Limits of Accuracy

Every measurement is recorded to a finite precision. When a length is reported as "$12\,\text{cm}$ to the nearest cm", the actual value sits anywhere in $[11.5, 12.5)$.

The rule. A measurement rounded to the nearest $u$ has bounds: $x_{\min} = x - \dfrac{u}{2}, \quad x_{\max} = x + \dfrac{u}{2}.$

The lower bound is INCLUSIVE — that's the smallest value that would still round to the recorded value. The upper bound is technically not reached (the next unit up would round to a larger value).

Cambridge accepts both forms: $11.5 \le x < 12.5$ or $11.5 \le x \le 12.5$.

Worked. A length is given as $4.5\,\text{m}$ to 1 d.p. Find the bounds.

• Unit of accuracy: $0.1\,\text{m}$. Half: $0.05\,\text{m}$.
• Lower: $4.45\,\text{m}$. Upper: $4.55\,\text{m}$.

Worked. A mass is $250\,\text{g}$ to 2 s.f. Find the bounds.

• 2 s.f. → unit of accuracy is $10\,\text{g}$. Half: $5\,\text{g}$.
• Lower: $245\,\text{g}$. Upper: $255\,\text{g}$.

Time given to nearest minute. Unit is $1\,\text{min} = 60\,\text{s}$. Half-unit: $30\,\text{s}$. So "10 minutes" lies in $[9\,\text{min}\,30\,\text{s},\, 10\,\text{min}\,30\,\text{s})$.

When you combine two measured values, the resulting expression has bounds too — and they're often what Paper 4 asks for.

Addition: $a + b$.

• $\max(a + b) = a_{\max} + b_{\max}$.
• $\min(a + b) = a_{\min} + b_{\min}$.

Subtraction: $a - b$. Counterintuitive.

• $\max(a - b) = a_{\max} - b_{\min}$ (largest minus smallest).
• $\min(a - b) = a_{\min} - b_{\max}$ (smallest minus largest).

Multiplication (positive values): $a \times b$.

• $\max(a \times b) = a_{\max} \times b_{\max}$.
• $\min(a \times b) = a_{\min} \times b_{\min}$.

Division (positive values): $a / b$. Like subtraction — flip the bound on the divisor.

• $\max(a / b) = a_{\max} / b_{\min}$ (largest divided by smallest).
• $\min(a / b) = a_{\min} / b_{\max}$ (smallest divided by largest).

Worked. A rectangle has length $12\,\text{cm}$ and width $5\,\text{cm}$, both to the nearest cm. Bounds for the area?

• Length: $11.5 \le L < 12.5$.
• Width: $4.5 \le W < 5.5$.
• $\max$ area: $12.5 \times 5.5 = 68.75\,\text{cm}^{2}$.
• $\min$ area: $11.5 \times 4.5 = 51.75\,\text{cm}^{2}$.

Worked (subtraction). Two heights are $1.85\,\text{m}$ and $1.62\,\text{m}$, both to 2 d.p. Largest possible DIFFERENCE?

• Heights: $1.845 \le h_{1} < 1.855$, $1.615 \le h_{2} < 1.625$.
• $\max(h_{1} - h_{2}) = h_{1,\max} - h_{2,\min} = 1.855 - 1.615 = 0.240\,\text{m}$.

Worked (division). Distance is $400\,\text{m}$ to 3 s.f., time is $50\,\text{s}$ to 2 s.f. Largest possible average speed?

• Distance: $399.5 \le d < 400.5$.
• Time: $49.5 \le t < 50.5$.
• $\max(d/t) = 400.5 / 49.5 \approx 8.09\,\text{m/s}$.

Cambridge wraps bounds questions in physical contexts where uncertainty matters.

Worked: door fitting. A door is measured at $2.05\,\text{m}$ tall (to 2 d.p.) and a frame is $2.10\,\text{m}$ (to 2 d.p.). Will the door definitely fit?

• Door: $2.045 \le d < 2.055$.
• Frame: $2.095 \le f < 2.105$.
• Worst case: door at max ($2.055$) vs frame at min ($2.095$). Gap: $2.095 - 2.055 = 0.04\,\text{m} = 4\,\text{cm}$. So even in the worst case, the door fits with $4\,\text{cm}$ to spare.

Worked: speed limit. "A car covers $84\,\text{m}$ (to nearest m) in $4\,\text{s}$ (to nearest s). The speed limit is $20\,\text{m/s}$. Is the driver definitely speeding?"

• Distance: $83.5 \le d < 84.5$.
• Time: $3.5 \le t < 4.5$.
• Smallest possible speed: $83.5 / 4.5 = 18.6\,\text{m/s}$ < limit. So we CAN'T say definitively the driver was speeding.

The wording matters: "definitely" requires the LOWER bound of speed to exceed the limit; "possibly" only requires the upper bound to.