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The trap question field guide

A trap is not a secret topic. It is a familiar method with one fragile condition. Before calculating, write three things: the requested quantity, its unit, and the constraint. That ten-second habit catches most of the damage below.

Download the PDFChecked August 2026About 9 minutes to read

Unit conversions: the exponent travels too

Prefixes are easy when a quantity is linear. They become dangerous when the converted length is squared or cubed. Convert the measurement first, then apply the geometry.

Worked example Current density in a thin wire

A wire carries a current of 2.4,mathrmA2.4\\,\\mathrm{A} and has diameter 0.40,mathrmmm0.40\\,\\mathrm{mm}. Estimate the current density J=I/AJ=I/A.

The radius is half the diameter, expressed in metres:

r=0.20mm=2.0×104m.r=0.20\,\mathrm{mm}=2.0\times10^{-4}\,\mathrm{m}.

Hence

A=πr23.14(2.0×104)2=1.26×107m2,A=\pi r^2\approx3.14(2.0\times10^{-4})^2=1.26\times10^{-7}\,\mathrm{m^2},

so

J=2.41.26×1071.9×107Am2.J=\frac{2.4}{1.26\times10^{-7}}\approx\boxed{1.9\times10^7\,\mathrm{A\,m^{-2}}}.

Correct: the factor 10310^{-3} was squared, and the diameter was halved before the area was found.

Trap: 1.9×1041.9\times10^4 treats square millimetres as though the conversion were linear. Using 0.40mm0.40\,\mathrm{mm} as the radius creates a separate factor-of-four error.

Boundary conditions: test the equality case

Words such as positive, distinct, inside, at least and no real roots decide whether an endpoint belongs. Solve the inequality, then put the boundary back into the original sentence.

Worked example Two distinct positive roots

For which real values of mm does

x2(m+1)x+m=0x^2-(m+1)x+m=0

have two distinct positive roots? Factorise before reaching for the discriminant:

x2(m+1)x+m=(x1)(xm).x^2-(m+1)x+m=(x-1)(x-m).

The roots are 11 and mm. Both are positive when m>0m>0, but they are distinct only when m1m\ne1. Therefore

m>0 and m1.\boxed{m>0\text{ and }m\ne1}.

Correct: m=0m=0 is excluded because zero is not positive; m=1m=1 is excluded because the roots coincide.

Trap: m0m\geq0 checks neither the strict word “positive” nor the word “distinct”.

Subtle wording: answer the quantity asked for

The arithmetic can be flawless and still answer the wrong question. Percentage increase is not the final percentage. Momentum is not kinetic energy. A statement about one fixed variable does not survive if that variable changes.

Worked example A ten per cent change that becomes twenty-one

A particle's momentum increases by 10% while its mass stays constant. By what percentage does its kinetic energy increase?

Use K=p2/(2m)K=p^2/(2m). If p=1.10pp'=1.10p and mm is fixed, then

KK=(1.10p)2/(2m)p2/(2m)=(1.10)2=1.21.\frac{K'}K=\frac{(1.10p)^2/(2m)}{p^2/(2m)}=(1.10)^2=1.21.

The final kinetic energy is 121% of the original, so the increase is

21%.\boxed{21\%}.

Correct: translate the wording into a multiplier first, then convert the final multiplier back into a change.

Trap: 121% answers “what percentage of the original?”; 10% assumes kinetic energy is proportional to momentum.

Sign traps: predict the direction before the algebra

A minus sign can mean a negative charge, an opposite direction, a fall in a scalar, or simply the coordinate convention you chose. Give each sign a job before multiplying them.

Worked example An electron crosses a potential difference

An electron moves from a point at 120V120\,\mathrm{V} to one at 20V20\,\mathrm{V}. Only the electric force does work. What happens to its kinetic energy?

ΔV=20120=100V.\Delta V=20-120=-100\,\mathrm{V}.

For an electron q=eq=-e, so

ΔU=qΔV=(e)(100V)=+100eV.\Delta U=q\Delta V=(-e)(-100\,\mathrm{V})=+100\,\mathrm{eV}.

Conservation of energy gives ΔK=ΔU\Delta K=-\Delta U, hence

ΔK=100eV.\boxed{\Delta K=-100\,\mathrm{eV}}.

Its kinetic energy decreases by 100eV100\,\mathrm{eV}.

Correct: the negative charge reverses the usual positive-charge intuition. The energy check confirms the direction.

Trap: saying “potential fell, so kinetic energy rose” forgets that electric potential and electric potential energy are different quantities.

The 12-second trap scan

Before line one of working, run A-U-B-S:

  1. Ask: underline exactly what the question wants.
  2. Units: convert prefixes before powers, areas or volumes.
  3. Boundary: test equality, zero, domain and limiting cases.
  4. Sign: predict direction or change before calculation.

Do not hunt for tricks in every sentence. Most questions are direct. The scan exists so that a direct method does not become a wrong option through one silent assumption.

Sources

UAT-UK ESAT Content Specification: the assessed mathematics and science content, including compound units, inequalities, percentage change, momentum, energy and electrical quantities.

October 2025 post-sitting synthesis source: candidate reports cited in this guide repeatedly mention boundary conditions, unit conversions and subtle option wording. These are qualitative experience signals, not a representative survey.

All examples in this guide are original. UniGenius is independent and is not affiliated with UAT-UK or Pearson VUE.

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