Understanding Accuracy and Measurement Revision Notes | GCSE AQA Higher Maths

1. Why Measurements Are Never Perfectly Exact

Every measurement you take in real life is rounded. A ruler marked in millimetres cannot tell you whether a pencil is 14.3 mm or 14.32 mm — it gives you the nearest millimetre. That rounding creates a gap between the number written down and the true value hiding behind it.

This matters enormously in engineering, medicine, and construction. If a bridge beam is measured as 12 m to the nearest metre, the true length could be anywhere from 11.5 m to just under 12.5 m — a full metre of uncertainty. Knowing how to capture that uncertainty precisely is what this topic is all about.

Before finding any bounds, you must identify the degree of accuracy. That single piece of information determines everything else. Half of the degree of accuracy is the key quantity — it tells you how far the true value can stray either side of the rounded figure.

2. Upper and Lower Bounds

Picture a number line. You have a rounded value sitting in the middle. The true value could be anywhere in a symmetric interval around it. The edges of that interval are the lower bound and the upper bound.

Notice that the lower bound uses a closed inequality ($\geq$) — the true value can equal the lower bound. The upper bound uses a strict inequality ($<$) — the true value cannot equal the upper bound, because if it did, it would round up to the next value.

Worked example: Finding the Error Interval for a Rounded Measurement

A length $l$ is measured as 8.4 cm, correct to 1 decimal place. Write down the error interval for $l$.

1. Rounded to 1 decimal place $\Rightarrow$ degree of accuracy $= 0.1$ cm — 1 decimal place means the measurement is given to the nearest 0.1. This is your starting point every time. [M1]
2. $\frac{0.1}{2} = 0.05$ cm — This is the maximum amount the true value can differ from the rounded value in either direction.
3. $8.4 - 0.05 = 8.35$ cm — Subtract half the degree of accuracy from the rounded value. [M1]
4. $8.4 + 0.05 = 8.45$ cm — Add half the degree of accuracy to the rounded value.
5. $8.35 \leq l < 8.45$ — Use $\leq$ for the lower bound (included) and $<$ for the upper bound (excluded). [A1]

Answer: $8.35 \leq l < 8.45$

3. Bounds with Significant Figures

Significant figures add a layer of care because the degree of accuracy changes depending on the size of the number. A value given to 2 significant figures has a different degree of accuracy at 3,400 than at 0.0034.

The method is identical — identify what the last significant figure represents as a place value, treat that as your degree of accuracy, then halve it. The worked example below shows this clearly. Understanding these key concepts solidifies your foundational knowledge and prepares you for more advanced problems.

Worked example: Error Interval for a Value Given to Significant Figures

A population $P$ is recorded as 47,000, given to 2 significant figures. Write the error interval for $P$.

1. $47{,}000$ to 2 s.f. $\Rightarrow$ the second significant figure is the digit 7, which is in the thousands column. — The degree of accuracy is therefore 1,000 — the value is rounded to the nearest thousand. [M1]
2. $\frac{1000}{2} = 500$ — This is the maximum deviation from the stated value.
3. Lower bound: $47{,}000 - 500 = 46{,}500$ Upper bound: $47{,}000 + 500 = 47{,}500$ — Subtract and add 500 respectively. [M1]
4. $46{,}500 \leq P < 47{,}500$ — Standard error interval notation. [A1]

Answer: $46{,}500 \leq P < 47{,}500$

4. Bounds with Calculated Values and Techniques

Here is where the topic becomes genuinely powerful — and where exam questions get harder. When you combine two rounded measurements in a calculation, the result also has a range of possible values. You need to think carefully about which combination of bounds gives the maximum result and which gives the minimum.

The rules below are the key to getting this right every time. Memorise the logic, not just the rule — understanding why each combination works will save you in unfamiliar questions.

The division rule is the one that trips students up most often. To make a fraction as large as possible, you want the numerator as big as possible and the denominator as small as possible. Think of it like this: $\frac{10}{2} = 5$ but $\frac{10}{1} = 10$ — a smaller denominator gives a bigger result.

Worked example: Maximum and Minimum of a Calculated Area

A rectangle has length $l = 12$ cm (to the nearest cm) and width $w = 7$ cm (to the nearest cm). Calculate the upper bound and lower bound of the area of the rectangle.

1. Degree of accuracy $= 1$ cm, so half $= 0.5$ cm $\text{LB}(l) = 12 - 0.5 = 11.5$ cm $\text{UB}(l) = 12 + 0.5 = 12.5$ cm — Standard bounds calculation for $l$. [M1]
2. Degree of accuracy $= 1$ cm, so half $= 0.5$ cm $\text{LB}(w) = 7 - 0.5 = 6.5$ cm $\text{UB}(w) = 7 + 0.5 = 7.5$ cm — Standard bounds calculation for $w$. [M1]
3. $\text{UB}(A) = \text{UB}(l) \times \text{UB}(w) = 12.5 \times 7.5$ $= 12.5 \times 7 + 12.5 \times 0.5 = 87.5 + 6.25 = 93.75$ cm$^2$ — Maximum area uses both upper bounds. Arithmetic verified: $12.5 \times 7.5 = 93.75$. [M1]
4. $\text{LB}(A) = \text{LB}(l) \times \text{LB}(w) = 11.5 \times 6.5$ $= 11.5 \times 6 + 11.5 \times 0.5 = 69 + 5.75 = 74.75$ cm$^2$ — Minimum area uses both lower bounds. Arithmetic verified: $11.5 \times 6.5 = 74.75$. [A1]

Answer: Upper bound of area $= 93.75$ cm$^2$; Lower bound of area $= 74.75$ cm$^2$

Worked example: Upper Bound of a Calculated Speed

Priya runs a distance $d = 400$ m, measured to the nearest 10 m, in a time $t = 52$ s, measured to the nearest second. Calculate the upper bound of her average speed. Give your answer to 3 significant figures.

1. Degree of accuracy $= 10$ m, half $= 5$ m $\text{LB}(d) = 400 - 5 = 395$ m $\text{UB}(d) = 400 + 5 = 405$ m — Nearest 10 m gives a half-interval of 5 m. [M1]
2. Degree of accuracy $= 1$ s, half $= 0.5$ s $\text{LB}(t) = 52 - 0.5 = 51.5$ s $\text{UB}(t) = 52 + 0.5 = 52.5$ s — Nearest second gives a half-interval of 0.5 s. [M1]
3. $\text{Speed} = \frac{d}{t}$, so maximum speed $= \frac{\text{UB}(d)}{\text{LB}(t)} = \frac{405}{51.5}$ — To maximise a fraction, use the largest numerator and the smallest denominator. [M1]
4. $\frac{405}{51.5} = 7.864...$ m/s — Carry out the division without rounding mid-calculation. Keep the full decimal for now.
5. $7.864... \approx 7.86$ m/s (3 s.f.) — The fourth significant figure is 4, so round down. [A1]

Answer: Upper bound of speed $\approx 7.86$ m/s (3 s.f.)

5. Truncation vs Rounding

Rounding is not the only way a value gets approximated. Truncation cuts off digits after a certain point without looking at what comes next. A calculator display that shows 3.14 for $\pi$ has truncated, not rounded — the true value is 3.14159..., which would round to 3.14 anyway, but truncation always discards, never rounds up.

This changes the error interval. When a value is truncated to, say, 2 decimal places, the lower bound is the truncated value itself (the true value is at least that big), and the upper bound is the truncated value plus the degree of accuracy. The interval shifts entirely above the stated value.

The table makes the contrast clear. Rounding creates a symmetric interval centred on the stated value. Truncation creates an interval that starts at the stated value and extends upward by one full degree of accuracy. Examiners test this distinction directly, so it is worth fixing firmly in your mind.

6. Deliberate Practice

These problems cover the full range of techniques from this lesson. Work through each one before checking the solution — the act of attempting it first is what builds the skill.

7. Exam Technique and Common Mistakes

Bounds questions appear on both calculator-allowed questions and questions without a calculator. In questions without a calculator, you must show all arithmetic working clearly — you cannot rely on a calculator to verify your multiplication. On calculator-allowed exam questions , you can compute the final division or product directly, but you still need to show which bounds you selected and why.

The mark scheme for a bounds calculation question typically awards one mark for correct bounds on each variable and one mark for selecting the correct combination. The final answer mark is only awarded if the combination is right. This means you can still earn method marks even if your arithmetic slips — but only if your working is visible.