HookThe units mix-up that destroyed a Mars orbiter
On 23 September 1999, after a nine-month cruise, NASA's Mars Climate Orbiter fired its engine to slip into orbit around Mars — and was never heard from again. The inquiry found a cause almost too simple to believe: one team, at Lockheed Martin, had written the thruster software in imperial units (pound-force seconds), while NASA's navigation software read the numbers as metric (newton-seconds). Since one pound-force is about \(4.45\) newtons, every instruction was out by a factor of roughly \(4.45\), the orbiter came in far too low, and it broke up in the atmosphere. The orbiter alone cost around 125 million dollars, lost because two numbers meant different things.
That disaster is this whole section in miniature: a measurement is meaningless without its unit, a rounded figure is not the true value, and errors do not politely stay put — they combine and grow. Higher GCSE turns each of those into a skill: converting between standard and compound units, estimating to sanity-check an answer, stating the interval a rounded measurement really lies in, and — the flagship Higher technique — carrying upper and lower bounds through a calculation to find the range the true answer must occupy. Get these right and you never trust a wrong answer; get them wrong and, like the orbiter, you can be perfectly precise and completely lost.
ModelStandard and compound units
Standard units come in families linked by powers of ten — \(1\) km \(= 1000\) m, \(1\) m \(= 100\) cm, \(1\) kg \(= 1000\) g, \(1\) litre \(= 1000\) ml — so converting between them is multiplying or dividing by the right power of ten, with the decimal point doing the moving. Compound units combine two measures: speed in metres per second (\(\text{m/s}\)), density in grams per cubic centimetre (\(\text{g/cm}^3\)), pressure in newtons per square metre, unit prices in pence per gram. Converting these needs care, because both parts change: turning \(\text{m/s}\) into \(\text{km/h}\) means dealing with the metres and the seconds at once.
The reliable method is to convert one quantity at a time and keep the units written down as you go — the units themselves tell you whether to multiply or divide. This is precisely the discipline the Mars orbiter's software skipped: a number with no unit attached is an accident waiting to happen.
Convert \(90\) km/h into m/s. In one hour a \(90\) km/h vehicle travels \(90\) km \(= 90000\) m, and one hour is \(3600\) seconds, so the speed is \(\frac{90000}{3600} = 25\) m/s. (The shortcut '\(\div 3.6\)' is exactly this, since \(\frac{1000}{3600} = \frac{1}{3.6}\).) Now a best-value comparison, which is really a compound-unit question: \(750\) g of pasta for £2.10 costs \(\frac{210}{750} = 0.28\) pence per gram, while \(1.2\) kg for £3.48 costs \(\frac{348}{1200} = 0.29\) pence per gram — so, surprisingly, the smaller bag is the better deal per gram.
MechanismEstimation — the answer's smell test
Estimation is not guessing; it is a deliberate calculation with every number rounded to \(1\) significant figure, done to check that a real answer is the right size. Rounding \(28.6\) to \(30\), \(4.9\) to \(5\) and \(0.196\) to \(0.2\) turns an ugly sum into one you can do in your head, and the estimate it gives should land in the same ballpark as the calculator's answer. If they disagree wildly, you have mistyped something — a misplaced decimal point or a dropped digit — and the estimate has just saved the marks. AQA now asks this even of 'answers obtained using technology'.
The move that trips people is dividing by a number less than \(1\): it makes the answer bigger, not smaller, because you are asking how many small pieces fit inside. Estimation questions on the calculator papers exist largely to test whether you believe that.
Estimate \(\frac{28.6 \times 4.9}{0.196}\). Round each number to \(1\) significant figure: \(\frac{30 \times 5}{0.2} = \frac{150}{0.2} = 750\). The true value is about \(715\), so the estimate is close — and, crucially, both are near \(700\)–\(800\), not \(70\) or \(7000\). Had a calculator shown \(71.5\) because \(0.196\) was keyed as \(1.96\), the estimate of \(750\) would expose the slip instantly. Notice how dividing by \(0.2\) multiplied by five: \(150 \div 0.2 = 750\), the same as \(150 \times 5\).
ModelRounding and error intervals
Rounding replaces a number with a nearby, simpler one to a stated accuracy — a number of decimal places, or of significant figures (the digits that carry weight, counting from the first non-zero one). But a rounded measurement is a range in disguise. A length recorded as \(4.7\) cm 'to \(1\) decimal place' is really anything that rounds to \(4.7\): from \(4.65\) cm up to, but not including, \(4.75\) cm. In inequality notation, \(4.65 \le L \lt 4.75\) — and the \(\le\) on the bottom with the \(\lt\) on top is deliberate, because \(4.75\) would itself round up to \(4.8\).
Truncation is the blunter cousin: chopping a number off rather than rounding it. A value truncated to \(4.7\) could be anything from \(4.7\) up to \(4.8\) — that is \(4.7 \le x \lt 4.8\) — because truncation only ever throws digits away, never rounds up. Reading which process produced a figure decides which interval you write.
A crowd is reported as \(37000\) 'to the nearest thousand'. Write the error interval. The halfway points on either side are \(36500\) and \(37500\), so the true figure \(n\) satisfies \(36500 \le n \lt 37500\): as few as \(36500\) people (which rounds up) or as many as \(37499\), but not \(37500\) (which rounds to \(38000\)). Contrast a value given as \(3.6\) after truncation to \(1\) decimal place: there the interval is \(3.6 \le x \lt 3.7\), since \(3.699\ldots\) truncates to \(3.6\) but \(3.7\) does not.
DataUpper and lower bounds in calculations
The Higher tier's demand is to carry those intervals through a calculation. Each measurement has a lower bound (LB) and an upper bound (UB), and the true answer lies between bounds you build from them — the subtlety is which bound of each input to use. For a sum, the biggest total uses both upper bounds. For a difference, the biggest result uses the upper bound of the first and the lower bound of the second, \(\text{UB} - \text{LB}\). For a quotient \(\frac{a}{b}\), the biggest result comes from the largest top over the smallest bottom, \(\frac{\text{UB}_a}{\text{LB}_b}\), and the smallest from \(\frac{\text{LB}_a}{\text{UB}_b}\).
The payoff is a genuine, examinable conclusion: if the upper and lower bounds of your answer agree when rounded to some accuracy, you may quote the answer to that accuracy and defend it. If they do not, you have found the honest limit of what your measurements can tell you — which is exactly the judgement the Mars mission needed and lacked.
A sprinter runs \(100\) m, measured to the nearest metre, in \(12.5\) s, measured to the nearest \(0.1\) s. Find the bounds for her average speed. The distance lies in \(99.5 \le d \lt 100.5\) and the time in \(12.45 \le t \lt 12.55\). Speed is \(\frac{d}{t}\), so the maximum speed uses the largest distance over the smallest time: \(\frac{100.5}{12.45} = 8.072\ldots\) m/s. The minimum uses the smallest distance over the largest time: \(\frac{99.5}{12.55} = 7.928\ldots\) m/s. Both round to \(8\) m/s to \(1\) significant figure, so you may safely state the speed as \(8\) m/s — but no more precisely, because the second figure is already uncertain.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
State units on every line, and never quote a compound measure without them — 'speed \(= 25\)' is worth less than 'speed \(= 25\) m/s', and on a conversion the units are what tell you whether to multiply or divide. When a question says 'estimate', it means round each number to \(1\) significant figure and show that rounding; a precise answer to an 'estimate' question can actually lose the mark, because it proves you did not estimate.
Error intervals live or die on the choice of \(\le\) versus \(\lt\): write the lower bound with \(\le\) and the upper bound with \(\lt\), and take each bound as the exact halfway value for rounding — \(36500\), not \(36499\). For bounds calculations, write down LB and UB for every measurement first, then reason about which to use: the maximum of a division is largest-over-smallest, the minimum is smallest-over-largest, and a subtraction flips the second bound.
The final flourish AQA rewards is the conclusion — 'both bounds round to \(8\) m/s, so the answer is \(8\) m/s to \(1\) s.f.' — because that sentence shows you understand what a bound is for, not merely how to compute one. Marks in this section are lost far more often to a missing unit or the wrong inequality sign than to the arithmetic itself.