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.Subtract, giving your answer as a single fraction in its simplest form: 4/(x − 1) − 2/(x + 3)
- 2.A semicircle has a diameter of 8 cm. Work out the exact area of the semicircle, in terms of π.
- 3.A set of kitchen scales displays the mass of a bag of sugar as 0.63 recurring kilograms, meaning 0.636363... kg with the block '63' repeating forever. Convert this mass to a fraction of a kilogram, then work out the mass in grams, giving your answer to the nearest gram.
- 4.Write 7/12 as a decimal, showing clearly which digit is recurring.
- 5.The decimal 0.2333... has one non-recurring digit (the 2) followed by a single recurring digit (the 3), so it can be written as 0.2 recurring 3. Let x = 0.2333... . Work out x as a fraction in its simplest form.
- 6.A square tile has an area of 72 cm². Work out the exact perimeter of the tile, giving your answer in the form k√2 cm.
- 7.Simplify √8 + √18, giving your answer in the form k√2.
- 8.Rationalise the denominator of 10/(4 − √6), giving your answer in its simplest form.
- 9.Work out the value of the discriminant b² − 4ac for the equation 2x² + 3x − 2 = 0.
- 10.A tent's sloping side makes an angle of 45° with the ground. The sloping side is 3√2 m long. Using the exact value of sin 45°, work out the exact height of the tent.
- 11.Work out the exact value of cos 0° − sin 30°.
- 12.Expand and simplify (2 + √3)², giving your answer in the form a + b√3.
- 13.Solve x² − 2x − 24 = 0.
- 14.3/5 of a 5/6 litre bottle of juice is poured out. Work out the exact volume poured out, in litres.
- 15.Chloe wants to work out 3 1/4 − 1 2/3. She converts both mixed numbers to twelfths, then subtracts the whole numbers and the fraction parts separately, without checking whether she needs to exchange first. Work out the correct value of 3 1/4 − 1 2/3, giving your answer as a mixed number in its simplest form.
- 16.Write 0.36 as a fraction in its simplest form.
- 17.One solution of the equation x² − (k + 1)x + k = 0 is x = 3. Work out the value of k.
- 18.Work out √3 × √12, giving your answer as an integer.
- 19.Work out the exact value of (sin 30°)² + (cos 30°)².
- 20.Factorise fully 12x² + 18x.
Answer key
- (b) (2x + 14)/((x − 1)(x + 3)) — To subtract these fractions, write them over the common denominator (x − 1)(x + 3): the numerator becomes 4(x + 3) − 2(x − 1). Expanding gives 4x + 12 − 2x + 2, which simplifies to 2x + 14, so the answer is (2x + 14)/((x − 1)(x + 3)). Writing (2x + 11)/((x − 1)(x + 3)) comes from not distributing the minus sign fully across the second bracket, treating −2(x − 1) as −2x − 1 instead of −2x + 2. Writing 2/((x − 1)(x + 3)) comes from subtracting the two original numerators directly, 4 − 2 = 2, the same mistake as subtracting fractions without a common denominator. Writing (2x − 10)/((x − 1)(x + 3)) comes from swapping which numerator multiplies which bracket, forming 4(x − 1) − 2(x + 3) instead of 4(x + 3) − 2(x − 1).
- (b) 8π cm² — A diameter of 8 cm gives a radius of 4 cm. The area of a full circle would be π × r² = π × 4² = 16π cm², and a semicircle is exactly half of this, giving 16π ÷ 2 = 8π cm². Forgetting to halve the area for the semicircle gives 16π cm², the area of the whole circle. Halving the diameter twice, using a radius of 2 instead of 4, gives π × 2² = 4π cm². Using the diameter itself as the radius, so π × 8² = 64π, and then halving that for the semicircle gives 32π cm².
- (d) 636 g — Let x = 0.636363... . Since two digits repeat, multiply by 100: 100x = 63.636363... . Subtracting removes the recurring part exactly: 100x − x = 63.636363... − 0.636363... = 63, so 99x = 63, giving x = 63/99 = 7/11 kg. Converting to grams: 7/11 × 1000 = 7000/11 = 636.3636... g, which rounds to 636 g. Treating the decimal as if it terminated, writing 0.63 as 63/100 kg, gives 630 g when multiplied by 1000 — this drops the recurring part entirely. Subtracting 10x instead of x, using 100x − 10x = 90x = 63, is the wrong power of ten for a two-digit block, giving x = 63/90 = 7/10 kg, which is 700 g. A numerator slip in the subtraction, 63 − 1 = 62 instead of 63, gives x = 62/99 kg, which is 62000/99 = 626.26... g, rounding to 626 g.
- (b) 0.58333... — Divide 7 by 12 using long division. 70 ÷ 12 = 5 remainder 10, so the first decimal digit is 5. Bring down a 0 to make 100: 100 ÷ 12 = 8 remainder 4, so the second digit is 8. Bring down a 0 to make 40: 40 ÷ 12 = 3 remainder 4, so the third digit is 3. Bring down a 0 to make 40 again — the remainder 4 has reappeared, so the digit 3 repeats forever from here. This gives 7/12 = 0.58333... . Stopping the division after two digits and writing 0.58 treats it as if it terminated, when the remainder is not yet zero. Misreading the pattern as a two-digit repeating block, '58', gives 0.585858..., which wrongly makes the 5 recur as well as the 3. A slip in the long division that carries the wrong remainder forward can make the second digit itself appear to repeat instead of the third, giving 0.588888... .
- (c) 7/30 — Let x = 0.2333... . Because only the 3 recurs, use two multiples of x that line up the recurring part exactly: 10x = 2.333... and 100x = 23.333... . Subtracting removes the recurring tail completely: 100x − 10x = 23.333... − 2.333... = 21, so 90x = 21, giving x = 21/90 = 7/30. Treating the decimal as if it terminated after two places, writing 0.23 as 23/100, ignores that the 3 carries on forever. Misreading which digits recur — treating 0.2333... as if the block '23' repeated, giving 0.232323... — leads to x = 23/99, which is a different, larger recurring decimal from the one given. A numerator slip in the subtraction, computing 22 instead of 21, gives x = 22/90 = 11/45.
- (b) 24√2 — The side length of the tile is √72. Since 72 = 36 × 2, √72 = √36 × √2 = 6√2 cm. A square has four equal sides, so the perimeter is 4 × 6√2 = 24√2 cm. Simplifying √72 by writing the perfect-square factor itself as the coefficient instead of its root, 36√2 instead of 6√2, and then multiplying by 4 lands on 144√2. Working out the correct side length, 6√2 cm, but then giving that as the final answer without multiplying by 4 for the perimeter gives 6√2. Doubling the side length instead of quadrupling it, as if the perimeter were 2 × 6√2 rather than 4 × 6√2, gives 12√2.
- (a) 5√2 — Simplify each surd first: √8 = √4 × √2 = 2√2, and √18 = √9 × √2 = 3√2. Both terms are now multiples of √2, so they are like terms: 2√2 + 3√2 = 5√2. Adding the numbers under the two roots first, 8 + 18 = 26, and writing √26 treats unlike surds as if they combine under one root — they only combine once they share the same radicand, which is not how addition of surds works. Writing 9√2 for √18 instead of 3√2 (forgetting to root the 9) and then adding gives 2√2 + 9√2 = 11√2. Writing 4√2 for √8 instead of 2√2 (forgetting to root the 4) and adding gives 4√2 + 3√2 = 7√2.
- (a) 4 + √6 — Multiply top and bottom by the conjugate, 4 + √6. The denominator becomes (4 − √6)(4 + √6) = 4² − (√6)² = 16 − 6 = 10. The numerator becomes 10 × (4 + √6) = 40 + 10√6. So the fraction is (40 + 10√6)/10 = 4 + √6, since both terms in the numerator divide by 10. Distributing the conjugate to only the whole-number term of the numerator, and forgetting the surd term entirely, leaves just 4. Rationalising by multiplying the numerator by the conjugate but leaving the ORIGINAL denominator's sign unchanged instead of squaring it lands on 4 − √6, with the surd's sign never actually flipping to positive. Dividing only the whole-number part of the numerator by 10 and forgetting to divide the surd term too leaves 4 + 10√6.
- (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) 3 m — Height = sloping length × sin 45° = 3√2 × √2/2 = (3 × 2)/2 = 3 m, since √2 × √2 = 2. 3√2 m comes from forgetting to multiply by sin 45° at all. 3√2/2 m comes from using sin 30° = 1/2 instead of sin 45° = √2/2. 6 m comes from using √2 instead of √2/2 for sin 45°, dropping the denominator of the exact value: 3√2 × √2 = 6.
- (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.
- (c) 7 + 4√3 — Expand the brackets fully: (2 + √3)² = 2² + 2 × 2 × √3 + (√3)² = 4 + 4√3 + 3. Adding the two whole-number terms, 4 + 3 = 7, gives 7 + 4√3. Using (a + b)² = a² + b² and skipping the middle cross term entirely gives just 4 + 3 = 7, with no surd term at all. Treating (√3)² as if it stayed √3 rather than becoming 3, then merging it with the existing surd term, gives 4 + 5√3. Squaring only the surd term correctly but carrying the whole-number term as 2 instead of squaring it to 4 gives 2 + 3 + 4√3 = 5 + 4√3.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (a) 1/2 — To find a fraction of an amount, multiply the fractions together: 3/5 × 5/6 = 15/30, which simplifies to 1/2 litre. Adding the fractions instead of multiplying them, using a common denominator of 30, gives 18/30 + 25/30 = 43/30, a value greater than the whole bottle. Dividing by 5/6 instead of multiplying by it, using its reciprocal 6/5, gives 3/5 × 6/5 = 18/25. Multiplying 5/6 by itself instead of by 3/5 gives 25/36.
- (d) 1 7/12 — Method: convert both mixed numbers to improper fractions with a common denominator, then subtract. Working: 3 1/4 = 39/12 and 1 2/3 = 20/12, so 39/12 − 20/12 = 19/12 = 1 7/12. Answer: 1 7/12. Chloe's method, subtracting whole numbers (3−1=2) and fraction parts (2/3−1/4=5/12) separately without exchanging, gives 2 5/12. 1 3/4 comes from converting 2/3 to twelfths incorrectly as 6/12 instead of 8/12, then subtracting. 8 comes from converting both mixed numbers to improper fractions correctly (13/4 and 5/3) but then subtracting numerators and denominators separately: (13−5)/(4−3) = 8/1.
- (b) 9/25 — Method: write the decimal over the matching power of ten, then divide the numerator and the denominator by their highest common factor. Working: 0.36 has two digits after the point, so 0.36 = 36/100; the highest common factor of 36 and 100 is 4, and 36 ÷ 4 = 9 with 100 ÷ 4 = 25; since 9 and 25 share no factor greater than 1, the fraction is fully cancelled. Answer: 9/25. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 9/50 comes from dividing the numerator by 4 but the denominator by only 2; 36/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (a) k = 3 — Method: a solution of an equation makes both sides balance, so substitute it in and solve the equation in k that is left. Working: putting x = 3 gives 3² − (k + 1) × 3 + k = 0, that is 9 − 3k − 3 + k = 0, so 6 − 2k = 0 and k = 3; the equation is then x² − 4x + 3 = 0, whose solutions are 3 and 1. Answer: k = 3. The distractors: k = −3 comes from solving 6 − 2k = 0 as though it gave 2k = −6; k = 4 comes from expanding −3(k + 1) as −3k − 1, multiplying only the k by 3; k = 1.5 comes from working 3² as 3 × 2 = 6, which leaves 3 − 2k = 0.
- (a) 6 — Use √a × √b = √(ab): √3 × √12 = √(3 × 12) = √36 = 6. Adding the numbers under the roots instead of multiplying them, 3 + 12 = 15, gives √15 — that comes from applying the rule for adding surds to a multiplication question. Multiplying the two numbers under the roots but then forgetting to take the square root at the end leaves 36. Simplifying only √12 to 2√3 and then dropping the other √3 factor entirely gives 2√3.
- (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.
- (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
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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