Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Algebra worksheet — GCSE Higher
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- 1.A delivery van's distance-time graph shows the following: it travels 9 km in the first 15 minutes at a steady speed, stops for 45 minutes to unload, then travels a further 21 km in the next 30 minutes, also at a steady speed. Work out the average speed, in km/h, for the whole journey.
- 2.A box holds n pencils. A shop has 6 full boxes and 4 loose pencils. All of these pencils are shared equally between 2 classes. Write down the expression for the number of pencils each class receives.
- 3.A student is asked to simplify (6x³ − 15x²)/(4x² − 10x) fully. Four attempts are shown below. Which one is correct?
- 4.A student claims: 'For every positive integer n, n² + n + 1 is a prime number.' Which value of n shows that this claim is false?
- 5.Expand and simplify 2(a + 6) − 2(a − 7)
- 6.Solve x² − 8x + 3 = 0 by completing the square. Give your answers in surd form.
- 7.Solve 6x − 5 = 3x + 13.
- 8.The numbers x and y satisfy x + y = 10 and xy = 21. Work out the pair of values.
- 9.Three consecutive integers add up to 72. Using n for the smallest integer, form an equation and solve it to find the smallest of the three integers.
- 10.Ben writes n + 4 ≤ 9. Which statement describes what Ben has written? Give a reason for your answer.
- 11.Write down the expression that means the same as (x + 7) ÷ 4.
- 12.The point (4, 2) lies on the circle x² + y² = 20. Work out the equation of the tangent to the circle at (4, 2).
- 13.A proof that (n + 3)² − (n − 3)² is always a multiple of a certain number begins: Line 1: (n + 3)² − (n − 3)² = (n² + 6n + 9) − (n² − 6n + 9). Which expression correctly completes Line 2?
- 14.A hot air balloon is at a height of 20 m and rises at a steady rate. After 4 minutes it is at a height of 52 m. Work out the rate at which the balloon rises, in metres per minute.
- 15.A taxi fare in pounds is given by F = 2.50 + 1.20m, where m is the number of miles travelled. Sam takes a taxi for a journey of 6 miles. Work out the fare.
Answer key
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (c) (6n + 4)/2 — Six full boxes hold 6 lots of n pencils, which is 6n, and the 4 loose pencils are added on, so the shop has 6n + 4 pencils altogether. Sharing them equally between 2 classes divides that whole total by 2, and brackets are what show that the division applies to all of it: (6n + 4)/2. Without the brackets, 6n + 4/2 halves only the loose pencils; 6(n + 4)/2 adds the loose pencils to every box before the division; 2(6n + 4) doubles the total instead of halving it.
- (a) 3x/2 — Factorise top and bottom first: 6x³ − 15x² = 3x²(2x − 5), and 4x² − 10x = 2x(2x − 5). The bracket (2x − 5) is common to both, so it cancels, leaving 3x²/(2x); dividing the power of x, 3x² ÷ x = 3x, gives 3x/2. Writing 3x²/2 cancels the (2x − 5) correctly and removes the denominator's x, but never reduces the power of x left in the numerator — 3x² ÷ x should give 3x, not stay as 3x². Writing 3/2 cancels the (2x − 5) correctly but then drops the x from the numerator altogether, treating 3x² over x as if it cancelled completely to 3 instead of reducing to 3x. Writing −3x/2 comes from factorising the denominator with the wrong sign, as 2x(5 − 2x) instead of 2x(2x − 5); cancelling (5 − 2x) against the numerator's (2x − 5) then needs an extra minus sign, which flips the answer to −3x/2.
- (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
- (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
- (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
- (d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
- (a) 3 and 7 — Method: one equation is linear and the other is not, so rearrange the linear equation and substitute it into the other to leave a single quadratic in one letter. Working: x + y = 10 gives y = 10 − x, so xy = 21 becomes x(10 − x) = 21, which rearranges to x² − 10x + 21 = 0; factorising gives (x − 3)(x − 7) = 0, so the two values are 3 and 7, and 3 + 7 = 10 with 3 × 7 = 21. Answer: 3 and 7. The distractors: 1 and 21 comes from using only the product and taking the first factor pair of 21; −3 and −7 comes from factorising as (x + 3)(x + 7), which reverses the sign of both values; 5 and 5 comes from using only the sum and splitting 10 into two equal parts.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
- (d) An inequality, because ≤ compares the two sides — The symbol ≤ means 'is less than or equal to', so the statement compares the sizes of the two sides instead of saying they are equal: that makes it an inequality. Solving it gives n ≤ 5, a whole range of values rather than the single value an equation would give. An identity has to be true for every value of the letter, and this fails at n = 6, so it is not one. A formula works one quantity out from another, and there is only one letter here.
- (c) (x + 7)/4 — Method: dividing an expression by a number is written as a fraction, with the whole expression on top. Working: (x + 7) ÷ 4 = (x + 7)/4. Answer: (x + 7)/4. 4/(x + 7) comes from writing the numbers the wrong way round, putting 4 on top. 4(x + 7) comes from reading ÷ as ×, multiplying instead of dividing. (x + 7) − 4 comes from reading ÷ as −, subtracting instead of dividing.
- (d) y = −2x + 10 — Method: a tangent is perpendicular to the radius drawn to the point where it touches, so work out the gradient of that radius, take its negative reciprocal for the tangent, then substitute into y − y₁ = m(x − x₁). Working: the radius joins (0, 0) to (4, 2), so its gradient is 2 ÷ 4 = 1/2; turning 1/2 upside down gives 2 and changing the sign gives −2. Substituting into y − 2 = −2(x − 4) gives y − 2 = −2x + 8, so y = −2x + 10. Answer: y = −2x + 10. The distractors: y = −0.5x + 4 changes the sign of the radius gradient but never turns it upside down, using −1/2 where −2 belongs; y = 2x − 6 turns the gradient upside down but leaves it positive, using 2 where −2 belongs; y = −2x − 10 has the correct gradient but substitutes the point with both signs reversed, writing y + 2 = −2(x + 4) instead of y − 2 = −2(x − 4).
- (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
- (a) 8 — Method: rate = height gained ÷ time taken. Working: height gained = 52 − 20 = 32 m, time taken = 4 minutes, so rate = 32 ÷ 4 = 8 metres per minute. Answer: the rate is 8 metres per minute. 32 comes from working out the height gained but forgetting to divide by the time taken. 13 comes from dividing the final height, 52, by the time taken instead of the height gained. 5 comes from dividing the starting height, 20, by the time taken instead of the height gained.
- (c) £9.70 — F = 2.50 + 1.20 × 6 = 2.50 + 7.20 = £9.70. £7.20 comes from forgetting to add the £2.50 fixed charge. £10.20 comes from adding the 2.50 to the number of miles before multiplying by 1.20, 1.20 × (6 + 2.50). £22.20 comes from adding the fixed charge and the rate together first and then multiplying by the miles, (2.50 + 1.20) × 6.
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