The Hardest GCSE Maths Questions

What this is testing. Whether you can turn a statement about two ratios into an equation, and whether "exact form" makes you reach for the quadratic formula rather than a calculator.

1. Equal ratios are equal fractions, so $\frac{x - 1}{x + 2} = \frac{x + 2}{2x + 7}$, which cross-multiplies to $(x - 1)(2x + 7) = (x + 2)^2$.
   M1 - forming an equation from the two ratios
2. Expand both sides: $2x^2 + 5x - 7 = x^2 + 4x + 4$.
   M1 - expanding both brackets
3. Collect everything on one side: $x^2 + x - 11 = 0$.
   A1 - correct quadratic
4. Nothing multiplies to $-11$ and adds to $1$, so use the formula with $a = 1$, $b = 1$, $c = -11$: $x = \frac{-1 \pm \sqrt{1 + 44}}{2}$.
   M1 - quadratic formula with correct substitution
5. $\sqrt{45} = 3\sqrt{5}$, so $x = \frac{-1 \pm 3\sqrt{5}}{2}$.
   A1 - both solutions, surd simplified

Answer. $x = \frac{-1 + 3\sqrt{5}}{2}$ or $x = \frac{-1 - 3\sqrt{5}}{2}$

Where students lose marks. Two places. Expanding $(x + 2)^2$ as $x^2 + 4$ kills the question at the second line. And "exact form" means the surd stays: leaving the answer as $\frac{-1 \pm \sqrt{45}}{2}$, or rounding to $2.85$, drops the final mark even though the working is right.

What this is testing. Whether you can pair bounds to make a quantity as large or as small as possible, and whether you can then say what that pair of numbers licenses you to claim.

1. Distance: $35.95 \le d < 36.05$.
   B1 - bounds for the distance
2. Time: $7.15 \le t < 7.25$.
   B1 - bounds for the time
3. Speed is largest when the distance is largest and the time is smallest: $36.05 \div 7.15 = 5.0419...$ m/s.
   M1 - upper bound, correctly paired
4. Speed is smallest the other way round: $35.95 \div 7.25 = 4.9586...$ m/s.
   A1 - lower bound, correctly paired
5. Both bounds round to $5.0$ to 2 significant figures but differ in the third, so 2 significant figures is the greatest accuracy the data supports.
   C1 - comparison and conclusion

Answer. $5.0$ m/s, because the upper and lower bounds agree to 2 significant figures and no further

Where students lose marks. Pairing the bounds the wrong way (largest distance with largest time) gives two numbers that are neither bound, and the whole method mark goes. The other loss is arithmetic without argument: a student who calculates both bounds correctly and then writes "5.0 m/s" with no comparison sentence has not answered "you must justify your answer".

What this is testing. Whether you spot that rationalising three ugly fractions makes them cancel each other. "Show that" means the printed answer is not the goal, the route to it is.

1. Multiply each fraction by the conjugate of its own denominator over itself. The denominators then become differences of two squares.
   M1 - conjugate method on at least one term
2. $\frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$
   A1 - first term
3. $\frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}$
   A1 - second term
4. $\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$
   A1 - third term
5. Add them: $(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (2 - \sqrt{3})$. The $\sqrt{2}$ terms cancel, the $\sqrt{3}$ terms cancel, and $-1 + 2 = 1$ is left, as required.
   A1 - complete cancellation and conclusion

Answer. The three terms telescope to $-1 + 2 = 1$

Where students lose marks. Reaching for a common denominator of $(\sqrt{2} + 1)(\sqrt{3} + \sqrt{2})(2 + \sqrt{3})$ is legal and almost never finishes inside the time. The other loss is stopping at three rationalised terms without adding them: a "show that" is only complete when the last line is the printed result.

What this is testing. Substitution into a quadratic, and whether you remember that a line cutting a circle gives two points, so each $x$ needs its own $y$.

1. Substitute $y = x + 1$ into the circle: $x^2 + (x + 1)^2 = 25$.
   M1 - substitution
2. Expand the bracket: $x^2 + x^2 + 2x + 1 = 25$, so $2x^2 + 2x + 1 = 25$.
   A1 - correct expansion
3. Rearrange and divide by 2: $2x^2 + 2x - 24 = 0$, so $x^2 + x - 12 = 0$.
   A1 - correct quadratic
4. Factorise: $(x + 4)(x - 3) = 0$, so $x = -4$ or $x = 3$.
   M1 - solving the quadratic
5. Put each $x$ back into $y = x + 1$: $x = 3$ gives $y = 4$, and $x = -4$ gives $y = -3$.
   A1 - both pairs, correctly matched

Answer. $x = 3$, $y = 4$ and $x = -4$, $y = -3$

Where students lose marks. Expanding $(x + 1)^2$ as $x^2 + 1$ is the classic, and it produces a quadratic that still factorises, so nothing looks wrong. The second loss is presentational: listing $x = 3$, $x = -4$, $y = 4$, $y = -3$ as four numbers instead of two pairs. The mark is for the pairing.

What this is testing. Whether you can see one base hiding inside four different numbers. Once everything is a power of 3 the question is a one-line linear equation.

1. Write every term as a power of 3: $27 = 3^3$, $9 = 3^2$ and $\sqrt{3} = 3^{\frac{1}{2}}$.
   M1 - common base
2. Left side: $3^{6x} \div 3^{2x - 2} = 3^{6x - (2x - 2)} = 3^{4x + 2}$.
   A1 - left side simplified
3. Right side: $3^{x + 5} \times 3^{\frac{1}{2}} = 3^{x + \frac{11}{2}}$.
   A1 - right side simplified
4. Equal powers of the same base means equal indices: $4x + 2 = x + \frac{11}{2}$, so $3x = \frac{7}{2}$ and $x = \frac{7}{6}$.
   A1 - solving the linear equation

Answer. $x = \frac{7}{6}$

Where students lose marks. Subtracting the index of a negative term: $6x - (2x - 2)$ becomes $4x - 2$ for a lot of students, and the bracket is the whole difficulty. The other is dropping $\sqrt{3}$ because it looks decorative. It contributes the $\frac{1}{2}$ that makes the answer a fraction rather than a whole number.

What this is testing. Clearing fractions without losing a bracket, then recognising a quadratic that will not factorise.

1. Multiply every term by $(x + 1)(x - 2)$.
   M1 - clearing the denominators
2. This gives $3(x - 2) + 2(x + 1) = (x + 1)(x - 2)$.
   A1 - correct equation
3. Expand: $3x - 6 + 2x + 2 = x^2 - x - 2$, so $5x - 4 = x^2 - x - 2$.
   A1 - both sides expanded
4. Rearrange: $x^2 - 6x + 2 = 0$.
   A1 - correct quadratic
5. Formula: $x = \frac{6 \pm \sqrt{36 - 8}}{2} = \frac{6 \pm 2\sqrt{7}}{2} = 3 \pm \sqrt{7}$.
   A1 - both exact solutions, simplified

Answer. $x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$

Where students lose marks. Multiplying the right-hand side by only one bracket, so the $1$ becomes $(x + 1)$ rather than $(x + 1)(x - 2)$. And at the end, $\frac{6 \pm \sqrt{28}}{2}$ is exact but not simplified: $\sqrt{28} = 2\sqrt{7}$, and the $2$s then cancel.

What this is testing. Whether you count both orders, and whether you check every root of the quadratic instead of assuming the tidier one is the answer.

1. There are $n + 3$ counters. Red then blue has probability $\frac{3}{n + 3} \times \frac{n}{n + 2}$.
   M1 - one branch, denominator reduced for the second pick
2. Blue then red has the same probability, so P(different) $= \frac{6n}{(n + 3)(n + 2)}$.
   M1 A1 - both orders added
3. Set this equal to $\frac{1}{2}$ and cross-multiply: $12n = (n + 3)(n + 2)$.
   M1 - forming the equation
4. Expand and collect: $n^2 + 5n + 6 - 12n = 0$, so $n^2 - 7n + 6 = 0$.
   A1 - correct quadratic
5. Factorise: $(n - 1)(n - 6) = 0$. Both roots are positive integers, so both are possible: check $n = 1$ gives $\frac{6}{12} = \frac{1}{2}$ and $n = 6$ gives $\frac{36}{72} = \frac{1}{2}$.
   A1 - both values, both checked

Answer. $n = 1$ or $n = 6$

Where students lose marks. Forgetting that "different colours" happens two ways, which halves the probability and produces a quadratic with no whole-number roots. Then, having got there, discarding $n = 1$ because a bag with one blue counter feels wrong. It is not wrong, and this question is set precisely to see who checks.

What this is testing. That the cut piece is similar to the whole cone, and that the frustum is a subtraction, not a formula you are expected to remember.

1. The cut is $8$ cm above the base, so the small cone on top has height $12 - 8 = 4$ cm.
   B1 - height of the small cone
2. The small cone is similar to the whole cone with length scale factor $\frac{4}{12} = \frac{1}{3}$, so its radius is $\frac{1}{3} \times 6 = 2$ cm.
   M1 A1 - scale factor and radius
3. Volume of the small cone: $\frac{1}{3}\pi(2)^2(4) = \frac{16\pi}{3}$.
   A1 - small cone volume
4. Volume of the whole cone: $\frac{1}{3}\pi(6)^2(12) = 144\pi$. Subtract: $144\pi - \frac{16\pi}{3} = \frac{432\pi - 16\pi}{3} = \frac{416\pi}{3}$.
   A1 - frustum volume

Answer. $\frac{416\pi}{3}$ cm$^3$

Where students lose marks. Using $8$ as the height of the small cone. The cut is measured from the base, the small cone sits above it, and the two numbers are not interchangeable. The other error is scaling the volume by $\frac{1}{3}$: volume scales by the cube of the length scale factor, so the small cone is $\frac{1}{27}$ of the whole one.

What this is testing. Two theorems in one figure, and whether you can name them. Half the marks here are for reasons, not numbers.

1. (a) $ABCD$ is a cyclic quadrilateral, so opposite angles add to $180^\circ$: $(4x - 10) + (2x + 16) = 180$.
   M1 - opposite angles of a cyclic quadrilateral
2. $6x + 6 = 180$, so $x = 29$.
   A1
3. (b) Angle $ADC = 4(29) - 10 = 106^\circ$.
   B1 - substituting into the right expression
4. By the alternate segment theorem, the angle between tangent $AT$ and chord $AD$ equals the angle in the alternate segment: angle $ACD =$ angle $TAD = 58^\circ$.
   M1 A1 - theorem named and applied
5. Angles in triangle $ACD$ add to $180^\circ$: angle $CAD = 180 - 106 - 58 = 16^\circ$.
   A1

Answer. (a) $x = 29$. (b) Angle $CAD = 16^\circ$

Where students lose marks. Reasons. "Angles in the same segment" and "the alternate segment theorem" are different theorems, and an examiner marking a "give a reason" question will not accept the wrong name or a vague "circle theorem". The number error worth watching is substituting $x = 29$ into $2x + 16$ when the triangle needs angle $ADC$.

What this is testing. Whether "parallel to" is something you can write down algebraically. Parallel means one vector is a multiple of the other, which here forces a coefficient to be zero.

1. $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, so $\overrightarrow{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}(\mathbf{a} + \mathbf{b})$.
   M1 A1 - route to the midpoint
2. $P$ is on line $OM$, so $\overrightarrow{OP} = k\overrightarrow{OM} = \frac{k}{2}\mathbf{a} + \frac{k}{2}\mathbf{b}$ for some number $k$.
   M1 - introducing a scalar multiple
3. $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \left(\frac{k}{2} - 1\right)\mathbf{a} + \frac{k}{2}\mathbf{b}$.
   M1 - expression for AP
4. $AP$ is parallel to $OB$, so $\overrightarrow{AP}$ must be a multiple of $\mathbf{b}$ alone. The $\mathbf{a}$ term is therefore zero: $\frac{k}{2} - 1 = 0$, so $k = 2$ and $\overrightarrow{OP} = \mathbf{a} + \mathbf{b}$, as required.
   A1 - complete argument

Answer. $k = 2$, giving $\overrightarrow{OP} = \mathbf{a} + \mathbf{b}$

Where students lose marks. Writing $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$. Travel from $A$ to $B$ is back along $\mathbf{a}$ and then out along $\mathbf{b}$, so it is $\mathbf{b} - \mathbf{a}$, and getting it backwards makes $M$ land outside the triangle. The other loss is asserting the result: measuring the diagram, or saying "$P$ looks like $\mathbf{a} + \mathbf{b}$", earns nothing on a show-that question.

What this is testing. Whether you can turn an expression into a statement about consecutive integers. Numerical examples are not a proof, and this question is designed to catch students who offer them.

1. Factorise: $n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1)$.
   M1 A1 - full factorisation, including the difference of two squares
2. Written in order this is $(n - 1)n(n + 1)$: the product of three consecutive integers.
   B1 - identifying consecutive integers
3. Any two consecutive integers include an even one, so the product is a multiple of 2.
   B1 - divisibility by 2
4. Exactly one of any three consecutive integers is a multiple of 3, so the product is also a multiple of 3. As 2 and 3 are different primes, the product is a multiple of $2 \times 3 = 6$.
   B1 - divisibility by 3 and conclusion

Answer. $n^3 - n = (n - 1)n(n + 1)$, a product of three consecutive integers, which is always a multiple of both 2 and 3, so a multiple of 6

Where students lose marks. Testing $n = 1, 2, 3, 4$ and writing "so it works". That is verification, not proof, and it scores zero for the reasoning marks however many cases you list. The second gap is stopping at $n(n - 1)(n + 1)$ without ever saying the words "three consecutive integers": the factorisation alone is not an argument.

What this is testing. Three separate skills stacked: reading a surd ratio as an algebraic length, rationalising to get a usable width, then expanding a product of surds without losing a term.

1. Let the width be $w$, so the length is $w(1 + \sqrt{2})$. Perimeter: $2\big(w + w(1 + \sqrt{2})\big) = 20$, so $w(2 + \sqrt{2}) = 10$.
   M1 A1 - perimeter equation in one unknown
2. $w = \frac{10}{2 + \sqrt{2}}$. Rationalise: $\frac{10(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{10(2 - \sqrt{2})}{2}$.
   M1 - rationalising the denominator
3. So $w = 5(2 - \sqrt{2}) = 10 - 5\sqrt{2}$.
   A1 - exact width
4. Area $= w \times w(1 + \sqrt{2}) = w^2(1 + \sqrt{2})$, and $w^2 = (10 - 5\sqrt{2})^2 = 100 - 100\sqrt{2} + 50 = 150 - 100\sqrt{2}$.
   M1 - area in terms of w, w squared expanded
5. $(150 - 100\sqrt{2})(1 + \sqrt{2}) = 150 + 150\sqrt{2} - 100\sqrt{2} - 200 = 50\sqrt{2} - 50 = 50(\sqrt{2} - 1)$.
   A1 - required form

Answer. $50(\sqrt{2} - 1)$ cm$^2$, which is about $20.7$ cm$^2$

Where students lose marks. Squaring termwise: $(10 - 5\sqrt{2})^2$ is not $100 - 50$. The middle term $-100\sqrt{2}$ is the one the answer depends on. Before that, halving the perimeter incorrectly is common: the width appears twice and the length appears twice, so it is $2(w + \text{length}) = 20$, not $w + \text{length} = 20$.