Why Rounding Too Early Costs You Marks

One question, worked twice

Take the kind of two-stage question that appears on every Higher and Extended paper. The sine rule gives a length in the first triangle. That length is then a side of a second triangle, and the cosine rule gives an angle.

Stage one: a side of 7.3 cm sits opposite an angle of 41°, and the side you want, b, sits opposite an angle of 88°. The sine rule gives b = 7.3 × sin 88° ÷ sin 41° = 11.1202... cm.

Stage two: that side b is opposite the angle you want in a second triangle whose other two sides are 9.6 cm and 6.4 cm. Rearranging the cosine rule gives cos C = (9.6² + 6.4² - b²) ÷ (2 × 9.6 × 6.4). Now run it three times, with b carried at three different accuracies.

The middle line is the one to look at. Rounding to 3 significant figures is not carelessness - it is rounding to exactly the accuracy the final answer is being asked for. It still moved the answer by 0.2°, which is outside any tolerance, and the two-significant-figure version is out by more than a whole degree.

• Keeping b = 11.1202... → C = 85.58...° → 85.6° to 3 significant figures
• Rounding b to 11.1 first → C = 85.37...° → 85.4°
• Rounding b to 11 first → C = 84.34...° → 84.3°

Why a small error does not stay small

Rounding 11.1202 to 11.1 is an error of about 0.02, which is roughly two parts in a thousand. In an addition that would be harmless. What happens here is worse, and it is worth seeing exactly where.

First the value is squared, which roughly doubles the relative error: b² moves from 123.66 to 123.21. Then comes the damaging step. That figure is subtracted from 9.6² + 6.4² = 133.12, and the two numbers are close together, so the difference is small - about 9.46. An absolute error of 0.45 in a quantity of 123 is invisible; the same 0.45 in a quantity of 9.46 is nearly five per cent. The subtraction has not created error, it has magnified the share the error represents.

That is the general pattern and it is worth learning as a warning sign. Subtracting two numbers of similar size is the operation that punishes early rounding hardest, and the cosine rule for an angle does exactly that every time. Multiplying and dividing carry relative error through roughly unchanged. Adding and subtracting numbers of very different sizes are harmless. The trouble appears when a calculation strings several stages together, which is what a multi-stage question is.

What to do instead, on the calculator you already have

The written record and the stored value are two different things, and students routinely assume that writing 11.1 obliges them to use 11.1. It does not. Write the fuller value on the page so the method can be followed, keep the exact value inside the machine, and round once at the end.

Every scientific calculator permitted in GCSE and IGCSE has an answer key and at least one memory. Finding out where they are is a five-minute job, and it protects more marks per minute than almost any other five minutes of revision.

• Do not write an intermediate value down and then retype it. Use the answer key, or store it in a memory.
• Where a value is needed more than once, store it once rather than retyping it - retyping is also where transcription errors get in.
• Write the intermediate value on the page to more figures than the final answer needs, so your method is visible.
• Round once, at the very end, to the accuracy the question asked for.

How the mark is actually lost

Method marks are usually safe. Working that shows the right sine rule and the right cosine rule has earned those marks whether or not the arithmetic finishes in the right place. What goes is the accuracy mark, which is awarded for the value alone and is often the last mark on a long question.

There is a second cost that never appears in a mark scheme. A student who rounds early and arrives at 85.4 has no way of noticing anything is wrong, because the answer looks entirely sensible. Errors that produce absurd answers get caught by the person making them; errors that produce plausible ones do not.

And in a question with parts that follow on, an early-rounded answer becomes the input to the next part, where the same magnification happens again to a value that was already wrong.