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 (√5)⁴
- 2.A circle has a circumference of 8π cm. Work out the exact area of the circle in terms of π.
- 3.Simplify (3a²)² × (2a)³
- 4.Work out the exact value of (cos 45°)² + (sin 45°)².
- 5.Factorise fully 56x − 24
- 6.Write 0.35 as a fraction in its lowest terms.
- 7.Work out the value of .
- 8.Work out 1 1/2 ÷ 3/4 exactly, giving your answer in its simplest form.
- 9.Which of these values of x is a solution of x² + 2x − 15 = 0?
- 10.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.
- 11.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.
- 12.Work out the value of .
- 13.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
- 14.When a number is added to its square the result is 30. Work out the possible values of the number.
- 15.Work out the value of .
- 16.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.
- 17.Expand and simplify 4(2x − 1) − 3(x + 2) − 5x
- 18.Sam compares 0.6 and 5/8 by comparing the digit 6 with the digit 5, and says that 0.6 is the larger number. Convert 5/8 to a decimal to find the correct larger value.
- 19.Expand and simplify (x − 3)²
- 20.Decide which of 2³⁰ and 3²⁰ is the larger number, and write down the correct statement.
Answer key
- (a) 25 — Method: squaring a square root removes the root, so a square root raised to the power 4 can be squared in two stages. Working: (√5)⁴ = ((√5)²)² = 5² = 5 × 5 = 25. Answer: 25. The distractors: 5 comes from squaring once and stopping, treating the fourth power as a square; 625 comes from raising 5 to the power 4 and ignoring the root sign altogether; 20 comes from multiplying 5 by the index 4 instead of raising 5 to that power.
- (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.
- (d) 72a⁷ — Method: a power outside brackets applies to every factor inside them, and multiplying two powers of the same letter adds their indices. Working: (3a²)² = 3² × a⁴ = 9a⁴, and (2a)³ = 2³ × a³ = 8a³. Multiplying the two results gives 9 × 8 = 72 for the number and 4 + 3 = 7 for the index of a. Answer: 72a⁷. The distractors: 36a⁷ comes from squaring the 2 in (2a)³ instead of cubing it, giving 4a³ and then 9 × 4; 72a¹² comes from multiplying the indices 4 and 3 when the two terms are multiplied, instead of adding them; 17a⁷ comes from adding the coefficients 9 and 8 rather than multiplying them.
- (c) 1 — cos 45° = √2/2 and sin 45° = √2/2. Squaring each gives (√2/2)² = 2/4 = 1/2, so (cos 45°)² + (sin 45°)² = 1/2 + 1/2 = 1. The distractor √2 comes from adding cos 45° + sin 45° directly without squaring first (√2/2 + √2/2 = √2). The distractor 2 comes from squaring the top of the fraction, (√2)² = 2, but then dividing by 2 instead of 4 for each term, giving 1 + 1 = 2. The distractor 1/2 comes from squaring only cos 45° and forgetting to add the sin 45° term.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (d) 7/20 — Method: write the decimal over the power of ten that matches the number of digits after the point, then divide the numerator and the denominator by their highest common factor. Working: 0.35 has two digits after the point, so it is 35 hundredths and can be written as 35/100; the highest common factor of 35 and 100 is 5, and 35 ÷ 5 = 7 with 100 ÷ 5 = 20. Answer: 7/20. The distractors: 3/10 comes from reading only the first digit after the point and converting 0.3; 7/25 comes from dividing the numerator by 5 but the denominator by 4, using a different factor on the top and on the bottom; 35/10 comes from counting one decimal place instead of two and writing the digits over 10.
- (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.
- (c) 2 — First write 1 1/2 as an improper fraction, 3/2. To divide by 3/4, multiply by its reciprocal, 4/3: 3/2 × 4/3 = 12/6 = 2. Dropping the whole number and dividing only the fractional part, 1/2 ÷ 3/4 = 1/2 × 4/3, gives 2/3. Multiplying by 3/4 directly instead of using its reciprocal, 3/2 × 3/4, gives 9/8. Using the reciprocal of the first fraction instead of the second, 2/3 × 3/4, gives 1/2.
- (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.
- (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.
- (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.
- (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) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
- (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.
- (d) 1/4 — Method: deal with the fractional index first, then the negative sign. Working: $8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$. A negative index means take the reciprocal of that result, so $8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{4}$. Answer: 1/4. A candidate who evaluates $8^{2/3}$ correctly but forgets the negative sign entirely gets 4 — they have dropped the instruction to take a reciprocal. A candidate who takes the reciprocal step but applies it as a sign change to the finished number instead of inverting it gets −4. A candidate who multiplies 8 by −2/3, treating the index as an ordinary factor rather than a power, gets −16/3.
- (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) −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.
- (a) 0.625 — Method: convert the fraction to a decimal so it can be compared properly with 0.6. Working: 5/8 = 0.625, and since 0.625 > 0.6, the larger value is 0.625. Answer: 0.625. 0.6 repeats Sam's incorrect claim, made by comparing single digits rather than full place value. 0.58 comes from converting 5/8 incorrectly, treating it as if it read 5 tenths and 8 hundredths. 0.85 comes from turning the fraction upside down and writing its digits straight after the decimal point, 8 then 5, instead of dividing.
- (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).
- (b) 3²⁰ is larger — Method: two powers with different bases and different indices can be compared once they are rewritten with a common index, which is possible whenever the indices share a factor. Working: 30 and 20 have a highest common factor of 10, so 2³⁰ = (2³)¹⁰ = 8¹⁰ and 3²⁰ = (3²)¹⁰ = 9¹⁰. Both are now tenth powers, and since 9 is larger than 8, 9¹⁰ is larger than 8¹⁰. Answer: 3²⁰ is larger. The distractors: 2³⁰ is larger comes from comparing only the indices and choosing the power with the bigger index; They are equal comes from multiplying base by index, 2 × 30 and 3 × 20, and finding 60 each time; They cannot be compared without a calculator comes from assuming that powers this large can only be ranked by evaluating them in full.
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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