimate, then decide if it is an overestimate [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• estimation, rounding
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Question
(a) Work out an estimate for the value of $\dfrac{6.82 \times 18.7}{0.0413}$

Show your working clearly. [2]

(b) Is your answer to part (a) an overestimate or an underestimate? Give a reason for your answer. [1]
Step-by-step solution
1. Step 1
  M1 (a) Round each number to $1$ significant figure and write the rounded values down.
  $\frac{7 \times 20}{0.04}$
2. Step 2
  A1
  $\frac{140}{0.04} = 3500$
3. Step 3
  B1 (b) Both numbers on top were rounded up ( $7 > 6.82$ , $20 > 18.7$ ) and the number underneath was rounded down ( $0.04 < 0.0413$ ). A bigger numerator and a smaller denominator both make the answer bigger, so $3500$ is an overestimate.
Answer
(a) $3500$ (b) Overestimate, because the numerator values were rounded up and the denominator was rounded down.
Examiner tip
Examiner tip: In (b), comment on every rounding. The direction of the denominator's rounding has the opposite effect to the numerator's.

Common mistake: Saying "overestimate because the numbers were rounded up". That ignores $0.0413 \to 0.04$ , which was rounded down, and does not score.
9Lower bound of a hypotenuse [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• bounds, pythagoras
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Question
A right-angled triangle has shorter sides of length $6.2$ cm and $8.5$ cm, both correct to $1$ decimal place.

Calculate the lower bound for the length of the hypotenuse.

Give your answer correct to $2$ decimal places. [4]
Step-by-step solution
1. Step 1
  B1 The lower bounds of the two sides are $6.15$ cm and $8.45$ cm.
2. Step 2
  M1 Use Pythagoras with both lower bounds.
  $h^{2} = 6.15^{2} + 8.45^{2} = 37.8225 + 71.4025$
3. Step 3
  M1
  $h = \sqrt{109.225}$
4. Step 4
  A1
  $h = 10.4510\ldots \approx 10.45\ \text{cm}$
Answer
$10.45$ cm
Examiner tip
Examiner tip: The hypotenuse gets longer as either side gets longer, so the lower bound uses both lower bounds.

Common mistake: Working out $\sqrt{6.2^2 + 8.5^2} = 10.52$ and then subtracting $0.05$ . Bounds must go into the calculation, not be applied to the answer.
10Lower bound of average speed over two stages [5 marks]
Stretch• Paper 1H/2H style — 5 marks• bounds, speed, compound measures
▼
Question
A cyclist rides two stages.

Stage 1: $42$ km in $35$ minutes.

Stage 2: $18$ km in $20$ minutes.

Each distance is correct to the nearest kilometre. Each time is correct to the nearest minute.

Work out the lower bound for the average speed of the cyclist over the whole journey. Give your answer in km/h correct to $3$ significant figures. [5]
Step-by-step solution
1. Step 1
  B1 Lower bounds of the distances: $41.5$ km and $17.5$ km.
2. Step 2
  B1 Upper bounds of the times: $35.5$ min and $20.5$ min.
3. Step 3
  M1 Least total distance and greatest total time.
  $41.5 + 17.5 = 59\ \text{km}, \qquad 35.5 + 20.5 = 56\ \text{min}$
4. Step 4
  M1 Convert the time to hours and divide.
  $59 \div \frac{56}{60}$
5. Step 5
  A1
  $= 63.214\ldots \approx 63.2\ \text{km/h}$
Answer
$63.2$ km/h (awrt)
Examiner tip
Examiner tip: Average speed is total distance over total time. Add the bounds first, then divide once.

Common mistake: Averaging the two stage speeds. The stages take different times, so the mean of the two speeds is not the average speed. Also watch the units: $59 \div 56 = 1.05$ is km per minute, not km/h.

Question 1
Paper 1H/2H style — 1 mark1 mark
Write $0.0697$ correct to $2$ decimal places. [1]
Model answer
The second decimal place is $6$ and the next digit is $9$ , so the $6$ rounds up to $7$ .

Answer: $0.07$
Why this scores
Mark scheme:
- B1 for $0.07$ cao
Why this scores full marks: Decimal places are counted from the decimal point, zeros included, so the answer has exactly two decimal places. Writing $0.070$ , which is the answer to $2$ significant figures and has three decimal places, loses the mark.
Question 2
Paper 1H/2H style — 2 marks2 marks
The mass, $m$ kg, of a sack of rice is $25$ kg, correct to the nearest $0.5$ kg.

Complete the error interval for $m$ .

$\ldots\ldots \leqslant m < \ldots\ldots$ [2]
Model answer
Half of $0.5$ is $0.25$ .

$25 - 0.25 = 24.75, \qquad 25 + 0.25 = 25.25$

Answer: $24.75 \leqslant m < 25.25$
Why this scores
Mark scheme:
- B1 for $24.75$
- B1 for $25.25$
Why this scores full marks: The bounds are half of the rounding unit, $0.5$ , either side of $25$ . The most common wrong answer, $24.5 \leqslant m < 25.5$ , treats the mass as if it were to the nearest kilogram.
Question 3
Paper 1H/2H style — 2 marks2 marks
The total mass of $20$ identical coins is $150$ g, correct to the nearest $10$ g.

Work out the lower bound for the mass of one coin. [2]
Model answer
Lower bound of total mass $= 150 - 5 = 145$ g

$145 \div 20 = 7.25$

Answer: $7.25$ g
Why this scores
Mark scheme:
- M1 for $145 \div 20$
- A1 for $7.25$ cao
Why this scores full marks: The bound belongs to the measured total, not to each coin. The $20$ is an exact count, so it has no bounds. Weaker scripts work out $150 \div 20 = 7.5$ and then subtract $0.5$ or $0.05$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
Work out an estimate for the value of

$\frac{5.93^{2} + 3.12}{\sqrt{0.0386}}$

Show your working clearly. [3]
Model answer
Round each number to $1$ significant figure:

$\frac{6^{2} + 3}{\sqrt{0.04}}$

$= \frac{36 + 3}{0.2} = \frac{39}{0.2}$

$= 195$

Answer: $195$
Why this scores
Mark scheme:
- M1 for at least two of $6$ , $3$ , $0.04$ seen
- M1 for $\dfrac{39}{0.2}$ oe
- A1 for $195$
Why this scores full marks: Rounding $0.0386$ to $0.04$ makes the square root exact, $0.2$ . The rounded values are all written down before any arithmetic. A calculator value such as $194.9$ with no rounded values shown scores $0$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
The perimeter of a square is $36$ cm, correct to the nearest centimetre.

Work out the lower bound for the area of the square.

Give your answer correct to $3$ significant figures. [3]
Model answer
Lower bound of perimeter $= 35.5$ cm

Lower bound of side $= 35.5 \div 4 = 8.875$ cm

$8.875^{2} = 78.765\,625$

Answer: $78.8$ cm²
Why this scores
Mark scheme:
- B1 for $35.5$ seen
- M1 for $(35.5 \div 4)^{2}$ oe
- A1 for awrt $78.8$
Why this scores full marks: The bound is applied to the perimeter, which is the measured quantity. The side comes from it. Weaker scripts take the side as $9$ cm and use $8.5^2 = 72.25$ , putting a bound on a value that was never measured.
Question 6
Paper 1H/2H style — 3 marks3 marks
Each of $12$ tins has a mass of $450$ g, correct to $2$ significant figures.

Work out the lower bound for the total mass of the $12$ tins. Give your answer in kilograms. [3]
Model answer
$450$ to $2$ s.f. is to the nearest $10$ g, so the lower bound is $445$ g.

$12 \times 445 = 5340\ \text{g}$

Answer: $5.34$ kg
Why this scores
Mark scheme:
- B1 for $445$ seen
- M1 for $12 \times 445$
- A1 for $5.34$ cao
Why this scores full marks: The bound is applied to each tin and then multiplied. Applying it to the total ( $12 \times 450 = 5400$ , then $5395$ ) is the most common error. The answer is also given in kilograms, as asked.
Question 7
Paper 1H/2H style — 4 marks4 marks
A rectangular garden measures $14$ m by $9$ m, each correct to the nearest metre.

The garden contains a square pond of side $2.4$ m, correct to $1$ decimal place. The rest of the garden is lawn.

Work out the upper bound for the area of the lawn. [4]
Model answer
Upper bounds of garden: $14.5$ m and $9.5$ m. Lower bound of pond side: $2.35$ m.

$14.5 \times 9.5 = 137.75$

$2.35^{2} = 5.5225$

$137.75 - 5.5225 = 132.2275$

Answer: $132.2275$ m²
Why this scores
Mark scheme:
- B1 for $14.5$ and $9.5$ seen
- B1 for $2.35$ seen
- M1 for $14.5 \times 9.5 - 2.35^{2}$
- A1 for awrt $132$
Why this scores full marks: The lawn is largest when the garden is largest and the pond is smallest, so the candidate subtracts the lower bound of the pond. Weaker scripts use $2.45$ , the upper bound for every length, giving $131.7475$ .
Question 8
Paper 1H/2H style — 4 marks4 marks
A ramp rises a vertical height of $0.45$ m over a horizontal distance of $3.2$ m. Both measurements are correct to $2$ significant figures.

Calculate the upper bound for the angle the ramp makes with the horizontal.

Give your answer correct to $1$ decimal place. [4]
Model answer
Upper bound of height $= 0.455$ m. Lower bound of horizontal distance $= 3.15$ m.

$\tan \theta = \frac{0.455}{3.15} = 0.1444\ldots$

$\theta = \tan^{-1}(0.1444\ldots) = 8.219\ldots^{\circ}$

Answer: $8.2^{\circ}$
Why this scores
Mark scheme:
- B1 for $0.455$ seen
- B1 for $3.15$ seen
- M1 for $\tan^{-1}\left(\dfrac{0.455}{3.15}\right)$
- A1 for awrt $8.2$
Why this scores full marks: A steeper ramp has a larger $\tan \theta$ , so the candidate maximises the quotient: the upper bound of the height over the lower bound of the distance. The $2$ s.f. bounds are different for each: $\pm 0.005$ for $0.45$ and $\pm 0.05$ for $3.2$ .
Question 9
Paper 1H/2H style — 4 marks4 marks
$x = 5.8$ correct to $1$ decimal place.

(a) Work out the upper bound for the value of $\dfrac{x^{2}}{x + 2}$

Give your answer correct to $3$ significant figures. [3]

(b) Kiran works out $\dfrac{5.85^{2}}{7.75}$ for part (a).

Explain why Kiran is wrong. [1]
Model answer
(a) Upper bound of $x = 5.85$

$\frac{5.85^{2}}{5.85 + 2} = \frac{34.2225}{7.85} = 4.3595\ldots$

Answer: $4.36$

(b) $x$ cannot have two different values at the same time. Kiran uses $x = 5.85$ on the top and $x = 5.75$ on the bottom, but the same value of $x$ must be used in both places.
Why this scores
Mark scheme:
- B1 for $5.85$ seen
- M1 for $\dfrac{5.85^{2}}{5.85 + 2}$
- A1 for awrt $4.36$
- B1 for an explanation that the same value of $x$ must be used in the numerator and the denominator
Why this scores full marks: Bounds are mixed only when the quantities are different. Here there is one measurement, so the candidate substitutes one value. For $x > 0$ the expression increases as $x$ increases, so $x = 5.85$ gives the upper bound. Kiran's $4.4158\ldots$ is larger, but no value of $x$ produces it.
Question 10
Paper 1H/2H style — 5 marks5 marks
The braking distance, $s$ metres, of a car is given by

$s = \frac{v^{2}}{2a}$

$v = 17.3$ correct to $3$ significant figures.

$a = 4.25$ correct to $3$ significant figures.

By considering bounds, work out the value of $s$ to a suitable degree of accuracy.

You must show all your working and give a reason for your answer. [5]
Model answer
$17.25 \leqslant v < 17.35$ , $\quad 4.245 \leqslant a < 4.255$

$\text{UB of } s = \frac{17.35^{2}}{2 \times 4.245} = \frac{301.0225}{8.49} = 35.456\ldots$

$\text{LB of } s = \frac{17.25^{2}}{2 \times 4.255} = \frac{297.5625}{8.51} = 34.966\ldots$

To $3$ s.f.: $35.5$ and $35.0$ , which differ.

To $2$ s.f.: $35$ and $35$ , which agree.

Answer: $s = 35$ m, because both bounds round to $35$ to $2$ significant figures.
Why this scores
Mark scheme:
- B1 for any correct bound of $v$ or $a$
- M1 for $\dfrac{17.35^{2}}{2 \times 4.245}$
- M1 for $\dfrac{17.25^{2}}{2 \times 4.255}$
- A1 for $35.45\ldots$ and $34.96\ldots$
- B1 for $35$ with a correct reason, dep on both M marks
Why this scores full marks: Each bound pairs the right bound of $v$ with the opposite bound of $a$ , because $a$ is in the denominator. The candidate tests $3$ s.f. and then $2$ s.f., and states the reason. Weaker scripts give $35.2$ from the stated values, or $35.0$ , which claims more accuracy than the bounds support.

4MA1 Degree of Accuracy

At a glance

What you’ll learn

Rounding to d.p. and s.f.

Upper and lower bounds — single value

Bounds in calculations

Quick recap

Memorise this

How it’s examined

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

1Round to a given accuracy

2Find upper and lower bounds

3Bounds with multiplication

4Bounds with subtraction

5Truncate vs round

6Rounding a large number to 2 significant figures [1 mark]

7Error interval for a population to 2 significant figures [2 marks]

8Estimate, then decide if it is an overestimate [3 marks]

9Lower bound of a hypotenuse [4 marks]

10Lower bound of average speed over two stages [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