Limits of Accuracy Worked ExamplesIGCSE Maths 0580

Find the size of the rounding interval. To $1$ decimal place the interval is $0.1$, so the bounds sit half of that either side.

Subtract for the lower bound.

Add for the upper bound.

The lower bound is attainable ($27.35$ rounds up to $27.4$); the upper bound is not ($27.45$ rounds to $27.5$). So the inequality is strict on the right only.

Write the bounds for each side — half of $0.1$ either way.

For a product, the largest result comes from multiplying the two upper bounds.

The smallest result comes from multiplying the two lower bounds.

Write the bounds.

For a difference, the result is largest when you subtract as little as possible — so take the upper bound of $a$ and the lower bound of $b$.

Evaluate.

Bounds for distance: half of $10$ is $5$. Bounds for time: half of $1$ is $0.5$.

Speed $= \frac{d}{t}$. For a quotient, the result is largest when the numerator is largest and the denominator smallest.

Evaluate.

Round to $3$ significant figures as instructed.

(a) Half of $100$ is $50$.

(b) The first is within $\pm 50$ of $4000$; the second is within $\pm 500$ of $3000$.

The first interval spans $100$ people, the second spans $1000$ — ten times wider.

Bounds for one tile.

Multiply each bound by $20$ — repeated addition of the same measurement.

Compare the bounds digit by digit: they agree on the leading $30$, then differ ($303$ vs $305$).

Only the digits both bounds agree on can be trusted, so quote the length to $2$ significant figures.

(a) A filled circle means the value is included ($\leqslant$); an open circle means it is not ($<$).

(b) The rounded value is the midpoint of the interval, and the accuracy is the full width of the interval.

So Diagram 1 shows $6.5\text{ kg}$, correct to $1$ decimal place. Repeat for Diagram 2.

Diagram 2 shows $300\text{ m}$, correct to the nearest $100\text{ m}$ (equivalently, to $1$ significant figure).

(c) A mass of exactly $6.55\text{ kg}$ rounds up to $6.6\text{ kg}$, not $6.5\text{ kg}$. So $m$ cannot equal $6.55$, and the right-hand sign must be the strict $<$.

Write the bounds of each length — half of $0.1$ either side.

(a) Area $= \frac{1}{2}(a + b)h$. Every length is added or multiplied, so the largest area uses all three upper bounds.

(b) The smallest area uses all three lower bounds.

(c) One box covers between $24.5\text{ m}^2$ and $25.5\text{ m}^2$. To be certain, plan for the worst case: the largest lawn and the least coverage per box.

So $2$ boxes might not be enough. $3$ boxes cover at least $3 \times 24.5 = 73.5\text{ m}^2$, which is more than the largest possible area.

The nearest millimetre is the nearest $0.1\text{ cm}$, so each bound is $0.05\text{ cm}$ either side.

(a) By Pythagoras, $AB^2 = AC^2 - BC^2$. This is a subtraction, so $AB$ is largest with the upper bound of $AC$ and the lower bound of $BC$.

(b) $\sin x = \frac{BC}{AC}$. Between $0^{\circ}$ and $90^{\circ}$ a larger sine means a larger angle, so the upper bound of $x$ uses the largest $BC$ over the smallest $AC$.

The lower bound uses the smallest $BC$ over the largest $AC$.

(c) The true angle could be anywhere from about $35.18^{\circ}$ to $36.02^{\circ}$. To $1$ decimal place these round to $35.2^{\circ}$ and $36.0^{\circ}$, and to the nearest degree they round to $35^{\circ}$ and $36^{\circ}$. The bounds do not even agree to the nearest degree, so the first decimal place in $35.6^{\circ}$ cannot be trusted.

Write down the bounds of each value. 'Nearest $10\text{ m}^3$' means $\pm 5$; 'nearest metre' means $\pm 0.5$; '$1$ decimal place' means $\pm 0.05$.

Depth $= \dfrac{\text{volume}}{\text{surface area}}$. A fraction is smallest when its numerator is smallest and its denominator is largest, so use the lower bound of the volume with the upper bounds of both lengths.

Evaluate. The value does not terminate, so give it to $3$ significant figures.

(a) 'Nearest millimetre' means $\pm 0.05\text{ cm}$, so the diameter is at least $2.35\text{ cm}$. The radius is half the diameter.

(b) Density $= \dfrac{\text{mass}}{\text{volume}}$. Its greatest value uses the upper bound of the mass and the lower bound of the volume, which comes from the smallest radius.

Find the smallest volume, keeping the full calculator value.

Divide, then round to $3$ significant figures only at the end.