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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- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (b) −3 — Method: the gradient of a straight line is the change in y divided by the change in x, with the two coordinates taken in the same order in the numerator as in the denominator. Working: going from (−1, 5) to (3, −7), the change in y is −7 − 5 = −12 and the change in x is 3 − (−1) = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3. The distractors: 3 comes from subtracting the y-coordinates in one order and the x-coordinates in the other, giving 12 ÷ 4; −1/3 comes from dividing the change in x by the change in y instead of the other way round, giving 4 ÷ (−12); −6 comes from working out 3 − (−1) as 3 − 1 = 2, so that the change in y is divided by 2 rather than by 4.
- (d) P lies outside the circle — The squared distance from the origin to P is 8² + 9² = 64 + 81 = 145, which is greater than 100, so P lies outside the circle. Choosing 'P lies inside the circle' comes from adding the coordinates directly (8 + 9 = 17) instead of squaring and adding them, and comparing 17 with 100. Choosing 'P lies on the circle' comes from squaring 9 incorrectly as 36 (as if it were 6²) instead of 81, giving 64 + 36 = 100, equal to the right-hand side. Choosing 'Cannot be determined' wrongly assumes the equation of the line itself is needed, when the radius and the distance from the origin to P are already enough.
- (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.
- (d) 5 — Method: rearrange the formula to make m the subject, then substitute F = 15.50. Working: F = 3 + 2.5m, so subtracting 3 from both sides gives F − 3 = 2.5m, then dividing by 2.5 gives m = (F − 3) / 2.5. Substituting F = 15.50: m = (15.50 − 3) / 2.5 = 12.50 / 2.5 = 5. The value 6.2 comes from dividing 15.50 by 2.5 without subtracting the fixed £3 first. The value 3.2 comes from dividing first and subtracting afterwards, in the wrong order: (15.50 / 2.5) − 3 = 3.2. The value 7.4 comes from adding £3 instead of subtracting it before dividing: (15.50 + 3) / 2.5 = 7.4.
- (d) 7n + 5 — Method: find the rate charged per extra chair, then find the fixed part of the cost that fits hiring 1 chair. Working: the cost rises by £7 for each extra chair (19 − 12 = 7, 26 − 19 = 7, 33 − 26 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 12, so c = 5. Answer: the cost in pounds is 7n + 5. The value 7n comes from ignoring the fixed part of the charge entirely. The value 7n + 12 comes from using the cost of 1 chair as the fixed part directly, without subtracting the per-chair rate first. The value 12n + 7 comes from swapping the roles of the cost of hiring 1 chair, £12, and the rate per extra chair, £7 — using the total for one chair as the coefficient of n and the rate as the fixed part.
- (b) 60 km/h — Method: the gradient of a distance-time graph is the change in distance divided by the change in time, and for a journey at a steady rate that gradient is the speed. Working: the change in distance is 195 − 15 = 180 km and the change in time is 3 hours, so the gradient is 180 ÷ 3 = 60 km/h. Answer: 60 km/h. The distractors: 180 km/h comes from stopping at the change in distance and never dividing by the 3 hours; 195 km/h comes from reading the final marker as the rate instead of working with the change between the two markers; 3 km/h comes from quoting the time taken, which belongs on the bottom of the fraction, as though it were the value of the fraction itself.
- (a) (3, 7) — y = f(x) + 5 is a vertical translation of y = f(x) by 5 units up — the translation vector is (0, 5) — so only the y-coordinate of any point changes. Turning point (3, 2) → (3, 2 + 5) = (3, 7). Adding the 5 to the x-coordinate, or treating it as a horizontal shift like y = f(x + 5), moves the wrong coordinate — check first whether the number sits inside or outside the brackets.
- (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).
- (d) 2x/(x − 2) — Factorise x² − 4 as (x − 2)(x + 2) first — it's a difference of two squares. The (x + 2) in the numerator then cancels with the (x + 2) in the factorised denominator, and 6x² ÷ 3x simplifies to 2x, leaving 2x/(x − 2). Writing 2/(x − 2) comes from over-cancelling 6x² ÷ 3x as 2 instead of 2x, dropping the x that should remain. Writing 2x/(x + 2) comes from factorising x² − 4 as (x + 2)² instead of (x − 2)(x + 2), a difference of two squares always has one plus and one minus bracket. Writing 2x²/(x − 2) comes from simplifying 6x² ÷ 3x as 2x² instead of 2x, dividing the coefficients but not reducing the power of x.
- (b) x = −4, y = −3 and x = 3, y = 4 — Substitute y = x + 1 into x² + y² = 25: x² + (x + 1)² = 25, which expands to x² + x² + 2x + 1 = 25, giving 2x² + 2x − 24 = 0, or x² + x − 12 = 0. Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3. Using y = x + 1: when x = −4, y = −3; when x = 3, y = 4. Distractor routes: x = 4, y = 5 and x = −3, y = −2 comes from mis-factorising x² + x − 12 as (x − 4)(x + 3), reversing the sign of each root. x = −4, y = −3 alone stops after finding the first root of the quadratic and never goes back for the second pair. x = 12, y = 13 comes from misreading x² + y² = 25 as the linear equation x + y = 25 and solving that together with y = x + 1.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (b) x = −3, y = −7 — x² + 6x + 2 = (x + 3)² − 3² + 2 = (x + 3)² − 7. Substituting x = −3: (−3)² = 9, 6 × (−3) = −18, so 9 − 18 + 2 = −7, confirming the minimum value −7 at x = −3: turning point (−3, −7). Using 6 instead of half of 6 inside the bracket gives (x + 6)² − 34, turning point (−6, −34) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (3, −7) — wrong, because (x + 3)² is zero at x = −3, not x = 3. Computing 9 − 2 = 7 instead of 2 − 9 = −7 flips the sign of the constant, giving (−3, 7) — wrong, since the completed square's constant must be evaluated as 2 minus 9, not 9 minus 2. Always check a turning point by substituting its x-value back into the original equation.
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (d) 6th term — The terms are aₙ = 3 × (√3)ⁿ⁻¹: a₁ = 3, a₂ = 3√3 ≈ 5.196, a₃ = 9, a₄ = 9√3 ≈ 15.588, a₅ = 27, a₆ = 27√3 ≈ 46.765. Since a₅ = 27 is still below 30 and a₆ ≈ 46.8 is above 30, the 6th term is the first to exceed 30. 4th term comes from treating the common ratio as 3 instead of √3: 3, 9, 27, 81, … — under that wrong ratio, 81, the 4th term, is the first over 30. 5th term comes from misjudging a₅ = 27 as already greater than 30. 7th term comes from tracking only the whole-number terms (3, 9, 27, 81, …) and reporting the ORIGINAL position of 81 in the full sequence — 81 is indeed a₇, but a₆ ≈ 46.8 already exceeds 30 first.
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