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.Solve 5x² − 15x = 0.
- 2.The recurring decimal 0.181818... can be written as 0.18 recurring, where both digits repeat forever. Let x = 0.18 recurring. Work out x as a fraction in its simplest form.
- 3.Write 0.36 as a fraction in its simplest form.
- 4.A cake recipe needs 3/4 of a kilogram of sugar. Aisha wants to make half the recipe. Work out how much sugar she needs, giving your answer as a fraction of a kilogram in its simplest form.
- 5.Work out the value of .
- 6.Write 5/6 as a decimal, showing clearly which digit recurs.
- 7.A ramp rises at 30° to the horizontal. Its sloping surface is 4.8 m long. Safety rules say the vertical rise of a ramp must be no more than 2.5 m. Work out how far below that limit the rise of this ramp is.
- 8.Work out the value of √49 + ∛27
- 9.3/5 of a 5/6 litre bottle of juice is poured out. Work out the exact volume poured out, in litres.
- 10.Put sin 30°, tan 30° and cos 30° in order of size, starting with the smallest.
- 11.Which of these values of x is a solution of x² + 2x − 15 = 0?
- 12.Work out the value of 2⁻²
- 13.Factorise 6x² − 7x − 3.
- 14.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 15.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.
- 16.Work out the value of .
- 17.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.
- 18.When a number is added to its square the result is 30. Work out the possible values of the number.
- 19.Work out the value of .
- 20.Rationalise the denominator of 6/√3, giving your answer in its simplest form.
Answer key
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (a) 2/11 — Let x = 0.18 recurring, so x = 0.181818... . Since two digits repeat, multiply by 100: 100x = 18.181818... . Subtracting the original x removes the recurring part, because the digits line up exactly: 100x − x = 18.181818... − 0.181818... = 18, so 99x = 18, giving x = 18/99 = 2/11. Treating the decimal as if it terminated at two places gives 18/100 = 9/50, which is only 0.18 and drops the repeating part entirely. Subtracting 10x instead of x — using 100x − 10x = 90x = 18 — is the wrong power of ten for a two-digit repeating block, and gives x = 18/90 = 1/5. Making an arithmetic slip in the numerator, 18 − 1 = 17 instead of 18, gives 17/99.
- (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.
- (d) 3/8 — Method: making half the recipe means dividing the quantity of sugar by 2. Working: 3/4 ÷ 2 = 3/8. Answer: 3/8. 3/2 comes from multiplying by 2 instead of dividing, as if doubling the recipe. 5/4 comes from adding 1/2 to 3/4 instead of halving it, confusing "half of" with "plus a half". 3/4 comes from leaving the amount unchanged, forgetting to halve it for the smaller recipe.
- (d) 8 — Method: write $16^{3/4}$ as $(\sqrt[4]{16})^3$ — the denominator of the index gives the root, the numerator gives the power. Working: $\sqrt[4]{16} = 2$, so $16^{3/4} = 2^3 = 8$. Answer: 8. A candidate who multiplies 16 by 3/4 is treating the index as an ordinary factor and gets 12 — a fractional index is not a multiplier. A candidate who takes the square root instead of the fourth root and then cubes it works out $(\sqrt{16})^3 = 4^3$ and gets 64; the denominator 4 names a fourth root, not a square root. A candidate who takes the fourth root of 16 correctly but stops there, without cubing it, gets 2.
- (b) 0.83333... — Divide 5 by 6 using long division. 5.000... ÷ 6: 50 ÷ 6 = 8 remainder 2, giving the first decimal digit 8. Bring down a 0 to make 20, and 20 ÷ 6 = 3 remainder 2 — the remainder 2 has reappeared, so from here the digit 3 repeats forever. This gives 5/6 = 0.83333... . Stopping after two decimal places and writing 0.83 treats the division as if it terminated, when the remainder never reaches zero. Shifting the decimal point one place too far to the left gives 0.083333..., the same digits divided by an extra power of ten. A slip in the long division itself, misreading a remainder, can produce the wrong repeating digit, 0.85555... .
- (a) 0.1 m — Method: the sloping surface is the hypotenuse and the vertical rise is the side opposite the 30° angle, so rise = 4.8 × sin 30°; then compare that rise with the limit. Working: the exact value of sin 30° is one half, so the rise = 4.8 × 1/2 = 2.4 m. The limit is 2.5 m, and 2.5 − 2.4 = 0.1. Answer: the ramp is 0.1 m below the limit. Working out the rise and stopping there gives 2.4 m, which answers a question that was not asked. Dividing by sin 30° instead of multiplying gives 4.8 ÷ 0.5 = 9.6 and then 9.6 − 2.5 = 7.1 m. Treating sine as proportional to the angle, so that sin 30° is a third of sin 90°, gives 4.8 ÷ 3 = 1.6 and then 2.5 − 1.6 = 0.9 m.
- (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.
- (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) sin 30°, tan 30°, cos 30° — sin 30° = 1/2 = 0.5, tan 30° = √3/3 ≈ 0.577 and cos 30° = √3/2 ≈ 0.866, so the correct order from smallest to largest is sin 30°, tan 30°, cos 30°. 'sin 30°, cos 30°, tan 30°' swaps the last two, wrongly putting cos 30° before tan 30°. 'cos 30°, tan 30°, sin 30°' is the correct list written backwards, from largest to smallest. 'tan 30°, sin 30°, cos 30°' wrongly swaps sin 30° and tan 30° at the start.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (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.
- (b) (3x + 1)(2x − 3) — To factorise 6x² − 7x − 3, look for two numbers that multiply to 6 × (−3) = −18 and add to −7: these are −9 and 2. Splitting the middle term gives 6x² − 9x + 2x − 3, which groups to 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Writing (3x − 1)(2x + 3) comes from using the pair 9 and −2 instead — the right product but the wrong signs, giving +7x instead of −7x. Writing (6x − 1)(x + 3) comes from picking the factor pair 18 and −1: it multiplies to −18 correctly, but it adds to 17, not −7, so checking only the product splits the middle term as 6x² + 18x − x − 3 and pairs the wrong factors of 6 and 3 together. Writing (x − 3)(6x + 1) comes from that same unchecked pair used the other way round, splitting the middle term as 6x² + x − 18x − 3.
- (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.
- (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.
- (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.
- (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.
- (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) 0.001 — Method: a negative index means 'one over' the positive power, so $10^{-3}$ means one over $10^{3}$. Working: ten cubed is 1000, and one over 1000 is 0.001. 1000 comes from ignoring the negative sign and working out ten cubed instead of its reciprocal. −1000 comes from ignoring what the negative index does to the size, while still writing a negative sign on the large value. −0.001 comes from correctly finding the size, 0.001, but wrongly keeping a negative sign because the index was negative. Answer: 0.001.
- (a) 2√3 — Multiply the top and bottom of the fraction by √3, since √3 × √3 = 3: 6/√3 = (6 × √3)/(√3 × √3) = 6√3/3. Dividing 6 by 3 gives 2, so the fraction simplifies to 2√3. Multiplying only the numerator by √3 and then cancelling the surd in the denominator against it as if they were the same term, without properly squaring the denominator, leads to 6. Dividing 6 by 3 as 3 instead of 2 gives 3√3 — a slip in the final division. Simplifying 6√3/3 by cancelling the whole numerator's 3 with the denominator's 3, including the surd, gives 2, which loses the surd altogether.
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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