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.Work out the value of .
- 2.A shape is made by joining a rectangle measuring 10 cm by 6 cm to a semicircle of diameter 6 cm along one of the rectangle's shorter sides. Work out the exact area of the shape, in terms of π.
- 3.Work out the value of √49 + ∛27
- 4.A ladder rests against a wall, making an angle of 60° with the ground. The ladder is 2 m long. Using the exact value of sin 60°, work out how high up the wall the ladder reaches, giving your answer in centimetres.
- 5.Expand and simplify 4(2x − 1) − 3(x + 2) − 5x
- 6.Solve x² − 5x + 6 = 0 by factorising.
- 7.Solve 4x² − 9 = 0.
- 8.Work out the exact value of cos 0° − sin 30°.
- 9.Expand and simplify √3(2 + √12).
- 10.A recipe uses 0.625 kg of flour. Write this mass as a fraction of a kilogram, in its simplest form.
- 11.A circle has a circumference of 8π cm. Work out the exact area of the circle in terms of π.
- 12.Solve x² + 7x = 0.
- 13.A pizza is cut into 12 equal slices. Ben eats 5 slices and Mia eats 3 slices. What fraction of the pizza is left, giving your answer in its simplest form?
- 14.Work out the exact value of sin 45° × cos 45°.
- 15.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 16.Write ∛(x²) as a single power of x.
- 17.Work out the value of 2⁻²
- 18.A right-angled triangle has a hypotenuse of 10 cm. One of its other angles is 45°. Work out the exact length of one of the two shorter sides.
- 19.Triangle DEF has a right angle at E. Angle DFE = 60° and EF = 9 cm. Work out the exact length of DE.
- 20.A quarter-circle has a radius of 6 cm. Work out the exact perimeter of the quarter-circle, giving your answer in terms of π.
Answer key
- (a) 17 — Method: work out each power separately, then combine them as the question asks. Working: $2^3 = 8$ and $3^2 = 9$, and 8 + 9 = 17. 72 comes from working out 8 × 9 = 72, multiplying the two powers instead of adding them. 12 comes from misreading the powers as repeated multiplication of the base by the index, 2 × 3 + 3 × 2 = 6 + 6 = 12. −1 comes from working out 8 − 9 = −1, subtracting the powers instead of adding them. Answer: 17.
- (b) (60 + 4.5π) cm² — Method: find the area of the rectangle and the area of the semicircle separately, then add them. Working: the rectangle has area 10 × 6 = 60 cm². The semicircle has radius 3 cm, so its area is half of π × 3² = half of 9π = 4.5π cm². Total area = (60 + 4.5π) cm². Answer: (60 + 4.5π) cm². (60 + 18π) cm² comes from using the diameter (6 cm) as the radius in the semicircle area formula: half of π × 6² = 18π. (60 + 9π) cm² comes from forgetting to halve the full circle's area: π × 3² = 9π. (60 + 3π) cm² comes from finding the semicircle's arc length instead of its area: half of 2 × π × 3 = 3π.
- (a) 10 — √49 = 7 and ∛27 = 3, so √49 + ∛27 = 7 + 3 = 10. Treating the cube root as dividing by 3 instead of finding the cube root gives 27 ÷ 3 = 9, then 7 + 9 = 16. Multiplying the two roots instead of adding them gives 7 × 3 = 21. Ignoring the cube root symbol and using 27 as it stands gives 7 + 27 = 34.
- (c) 100√3 cm — The height is opposite the 60° angle, so height = 2 × sin 60° = 2 × √3/2 = √3 m. Converting to centimetres: √3 m = 100√3 cm. '√3 cm' forgets to convert the answer from metres to centimetres. '200√3 cm' comes from mis-recalling sin 60° as √3 instead of √3/2, dropping the denominator of the exact value: 2 × √3 = 2√3 m = 200√3 cm. '50√3 cm' comes from halving the ladder's length before multiplying by sin 60°, instead of using the full 2 m.
- (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.
- (c) x = 2 or x = 3 — Method: factorise into two brackets whose numbers multiply to the constant term and add to the coefficient of x, then set each bracket equal to zero. Working: two numbers that multiply to 6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0; then x − 2 = 0 gives x = 2 and x − 3 = 0 gives x = 3. Answer: x = 2 or x = 3. The distractors: x = −2 or x = −3 comes from reading the numbers inside the brackets as the solutions instead of changing their signs; x = 1 or x = 6 comes from taking the first factor pair of 6 without checking that the pair adds to −5; x = 5 or x = 6 comes from reading the solutions straight off the 5 and the 6 in the equation.
- (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.
- (a) 1/2 — cos 0° = 1 and sin 30° = 1/2, so cos 0° − sin 30° = 1 − 1/2 = 1/2. 1 comes from writing down cos 0° alone and forgetting to subtract sin 30°. 3/2 comes from adding the two values instead of subtracting. −1/2 comes from working out sin 30° − cos 0°, the two terms the wrong way round.
- (d) 6 + 2√3 — Multiply √3 by each term in the bracket separately. First term: √3 × 2 = 2√3. Second term: √3 × √12 = √(3 × 12) = √36 = 6. Adding the two results in the order they were found, and writing the whole-number term first, gives 6 + 2√3. Adding the numbers under the root for the second term instead of multiplying them (3 + 12 = 15) gives √15 in place of 6, leading to √15 + 2√3. Multiplying √3 by the 2 but never distributing to the √12 term at all leaves just 2√3. Treating √3 × 2 as if the 3 were multiplied by the 2 inside the root, √3 × 2 → √6, while still getting the second term correct, gives 6 + √6.
- (c) 5/8 — Method: write the decimal over 1000 using its three decimal places, then simplify. Working: 0.625 = 625/1000 = 5/8 (dividing both numerator and denominator by 125). Answer: 5/8. 25/4 comes from writing the decimal over 100 instead of 1000, as if there were only two decimal places. 31/50 comes from rounding 0.625 to 0.62 before converting. 8/5 comes from simplifying correctly to 5/8 and then writing the fraction upside down.
- (b) 16π — Use the circumference formula C = 2πr to find the radius: 8π = 2πr, so r = 8π ÷ 2π = 4 cm. Then use the area formula A = πr²: A = π × 4² = 16π cm². Using C = πr instead of C = 2πr gives r = 8, and squaring that gives 64π. Finding r = 4 correctly but then substituting it back into the circumference formula instead of the area formula gives 2π × 4 = 8π. Finding r = 4 correctly but forgetting to square it in the area formula, using A = πr instead of A = πr², gives 4π.
- (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.
- (a) 1/3 — Method: find the total fraction eaten, then subtract it from the whole pizza. Working: together they eat 5/12 + 3/12 = 8/12, so the fraction left is 12/12 − 8/12 = 4/12 = 1/3. Answer: 1/3. 2/3 comes from giving the fraction eaten instead of the fraction left. 7/12 comes from only subtracting Ben's slices and forgetting Mia's. 1/2 comes from comparing the 4 slices left with the 8 slices eaten, 4/8, a part-to-part comparison instead of comparing with the whole pizza of 12 slices.
- (c) 1/2 — sin 45° = √2/2 and cos 45° = √2/2, so sin 45° × cos 45° = √2/2 × √2/2 = 2/4 = 1/2. √2/2 comes from writing down only one of the two factors and forgetting to multiply by the other. √2 comes from adding the two exact values instead of multiplying them: √2/2 + √2/2 = √2. 1 comes from wrongly treating sin 45° × cos 45° as sin(45° + 45°) = sin 90° = 1 — multiplying two ratios is not the same as adding their angles.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (c) x⁽²⁄³⁾ — Method: a root can be written as a fractional index, with the root's index as the denominator and the power inside the root as the numerator. Working: the cube root gives a denominator of 3 and the square inside gives a numerator of 2, so ∛(x²) = x⁽²⁄³⁾. Answer: x⁽²⁄³⁾. The distractors: x⁽³⁄²⁾ comes from writing the fraction upside down, with the root's index on top; x⁽¹⁄⁶⁾ comes from treating the square as a second root and multiplying 1/3 by 1/2; x⁶ comes from multiplying the root's index by the power, 3 × 2, and keeping the result as a whole-number index.
- (c) 1/4 — A negative index means the reciprocal of the positive power, so 2⁻² = 1 ÷ 2² = 1/4. Treating the negative sign as making the answer negative instead gives −(2²) = −4. Ignoring the negative sign altogether gives just 2² = 4. Finding the reciprocal correctly but then also applying a negative sign gives −1/4.
- (a) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (d) 9√3 cm — DE is opposite the 60° angle at F, and EF is adjacent to it, so DE = EF × tan 60° = 9 × √3 = 9√3 cm. 9√3/2 cm comes from using sin 60° = √3/2 instead of tan 60°. 3√3 cm comes from using tan 30° = 1/√3 instead of tan 60° (9 × 1/√3 = 9/√3 = 3√3). 18 cm is the hypotenuse DF, not DE: it comes from using cos 60° = 1/2 and working out 9 ÷ 1/2 = 18, which finds the wrong side of the triangle.
- (c) (12 + 3π) cm — The perimeter of a quarter-circle is made up of two straight radii plus a quarter of the circumference. The two radii give 2 × 6 = 12 cm, and a quarter of the circumference is (1/4) × 2 × π × 6 = 3π cm, so the total perimeter is (12 + 3π) cm. Giving only the curved part, 3π cm, forgets the two straight edges entirely. Using the full circumference, 2 × π × 6 = 12π, instead of a quarter of it gives (12 + 12π) cm. Including only one radius instead of two gives (6 + 3π) cm.
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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