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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Answer key: Algebra worksheet — GCSE Higher
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- (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.
- (b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (d) 3m³ — Method: multiply the powers of m by adding their indices, then bring the number coefficient to the front. Working: m² × m has indices 2 and 1; add them to get 3, giving m³, then × 3 gives 3m³. Answer: 3m³. 3m² comes from multiplying the indices instead of adding them: 2 × 1 = 2, giving m², then × 3 = 3m². m³ comes from correctly combining the m's but dropping the coefficient 3. m⁶ comes from multiplying the index by the coefficient instead of writing the coefficient in front: taking the 2 in m² and the 3 to give m raised to the power 2 × 3, which is m⁶, with the lone m left out.
- (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
- (c) 7n − 1 — Method: find the common difference, then find the constant that fits the first term. Working: 13 − 6 = 7, 20 − 13 = 7, 27 − 20 = 7, so the terms increase by 7 each time and the nth term has the form 7n + c. Substituting n = 1: 7(1) + c = 6, so c = −1. Answer: the nth term is 7n − 1. The value 7n comes from leaving out the constant altogether. The value 7n + 6 comes from using the first term as the constant directly, without subtracting the common difference first. The value 6n + 1 comes from using the first term, 6, as the coefficient of n instead of the common difference, and then attaching +1 with the sign of the constant flipped.
- (c) 1 — At x = 0, f(0) = 4. Applying the transformations in order — reflect in the x-axis first, then translate up by 5 — gives −f(0) + 5 = −4 + 5 = 1. Applying the translation but forgetting the reflection gives f(0) + 5 = 9. Applying the reflection to the whole expression, including the +5, gives −f(0) − 5 = −9. Applying the reflection but forgetting the translation gives −f(0) = −4.
- (c) 6 — Method: substitute the value, work out the top of the fraction first, then the division, and add the 3 last. Working: the top gives 10 − 4 = 6, dividing by 2 gives 6 ÷ 2 = 3, and adding 3 gives 3 + 3 = 6. Answer: 6. The distractors: 11 comes from dividing only the 4 by 2 instead of the whole of the top, giving 10 − 2 + 3; 4.5 comes from dividing the + 3 by 2 as well, giving (10 − 4 + 3) ÷ 2; 0 comes from subtracting the wrong way round on the top, giving (4 − 10) ÷ 2 = −3 and then −3 + 3.
- (d) 9 — Method: expand the brackets fully first, then collect the x-terms and constants before dividing. Working: 4(x − 3) = 4x − 12, so 4x − 12 = 2x + 6; 4x − 2x = 6 + 12; 2x = 18; x = 9. Answer: x = 9. 4.5 comes from expanding only the x-term in the bracket and forgetting to multiply the 3, giving 4x − 3 = 2x + 6, then 2x = 9, x = 4.5. −3 comes from a sign error when expanding, giving 4x + 12 = 2x + 6, then 2x = −6, x = −3. 3 comes from a sign error collecting the x-terms, adding instead of subtracting: 4x + 2x = 6 + 12, so 6x = 18, x = 3.
- (a) x = 4, y = −3 — Method: both equations contain +y with the same coefficient, so subtracting one equation from the other removes y. Working: (2x + y) − (x + y) = 5 − 1 gives x = 4; substituting x = 4 into x + y = 1 gives 4 + y = 1, so y = −3. Answer: x = 4, y = −3, which also satisfies 8 − 3 = 5. The distractors: x = 4, y = 5 comes from rearranging x + y = 1 as y = 1 + x; x = −2, y = 3 comes from eliminating x by doubling the first equation and then reading −y = 3 as y = 3; x = 6, y = −5 comes from adding the constants instead of subtracting them while eliminating y, taking x as 5 + 1.
- (c) 6 — Method: rate = amount of fuel used ÷ time taken. Working: fuel used = 50 − 38 = 12 litres, time taken = 2 hours, so rate = 12 ÷ 2 = 6 litres per hour. Answer: the rate is 6 litres per hour. 44 comes from finding the average of the two fuel amounts, (50 + 38) ÷ 2, instead of the fuel used. 12 comes from working out the fuel used but forgetting to divide by the time taken. 19 comes from dividing the remaining fuel, 38, by the time taken instead of the fuel used.
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (b) y ≥ 2, y ≤ x, x ≤ 6 — "On or above the line y = 2" means y ≥ 2. "On or below the line y = x" means y ≤ x. "On or to the left of the line x = 6" means x ≤ 6. Together these give y ≥ 2, y ≤ x, x ≤ 6. Distractor routes: y ≤ 2, y ≤ x, x ≤ 6 flips the first inequality, describing "on or below" y = 2 instead of "on or above". y ≥ 2, y ≥ x, x ≤ 6 flips the second, describing "on or above" y = x instead of "on or below". y ≥ 2, y ≤ x, x ≥ 6 flips the third, describing "on or to the right of" x = 6 instead of "on or to the left".
- (c) y = 3x + 2 — Method: divide the change in the outputs by the change in the inputs to find the multiplier, then put one pair of values into the rule to find the number added on. Working: the output rises by 11 − 5 = 6 while the input rises by 3 − 1 = 2, so the multiplier is 6 ÷ 2 = 3; with an input of 1, 3 × 1 = 3 and the output is 5, so 2 is added. Answer: y = 3x + 2, checked against the second pair by 3 × 3 + 2 = 11. The distractors: y = 3x − 2 comes from finding the multiplier 3 and then subtracting the 2 instead of adding it; y = 2x + 3 comes from swapping the multiplier and the number added on; y = x + 4 comes from assuming the input is multiplied by 1 and using 5 − 1 = 4 as the number added on.
- (c) 37 — The first differences are 3, 5, 7, 9 — they increase by 2 each time (the second difference), so the next first difference is 11, giving 26+11=37. A candidate who repeats the last first difference (9) instead of increasing it would reach 26+9=35. A candidate who increases the difference by 4 instead of 2 would reach 26+13=39. A candidate who adds only the second difference (2) to the last term, instead of the next first difference, would reach 26+2=28.
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