Estimation and Rounding Numbers Worked ExamplesIGCSE Maths 0580

(a) The 3rd decimal place is $1$; the next digit is $5$, so round up.

(b) Leading zeros are not significant. The 1st significant figure is $4$, the 2nd is $9$; the next digit is $8$, so round up.

(c) The 3rd significant figure is $7$; the next digit is $3$, so round down. Keep place value with zeros.

Round every number to $1$ significant figure and write each rounded value down — these carry the method mark.

Evaluate the root and the power.

Substitute.

Dividing by $0.02$ is multiplying by $50$.

For a square root, round to a nearby value whose root is exact rather than blindly to 1 s.f. — $80.6 \to 81$ because $\sqrt{81} = 9$.

Round the other value to 1 significant figure.

Add.

(a) Take each rounding in turn. $61.4 \to 60$ rounds down, which makes the numerator smaller.

$3.8 \to 4$ rounds up, which makes the numerator larger. These two partly cancel.

$0.49 \to 0.5$ rounds up, and a larger denominator makes the whole fraction smaller.

Numerator: $60 \times 4 = 240$ against the true $61.4 \times 3.8 = 233.32$, so the numerator was rounded up overall. That effect outweighs the denominator, so the estimate is larger than the true value.

(b) Exact value.

Percentage error uses the true value as the denominator.

A ruler marked in millimetres can measure to the nearest $1\text{ mm} = 0.001\text{ m}$, so at most $3$ decimal places in metres.

The fourth decimal place ($7$, representing $0.1\text{ mm}$) claims precision the instrument cannot provide.

Round to the millimetre.

Round each value to the nearest whole number — enough precision for an estimate.

Add the rounded values.

Divide by how many values there are.

(a) Round each length to $1$ significant figure.

Estimate the area of the rectangle.

The triangle's base is the $21.4\text{ m}$ side it shares with the rectangle, and its perpendicular height is $9.8\text{ m}$.

Add the two parts.

(b) Round $52$ to $1$ significant figure and divide.

(a) Subtract the window from the whole wall.

(b) Work in the same units. $30\text{ cm} = 0.3\text{ m}$.

Divide the wall area by the tile area.

A part-tile is still needed to finish the wall, so round up.

(c) Packs also round up — $9$ packs hold only $108$ tiles.

Cost of $10$ packs, written to the nearest cent: $239.50.

(d) Test Anya's answer: $117$ tiles cover less than the wall, so part of the wall would be left bare.

(a) The two semicircles together make one full circle of diameter $95.3\text{ m}$, so one lap is two straights plus one circumference.

Round $52.8 \to 50$, $95.3 \to 100$ and $\pi \to 3$.

(b) Ten laps is about $4000\text{ m} = 4\text{ km}$. The time is about $30$ minutes, which is $0.5$ hours.

(c) Now use a calculator for the exact lap, keeping all the digits.

Divide the race distance by one lap and keep the whole number of laps.

Subtract twelve full laps from the race distance.

Round each number to $1$ significant figure. Look at the second significant figure to decide: $19.6$ rounds up to $20$, and in $0.0987$ the first significant figure is $9$, followed by $8$, so it rounds up to $0.1$.

Substitute the rounded values and simplify the fraction under the root.

The rounded values were chosen by the rule, and here they lead to a perfect square.