Approximation and error · Approximation & error

Most measurements and many answers can’t be exact, so they are rounded. This lesson covers how to round, when to round, and how big an error is.

Rounding

To round, find the last digit you’re keeping and look at the next digit. If it is 5 or more, round up. If it is less than 5, leave the kept digit as it is.

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 halfway 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
- ${27.63 - 12.37 = 15.26}$ and ${27.63 + 12.37 = 40}$ .
- So the expression is ${15.26 \times 40}$ .
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: a question may say “reduce each number to two significant figures, then evaluate”. That 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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