Upper and lower boundsIB MYP Maths Extended: Revision notes

A measurement that has been rounded could have been any value in a small range. The lower bound is the smallest value that rounds to it and the upper bound is the largest (the point where it would round up). To find them, take half the rounding unit either side. A length of $48$ m to the nearest metre has $47.5\le L<48.5$, so lower bound $47.5$ and upper bound $48.5$. A time of $12.4$ s to the nearest $0.1$ s has bounds $12.35$ and $12.45$. A mass of $450$ g to the nearest $10$ g has bounds $445$ and $455$. In calculations the upper bound is used as $48.5$, even though $48.5$ itself rounds up.

For a sum $a+b$, the largest result uses both upper bounds and the smallest uses both lower bounds: $(a+b)_{\max}=a_{\max}+b_{\max},\qquad(a+b)_{\min}=a_{\min}+b_{\min}.$ For a difference $a-b$ the largest result uses the upper bound of $a$ and the lower bound of $b$: $(a-b)_{\max}=a_{\max}-b_{\min},\qquad(a-b)_{\min}=a_{\min}-b_{\max}.$ Example: perimeter of the plot, upper bound $=2(48.5+35.5)=168$ m.

For a product $a\times b$ (all values positive) use two upper bounds for the maximum and two lower bounds for the minimum. Example: lower bound of area $=47.5\times34.5=1638.75$ m$^2$. For a quotient $\frac ab$ the maximum is $\frac{a_{\max}}{b_{\min}}$ and the minimum is $\frac{a_{\min}}{b_{\max}}$, because dividing by a smaller number gives a bigger answer. Example: speed $=\frac{\text{distance}}{\text{time}}$ has upper bound $\frac{100.5}{12.35}=8.14$ m/s and lower bound $\frac{99.5}{12.45}=7.99$ m/s.

After finding both bounds of an answer, round them to the same number of significant figures. If they match, that is the accuracy you can give. If they differ, there are too many figures. Example: bounds $3.82$ kg and $3.98$ kg give $3.8$ and $4.0$ to 2 s.f. (different), but both give $4$ to 1 s.f. So the mass is $4$ kg to 1 s.f. Bounds of $8.46$ and $8.54$ give $8.5$ to 2 s.f., so the answer is $8.5$.

A block has mass $540$ g (nearest $10$ g) and volume $116.3$ to $123.7$ cm$^3$ from its measured sides. Density $=\frac{\text{mass}}{\text{volume}}$. Upper bound $=\frac{545}{116.34}=4.68$ g/cm$^3$ and lower bound $=\frac{535}{123.74}=4.32$ g/cm$^3$. If the block is claimed to be titanium (density $4.51$ g/cm$^3$), that lies between the bounds, so the claim is possible. Steel at $7.8$ g/cm$^3$ lies outside, so it is impossible.

Bounds let you decide whether a claim can be trusted. If a value lies outside the range between the bounds, the claim is impossible. If it lies inside, the claim is only possible, not proven. In real problems such as limits for luggage, speed checks or materials, use the bound that makes the claim hardest to meet: the upper bound of a mass against a limit, for example. State your conclusion in words and give units.