Rounding, Bounds and Error

Every measurement is rounded. A ruler, a scale or a stopwatch can only give a number to a certain accuracy, so the true value is hidden somewhere in a small interval around the reading. This page shows how to round sensibly, how to find the bounds that the true value must lie between, how those bounds carry through a calculation, and how to measure the size of an error as a percentage.

Key ideas

Decimal places and significant figures

• Decimal places (d.p.) count digits after the decimal point. $3.14159$ to $2$ d.p. is $3.14$.
• Significant figures (s.f.) count digits from the first non-zero digit. Leading zeros are never significant; zeros between non-zero digits always are; trailing zeros after the decimal point are significant (they show the accuracy).

Number	To 3 s.f.	Why
$52\,847$	$52\,800$	the zeros are placeholders, keeping the size of the number
$0.004\,037\,2$	$0.004\,04$	the leading zeros don’t count; start at the $4$
$3.0496$	$3.05$	the $0$ between $3$ and $4$ counts
$2.0004$	$2.00$	the trailing zeros show the answer is accurate to $3$ s.f.

To round, look at the next digit: $5$ or more rounds up, $4$ or less leaves the digit alone.

In IB exams, give answers exactly or to 3 s.f. unless the question says otherwise. More generally, choose an accuracy that matches the data: if lengths were measured to the nearest centimetre, an area to $8$ s.f. claims an accuracy you don’t have. Keep full calculator values during working and round only at the end.

Upper and lower bounds

If a value is given to some accuracy, the true value can be up to half a unit of that accuracy either side. For $x = 7.3$ to $1$ d.p., the unit is $0.1$, so half a unit is $0.05$:

$$7.25 \le x \lt 7.35$$

$7.25$ is the lower bound (it rounds up to $7.3$, so it’s included) and $7.35$ is the upper bound (it would round to $7.4$, so it’s excluded).

Every number from $7.25$ up to (but not including) $7.35$ rounds to $7.3$.

Given as	Half a unit	Bounds
$64$ kg to the nearest kg	$0.5$ kg	$63.5 \le m \lt 64.5$
$3.70$ m to $2$ d.p.	$0.005$ m	$3.695 \le L \lt 3.705$
$2400$ to the nearest $100$	$50$	$2350 \le n \lt 2450$

Bounds of a calculation

To find the largest or smallest possible result, choose the bound of each input that pushes the answer the right way:

Calculation	Upper bound	Lower bound
$a + b$	$a_{\max} + b_{\max}$	$a_{\min} + b_{\min}$
$a - b$	$a_{\max} - b_{\min}$	$a_{\min} - b_{\max}$
$a \times b$ (positive)	$a_{\max} \times b_{\max}$	$a_{\min} \times b_{\min}$
$a \div b$ (positive)	$a_{\max} \div b_{\min}$	$a_{\min} \div b_{\max}$

Subtracting or dividing by the smallest possible number gives the biggest answer. That’s the one people get wrong.

Percentage error

The percentage error compares an approximate value $v_A$ with the exact value $v_E$:

$$\varepsilon = \left|\frac{v_A - v_E}{v_E}\right| \times 100\%$$

Divide by the exact value. For example, using $\dfrac{22}{7}$ for $\pi$ gives

$$\varepsilon = \left|\frac{\frac{22}{7} - \pi}{\pi}\right| \times 100\% \approx 0.0402\%$$

When a value is measured, the exact value is unknown but lies between the bounds. The maximum percentage error is the biggest percentage error over that interval. Work it out with both bounds as $v_E$ and take the larger. (It’s usually the lower bound, because you divide by a smaller number.)

Estimation and reasonableness

Before trusting a calculator, estimate: round each number to $1$ s.f. and work it out in your head. If your answer is wildly different, look for a slip. Also ask whether the answer makes sense: lengths, areas and times can’t be negative, a probability can’t be more than $1$, and a person isn’t $17$ m tall.

Worked examples

Example 1: Rounding and estimating

• (a) Round $0.006\,195$ to $2$ s.f. and $3.849\,7$ to $2$ d.p.
• (b) Estimate $\dfrac{48.7 \times 0.212}{9.8}$ by rounding each number to $1$ s.f. Then compare with the calculator value.

Solution.

(a) In $0.006\,195$ the first significant figure is the $6$. Two significant figures: $6$ and $1$, and the next digit is $9$, so round up: $0.006\,2$. For $3.849\,7$: the third decimal digit is $9$, so the $4$ rounds up to $5$: $3.85$.

(b) Round to $1$ s.f.: $48.7 \approx 50$, $0.212 \approx 0.2$, $9.8 \approx 10$.

$$\frac{50 \times 0.2}{10} = \frac{10}{10} = 1$$

The calculator gives $1.053\,51\ldots \approx 1.05$ (3 s.f.). The estimate is close, so the calculator answer is believable.

Example 2: Bounds of a perimeter and an area

A rectangular garden is $18$ m long and $11$ m wide, each to the nearest metre. Find the lower and upper bounds of its perimeter and of its area.

Solution. The bounds of the sides are $17.5 \le L \lt 18.5$ and $10.5 \le W \lt 11.5$.

Perimeter $P = 2(L + W)$:

$$\begin{aligned} P_{\min} &= 2(17.5 + 10.5) = 56 \text{ m} \\ P_{\max} &= 2(18.5 + 11.5) = 60 \text{ m} \end{aligned}$$

Area $A = LW$:

$$\begin{aligned} A_{\min} &= 17.5 \times 10.5 = 183.75 \text{ m}^2 \\ A_{\max} &= 18.5 \times 11.5 = 212.75 \text{ m}^2 \end{aligned}$$

Notice how wide the range of areas is: the “area” $18 \times 11 = 198$ m² could really be anywhere from about $184$ to $213$ m².

Example 3: Bounds of a speed

A cyclist rides $120$ km, to the nearest $10$ km, in $4.5$ hours, to the nearest $0.1$ hour. Find the bounds of the cyclist’s average speed.

Solution. Bounds: $115 \le d \lt 125$ km and $4.45 \le t \lt 4.55$ h.

Speed is $d \div t$. For the biggest speed, use the biggest distance and the smallest time:

$$v_{\max} = \frac{125}{4.45} = 28.089\ldots \approx 28.1 \text{ km/h}$$

For the smallest speed, use the smallest distance and the biggest time:

$$v_{\min} = \frac{115}{4.55} = 25.274\ldots \approx 25.3 \text{ km/h}$$

The average speed is between about $25.3$ and $28.1$ km/h (3 s.f.).

Example 4: Maximum percentage error in an area

The side of a square tile is measured as $6.4$ cm, to $1$ d.p. Find the maximum percentage error in the area calculated from this measurement.

Solution. The calculated (approximate) area is $v_A = 6.4^2 = 40.96$ cm².

The true side lies in $6.35 \le s \lt 6.45$, so the true area lies between

$$6.35^2 = 40.3225 \qquad \text{and} \qquad 6.45^2 = 41.6025$$

Try each bound as the exact value:

$$\begin{aligned} \text{lower: } &\left|\frac{40.96 - 40.3225}{40.3225}\right| \times 100\% = 1.581\ldots\% \\ \text{upper: } &\left|\frac{40.96 - 41.6025}{41.6025}\right| \times 100\% = 1.544\ldots\% \end{aligned}$$

The maximum percentage error is about $1.58\%$ (3 s.f.). A rounding error of at most $0.05$ cm in the side (under $1\%$) becomes about $1.6\%$ in the area, because the side is squared.

Common mistakes

Dropping the placeholder zeros. $52\,847$ to $3$ s.f. is $52\,800$, not $528$. Rounding must keep the size of the number.

Making the upper bound too small. For $64$ kg to the nearest kg, the upper bound is $64.5$, not $64.4$ or $64.49$. Use $64.5$ in calculations, even though the true value is strictly less than it.

Using max with max for a difference or quotient. The largest value of $a - b$ is $a_{\max} - b_{\min}$, and the largest value of $a \div b$ is $a_{\max} \div b_{\min}$. Ask yourself, “which way does each input push the answer?”

Dividing by the approximate value in percentage error. The formula divides by the exact value $v_E$. With $v_A = 40.96$ and $v_E = 40.3225$, dividing by $40.96$ instead gives $1.56\%$, which is wrong.

Rounding in the middle of a calculation. Rounding early adds extra error that can change your 3 s.f. answer. Store full values in your GDC and round once, at the end.

Not checking whether an answer is sensible. A negative length, a speed of $2810$ km/h for a cyclist, or a percentage error of $150\%$ from a careful measurement are all signs of a slip. Estimate first, then compare.

Practice

1. (Warm-up) Write $0.030\,695\,1$:

• (a) to $3$ s.f.
• (b) to $2$ d.p.
• (c) to $4$ s.f.

Solution

(a) The first significant figure is the $3$. Three figures: $3, 0, 6$, and the next digit is $9$, so round up: $0.030\,7$.

(b) Two decimal places: $0.03$ (the next digit is $0$).

(c) Four figures: $3, 0, 6, 9$, and the next digit is $5$, so round up: $3069 \to 3070$, giving $0.030\,70$. Keep the final zero: it shows the answer is accurate to $4$ s.f.

2. (Warm-up) Write down the lower and upper bounds of each value.

• (a) A runner’s time is $12.7$ s, to $1$ d.p.
• (b) A bridge is $340$ m long, to the nearest $10$ m.
• (c) A coin has mass $6.27$ g, to $3$ s.f.

Solution

(a) Half of $0.1$ is $0.05$: $12.65 \le t \lt 12.75$ s.

(b) Half of $10$ is $5$: $335 \le L \lt 345$ m.

(c) The last significant figure is in the hundredths place, so half a unit is $0.005$: $6.265 \le m \lt 6.275$ g.

3. (Warm-up) Estimate $\dfrac{612 \times 0.0487}{2.93}$ by rounding each number to $1$ s.f. Then find the value to 3 s.f. with a calculator.

Solution

$612 \approx 600$, $0.0487 \approx 0.05$, $2.93 \approx 3$:

$$\frac{600 \times 0.05}{3} = \frac{30}{3} = 10$$

Calculator: $10.172\ldots \approx 10.2$ (3 s.f.). The estimate agrees.

4. (Core) Two rods have lengths $8.3$ cm and $5.6$ cm, each to $1$ d.p. Find the lower and upper bounds of:

• (a) their total length
• (b) the difference in their lengths

Solution

Bounds: $8.25 \le a \lt 8.35$ and $5.55 \le b \lt 5.65$.

(a) Lower bound $8.25 + 5.55 = 13.8$ cm; upper bound $8.35 + 5.65 = 14.0$ cm.

(b) Lower bound $a_{\min} - b_{\max} = 8.25 - 5.65 = 2.6$ cm; upper bound $a_{\max} - b_{\min} = 8.35 - 5.55 = 2.8$ cm.

5. (Core) A cylindrical water tank has radius $0.8$ m and height $1.5$ m, each to $1$ d.p. Find the lower and upper bounds of its volume, $V = \pi r^2 h$, to 3 s.f.

Solution

Bounds: $0.75 \le r \lt 0.85$ and $1.45 \le h \lt 1.55$.

$$\begin{aligned} V_{\min} &= \pi (0.75)^2 (1.45) = 2.562\ldots \approx 2.56 \text{ m}^3 \\ V_{\max} &= \pi (0.85)^2 (1.55) = 3.518\ldots \approx 3.52 \text{ m}^3 \end{aligned}$$

(The calculated volume $\pi(0.8)^2(1.5) \approx 3.02$ m³ sits between them.)

6. (Core) A car travels $450$ km, to the nearest $10$ km, using $32$ litres of fuel, to the nearest litre. Fuel consumption is measured in litres per $100$ km. Find its lower and upper bounds, to 3 s.f.

Solution

Consumption $= \dfrac{\text{fuel}}{\text{distance}} \times 100$. Bounds: $31.5 \le F \lt 32.5$ L and $445 \le d \lt 455$ km.

$$\begin{aligned} \text{upper} &= \frac{32.5}{445} \times 100 = 7.303\ldots \approx 7.30 \text{ L/100 km} \\ \text{lower} &= \frac{31.5}{455} \times 100 = 6.923\ldots \approx 6.92 \text{ L/100 km} \end{aligned}$$

7. (Core) The edge of a cube is measured as $4.2$ cm, to $1$ d.p. Find the maximum percentage error in the volume calculated from this measurement.

Solution

Calculated volume: $v_A = 4.2^3 = 74.088$ cm³. The true edge is in $4.15 \le s \lt 4.25$, so the true volume is between $4.15^3 = 71.473\,375$ and $4.25^3 = 76.765\,625$.

$$\begin{aligned} \text{lower: } &\left|\frac{74.088 - 71.473\,375}{71.473\,375}\right| \times 100\% = 3.658\ldots\% \\ \text{upper: } &\left|\frac{74.088 - 76.765\,625}{76.765\,625}\right| \times 100\% = 3.488\ldots\% \end{aligned}$$

Maximum percentage error $\approx 3.66\%$ (3 s.f.).

8. (Challenge) A ruler measures lengths to the nearest millimetre. What is the shortest length you can measure with it so that the percentage error from rounding is at most $1\%$? Give your answer in centimetres.

Solution

To the nearest millimetre, the rounding error is at most $0.5$ mm $= 0.05$ cm. Percentage error divides by the exact value, and the worst case is when the true length is as small as possible. If the ruler reads $L$ cm, the true length could be as small as $L - 0.05$ cm, so we need

$$\frac{0.05}{L - 0.05} \times 100 \le 1 \quad\Rightarrow\quad 5 \le L - 0.05 \quad\Rightarrow\quad L \ge 5.05$$

Readings go up in steps of $0.1$ cm, so the shortest reading that works is $5.1$ cm. Check: a reading of $5.0$ cm could be a true length of $4.95$ cm, giving $\dfrac{0.05}{4.95} \times 100\% \approx 1.01\%$, just over $1\%$; a reading of $5.1$ cm gives at most $\dfrac{0.05}{5.05} \times 100\% \approx 0.990\%$. (Shorter objects need a more precise instrument, such as calipers.)

9. (Challenge) $p = 5.0$ and $q = 2.4$, each to $1$ d.p.

• (a) Explain why $\dfrac{5.05}{4.95 - 2.45}$ is not the upper bound of $\dfrac{p}{p - q}$.
• (b) By writing $\dfrac{p}{p - q} = \dfrac{1}{1 - \frac{q}{p}}$, find the lower and upper bounds of $\dfrac{p}{p - q}$, to 3 s.f.

Solution

(a) That calculation uses $p = 5.05$ in the numerator and $p = 4.95$ in the denominator at the same time. But $p$ has only one true value, so it can’t be both. The value $\dfrac{5.05}{2.5} = 2.02$ can never actually happen.

(b) Bounds: $4.95 \le p \lt 5.05$ and $2.35 \le q \lt 2.45$. The expression $\dfrac{1}{1 - \frac{q}{p}}$ gets bigger as $\dfrac{q}{p}$ gets bigger (the denominator shrinks). So use the biggest and smallest possible values of $\dfrac{q}{p}$.

Upper bound: biggest $q$, smallest $p$, so $p = 4.95$, $q = 2.45$:

$$\frac{4.95}{4.95 - 2.45} = \frac{4.95}{2.5} = 1.98$$

Lower bound: smallest $q$, biggest $p$, so $p = 5.05$, $q = 2.35$:

$$\frac{5.05}{5.05 - 2.35} = \frac{5.05}{2.7} = 1.870\ldots \approx 1.87$$

So $\dfrac{p}{p - q}$ lies between about $1.87$ and $1.98$ (3 s.f.).