20 exact-answer questions with no calculator: surds, indices, exact trigonometric values, algebraic manipulation and fractions.
🧮 Paper 1 non-calculator warm-up — Higher
On Higher, Paper 1 asks for answers in exact form — a surd left as a surd, sin 60° left as √3/2, a fraction left as a fraction — and a rounded decimal scores nothing. This sheet drills exactly that. Twenty questions across simplifying surds and rationalising denominators, the index laws including fractional and negative indices, exact values of sin, cos and tan at the standard angles, expanding and factorising, and arithmetic with fractions and mixed numbers. Give yourself twenty-five minutes and no calculator, and write every answer in its exact form even when a decimal feels more natural. If you find yourself reaching for the calculator out of habit, that is the habit this sheet exists to break.
- 1.Expand and simplify (x − 3)²
- 2.The two shorter sides of a right-angled triangle are √12 cm and √24 cm. Work out the exact length of the hypotenuse.
- 3.Simplify 2³ × 2⁴, giving your answer as a single power of 2.
- 4.A zip-wire is fixed at an angle of 60° to the horizontal ground. The zip-wire is 12 m long. Using the exact value of cos 60°, work out the exact horizontal distance it covers.
- 5.In a school, 3/5 of the students study French. Of these students, 2/3 also study Spanish. What fraction of all the students study both French and Spanish?
- 6.A student works out the exact area of a circle with radius 4 cm by squaring the radius but forgetting to multiply by π. Work out the correct exact area of the circle, in terms of π.
- 7.The equation x² − 4x + k = 0 has two different real solutions. Work out the range of values of k.
- 8.Solve x² + 3x − 10 = 0.
- 9.Solve 4x² − 9 = 0.
- 10.The graph of y = x² − 2x − 8 takes these values: when x = −2, y = 0; when x = −1, y = −5; when x = 0, y = −8; when x = 3, y = −5; when x = 4, y = 0. Use these values to write down the two solutions of x² − 2x − 8 = 0.y = x² − 2x − 8
- 11.Work out the exact value of (sin 30°)² + (cos 30°)².
- 12.Simplify (2x² + 7x + 3) ÷ (x² − 9), giving your answer as a single fraction.
- 13.Solve x² + 7x = 0.
- 14.A right-angled triangle has a hypotenuse of 14 cm. One of the other angles is 30°. Work out the exact length of the side opposite the 30° angle.
- 15.When a number is added to its square the result is 30. Work out the possible values of the number.
- 16.Work out the value of the discriminant b² − 4ac for the equation 2x² + 3x − 2 = 0.
- 17.Expand and simplify 4(2x − 1) − 3(x + 2) − 5x
- 18.Simplify x⁽³⁄⁴⁾ ÷ x⁽¹⁄⁴⁾
- 19.A student writes 0.08 as the fraction 8/10, reading the 8 as if it stood in the tenths column and ignoring the zero. Work out the correct fraction that 0.08 is equal to, giving your answer in its simplest form.
- 20.A pop-up canopy has two sloping supports that meet at the top. Each support makes an angle of 30° with the ground and is 6 m long. Using the exact value of cos 30°, work out the total width of the canopy's base.
Answer key
- (d) x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3).
- (d) 6 — By Pythagoras' theorem, the square of the hypotenuse equals the sum of the squares of the other two sides: (√12)² + (√24)² = 12 + 24 = 36. The hypotenuse is √36 = 6 cm. Adding the two side lengths directly instead of squaring them first, treating the theorem as if it were a straight sum of the sides, gives √12 + √24 = 2√3 + 2√6. Multiplying the two squared values, 12 × 24 = 288, instead of adding them, then taking the root, gives √288 = 12√2. Adding the squares correctly to get 36 but forgetting to take the square root at the end leaves 36 as the answer instead of the hypotenuse itself.
- (d) 2⁷ — When multiplying powers of the same base, the indices add: 3 + 4 = 7, so 2³ × 2⁴ = 2⁷. Multiplying the indices instead of adding them gives 3 × 4 = 12, so 2¹². Subtracting the indices instead of adding them gives 4 − 3 = 1, so 2¹. Multiplying the bases together as well as adding the indices gives 2 × 2 = 4, so 4⁷.
- (d) 6 m — The horizontal distance is adjacent to the 60° angle and the zip-wire is the hypotenuse, so horizontal distance = 12 × cos 60° = 12 × 1/2 = 6 m. 6√3 m comes from using sin 60° = √3/2 instead of cos 60°, which would give the vertical drop, not the horizontal distance. 4√3 m comes from treating 12 as the side adjacent to a tangent ratio and dividing by tan 60° = √3. 24 m comes from dividing 12 by cos 60° instead of multiplying by it.
- (c) 2/5 — To find a fraction of a fraction, multiply them together: 3/5 × 2/3 = 6/15, which simplifies to 2/5. Multiplying only the numerators, 3 × 2 = 6, but adding the denominators, 5 + 3 = 8, instead of multiplying them gives 6/8, which simplifies to 3/4. Using only the fraction who study French, 3/5, and ignoring that a further fraction of them also study Spanish gives 3/5. Dividing by 2/3 instead of multiplying by it, using its reciprocal 3/2, gives 3/5 × 3/2 = 9/10.
- (d) 16π cm² — Method: for a circle, area = π × radius². Working: area = π × 4² = π × 16 = 16π cm². Answer: 16π cm². The student squared the radius but left out the π, which is why 16 cm² is not the exact area. 4π cm² comes from multiplying by the radius once instead of squaring it: π × 4 = 4π. 8π cm² comes from using the circumference formula 2 × π × radius instead of the area formula: 2 × π × 4 = 8π. 64π cm² comes from using the diameter (8 cm) as the radius in the area formula: π × 8² = 64π.
- (d) k < 4 — Method: the number of real solutions of ax² + bx + c = 0 is decided by the discriminant b² − 4ac, and two different real solutions need it to be greater than zero. Working: here a = 1, b = −4 and c = k, so b² − 4ac = 16 − 4k; the condition is 16 − 4k > 0, which gives 16 > 4k and then k < 4. Answer: k < 4; for example k = 3 gives x² − 4x + 3 = 0, whose solutions are 1 and 3. The distractors: k ≤ 4 comes from using b² − 4ac ≥ 0, which also allows the single repeated solution at k = 4; k > 4 comes from dividing −4k > −16 by −4 without reversing the inequality sign; k < 16 comes from leaving the factor 4 out of 4ac and solving 16 − k > 0.
- (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
- (a) x = 3/2 or x = −3/2 — Method: the equation has an x² term and a number but no x term, so make x² the subject and then take the square root of both sides, keeping the negative root as well as the positive one. Working: adding 9 to both sides of 4x² − 9 = 0 gives 4x² = 9, and dividing both sides by 4 gives x² = 9/4. Square-rooting the top and the bottom of 9/4 gives 3/2, so x = 3/2 or x = −3/2, and each value checks out because 4 × 9/4 − 9 = 0. Answer: x = 3/2 or x = −3/2. The distractors: x = 3 or x = −3 comes from square-rooting both sides of 4x² = 9 without first dividing by the 4, so the coefficient of x² is ignored; x = 9/4 or x = −9/4 comes from stopping at x² = 9/4 and writing that value down as x, leaving the square root undone; x = 3/2 only comes from taking the positive square root of 9/4 and losing the negative solution.
- (c) x = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
- (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
- (b) (2x + 1)/(x − 3) — Factorise both the numerator and the denominator before you cancel anything. The numerator 2x² + 7x + 3 factorises to (2x + 1)(x + 3), and the denominator x² − 9 is a difference of two squares, factorising to (x − 3)(x + 3). The (x + 3) factor is common to both, so it cancels, leaving (2x + 1)/(x − 3). Writing (2x + 1)/(x + 3) comes from factorising x² − 9 as (x + 3)² instead of (x − 3)(x + 3) — a difference of two squares always has one plus and one minus bracket. Writing (2x + 3)/(x − 3) comes from mis-factorising the numerator as (2x + 3)(x + 1) and then wrongly cancelling the (x + 1) against the denominator's (x + 3) as though they were the same bracket. Writing 2x + 1 comes from cancelling the (x + 3) factor correctly but then dropping the remaining (x − 3) on the denominator altogether.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (b) 7 cm — The side opposite the 30° angle is found using sin 30° = opposite/hypotenuse, so opposite = 14 × sin 30° = 14 × 1/2 = 7 cm. 7√3 cm comes from using cos 30° = √3/2 instead of sin 30° (mixing up the opposite and adjacent sides). 14/√3 cm comes from using tan 30° = 1/√3 instead of sin 30°. 28 cm comes from dividing 14 by sin 30° instead of multiplying by it.
- (b) 5 and −6 — Method: turn the sentence into an equation in one letter, collect every term on one side so it reads as a quadratic equal to zero, then factorise. Working: if the number is x then x² + x = 30, which rearranges to x² + x − 30 = 0; two numbers that multiply to −30 and add to 1 are 6 and −5, so (x + 6)(x − 5) = 0 and x = −6 or x = 5. Both values work: 5 + 25 = 30 and −6 + 36 = 30. Answer: 5 and −6. The distractors: 6 and −5 comes from reading the numbers inside the brackets as the solutions without changing their signs; 5 only comes from discarding the negative solution, although nothing in the question rules out a negative number; 5 and 6 comes from hunting for a factor pair of 30 instead of forming and solving the quadratic.
- (c) 25 — Method: read off a, b and c with their signs and substitute them into b² − 4ac. Working: for 2x² + 3x − 2 = 0, a = 2, b = 3 and c = −2, so b² − 4ac = 3² − 4 × 2 × (−2) = 9 − (−16) = 9 + 16 = 25. Answer: 25. The distractors: −7 comes from taking c as +2, which gives 9 − 16; 22 comes from working b² as 2 × 3 = 6 and then 6 + 16; 13 comes from leaving the 4 out of 4ac and working 9 − 2 × (−2).
- (a) −10 — Method: multiply each bracket out, treating the second bracket as being multiplied by −3 because it is subtracted, then collect like terms. Working: 4(2x − 1) = 8x − 4 and −3(x + 2) = −3x − 6, so the expression becomes 8x − 4 − 3x − 6 − 5x; the x terms give 8x − 3x − 5x = 0, so no term in x survives, and the numbers give −4 − 6 = −10. Answer: −10. The distractors: 2 comes from expanding −3(x + 2) as −3x + 6, leaving the numbers −4 + 6; 5x − 10 comes from forgetting the final −5x, so the x terms give 8x − 3x = 5x; −7 comes from multiplying the 4 over only the first term of its bracket, giving 8x − 1 and so the numbers −1 − 6.
- (d) x⁽¹⁄²⁾ — Method: dividing two powers of the same letter subtracts the index of the divisor from the index of the term being divided, and fractional indices are subtracted like any other fractions. Working: 3/4 − 1/4 = 2/4, which simplifies to 1/2, so the result is x⁽¹⁄²⁾. Answer: x⁽¹⁄²⁾. The distractors: x comes from adding the indices, 3/4 + 1/4 = 1, as though the powers were being multiplied; x³ comes from dividing the indices, so that 3/4 divided by 1/4 gives 3; x⁽³⁄¹⁶⁾ comes from multiplying the indices, 3/4 × 1/4.
- (d) 2/25 — Method: write the decimal over 100 using its two decimal places, then simplify. Working: 0.08 = 8/100 = 2/25 (dividing both numerator and denominator by 4). Answer: 2/25. The student's fraction, 8/10, comes from ignoring the zero in the tenths column and reading 0.08 as though it were 0.8; it simplifies to 4/5. 1/125 comes from writing the decimal over 1000 instead of 100, as if there were three decimal places. 25/2 comes from flipping the correct fraction upside down.
- (b) 6√3 m — The horizontal distance covered by one support is adjacent to the 30° angle, so it equals 6 × cos 30° = 6 × √3/2 = 3√3 m. The total base width is made up of both supports, so it is 2 × 3√3 = 6√3 m. '3√3 m' gives only one support's horizontal distance and forgets to double it for the total width. '6 m' comes from using sin 30° instead of cos 30° for the horizontal distance (6 × sin 30° = 3, doubled to 6). '12 m' comes from doubling the full sloping length of 6 m without using any trigonometry at all.
What is on this worksheet?
The sheet holds 20 questions drawn from the MathsUK bank — the content areas covered: Number, Algebra, Geometry and measures (statements N7, N8, N10, A4, A18, G21). It is pitched at GCSE Higher and takes about 25 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.
How to use the sheet well
- Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
- Do all 20 questions before checking — about 25 minutes is the guide, but there is no time pressure.
- Check the answers — press “Show answers” or print the answer page separately.
- Redo the questions you got wrong — twice as effective as doing 20 fresh ones.
- “New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.
Why this sheet helps
MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.
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