The 5 Calculations Every IGCSE Physics Student Gets Wrong (CIE 0625)

I am a PhD physicist and I tutor CIE 0625 IGCSE Physics every single year, mostly with students at international schools in Milan and online around the world. After marking hundreds of practice papers, I can tell you something that surprises parents: the students who lose calculation marks are usually not the ones who do not understand physics. They understand it fine. They lose marks on small, repeatable mechanical mistakes that have nothing to do with how clever they are.

The good news is that this is the easiest kind of mark to win back. Conceptual gaps take weeks to close. Calculation habits can be fixed in an afternoon, because the same five mistakes appear again and again. Below are the five I see most often, each with a worked example showing the common wrong answer next to the correct method, plus the exact habit that fixes it.

One thing to keep in mind about CIE 0625 marking before we start: numerical answers in Paper 4 are usually worth more than one mark. There is typically a method mark (sometimes called an ECF or "error carried forward" mark) for the correct working and substitution, and a separate mark for the final answer with the correct unit. This matters enormously. It means that if your method is laid out clearly you can still earn marks even when your final number is wrong, and it means a perfect number with no unit, or the wrong unit, can still drop a mark. Almost every mistake below is really a fight over those method and unit marks.

This is the single most common calculation mistake I see, and it is also the most expensive because it can sink an otherwise perfect answer. A value is given with a prefix, the student types the number straight into the formula, and the prefix is silently ignored.

Why it happens: the prefix is part of the unit, not the number, so on the page "5.0 kV" looks like it is mostly about the digits. Under exam pressure the brain reads the 5.0 and drops the k.

Worked example.

A 2.0 microfarad capacitor is charged to 5.0 kV. Find the charge using Q = C V.

The common wrong answer:

Q = 2.0 x 5.0 = 10. The student writes 10 C, or maybe 10 microcoulombs, with no clear conversion. Both the number and the unit are wrong.

The correct method:

convert everything to base SI units first. C = 2.0 microfarad = 2.0 x 10^-6 F. V = 5.0 kV = 5.0 x 10^3 V. Then Q = C V = (2.0 x 10^-6) x (5.0 x 10^3) = 1.0 x 10^-2 C, which is 0.010 C.

The fix:

before you touch the formula, rewrite every given quantity in base SI units on a separate line. milli is 10^-3, micro is 10^-6, nano is 10^-9, kilo is 10^3, mega is 10^6. For length, 1 cm = 0.01 m and 1 mm = 0.001 m. Do this conversion as a visible step so the examiner sees it, then substitute. That one extra line is the difference between full marks and a careless zero.

Students are often taught "substitute then rearrange", and for simple cases it works. But the moment the unknown is on the bottom of a fraction or buried inside a product, substituting first turns a clean piece of algebra into a messy one, and that is where slips happen.

Why it happens: numbers feel safer than letters, so students rush to get rid of the symbols. The problem is that rearranging an equation full of digits is harder, not easier, and you cannot see the structure any more.

Worked example.

The resistance is R = 12 ohm and the power dissipated is P = 48 W. Find the current, given P = I^2 R.

The common wrong answer:

the student writes 48 = I^2 x 12, then divides wrong or forgets the square root entirely, ending up with I = 4 A (forgetting the root) or dividing by the wrong quantity. The structure I^2 gets lost in the digits.

The correct method:

rearrange in symbols first. From P = I^2 R we get I^2 = P / R, so I = square root of (P / R). Now substitute: I = square root of (48 / 12) = square root of 4 = 2.0 A. Clean, and the square root is impossible to forget because it is written before any number appears.

The fix:

rearrange the formula completely in letters, write the rearranged version on its own line, and only then put the numbers in. This also protects your method mark: the examiner can see the correct rearranged equation even if you mistype a digit on the calculator.

Two related habits cost marks here. The first is rounding partway through and then continuing the calculation with the rounded number, so the final answer drifts away from the true value. The second is giving a final answer to a silly number of figures, such as copying all eight digits off the calculator screen.

Why it happens: students round to "tidy up" as they go, not realising that errors compound. And nobody ever told them the simple rule the examiner uses.

Worked example.

A car travels 100 m in 7.0 s, then we want the kinetic energy related quantity, but take a cleaner case: average speed v = distance / time, then later square it. Distance = 100 m, time = 7.0 s.

The common wrong answer:

v = 100 / 7.0 = 14.28571429, rounded immediately to 14 m/s, then used as 14 in the next step. Or the opposite: the final answer written as 14.28571429 m/s, copied straight off the screen. The first introduces error; the second looks careless and can lose the significant figures mark.

The correct method:

carry the full unrounded value 14.2857... in your calculator (use the memory or "Ans" button) through every step, and round only at the very end. Then quote the final answer to a sensible number of significant figures: v = 14 m/s to 2 s.f., matching the 2 significant figures in the data (7.0 s).

The fix:

never write a rounded intermediate down and then retype it. Keep full precision in the calculator, round once at the end, and as a rule of thumb give your answer to the same number of significant figures as the data (usually 2 or 3 in 0625). Quoting 3 s.f. is almost always safe.

Energy and power questions are a minefield of mismatched units. The classic trap is putting time in minutes into P = E / t, which is built for seconds. The other classic is efficiency: students forget it is a ratio, multiply by 100 twice, or quote a value bigger than 1 (or bigger than 100 percent), which is physically impossible.

Why it happens: the question deliberately gives time in minutes or energy in kWh because that is realistic, and the student substitutes the friendly-looking number without converting to the SI base unit the formula expects.

Worked example.

A heater transfers 90 000 J of energy in 2.0 minutes. Find its power output, using P = E / t.

The common wrong answer:

P = 90 000 / 2.0 = 45 000 W. The time was left in minutes, so the answer is 60 times too big and the unit (watts requires seconds) is effectively wrong.

The correct method:

convert the time first: t = 2.0 minutes = 120 s. Then P = E / t = 90 000 / 120 = 750 W. Now the unit watts is honestly earned.

Efficiency check.

If that heater is supplied 100 000 J and usefully outputs 90 000 J, efficiency = useful output / total input = 90 000 / 100 000 = 0.90, which is 90 percent. It must be a value between 0 and 1 (or 0 and 100 percent). If you ever get an efficiency above 100 percent, stop, you have divided the wrong way round.

The fix:

for any P = E / t problem, force time into seconds. For efficiency, write it as the fraction useful / total, sanity check that it is below 1, and only then multiply by 100 if a percentage is asked for.

The fifth mistake is conceptual but it shows up as a number error. Students add forces or velocities as if direction did not matter, and in moments problems they multiply by the wrong distance.

Why it happens: on the page everything is just numbers, so two forces of 3 N and 4 N feel like they should give 7 N regardless of direction. And for moments, the perpendicular distance from the pivot is easy to confuse with whatever length the diagram happens to show.

Worked example A, resultant force.

Two forces act on a point: 3.0 N to the right and 4.0 N upward, at right angles. Find the resultant.

The common wrong answer:

3.0 + 4.0 = 7.0 N. The student added the magnitudes and ignored that they point in different directions.

The correct method:

for perpendicular vectors use Pythagoras: resultant = square root of (3.0^2 + 4.0^2) = square root of (9 + 16) = square root of 25 = 5.0 N. Only forces along the same line add directly (and opposite directions subtract).

Worked example B, moments.

A 6.0 N force is applied at the end of a spanner, and the spanner is 0.20 m long but held at 30 degrees to the horizontal. Take the simple straight case: force 6.0 N, perpendicular distance from pivot 0.20 m. Moment = force x perpendicular distance = 6.0 x 0.20 = 1.2 N m. The error to avoid is using a slanted distance instead of the perpendicular one, or forgetting the unit is newton metre, not newton.

The fix:

ask "do these quantities have direction?" before combining them. Same line means add or subtract; right angles mean Pythagoras. For moments, always identify the perpendicular distance to the line of action, and write the unit as N m.

Here is the whole thing in one table. Print it, stick it above your desk, and glance at it before every practice paper. After a few weeks the right method becomes automatic and you stop bleeding marks on the easy parts of long questions.

None of these fixes require new physics. They are habits, and habits are built by repetition. Run through this checklist on every calculation in your next three practice papers and the improvement is usually visible within a fortnight. The marks you save here are the cheapest marks in the whole exam.