N16 - Upper and lower bounds in calculations

Rounded measurements can produce a range of calculated answers. Learn to find the extremes for sums, differences, products and quotients, interpret whether a boundary is included, and justify a reliable degree of accuracy for a compound measure.

What measurement bounds mean (Higher)

A journey is recorded as $120\,\mathrm{km}$ in $2.4$ hours. The calculation gives $50\,\mathrm{km\,h^{-1}}$ . Could the actual average speed be $51\,\mathrm{km\,h^{-1}}$ ? First find out what those measurements allow.

You already know

A positive rounded measurement has boundaries half a rounding step below and above it. With the usual positive midpoint rule, include the lower boundary and exclude the upper one.

For a length recorded as $6.4\,\mathrm{cm}$ to the nearest tenth, $6.35\leq l<6.45$ , with $l$ in centimetres. Its upper bound is $6.45$ , although the length cannot equal that boundary. A bound need not be an attainable measurement.

Lower and upper bounds

The lower and upper boundaries of the possible values. State endpoint inclusion separately: $a\leq x<b$ includes $a$ and excludes $b$ .

The stated accuracy matters as much as the displayed number.

Worked example 1: a mass rounded to tens

Question. A positive mass $m$ is recorded as $200\,\mathrm{g}$ to the nearest $10\,\mathrm{g}$ . Find its bounds and write the interval.

Plan. Find half the rounding step and apply it on each side.

\begin{aligned} \text{half-step}&=10\div2=5\,\mathrm{g} && \text{half the stated accuracy} \\ \text{lower bound}&=200-5=195\,\mathrm{g} && \text{subtract} \\ \text{upper bound}&=200+5=205\,\mathrm{g} && \text{add} \end{aligned}

Why? $205$ rounds to $210$ , so it is excluded; $195$ rounds to $200$ , so it is included.

Answer. $195\leq m<205$ , with $m$ in grams.

Check. $204.9\,\mathrm{g}$ fits and rounds to $200\,\mathrm{g}$ to the nearest ten.

Change only the stated accuracy, from the nearest ten grams to the nearest gram. The displayed value stays two hundred.

Worked example 2: a more precise measurement

Question. A positive mass $m$ is $200\,\mathrm{g}$ to the nearest gram. Find its bounds and interval.

Plan. Replace the half-step by half a gram.

\begin{aligned} \text{half-step}&=1\div2=0.5\,\mathrm{g} && \text{use the new accuracy} \\ \text{lower bound}&=200-0.5=199.5\,\mathrm{g} && \text{subtract} \\ \text{upper bound}&=200+0.5=200.5\,\mathrm{g} && \text{add} \end{aligned}

Why? The same printed mass can represent a narrower interval when its accuracy is stated more precisely.

Answer. $199.5\leq m<200.5$ , with $m$ in grams.

Check. The interval width is one gram, a tenth of Example 1's width.

Exactly two hundred packets is an exact count, so it has no half-packet rounding interval. Conversion factors such as $100\,\mathrm{cm}=1\,\mathrm{m}$ are exact too. For truncated inputs, use their truncation intervals first: $3.2\,\mathrm{km}$ truncated to one decimal place means $3.2\leq d<3.3$ , not $3.15\leq d<3.25$ .

Watch out

An upper bound is a boundary, not necessarily a maximum that can occur. $m=200.5$ is excluded in Example 2 even though $200.5$ is its upper bound.

Change only the reported mass in the second example.

Choosing extremes for a calculation (Higher)

Ask what increasing each input does to the answer. Increasing either positive input increases a sum or product. Increasing the subtracted quantity decreases a difference; increasing a positive denominator decreases a quotient.

Bounds for a positive quotient

\text{lower boundary of }\frac ab=\frac{a_{\mathrm{L}}}{b_{\mathrm{U}}},\qquad \text{upper boundary of }\frac ab=\frac{a_{\mathrm{U}}}{b_{\mathrm{L}}}

Here both input intervals are positive, with $b_{\mathrm{L}}>0$ . The subscripts identify lower and upper boundaries. Check endpoint inclusion separately.

Not on the formula sheet: derive it from the effect of increasing a positive numerator or denominator.

For differences, pair small minus large to find the lower boundary, and large minus small for the upper boundary. The next examples apply all four decisions to the same inputs.

Worked example 3: bounds for four operations

Question. Measurements satisfy $4.5\leq a<5.5$ and $2.5\leq b<3.5$ . Find intervals for $a+b$ , $a-b$ , $ab$ and $a/b$ .

Plan. Choose the input extremes separately for each operation.

\begin{aligned} 4.5+2.5&\leq a+b<5.5+3.5 && \text{both inputs increase the sum} \\ 7&\leq a+b<9 && \text{evaluate} \\ 4.5-3.5&<a-b<5.5-2.5 && \text{reverse the subtracted extremes} \\ 1&<a-b<3 && \text{evaluate} \\ 4.5\times2.5&\leq ab<5.5\times3.5 && \text{positive product} \\ 11.25&\leq ab<19.25 && \text{evaluate} \\ \frac{4.5}{3.5}&<\frac ab<\frac{5.5}{2.5} && \text{reverse denominator extremes} \end{aligned}

Why? Each difference or quotient boundary needs one excluded upper input, so neither boundary is attained.

Answer. The intervals above; quotient boundaries are exactly $\frac97$ and $\frac{11}{5}$ .

Check. Taking $a=5$ , $b=3$ gives $8$ , $2$ , $15$ and $\frac53$ , all inside their intervals.

Change only the interval for $b$ , moving it down by one. The same choices of extremes still work.

Worked example 4: varying one measurement interval

Question. Now $4.5\leq a<5.5$ and $1.5\leq b<2.5$ . Find the same four intervals.

Plan. Retain each operation's extreme pairing and substitute the new bounds of $b$ .

\begin{aligned} 4.5+1.5&\leq a+b<5.5+2.5 && \text{sum extremes} \\ 6&\leq a+b<8 && \text{evaluate} \\ 4.5-2.5&<a-b<5.5-1.5 && \text{difference extremes} \\ 2&<a-b<4 && \text{evaluate} \\ 4.5\times1.5&\leq ab<5.5\times2.5 && \text{positive product extremes} \\ 6.75&\leq ab<13.75 && \text{evaluate} \\ \frac{4.5}{2.5}&<\frac ab<\frac{5.5}{1.5} && \text{quotient extremes} \end{aligned}

Why? Decreasing the denominator increases the quotient, so the new upper boundary is larger.

Answer. The quotient interval is $\frac95<\frac ab<\frac{11}{3}$ ; the other three intervals are shown above.

Check. $a=5$ , $b=2$ gives $7$ , $3$ , $10$ and $2.5$ , inside the four intervals.

A rectangle with $6.15\leq l<6.25$ and $3.35\leq w<3.45$ , in centimetres, has $20.6025\leq lw<21.5625$ in square centimetres. Keep the unrounded boundaries.

Signed products need a sign check. If $-2\leq x\leq1$ and $3\leq y\leq4$ , the four endpoint products are $-6$ , $-8$ , $3$ and $4$ , giving $-8\leq xy\leq4$ . The smallest product did not use both lower endpoints. If a denominator interval contains zero, division may be undefined or unbounded; the positive-quotient rule cannot simply be reused.

Watch out

Wrong: the lower boundary of $a-b$ is always lower $a$ minus lower $b$ . Increasing $b$ decreases the difference, so use lower $a$ minus upper $b$ . Check signs before extending positive-product rules.

Change only $b$ 's interval again for the matched quotient question.

Bounds, compound measures and reliable rounding (Higher)

A calculated speed inherits uncertainty from its distance and time. A reliable rounded answer must agree for the entire possible interval, not just for the central-value calculation.

Average speed

v=\frac dt

Use compatible distance and time units, with positive journey time $t$ .

Not on the formula sheet: learn it.

Use exact boundary expressions until the final rounding decision. A rounded boundary can accidentally cross a rounding threshold.

Worked example 5: finding a reliable speed

Question. Distance is $120\,\mathrm{km}$ to the nearest kilometre and time is $2.4$ hours to the nearest tenth. Find the speed bounds and decide whether one or two significant figures gives a unique report.

Plan. Form both measurement intervals, bound the quotient, then compare it with rounding thresholds.

\begin{aligned} 119.5&\leq d<120.5 && \text{half a kilometre each side} \\ 2.35&\leq t<2.45 && \text{half a tenth each side} \\ \frac{119.5}{2.45}&<v<\frac{120.5}{2.35} && \text{small over large; large over small} \\ 48.775510\ldots&<v<51.276595\ldots && \text{evaluate in kilometres per hour} \end{aligned}

Why? The whole interval lies between $45$ and $55$ , but crosses several whole-number rounding boundaries.

Answer. One significant figure gives $50\,\mathrm{km\,h^{-1}}$ uniquely; two significant figures does not. Possible values can round to $49$ , $50$ or $51$ .

Check. At distance $120\,\mathrm{km}$ , times $120/49$ and $120/51$ hours both fit the time interval. Speeds of both $49$ and $51\,\mathrm{km\,h^{-1}}$ really are possible, despite the central calculation being fifty.

Change only the time accuracy: $2.40$ hours is now given to the nearest hundredth. The central calculation remains fifty, but its uncertainty changes.

Worked example 6: more accurate time, narrower speed interval

Question. Keep the distance from Example 5. Time is $2.40$ hours to the nearest hundredth. Find the speed bounds and decide whether the nearest whole kilometre per hour is unique.

Plan. Use the new time half-step and repeat the quotient calculation.

\begin{aligned} 2.395&\leq t<2.405 && \text{half-step }0.005 \\ \frac{119.5}{2.405}&<v<\frac{120.5}{2.395} && \text{pair opposite extremes} \\ 49.688149\ldots&<v<50.313152\ldots && \text{evaluate} \\ 49.5&<v<50.5 && \text{entire interval inside these thresholds} \end{aligned}

Why? Both boundaries lie within the rounding interval for fifty, so every allowed speed rounds to fifty.

Answer. Uniquely $50\,\mathrm{km\,h^{-1}}$ to the nearest whole number. One decimal place is not unique: $49.8$ and $50.2$ are both possible.

Check. The speed interval is narrower than Example 5's because the time interval is narrower.

The same quotient method handles density, mass per unit volume, and pressure, force per unit area.

Density and pressure

\rho=\frac mV,\qquad p=\frac FA

Use compatible units, positive volume $V$ and positive area $A$ . Here $m$ is mass and $F$ is force.

Not on the formula sheet: learn it.

For mass $295\leq m<305$ grams and volume $95\leq V<105$ cubic centimetres, density $\rho=m/V$ satisfies $\frac{295}{105}<\rho<\frac{305}{95}$ . The boundaries are approximately $2.809524$ and $3.210526$ in $\mathrm{g\,cm^{-3}}$ . Every value rounds to three to the nearest whole number; $2.9$ and $3.1$ show why a decimal place is not unique. These intervals arise from $300\,\mathrm{g}$ and $100\,\mathrm{cm}^3$ , each rounded to the nearest ten of its unit. For pressure, use the supplied relationship $p=F/A$ and compatible units in the same way.

Watch out

A long calculator display is not evidence of reliable digits. Calculate both boundaries and test the entire interval against the requested rounding thresholds.

Change only the reported time in Example 6 for the matched problem.

Quick Recap

Start with the correct input intervals; exact counts and conversion factors have no rounding uncertainty.
Choose extremes by how each input changes the output; check signs and endpoint inclusion.
For positive quotients, use lower over upper and upper over lower.
Reliable accuracy requires the whole result interval to round to the same value.

Yes, the opening journey could have average speed $51\,\mathrm{km\,h^{-1}}$ : its permitted distance and time intervals allow that speed.

Last Accessed

Papers 1, 2 and 3

N16 - Upper and lower bounds in calculations

What measurement bounds mean (Higher)

Choosing extremes for a calculation (Higher)

Bounds, compound measures and reliable rounding (Higher)

Quick Recap

Learn faster with Nova