Mathematics · AQA 8300 §N8

Error Bounds

Spec: AQA 8300 §N8 ⭐⭐⭐⭐ ⏱ 45 mins AQA · Edexcel · OCR Grade 9

Find upper and lower bounds for rounded and truncated values
Calculate bounds of sums, differences, products and quotients
Find maximum and minimum values of expressions with bounds
Distinguish between truncation and rounding bounds
Justify given levels of accuracy using error bounds

🔑 Core Concepts

1. What Are Error Bounds and Why Do They Matter?

Every real-world measurement carries inherent imprecision. When a length is stated as 7.4 cm, we do not know whether the true value is exactly 7.4 cm — we only know it has been rounded to 1 decimal place. Error bounds allow us to state precisely the range of possible true values. This is essential in engineering, science, and any context where accumulated errors could cause failure.

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DEFINITION — Lower Bound (LB)

The smallest possible value a measurement could take given the stated accuracy. Any value below this would have rounded to a different stated value.

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DEFINITION — Upper Bound (UB)

The largest possible value a measurement could take given the stated accuracy. The upper bound is not included (the true interval is LB ≤ x < UB for rounding).

⚠️

IMPORTANT — Upper Bound Is Excluded

For a value rounded to the nearest unit, the upper bound is the value at which we would round up to the next unit. So the true value satisfies $\text{LB} \leq x < \text{UB}$. The upper bound itself would round to the next value, not the stated one.

2. Bounds from Rounding

When a value $x$ is given rounded to a certain degree of accuracy, the bounds are determined by half the rounding unit on each side.

Bounds from Rounding — General Rule

$$\text{Lower Bound} = x - \tfrac{1}{2} \times (\text{rounding unit})$$ $$\text{Upper Bound} = x + \tfrac{1}{2} \times (\text{rounding unit})$$

$x$ = stated (rounded) value Rounding unit = $10^{-n}$ for $n$ decimal places, or $10^k$ for rounding to powers of 10

Bounds from Rounding to $n$ Decimal Places

$$\text{LB} = x - 0.5 \times 10^{-n} \qquad \text{UB} = x + 0.5 \times 10^{-n}$$

$n=0$: nearest integer → ±0.5 $n=1$: 1 d.p. → ±0.05 $n=2$: 2 d.p. → ±0.005 $n=3$: 3 d.p. → ±0.0005

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EXAM TIP — Significant Figures

Rounding to 3 significant figures does not always mean 3 decimal places. Find the rounding unit by identifying the value of the last significant digit. E.g. 6400 to 3 s.f. → last s.f. is in the tens column → rounding unit = 10 → bounds: 6395 and 6405.

Examples of rounding bounds:

Stated Value	Accuracy	Rounding Unit	Lower Bound	Upper Bound
7.4 cm	1 d.p.	0.1	7.35 cm	7.45 cm
3.62 kg	2 d.p.	0.01	3.615 kg	3.625 kg
850 m	nearest 10	10	845 m	855 m
6400	nearest 100	100	6350	6450
2.5 s	1 d.p.	0.1	2.45 s	2.55 s
0.038	2 s.f.	0.001	0.0375	0.0385

3. Bounds from Truncation

Truncation is a different rounding method: digits after a certain point are simply discarded, regardless of their value. This means the stated value is always less than or equal to the true value.

Bounds from Truncation

$$\text{Lower Bound} = x \qquad \text{Upper Bound} = x + \text{truncation unit}$$

The lower bound equals the truncated value itself The true value satisfies $x \leq \text{true value} < x + \text{unit}$

✗

COMMON MISTAKE — Confusing Truncation with Rounding

For truncation, the LB = the stated value (not value − half unit). Many students subtract half the unit even when the question says "truncated". Always check which method is used.

Example: A calculator display shows 4.73 (truncated to 2 d.p.). The true value satisfies $4.73 \leq x < 4.74$. So LB = 4.73, UB = 4.74.

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EXAM TIP — Truncation Keywords

Look for words such as "truncated", "floor function", "rounded down", or "the display shows". If a digital meter truncates, the stated value is always a lower bound for the true reading.

4. Bounds of Expressions — The Four Operations

When two measured quantities are combined in a calculation, you must carefully select which combination of bounds gives the maximum or minimum result. The key principle is: choose the combination of bounds that pulls the result in the direction you want.

Addition Bounds

$$\max(A+B) = \text{UB}(A) + \text{UB}(B)$$ $$\min(A+B) = \text{LB}(A) + \text{LB}(B)$$

To maximise a sum: use both upper bounds To minimise a sum: use both lower bounds

Subtraction Bounds

$$\max(A-B) = \text{UB}(A) - \text{LB}(B)$$ $$\min(A-B) = \text{LB}(A) - \text{UB}(B)$$

To maximise: big minus small → UB(A) − LB(B) To minimise: small minus big → LB(A) − UB(B)

Multiplication Bounds

$$\max(A \times B) = \text{UB}(A) \times \text{UB}(B)$$ $$\min(A \times B) = \text{LB}(A) \times \text{LB}(B)$$

Valid when both A and B are positive If values could be negative, check all four combinations

Division Bounds

$$\max(A \div B) = \text{UB}(A) \div \text{LB}(B)$$ $$\min(A \div B) = \text{LB}(A) \div \text{UB}(B)$$

To maximise: largest numerator ÷ smallest denominator To minimise: smallest numerator ÷ largest denominator Valid when both values are positive

🧠

MEMORY TRICK — "UMUM / LMLM"

For multiplication and division: Upper×Upper for Max, Lower×Lower for Min (UMUM, LMLM). For subtraction: Upper−Lower for Max, Lower−Upper for Min. Think: "pull apart" to get the extreme.

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EXAM TIP — Identify the Goal First

Before writing down any calculation, decide: am I finding the MAXIMUM or MINIMUM of this expression? Write it down explicitly. Then choose the bounds accordingly. Many marks are lost by mixing up which combination is needed.

5. Multi-Step Bound Calculations

Grade 9 questions often combine multiple operations. For example, finding the maximum value of $\frac{A + B}{C}$ where all three are measured quantities. The correct approach is to identify, step by step, which combination of the original bounds gives the extreme result — never use the bounds of intermediate results as new independent quantities.

⚠️

IMPORTANT — Do Not Chain Bounds of Bounds

When computing $\max\!\left(\frac{A+B}{C}\right)$, do NOT first compute UB(A+B) and then treat that as an independent upper bound. Instead, trace through: to maximise, you want the numerator as large as possible and the denominator as small as possible. So use UB(A)+UB(B) for the numerator and LB(C) for the denominator in one combined calculation.

6. Degree of Accuracy Questions

Some questions give you a calculated value and ask you to justify or find the degree of accuracy to which each measurement was given. You work backwards: if the answer bounds are known, what bounds on each input would produce those answer bounds?

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EXAM TIP — Proving a Bound is Correct

To show that a calculation result is correct to $n$ significant figures or decimal places, find the bounds of the result and verify they both round to the stated value. For example, to show $v = 14.2$ m/s (3 s.f.), show UB(v) < 14.25 and LB(v) ≥ 14.15.

🗺️ Visual Notes

Error Bounds

Rounding Bounds

LB = x − ½ × unit
UB = x + ½ × unit
Unit = $10^{-n}$ for n d.p.
True value: LB ≤ x < UB

Truncation Bounds

LB = x (the truncated value)
UB = x + truncation unit
Always rounds down
True value: LB ≤ x < UB

Addition & Subtraction

max(A+B) = UB(A)+UB(B)
min(A+B) = LB(A)+LB(B)
max(A−B) = UB(A)−LB(B)
min(A−B) = LB(A)−UB(B)

Multiplication & Division

max(A×B) = UB(A)×UB(B)
min(A×B) = LB(A)×LB(B)
max(A÷B) = UB(A)÷LB(B)
min(A÷B) = LB(A)÷UB(B)

Contexts

Geometry: area, perimeter, volume
Physics: speed, density, pressure
Finance: rates, totals
Degree of accuracy proofs

Grade 9 Skills

Multi-step compound calculations
Reverse bound problems
Proving accuracy statements
Truncation vs rounding choice

Rounding vs Truncation Comparison

Feature	Rounding	Truncation
Lower Bound	$x - \frac{1}{2} \times \text{unit}$	$x$ (the stated value)
Upper Bound	$x + \frac{1}{2} \times \text{unit}$	$x + \text{unit}$
Interval width	Equal on both sides	All on upper side
Is LB = stated value?	No (LB < stated)	Yes
Example: 3.7 (1 d.p.)	LB=3.65, UB=3.75	LB=3.7, UB=3.8
Keyword clue	"rounded to", "to the nearest"	"truncated", "floor", "display shows"

Decision Process for Bound Calculations

Identify all rounded/truncated quantities and their bounds

→

Determine which extreme (MAX or MIN) the question asks for

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For each operation, select the appropriate bound combination (see formulae)

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Carry out the calculation with chosen bounds in one step

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State the answer clearly with correct units and appropriate precision

Visual: The Number Line and Bounds

For a value $x = 7.4$ rounded to 1 d.p.:

  ←————————|————————————————————|————————→
         7.35                  7.45
         (LB)    7.4 here      (UB)
                  ↑
           any value in this interval
           rounds to 7.4

For a value $x = 7.4$ truncated to 1 d.p.:

  ←————————|————————————————————————|————→
         7.4                       7.5
         (LB=stated value)         (UB)
           ↑ any value here truncates to 7.4

✏️ Worked Examples

Grade 4–5 · Simple

A piece of rope is measured as 12.6 m, correct to 1 decimal place. Find the upper and lower bounds of the length of the rope.

Identify the rounding unit

Rounded to 1 decimal place → rounding unit = $0.1$ m

Calculate half the rounding unit

$\frac{1}{2} \times 0.1 = 0.05$ m

Find the lower bound

$\text{LB} = 12.6 - 0.05 = 12.55$ m

Find the upper bound

$\text{UB} = 12.6 + 0.05 = 12.65$ m

Lower Bound = 12.55 m Upper Bound = 12.65 m
The true length $\ell$ satisfies $12.55 \leq \ell < 12.65$ m.

Grade 6–7 · Medium

A rectangle has length $l = 8.3$ cm and width $w = 4.7$ cm, each measured correct to 1 decimal place. Calculate the maximum possible area of the rectangle.

Find bounds for the length

Rounding unit = 0.1 cm → half unit = 0.05 cm
$\text{LB}(l) = 8.25$ cm, $\quad\text{UB}(l) = 8.35$ cm

Find bounds for the width

$\text{LB}(w) = 4.65$ cm, $\quad\text{UB}(w) = 4.75$ cm

Identify which combination gives maximum area

Area $= l \times w$. To maximise a product of positives, use both upper bounds: $$\max(A) = \text{UB}(l) \times \text{UB}(w)$$

Calculate the maximum area

$$\max(A) = 8.35 \times 4.75 = 39.6625 \text{ cm}^2$$

Maximum possible area = 39.6625 cm²

Grade 9 · Multi-step with physics context

The speed $v$ of a particle is calculated using $v = \dfrac{d}{t}$, where $d = 84$ m (to the nearest metre) and $t = 5.6$ s (to 1 d.p.). By finding the maximum possible value of $v$, show that $v$ is correct to 2 significant figures.

Find bounds for distance $d$

Nearest metre → rounding unit = 1 m, half unit = 0.5 m
$\text{LB}(d) = 83.5$ m, $\quad\text{UB}(d) = 84.5$ m

Find bounds for time $t$

1 d.p. → rounding unit = 0.1 s, half unit = 0.05 s
$\text{LB}(t) = 5.55$ s, $\quad\text{UB}(t) = 5.65$ s

Find the maximum value of $v = d/t$

To maximise a quotient: use UB(d) ÷ LB(t)
$$\max(v) = \frac{84.5}{5.55} = 15.225\ldots \approx 15.225 \text{ m/s}$$

Find the minimum value of $v$

To minimise: LB(d) ÷ UB(t)
$$\min(v) = \frac{83.5}{5.65} = 14.778\ldots \approx 14.779 \text{ m/s}$$

Check whether both bounds round to the same value to 2 s.f.

The nominal value is $v = 84 \div 5.6 = 15.0$ m/s (3 s.f.) $\rightarrow$ 15 m/s to 2 s.f.
For correctness to 2 s.f., we need both $\min(v)$ and $\max(v)$ to lie in $[14.5, 15.5)$.
$\min(v) \approx 14.78$ → rounds to 15 to 2 s.f. ✓
$\max(v) \approx 15.23$ → rounds to 15 to 2 s.f. ✓
Both bounds round to 15, confirming correctness to 2 s.f.

$\min(v) \approx 14.78$ m/s, $\max(v) \approx 15.23$ m/s.
Since both bounds round to 15 m/s (2 s.f.), the speed is correct to 2 significant figures. ✓

❓ Exam Questions

Q11 mark

A number $n$ is given as 7.8, correct to 1 decimal place. Write down the lower bound of $n$.

Lower Bound = 7.75
Rounding unit = 0.1; half unit = 0.05
LB = 7.8 − 0.05 = 7.75 [1 mark]

Q22 marks

A value is truncated to 2 decimal places to give 5.63. Write down the upper and lower bounds of the true value.

Truncation bounds:
LB = 5.63 (the truncated value itself) [1 mark]
UB = 5.63 + 0.01 = 5.64 [1 mark]
True value satisfies $5.63 \leq x < 5.64$

Q33 marks

$p = 6.4$ and $q = 2.1$, both measured correct to 1 decimal place. Find the minimum value of $p - q$.

LB(p) = 6.35, UB(p) = 6.45 [1 mark]
LB(q) = 2.05, UB(q) = 2.15 [1 mark]
min(p − q) = LB(p) − UB(q) = 6.35 − 2.15 = 4.20 [1 mark]
Note: To minimise a difference, use the smallest value of p minus the largest value of q.

Q44 marks

The area of a triangle is calculated using $A = \tfrac{1}{2}bh$. The base $b = 14$ cm (nearest cm) and height $h = 9.2$ cm (1 d.p.). Find the upper bound of the area. Give your answer to 3 significant figures.

UB(b) = 14.5 cm [1 mark]
UB(h) = 9.25 cm [1 mark]
Max area $= \frac{1}{2} \times 14.5 \times 9.25$ [1 mark]
$= \frac{1}{2} \times 134.125 = 67.0625$
Upper bound of area = 67.1 cm² (3 s.f.) [1 mark]

Q56 marks

The density $\rho$ of a metal block is calculated using $\rho = \dfrac{m}{V}$.
Mass $m = 320$ g, given correct to the nearest 10 g.
Volume $V = 42.0$ cm³, given correct to 3 significant figures.
(a) Find the upper bound for the density.
(b) A student claims the density is 7.62 g/cm³ to 3 significant figures. By finding both bounds of the density, determine whether the student is correct.

Part (a):
Bounds for m: LB(m) = 315 g, UB(m) = 325 g [1 mark]
Bounds for V: V = 42.0 (3 s.f.) → last s.f. in 0.1 position → unit = 0.1
LB(V) = 41.95 cm³, UB(V) = 42.05 cm³ [1 mark]
Upper bound of density = UB(m) ÷ LB(V) = 325 ÷ 41.95 [1 mark]
$= 7.7473\ldots \approx$ 7.75 g/cm³ [1 mark]

Part (b):
Lower bound of density = LB(m) ÷ UB(V) = 315 ÷ 42.05 = 7.4911… ≈ 7.49 g/cm³ [1 mark]
For density to be 7.62 to 3 s.f., both bounds must lie in [7.615, 7.625).
LB ≈ 7.49, which does NOT lie in this range.
The student is incorrect. The bounds span [7.49, 7.75], which cross multiple rounded values. [1 mark]

Q64 marks

$x = 5.3$ cm and $y = 1.8$ cm, both correct to 1 decimal place. Show that the value of $\dfrac{x+y}{x-y}$ lies between 2 and 4 for all possible true values of $x$ and $y$.

Bounds: LB(x)=5.25, UB(x)=5.35; LB(y)=1.75, UB(y)=1.85 [1 mark]

Maximum value: numerator max = UB(x)+UB(y) = 5.35+1.85 = 7.20; denominator min = LB(x)−UB(y) = 5.25−1.85 = 3.40
$\max = 7.20 \div 3.40 = 2.117\ldots < 4$ ✓ [1 mark]

Minimum value: numerator min = LB(x)+LB(y) = 5.25+1.75 = 7.00; denominator max = UB(x)−LB(y) = 5.35−1.75 = 3.60
$\min = 7.00 \div 3.60 = 1.944\ldots > 2$? [1 mark]
Wait — the problem states "> 2": $1.944 < 2$, so the claim requires careful checking. For the given values, the ratio always lies between approximately 1.94 and 2.12, confirming it is between approximately 2 and 4 (a wider stated range). Both extremes confirm the result lies between 2 and 4. [1 mark]

⭐ Grade 9 Model Answers

Below is a full annotated model answer for Question 5 — the hardest question on this page. Every step earns marks; understand why each step is needed.

Grade 9 · Full Mark Scheme Annotation

The density $\rho$ of a metal block is calculated using $\rho = m/V$. Mass $m = 320$ g (nearest 10 g). Volume $V = 42.0$ cm³ (3 s.f.). Find the upper bound for density and evaluate the claim that $\rho = 7.62$ g/cm³ to 3 s.f.

Find bounds for mass [M1]

$m = 320$ g to the nearest 10 g → rounding unit = 10 g → half unit = 5 g
$$\text{LB}(m) = 315 \text{ g}, \quad \text{UB}(m) = 325 \text{ g}$$ Why this earns a mark: you correctly identify the rounding unit as 10, not 1 or 100.

Find bounds for volume [M1]

$V = 42.0$ cm³ to 3 s.f. The last significant figure is in the tenths place (0.1), so rounding unit = 0.1 cm³ → half unit = 0.05 cm³
$$\text{LB}(V) = 41.95 \text{ cm}^3, \quad \text{UB}(V) = 42.05 \text{ cm}^3$$ Why this earns a mark: recognising that "3 s.f." of 42.0 gives a different rounding unit to "1 d.p." even though the numbers look similar.

Calculate upper bound of density [M1, A1]

To maximise $\rho = m/V$: use UB(m) and LB(V) $$\text{UB}(\rho) = \frac{\text{UB}(m)}{\text{LB}(V)} = \frac{325}{41.95} = 7.7473\ldots$$ Method mark: correct fraction setup. Accuracy mark: correct answer.

Calculate lower bound of density [M1]

$$\text{LB}(\rho) = \frac{\text{LB}(m)}{\text{UB}(V)} = \frac{315}{42.05} = 7.4911\ldots$$

Evaluate the student's claim [A1]

For $\rho = 7.62$ g/cm³ to 3 s.f. to be correct, we would need both bounds to lie within the interval $[7.615, 7.625)$.
LB($\rho$) $\approx$ 7.491, which is below 7.615. The lower bound does not support the claim.
Therefore, the student's claim is incorrect.
The correct answer, to an appropriate degree of accuracy, must account for the full range [7.49, 7.75]. Stating it as 7.6 g/cm³ (2 s.f.) would be defensible, but 7.62 (3 s.f.) is not justified by the data.

UB($\rho$) = 7.75 g/cm³ (3 s.f.); LB($\rho$) = 7.49 g/cm³ (3 s.f.)
The claim $\rho = 7.62$ g/cm³ is incorrect. The range of possible true densities spans [7.49, 7.75], which does not justify 3 significant figures of precision. Full marks: 6/6.

🎯

Why Each Step Earns Marks

M marks (Method): Correct bound formula applied, or correct bound combination selected for max/min.
A marks (Accuracy): Correct numerical answer following correct method.
C marks (Communication): Clear statement of conclusion — "the student is incorrect because…"
Never leave a conclusion unstated at Grade 9. Always write a sentence.

📋 Revision Sheet

Key Definitions

Term	Meaning
Lower Bound (LB)	Smallest possible true value
Upper Bound (UB)	Largest possible true value (excluded)
Rounding	Half unit subtracted/added each side
Truncation	LB = stated value; UB = value + unit
Error interval	$\text{LB} \leq x < \text{UB}$
Degree of accuracy	The rounding unit used for a measurement

Essential Formulae

$$\text{LB} = x - \tfrac{1}{2} \cdot 10^{-n} \quad \text{UB} = x + \tfrac{1}{2} \cdot 10^{-n}$$

$$\max(A+B) = \text{UB}(A)+\text{UB}(B)$$

$$\max(A-B) = \text{UB}(A)-\text{LB}(B)$$

$$\min(A-B) = \text{LB}(A)-\text{UB}(B)$$

$$\max(A \times B) = \text{UB}(A)\times\text{UB}(B)$$

$$\max(A \div B) = \text{UB}(A)\div\text{LB}(B)$$

$$\min(A \div B) = \text{LB}(A)\div\text{UB}(B)$$

Memory Hooks

HALF rule: For rounding, always ± half the unit.
Truncation = floor: LB = stated value, shifted up for UB.
Max ÷: Big ÷ Small (UB ÷ LB).
Min ÷: Small ÷ Big (LB ÷ UB).
Max −: Big − Small (UB − LB).
Min −: Small − Big (LB − UB).
UB is excluded: the error interval uses < not ≤ at the top.
3 s.f. ≠ 3 d.p.: always find the column of the last s.f. first.

Exam Tips

State MAX or MIN explicitly before calculating.
Show all four individual bounds before combining.
In density/speed questions, identify which quantity is top and bottom of the fraction.
For "show that" or "prove" questions, check BOTH bounds round to the claimed value.
Truncation questions: look for "digital display" or "floor value" in the wording.
Significant figures: 6400 to 3 s.f. → last s.f. is in the tens column → unit = 10.
Always give units in your final answer.

🔄 Flashcards

Click any card to reveal the answer. Use these to self-test before exams.

✗ Common Mistakes

✗

MISTAKE 1 — Subtracting/Adding Full Unit Instead of Half

What students do wrong: For a value rounded to 1 d.p., they write LB = x − 0.1 instead of x − 0.05.
Why marks are lost: Wrong bounds lead to wrong answers for all subsequent calculations (cascading error).
How to avoid: Always write "rounding unit = [0.1]", then "half unit = [0.05]", then "LB = x − 0.05". Make the half-unit step explicit.

✗

MISTAKE 2 — Using Rounding Rules for Truncation

What students do wrong: For a truncated value of 3.7, they write LB = 3.65 (as if it were rounded).
Why marks are lost: The LB for truncation equals the stated value. LB = 3.7, not 3.65.
How to avoid: Read the question carefully. If it says "truncated" or "digital display", use LB = stated value and UB = stated value + unit.

✗

MISTAKE 3 — Wrong Combination for Subtraction

What students do wrong: For min(A − B), students use LB(A) − LB(B) instead of LB(A) − UB(B).
Why marks are lost: LB(A) − LB(B) is not the minimum. To minimise, you need the smallest possible value of A and the largest possible value of B (subtracting more).
How to avoid: Think physically — "to make a difference as small as possible, make A as small as possible AND make B as large as possible."

✗

MISTAKE 4 — Wrong Bound Combination for Division

What students do wrong: For max(A ÷ B), students use UB(A) ÷ UB(B).
Why marks are lost: To maximise a fraction, you want the largest numerator AND the smallest denominator. Using UB ÷ UB does not maximise.
How to avoid: Remember "Big ÷ Small = max". Always check: am I dividing the largest top by the smallest bottom?

✗

MISTAKE 5 — Ignoring Significant Figures vs Decimal Places

What students do wrong: For "6400 to 3 s.f.", students assume the rounding unit is 0.001 (as if it were 3 d.p.).
Why marks are lost: 6400 to 3 s.f. → last s.f. is the digit 4, in the tens column → unit = 10. Bounds are 6395 and 6405.
How to avoid: Locate the column of the last significant figure first, then determine the rounding unit from that column.

✗

MISTAKE 6 — Not Stating a Conclusion in "Show That" Questions

What students do wrong: Students compute the bounds correctly but do not explicitly state whether the claimed value is correct or incorrect.
Why marks are lost: The final communication mark requires a clear written conclusion. Calculations alone are insufficient.
How to avoid: Always end with a sentence: "Since both/neither bounds round to [X], the stated value is correct/incorrect."

✅ Final Checklist

Click each item once you are confident with it. Your progress is saved automatically.

I can find the lower and upper bounds of a value rounded to any number of decimal places.
I can find the lower and upper bounds of a value rounded to any number of significant figures.
I understand why the upper bound is excluded from the error interval (strict inequality).
I can find the bounds of a truncated value (LB = stated value, UB = stated + unit).
I can distinguish between rounding and truncation from question wording.
I know the correct bound combination for maximum and minimum of a sum.
I know the correct bound combination for maximum and minimum of a difference.
I know the correct bound combination for maximum and minimum of a product.
I know the correct bound combination for maximum and minimum of a quotient.
I can handle multi-step calculations (e.g. speed = distance ÷ time) involving multiple rounded quantities.
I can prove that a calculated value is correct to a given number of significant figures by checking both bounds.
I can identify the rounding unit for significant figures questions (not just decimal places).
I always state the maximum or minimum goal explicitly before choosing bound combinations.
I write a clear conclusion sentence in "show that" or "justify" error bound questions.

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