CourseCourseQ BankQuestion BankDocsDocuments

E1.10 Limits of accuracy

Syllabus
0580–2028–2029
Topic
E1.10
Level
Extended

Learning objectives

E1.10.1Give upper and lower bounds for data rounded to a specified accuracy.• Give upper and lower bounds for data rounded to a specified accuracy.E1.10.2Find upper and lower bounds of the results of calculations which have• Find upper and lower bounds of the results of calculations which have used data rounded to a specified accuracy. e.g. write down the upper bound of a length measured correct to the nearest metre. Example calculations include: • calculate the upper bound of the perimeter or the area of a rectangle given dimensions measured to the nearest centimetre • find the lower bound of the speed given rounded values of distance and time.

Turn a rounded value into an interval

A rounded measurement stands for a range of possible original values. Identify one unit at the stated accuracy, halve it, then subtract and add that half-unit to locate the two boundaries.

a−r2≤x<a+r2a-\frac{r}{2}\le x<a+\frac{r}{2}

Here aa is the reported value, rr is the rounding step and xx is the original value. For positive data rounded to the nearest value, the lower boundary is included but the upper boundary is excluded: the upper boundary would round to the next reported value.

Reported accuracy Step rr Half-step Possible original values
470 to the nearest 10 10 5 465≤x<475465\le x<475
6.2 to 1 decimal place 0.1 0.05 6.15≤x<6.256.15\le x<6.25
830 to 2 significant figures 10 5 825≤x<835825\le x<835

Keep the step in the same unit as the data before halving. A length of 3.6 m correct to the nearest 20 cm has step 0.20.2 m and half-step 0.10.1 m, so 3.5≤L<3.73.5\le L<3.7 m.

Bounds are exact boundary values, so do not round them again. Do not use the full rounding step on each side, and do not include the upper boundary with ≤\le.

Choose input bounds for an extreme result

To bound a result, first replace every rounded input by its interval. Then choose the endpoint combination that makes the required result as small or as large as possible; do not automatically choose all lower bounds or all upper bounds.

Positive-quantity calculation Smallest result uses Largest result uses
A+BA+B or A×BA\times B lower AA, lower BB upper AA, upper BB
A−BA-B lower AA, upper BB upper AA, lower BB
A÷BA\div B lower AA, upper BB upper AA, lower BB

A rectangle is reported as 8.4 cm by 5 cm, correct to the nearest 0.1 cm and nearest centimetre. Its area approaches its greatest value as both dimensions approach their upper boundaries, so the upper bound is 8.45×5.5=46.4758.45\times5.5=46.475 cm2^2; the actual area is less than this value.

For speed =distance÷time=\text{distance}\div\text{time}, the smallest speed uses the lower distance and upper time. If 84.6 km is correct to 0.1 km and 6 h is correct to the nearest hour, the lower bound is 84.55÷6.5=13.007…84.55\div6.5=13.007\ldots km/h.

Keep boundary values exact throughout and round only if the question requests a final degree of accuracy. State whether the result is a lower or upper bound and retain its units.

The table assumes positive quantities and expressions that change monotonically. With negative values, squares or a denominator whose interval crosses zero, analyse how the expression changes instead of applying the table mechanically.