Approximation and error · Approximation & error

Rounding

To round, find the last digit you’re keeping and look at the next digit: 5 or more rounds up, less than 5 leaves it.

Decimal places count digits after the point: $2.71828$ to 2 d.p. is $2.72$ .
Significant figures count from the first non-zero digit: $0.004826$ to 3 s.f. is $0.00483$ .
Nearest ten, hundred, …: $527.38$ to the nearest hundred is $500$ .

Which neighbour is nearer?Past the halfway mark, round up

Rounding on a number linePick a number and how to round it

0.004826the number0.00483to 3 significant figures

significant figures = 3

To 3 significant figures, the last digit kept is in the 0.00001s place. 0.004826 lies between 0.00482 and 0.00483, and the half-way point is 0.004825. It is at or past half-way, so it rounds up to 0.00483. The zeros after the point are not significant: counting starts at the first non-zero digit.

Every rounding is a choice between two neighbours. The number rounds to whichever is nearer, and the half-way point decides.

More: rounding

Round at the end

Keep full accuracy all the way through a calculation and round only the final answer. Rounding halfway through lets errors build up.

Worked example · WAEC 2019

WAEC 2019 · Paper 1 · Q4

Simplify, correct to three significant figures, $(27.63)^2 - (12.37)^2$ .

Use the difference of two squares
$(27.63 - 12.37)(27.63 + 12.37) = 15.26 \times 40$ .
Think first. $a^2 - b^2 = (a - b)(a + b)$
Multiply
$15.26 \times 40 = 610.4$ .
Round at the end
To three significant figures: $610$ . (D)

One exception: when a question says “reduce each number to two significant figures, then evaluate”, it wants an estimate, so round every number first, exactly as it says.

More: calculating to a given accuracy

Percentage error

When a measurement or estimate is wrong, the error is the difference between it and the true value, and

\text{percentage error} = \frac{\text{error}}{\text{actual value}} \times 100\%

Percentage errorCompare the error with the actual value

More: percentage error

Your turn

WAEC 2023 · Paper 1 · Q36

Approximation & error

A student measured the height of a pole as $5.98\text{ m}$ , which is less than the actual height. If the percentage error is $5\%$ , find, correct to two decimal places, the actual height of the pole.

Worked solution (try it first)

Let the actual height be $h$ .
The percentage error is taken of the actual height, so the error is $0.05h$ .
The measurement is too small, so $h - 0.05h = 5.98$ , that is $0.95h = 5.98$ .
Divide by 0.95: $h = 6.2947\ldots$ , which is 6.29 m to two decimal places, option C.

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