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Paper 1 · Set 01

Paper 1: Applications of Mathematical Knowledge. 20 questions, 75 minutes. Answer on the grid or click the options; nothing is marked until you press Submit.

75 minutes20 marksNo calculator
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1Ratio, proportion and percentages · spec M3.10 · Paper 1 style · ★☆☆

Two similar solids have surface areas in the ratio \(4 : 9\). Find the ratio of their volumes.

  1. A\(12 : 27\)
  2. B\(16 : 81\)
  3. C\(2 : 3\)
  4. D\(8 : 27\)
  5. E\(64 : 729\)
  6. F\(4 : 9\)
Worked solution

Length ratio \(2 : 3\), so volume ratio \(2^3 : 3^3 = 8 : 27\).

Full method — how and why
Similar solids: areas scale as the square of lengths, volumes as the cube. Work back to the length ratio first.
  1. Areas are in ratio \(4 : 9\), so the length ratio is \(\sqrt4 : \sqrt9 = 2 : 3\).
    Area scales with the square of length, so undo it with a square root.
  2. Volumes scale with the cube of length: \(2^3 : 3^3\).
    Every linear dimension is scaled, so three dimensions give the cube.
  3. So the volume ratio is \(8 : 27\).
    Check: the ratios \(2:3\), \(4:9\), \(8:27\) are the classic length/area/volume triple.
Where it goes wrong
  • Cubing the area ratio to get \(64 : 729\) - an offered option.
  • Leaving the answer as the length ratio \(2:3\).
Takeaway Length \(k\), area \(k^2\), volume \(k^3\): always return to the length ratio before converting.
Answer D · Ratio, proportion and percentages (M3.10) · question RAT-06
2Number, units and estimation · spec M2.9 · Paper 1 style · ★☆☆

Express the recurring decimal \(0.\dot{2}\dot{7}\) as a fraction in its lowest terms.

  1. A\(\dfrac{9}{33}\)
  2. B\(\dfrac{5}{18}\)
  3. C\(\dfrac{3}{11}\)
  4. D\(\dfrac{27}{99}\)
  5. E\(\dfrac{3}{10}\)
  6. F\(\dfrac{27}{100}\)
Worked solution

\(x = 0.2727\ldots\), \(100x = 27.2727\ldots\), so \(99x = 27\) and \(x = \frac{27}{99} = \frac{3}{11}\).

Full method — how and why
The standard algebraic trick for recurring decimals: multiply by a power of \(10\) that shifts one full period, then subtract.
  1. Let \(x = 0.272727\ldots\) The repeating block has two digits, so multiply by \(10^2 = 100\): \(100x = 27.272727\ldots\)
    Shifting by exactly one period makes the decimal tails identical, so they cancel.
  2. Subtract: \(100x - x = 27.2727\ldots - 0.2727\ldots = 27\), so \(99x = 27\).
    The infinite tails cancel exactly - this is the whole method.
  3. Solve: \(x = \dfrac{27}{99}\).
    The denominator is always \(10^k - 1\), i.e. a string of nines matching the period length.
  4. Simplify by \(9\): \(\dfrac{3}{11}\).
    The question demands lowest terms, and \(\frac{27}{99}\) is offered as a trap.
Where it goes wrong
  • Choosing \(\frac{27}{99}\) - correct in value but not in lowest terms.
  • Multiplying by \(10\) instead of \(100\), which does not align the tails.
Takeaway Multiply by \(10^{(\text{period length})}\), subtract, then simplify fully.
Answer C · Number, units and estimation (M2.9) · question NUM-03
3Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Given that \(\log_{10} 2 = p\) and \(\log_{10} 3 = q\), express \(\log_{10} 0.75\) in terms of \(p\) and \(q\).

  1. A\(3q - 4p\)
  2. B\(\dfrac{q}{2p}\)
  3. C\(2p - q\)
  4. D\(q + 2p\)
  5. E\(q - p\)
  6. F\(q - 2p\)
Worked solution

\(0.75 = \frac34\), so \(\log 0.75 = \log 3 - \log 4 = q - 2p\).

Full method — how and why
Express the number in terms of the given primes, then apply the log laws.
  1. Write \(0.75 = \dfrac34\).
    Decimals must become fractions of small numbers before the laws help.
  2. Split: \(\log_{10}\frac34 = \log_{10}3 - \log_{10}4\).
    Quotient law.
  3. Write \(4 = 2^2\): \(\log_{10}4 = 2\log_{10}2 = 2p\).
    Only \(\log 2\) and \(\log 3\) are available, so every number must be built from \(2\)s and \(3\)s.
  4. Combine: \(q - 2p\).
    Substituting \(\log_{10}3 = q\).
Where it goes wrong
  • Using \(\log 4 = p^2\) or \(= 2 + p\).
  • Getting the subtraction backwards and answering \(2p - q\).
Takeaway Factorise the argument into the primes you have been given, then apply product/quotient/power laws.
Answer F · Exponentials and logarithms (MM5.2) · question LOG-13
4Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★☆☆

How many real solutions does the equation \(4^{x} + 2^{x+1} - 8 = 0\) have?

  1. A\(1\)
  2. B\(2\)
  3. C\(0\)
  4. D\(3\)
  5. Einfinitely many
Worked solution

\(y = 2^x\): \(y^2 + 2y - 8 = (y + 4)(y - 2) = 0\). \(y = -4\) is impossible, \(y = 2\) gives \(x = 1\). One solution.

Full method — how and why
Quadratic in \(2^x\) again, but this time one root must be rejected - which changes the count.
  1. Rewrite: \(4^x = (2^2)^x = (2^x)^2\) and \(2^{x+1} = 2\cdot 2^x\).
    Get everything in terms of \(2^x\) before substituting.
  2. Let \(y = 2^x\): \(y^2 + 2y - 8 = 0\), which factorises as \((y+4)(y-2) = 0\).
    Standard quadratic.
  3. Reject \(y = -4\): \(2^x\) is positive for every real \(x\), so it can never equal a negative number.
    This is the crucial filter - exponentials are always strictly positive.
  4. From \(y = 2\): \(2^x = 2\), so \(x = 1\). Exactly one real solution.
    The question asks how many, so the count is the answer.
Where it goes wrong
  • Answering \(2\) because the quadratic has two roots.
  • Rejecting \(y = 2\) as well, and answering \(0\).
Takeaway After substituting \(y = a^x\), discard every root with \(y \leq 0\) - then count what remains.
Answer A · Exponentials and logarithms (MM5.3) · question LOG-10
5Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(AB\) is a diameter of a circle and \(C\) is a point on the circle with \(AC = 6\) and \(BC = 8\). Find the radius of the circle.

  1. A\(\sqrt{50}\)
  2. B\(4\)
  3. C\(10\)
  4. D\(7\)
  5. E\(5\)
  6. F\(14\)
Worked solution

The angle in a semicircle is a right angle, so \(AB = \sqrt{36 + 64} = 10\) and the radius is \(5\).

Full method — how and why
The angle in a semicircle is a right angle, so \(AB\) is the hypotenuse of a right-angled triangle.
  1. \(AB\) is a diameter and \(C\) is on the circle, so angle \(ACB = 90^\circ\).
    The angle in a semicircle is a right angle - the key theorem here.
  2. Apply Pythagoras: \(AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\).
    The right angle is at \(C\), so the diameter \(AB\) is the hypotenuse.
  3. Radius \(= \frac{10}{2} = 5\).
    The question asks for the radius, not the diameter.
Where it goes wrong
  • Answering \(10\) (the diameter).
  • Assuming the right angle is at \(A\) or \(B\).
Takeaway A triangle inscribed with one side a diameter is right-angled at the third vertex.
Answer E · Circles and circle theorems (MM3.3) · question CIR-14
6Binomial expansion · spec MM2.4 · Paper 1 style · ★☆☆

The coefficient of \(x^2\) in the expansion of \((1 + kx)^6\) is \(60\). Given that \(k > 0\), find \(k\).

  1. A\(2\)
  2. B\(4\)
  3. C\(3\)
  4. D\(\sqrt{10}\)
  5. E\(\dfrac{1}{2}\)
  6. F\(10\)
Worked solution

\(\binom62 k^2 = 15k^2 = 60\), so \(k^2 = 4\) and \(k = 2\).

Full method — how and why
A coefficient is given and a constant inside the bracket is unknown. Set up the coefficient equation and solve.
  1. General term of \((1 + kx)^6\): \(\binom6r (kx)^r\), so the \(x^2\) coefficient is \(\binom62 k^2\).
    The \(1^{6-r}\) factor is invisible, which is why this form is the easiest to work with.
  2. Compute \(\binom62 = 15\), so the equation is \(15k^2 = 60\).
    \(\binom62 = \frac{6 \times 5}{2} = 15\).
  3. Solve: \(k^2 = 4\), so \(k = \pm 2\); the condition \(k > 0\) gives \(k = 2\).
    Squaring means two roots; the stated condition selects one.
Where it goes wrong
  • Forgetting to square \(k\) and solving \(15k = 60\) to get \(k = 4\) - which is offered.
  • Using \(\binom62 = 30\) (that is \(6 \times 5\) without dividing by \(2\)).
Takeaway The \(x^r\) coefficient of \((1 + kx)^n\) is \(\binom nr k^r\) - the constant is raised to the same power as \(x\).
Answer A · Binomial expansion (MM2.4) · question BIN-02
7Sequences and series · spec MM2.1 · Paper 1 style · ★☆☆

The sequence \(u_n = n^2 - 10n + 30\) is defined for positive integers \(n\). What is the smallest value taken by any term of the sequence?

  1. A\(21\)
  2. B\(5\)
  3. C\(6\)
  4. D\(-5\)
  5. E\(0\)
  6. F\(30\)
Worked solution

\(u_n = (n-5)^2 + 5\), minimised at \(n = 5\) with value \(5\).

Full method — how and why
A quadratic sequence. Completing the square finds the minimum - but \(n\) must be a positive integer, so check that the optimum is attainable.
  1. Complete the square: \(n^2 - 10n + 30 = (n - 5)^2 + 5\).
    Half of \(-10\) is \(-5\); \(30 - 25 = 5\).
  2. The square is smallest when \(n = 5\), giving \(u_5 = 5\).
    \((n-5)^2 \geq 0\) with equality only at \(n = 5\).
  3. \(n = 5\) is a positive integer, so the minimum is attained: the smallest term is \(5\).
    If the vertex had been at a non-integer (say \(4.5\)), you would have to test the integers either side.
Where it goes wrong
  • Giving \(n = 5\) (the position) instead of \(5\) (the value) - here they coincide, which hides the error.
  • Taking \(u_1 = 21\) as the smallest by assuming sequences always increase.
Takeaway Minimise a quadratic sequence by completing the square, then check the nearest integer values of \(n\).
Answer B · Sequences and series (MM2.1) · question SEQ-14
8Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Solve \(9^{x} - 4\times 3^{x} + 3 = 0\).

  1. A\(x = 1\) only
  2. B\(x = 0\) or \(x = 1\)
  3. C\(x = -1\) or \(x = 0\)
  4. D\(x = 0\) only
  5. E\(x = \log_3 4\)
  6. F\(x = 1\) or \(x = 3\)
Worked solution

Let \(y = 3^x\): \(y^2 - 4y + 3 = (y-1)(y-3) = 0\), so \(3^x = 1\) or \(3^x = 3\), giving \(x = 0\) or \(x = 1\).

Full method — how and why
A hidden quadratic. The key observation is \(9^x = (3^2)^x = (3^x)^2\), so the equation is quadratic in \(3^x\).
  1. Write \(9^x = (3^x)^2\) and substitute \(y = 3^x\): \(y^2 - 4y + 3 = 0\).
    Spotting that one exponential is the square of the other is the whole question; without it there is no route in.
  2. Factorise: \((y - 1)(y - 3) = 0\), so \(y = 1\) or \(y = 3\).
    Ordinary quadratic technique once the substitution is made.
  3. Undo the substitution: \(3^x = 1 \Rightarrow x = 0\); \(3^x = 3 \Rightarrow x = 1\).
    \(a^0 = 1\) for any \(a\) - the root \(y = 1\) is easy to lose because it looks trivial.
Where it goes wrong
  • Stopping at \(y = 1, 3\) and offering those as the answers - the question asks for \(x\), not \(3^x\).
  • Rejecting \(y = 1\) as 'not a real solution'. Only non-positive values of \(y\) must be rejected, because \(3^x > 0\) always.
Takeaway If an equation contains \(a^{2x}\) and \(a^x\), substitute \(y = a^x\); afterwards discard only the roots with \(y \leq 0\).
Answer B · Indices and surds (MM1.1) · question IND-08
9Integration · spec MM7.3 · Paper 1 style · ★☆☆

\(f(x) = ax + b\) satisfies \(\displaystyle\int_{0}^{1} f(x)\,dx = 2\) and \(\displaystyle\int_{0}^{2} f(x)\,dx = 6\). Find \(f(1)\).

  1. A\(4\)
  2. B\(0\)
  3. C\(5\)
  4. D\(1\)
  5. E\(3\)
  6. F\(2\)
Worked solution

\(\frac a2 + b = 2\) and \(2a + 2b = 6\), so \(a + b = 3\); subtracting the first equation from this gives \(\frac a2 = 1\), so \(a = 2\), \(b = 1\) and \(f(1) = 3\).

Full method — how and why
Two integral conditions on a linear function give two equations in \(a\) and \(b\).
  1. Integrate in general: \(\displaystyle\int_0^k (ax+b)dx = \frac{ak^2}{2} + bk\).
    Doing it once with a general limit saves repeating the work.
  2. \(k = 1\): \(\frac a2 + b = 2\). \(k = 2\): \(2a + 2b = 6\), i.e. \(a + b = 3\).
    Substitute each limit and simplify the second by dividing by \(2\).
  3. Subtract the first from the second: \(\frac a2 = 1\), so \(a = 2\), and then \(b = 1\).
    Elimination: \(a - \frac a2 = \frac a2\) and \(3 - 2 = 1\).
  4. Compute \(f(1) = a + b = 3\).
    Conveniently, \(f(1) = a + b\), which the second equation already gave.
Where it goes wrong
  • Forgetting the \(\frac12\) in \(\frac{ak^2}{2}\) and getting a different pair.
  • Answering \(2\) (the value of the first integral) instead of \(f(1)\).
Takeaway Integrate once with a general limit, then substitute each condition to get simultaneous equations.
Answer E · Integration (MM7.3) · question INT-13
10Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★☆☆

For how many integers \(n\) is \(n^2 - 10n + 16 < 0\)?

  1. A\(5\)
  2. B\(3\)
  3. C\(4\)
  4. D\(8\)
  5. E\(7\)
  6. F\(6\)
Worked solution

\((n-2)(n-8) < 0 \Rightarrow 2 < n < 8\), so \(n = 3, 4, 5, 6, 7\): five integers.

Full method — how and why
Counting integers in the solution set of a quadratic inequality. The counting at the end is where the marks are won or lost.
  1. Factorise: \(n^2 - 10n + 16 = (n - 2)(n - 8)\).
    Two numbers multiplying to \(16\), adding to \(-10\): \(-2\) and \(-8\).
  2. Negative between the roots: \(2 < n < 8\).
    Upward parabola, so it dips below the axis strictly between its roots.
  3. List the integers strictly inside: \(3, 4, 5, 6, 7\) - five values.
    Strict inequality excludes \(2\) and \(8\), where the expression equals zero.
Where it goes wrong
  • Including the endpoints and answering \(7\).
  • Counting \(8 - 2 = 6\) as the number of integers - the count between \(a\) and \(b\) exclusive is \(b - a - 1\).
Takeaway Write the integers out; the number strictly between \(a\) and \(b\) is \(b - a - 1\), not \(b - a\).
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-05
11Trigonometric functions and equations · spec MM4.3 · Paper 1 style · ★☆☆

Find the value of \(\tan 30^\circ \tan 60^\circ + \sin^2 45^\circ\).

  1. A\(\dfrac{\sqrt2}{2}\)
  2. B\(1\)
  3. C\(\dfrac{3}{2}\)
  4. D\(2\)
  5. E\(1 + \sqrt3\)
  6. F\(\dfrac{1}{2}\)
Worked solution

\(\tan30^\circ\tan60^\circ = \frac{1}{\sqrt3}\cdot\sqrt3 = 1\), \(\sin^2 45^\circ = \frac12\). Total \(\frac32\).

Full method — how and why
Exact-value recall plus one squared value. The first product is designed to collapse to \(1\).
  1. \(\tan 30^\circ = \dfrac{1}{\sqrt3}\) and \(\tan 60^\circ = \sqrt3\), so their product is \(1\).
    \(30^\circ\) and \(60^\circ\) are complementary, and \(\tan\theta\tan(90^\circ - \theta) = 1\) always.
  2. \(\sin 45^\circ = \dfrac{\sqrt2}{2}\), so \(\sin^2 45^\circ = \dfrac{2}{4} = \dfrac12\).
    Square top and bottom: \((\sqrt2)^2 = 2\) and \(2^2 = 4\).
  3. Add: \(1 + \frac12 = \frac32\).
    Straightforward once both pieces are exact.
Where it goes wrong
  • Writing \(\sin^2 45^\circ = \frac{\sqrt2}{2}\) (forgetting to square), giving \(1 + \frac{\sqrt2}{2}\).
  • Using \(\tan 30^\circ = \sqrt3\), which makes the product \(3\).
Takeaway \(\tan\theta \tan(90^\circ - \theta) = 1\), and \(\sin^2 45^\circ = \cos^2 45^\circ = \frac12\).
Answer C · Trigonometric functions and equations (MM4.3) · question TRG-18
12GCSE algebra · spec M4.3 · Paper 1 style · ★☆☆

Given that \(v^2 = u^2 + 2as\), find \(s\) when \(v = 13\), \(u = 5\) and \(a = 4\).

  1. A\(18\)
  2. B\(21\)
  3. C\(36\)
  4. D\(12\)
  5. E\(9\)
  6. F\(24\)
Worked solution

\(169 = 25 + 8s \Rightarrow s = 18\).

Full method — how and why
Substitution into a formula. Evaluate the squares first, then solve the linear equation.
  1. Substitute: \(13^2 = 5^2 + 2(4)s\), i.e. \(169 = 25 + 8s\).
    Compute each square separately; \(2as = 2 \times 4 \times s = 8s\).
  2. Rearrange: \(8s = 144\).
    \(169 - 25 = 144\).
  3. Divide: \(s = 18\).
    \(144 \div 8 = 18\).
Where it goes wrong
  • Forgetting the factor \(2\) in \(2as\) and getting \(36\) - an offered option.
  • Computing \(13^2 = 139\) or \(5^2 = 10\).
Takeaway Substitute numbers for every letter, evaluate powers first, then solve.
Answer A · GCSE algebra (M4.3) · question ALG-06
13Triangles, radians and sectors · spec MM4.1 · Paper 1 style · ★☆☆

In triangle \(ABC\), \(AB = 5\), \(AC = 8\) and angle \(BAC = 60^\circ\). Find \(BC\).

  1. A\(\sqrt{129}\)
  2. B\(\sqrt{49 + 40\sqrt3}\)
  3. C\(\sqrt{89}\)
  4. D\(\sqrt{69}\)
  5. E\(9\)
  6. F\(7\)
Worked solution

Cosine rule: \(BC^2 = 25 + 64 - 2\cdot5\cdot8\cdot\frac12 = 89 - 40 = 49\), so \(BC = 7\).

Full method — how and why
Two sides and the angle between them: that is exactly the cosine rule's input.
  1. Choose the rule: you know two sides and the included angle, so use the cosine rule, not the sine rule.
    The sine rule needs a side opposite a known angle; here the known angle sits between the known sides.
  2. Write it with \(BC\) opposite the known angle \(A\): \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A\).
    The side being found must be opposite the angle being used.
  3. Substitute, using \(\cos 60^\circ = \frac12\): \(BC^2 = 25 + 64 - 2(5)(8)\left(\frac12\right) = 89 - 40 = 49\).
    \(\cos 60^\circ = \frac12\) is one of the exact values you must recall.
  4. Take the square root: \(BC = 7\).
    Lengths are positive, so only the positive root.
Where it goes wrong
  • Forgetting the \(-2bc\cos A\) term and answering \(\sqrt{89}\) - which is offered.
  • Using \(\cos 60^\circ = \frac{\sqrt3}{2}\) (that is \(\cos 30^\circ\)), giving \(\sqrt{89 - 40\sqrt3}\).
Takeaway Two sides plus the included angle \(\Rightarrow\) cosine rule; the unknown side is opposite the known angle.
Answer F · Triangles, radians and sectors (MM4.1) · question TRI-01
14Simultaneous equations and inequalities · spec MM1.5 · Paper 1 style · ★☆☆

Find the complete set of values of \(x\) for which \(2x^2 - 5x - 3 \geq 0\).

  1. A\(x \leq -\dfrac{1}{2}\) or \(x \geq 3\)
  2. B\(x \leq -\dfrac12\)
  3. C\(x \geq 3\)
  4. D\(x \leq -3\) or \(x \geq \dfrac{1}{2}\)
  5. E\(-3 \leq x \leq \dfrac12\)
  6. F\(-\dfrac{1}{2} \leq x \leq 3\)
Worked solution

\((2x+1)(x-3) \geq 0\): the roots are \(-\frac12\) and \(3\) and the parabola opens upwards, so \(x \leq -\frac12\) or \(x \geq 3\).

Full method — how and why
A quadratic inequality with a leading coefficient other than \(1\). Factorise, find the roots, then use the shape of the parabola.
  1. Factorise: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).
    Two numbers multiplying to \(2 \times (-3) = -6\) and adding to \(-5\): \(-6\) and \(1\), so split \(-5x\) as \(-6x + x\).
  2. Roots: \(x = -\frac12\) and \(x = 3\).
    Set each bracket to zero.
  3. The coefficient of \(x^2\) is positive, so the parabola opens upwards and is \(\geq 0\) outside the roots.
    A quick sketch is the safest way to convert a factorisation into an inequality answer.
  4. Answer: \(x \leq -\frac12\) or \(x \geq 3\), with \(\leq\) because equality is allowed.
    The question has \(\geq 0\), so the roots themselves are included.
Where it goes wrong
  • Giving \(-\frac12 \leq x \leq 3\), which is where the expression is negative.
  • Mixing up the roots as \(-3\) and \(\frac12\) - the root of \((2x+1)\) is \(-\frac12\), not \(-2\).
Takeaway Factorise, mark the roots, sketch the parabola: 'outside' for \(>0\), 'between' for \(<0\) when \(a > 0\).
Answer A · Simultaneous equations and inequalities (MM1.5) · question SIM-02
15Sequences and series · spec MM2.3 · Paper 1 style · ★★☆

A geometric series has first term \(a\) and common ratio \(r\) with \(|r| < 1\). Its sum to infinity is \(8\) and the sum to infinity of the squares of its terms is \(\dfrac{64}{3}\). Find \(r\).

  1. A\(-\dfrac12\)
  2. B\(\dfrac{1}{2}\)
  3. C\(\dfrac{1}{4}\)
  4. D\(\dfrac{2}{3}\)
  5. E\(\dfrac{3}{4}\)
  6. F\(\dfrac{1}{3}\)
Worked solution

\(\dfrac{a}{1-r} = 8\) and \(\dfrac{a^2}{1 - r^2} = \dfrac{64}{3}\). Dividing the second by the square of the first: \(\dfrac{1-r}{1+r} = \dfrac{64/3}{64} = \dfrac13\), so \(3 - 3r = 1 + r\) and \(r = \frac12\).

Full method — how and why
Two infinite sums, two unknowns. Dividing one equation by the (square of the) other eliminates \(a\) elegantly.
  1. Write the two facts: \(\dfrac{a}{1 - r} = 8\) and, for the squares (first term \(a^2\), ratio \(r^2\)), \(\dfrac{a^2}{1 - r^2} = \dfrac{64}{3}\).
    Squaring every term of a geometric series gives another geometric series with first term \(a^2\) and ratio \(r^2\).
  2. Square the first: \(\dfrac{a^2}{(1-r)^2} = 64\).
    Squaring makes the numerators match, so they can be eliminated by division.
  3. Divide the second equation by this: \(\dfrac{(1-r)^2}{1 - r^2} = \dfrac{64/3}{64} = \dfrac13\).
    Dividing kills \(a^2\) entirely - the standard move when two equations share a factor.
  4. Simplify using \(1 - r^2 = (1-r)(1+r)\): \(\dfrac{1 - r}{1 + r} = \dfrac13\).
    One factor of \((1-r)\) cancels.
  5. Cross-multiply: \(3 - 3r = 1 + r \Rightarrow 4r = 2 \Rightarrow r = \frac12\).
    Check: \(a = 4\), and \(\frac{16}{1 - 1/4} = \frac{64}{3}\) ✓.
Where it goes wrong
  • Using ratio \(r\) (not \(r^2\)) for the series of squares.
  • Forgetting to square the first equation before dividing, which leaves an \(a\) behind.
Takeaway Squaring a geometric series squares both \(a\) and \(r\); divide equations to eliminate the unknown you do not want.
Answer B · Sequences and series (MM2.3) · question SEQ-05
16Binomial expansion · spec MM2.4 · Paper 1 style · ★★☆

Find the term independent of \(x\) in the expansion of \(\left(x^2 + \dfrac{2}{x}\right)^6\).

  1. A\(60\)
  2. B\(15\)
  3. C\(160\)
  4. D\(480\)
  5. E\(240\)
  6. F\(120\)
Worked solution

General term \(\binom6r (x^2)^{6-r} \left(\frac2x\right)^r = \binom6r 2^r x^{12 - 3r}\). Need \(12 - 3r = 0\), \(r = 4\): \(\binom64 \times 16 = 15 \times 16 = 240\).

Full method — how and why
'Term independent of \(x\)' means the power of \(x\) is zero. Track the power of \(x\) in the general term and solve for \(r\).
  1. General term: \(\binom6r (x^2)^{6-r}\left(\frac2x\right)^r = \binom6r 2^r x^{12 - 2r - r}\).
    \((x^2)^{6-r}\) gives \(x^{12-2r}\) and \(\left(\frac2x\right)^r\) gives \(2^r x^{-r}\); combine the powers.
  2. So the power of \(x\) is \(12 - 3r\).
    Getting this expression right is the whole question.
  3. Set it to zero: \(12 - 3r = 0\), so \(r = 4\).
    'Independent of \(x\)' means \(x^0\).
  4. Evaluate the term: \(\binom64 \times 2^4 = 15 \times 16 = 240\).
    \(\binom64 = \binom62 = 15\) by symmetry.
Where it goes wrong
  • Using the power \(12 - 2r\) and getting \(r = 6\), giving \(2^6 = 64\).
  • Forgetting the \(2^r\) factor and answering \(15\).
Takeaway Write the power of \(x\) in the general term as a linear expression in \(r\), then set it to the value you need.
Answer E · Binomial expansion (MM2.4) · question BIN-03
17Straight lines · spec MM3.1 · Paper 1 style · ★★☆

The line \(L\) passes through \((4, 0)\) and \((0, 6)\). Find the perpendicular distance from the origin to \(L\).

  1. A\(5\)
  2. B\(\dfrac{24}{13}\)
  3. C\(\dfrac{12}{13}\)
  4. D\(\dfrac{12}{\sqrt{13}}\)
  5. E\(2\sqrt{13}\)
  6. F\(\sqrt{13}\)
Worked solution

Area of triangle \(= \frac12 \cdot 4 \cdot 6 = 12\). Hypotenuse \(= \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\). Distance \(d\) satisfies \(\frac12 \cdot 2\sqrt{13} \cdot d = 12\), so \(d = \frac{12}{\sqrt{13}}\).

Full method — how and why
Perpendicular distance from a point to a line. The 'area two ways' method avoids memorising the distance formula.
  1. The line cuts the axes at \((4,0)\) and \((0,6)\), so with the origin it forms a right-angled triangle of area \(\frac12 \times 4 \times 6 = 12\).
    The two intercepts and the origin make a triangle whose legs lie along the axes - its area is easy.
  2. The hypotenuse (the segment of \(L\)) has length \(\sqrt{4^2 + 6^2} = \sqrt{52} = 2\sqrt{13}\).
    Pythagoras on the same triangle.
  3. Compute the area again using the hypotenuse as base and the required distance \(d\) as height: \(\frac12 \times 2\sqrt{13} \times d = 12\).
    The perpendicular distance from the origin to \(L\) is the height of that triangle on that base.
  4. Solve: \(d = \dfrac{12}{\sqrt{13}}\).
    Equivalently, the formula \(\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\) with \(3x + 2y = 12\) gives \(\frac{12}{\sqrt{13}}\).
Where it goes wrong
  • Using the distance to one of the intercepts (\(4\), \(6\) or \(2\sqrt{13}\)) instead of the perpendicular distance.
  • Forgetting the square root and answering \(\frac{12}{13}\).
Takeaway Area computed two ways turns a distance problem into arithmetic; or use \(\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\).
Answer D · Straight lines (MM3.1) · question LIN-03
18Graphs of functions · spec MM8.5 · Paper 1 style · ★★☆

The graph of \(y = f(x)\) has \(f'(x) > 0\) for \(x < 1\), \(f'(1) = 0\), \(f'(x) > 0\) for \(1 < x < 3\), \(f'(x) < 0\) for \(x > 3\). Which of the following is true?

  1. A\(f\) has local maxima at \(x = 1\) and \(x = 3\)
  2. B\(f\) has a local minimum at \(x = 1\) and a local maximum at \(x = 3\)
  3. C\(f\) has a local maximum at \(x = 3\) and a stationary point of inflexion at \(x = 1\)
  4. D\(f\) has a local maximum at \(x = 1\) and a local minimum at \(x = 3\)
  5. E\(f\) is increasing for all \(x\)
Worked solution

At \(x = 1\) the derivative is zero but does not change sign: a stationary point that is not a turning point (an inflexion). At \(x = 3\) the derivative changes from positive to negative: local maximum.

Full method — how and why
Reading a function's shape from the sign of its derivative. A zero derivative without a sign change is an inflexion, not a turning point.
  1. Before and after \(x = 1\): \(f' > 0\) on both sides, with \(f'(1) = 0\).
    The gradient dips to zero and comes back positive - the curve flattens momentarily but keeps rising.
  2. So \(x = 1\) is a stationary point that is not a turning point: a stationary point of inflexion.
    Turning points require a change of sign in \(f'\).
  3. At \(x = 3\): \(f' > 0\) just before and \(f' < 0\) just after, so the curve rises then falls - a local maximum.
    A \(+ \to -\) sign change is exactly a local maximum.
  4. So: local maximum at \(x = 3\), stationary inflexion at \(x = 1\).
    Match to the options.
Where it goes wrong
  • Assuming \(f'(1) = 0\) must be a maximum or minimum.
  • Reading the \(+ \to -\) change at \(x = 3\) as a minimum.
Takeaway \(f' = 0\) with no sign change \(=\) inflexion; \(+ \to -\) is a maximum, \(- \to +\) is a minimum.
Answer C · Graphs of functions (MM8.5) · question GRA-07
19Indices and surds · spec MM1.2 · Paper 1 style · ★★☆

Find the value of \(\dfrac{1}{\sqrt{2}+1} + \dfrac{1}{\sqrt{3}+\sqrt{2}} + \dfrac{1}{2+\sqrt{3}}\).

  1. A\(2\)
  2. B\(2 - \sqrt{3}\)
  3. C\(\sqrt{3} + 1\)
  4. D\(1\)
  5. E\(\sqrt{3}\)
  6. F\(\sqrt{2}\)
Worked solution

Rationalising each term gives \((\sqrt2 - 1) + (\sqrt3 - \sqrt2) + (2 - \sqrt3)\), which telescopes to \(1\).

Full method — how and why
A telescoping sum. Each term rationalises to a difference of two surds, and almost everything cancels - so do not compute decimals.
  1. Rationalise each term: \(\dfrac{1}{\sqrt2+1} = \sqrt2 - 1\), \(\dfrac{1}{\sqrt3+\sqrt2} = \sqrt3 - \sqrt2\), \(\dfrac{1}{2+\sqrt3} = 2 - \sqrt3\).
    Each denominator has the form \(\sqrt{n+1} + \sqrt n\) whose conjugate gives \((n+1) - n = 1\) - the denominators all become \(1\), which is the point of the construction.
  2. Add them: \((\sqrt2 - 1) + (\sqrt3 - \sqrt2) + (2 - \sqrt3)\).
    Write the terms in a line so the cancelling pairs are visible.
  3. Cancel: \(\sqrt2\) with \(-\sqrt2\), \(\sqrt3\) with \(-\sqrt3\), leaving \(-1 + 2 = 1\).
    This is a telescoping sum - the surds annihilate in pairs and only the two ends survive.
Where it goes wrong
  • Approximating numerically and choosing \(\sqrt2\) or \(\sqrt3\) because the total 'looks about right'.
  • Rationalising with the wrong conjugate sign, which flips one term and destroys the cancellation.
Takeaway \(\dfrac{1}{\sqrt{n+1}+\sqrt n} = \sqrt{n+1} - \sqrt n\): recognise this and long chains of such fractions collapse instantly.
Answer D · Indices and surds (MM1.2) · question IND-09
20Probability · spec M7.7 · Paper 1 style · ★★☆

A box contains \(n\) white balls and \(3\) black balls. Two balls are drawn without replacement. The probability that both are white is \(\dfrac{5}{12}\). Find \(n\).

  1. A\(6\)
  2. B\(4\)
  3. C\(5\)
  4. D\(3\)
  5. E\(7\)
  6. F\(9\)
Worked solution

\(\dfrac{n}{n+3}\cdot\dfrac{n-1}{n+2} = \dfrac{5}{12} \Rightarrow 12n^2 - 12n = 5n^2 + 25n + 30 \Rightarrow 7n^2 - 37n - 30 = 0 \Rightarrow (7n + 5)(n - 6) = 0\), so \(n = 6\).

Full method — how and why
Set up the without-replacement probability in terms of \(n\), then solve the resulting quadratic.
  1. There are \(n + 3\) balls in total. \(P(\text{both white}) = \dfrac{n}{n+3} \times \dfrac{n-1}{n+2}\).
    After one white is removed, \(n-1\) whites remain out of \(n+2\).
  2. Set equal to \(\frac{5}{12}\) and cross-multiply: \(12n(n-1) = 5(n+3)(n+2)\).
    Cross-multiplying is safe since all denominators are positive.
  3. Expand: \(12n^2 - 12n = 5(n^2 + 5n + 6) = 5n^2 + 25n + 30\).
    Expand the right side carefully: \((n+3)(n+2) = n^2 + 5n + 6\).
  4. Collect: \(7n^2 - 37n - 30 = 0\), which factorises as \((7n + 5)(n - 6) = 0\).
    Check the middle term: \(-42n + 5n = -37n\) ✓.
  5. \(n = -\frac57\) is impossible for a count, so \(n = 6\).
    Verify: \(\frac{6}{9}\times\frac58 = \frac{30}{72} = \frac{5}{12}\) ✓.
Where it goes wrong
  • Writing \(\left(\frac{n}{n+3}\right)^2\), the with-replacement version.
  • Accepting the negative root instead of rejecting it on contextual grounds.
Takeaway Model without replacement by decreasing both numerator and denominator, then reject roots that cannot be counts.
Answer A · Probability (M7.7) · question PRO-08
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