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.
Answer key: Algebra worksheet — GCSE Higher
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- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (d) x = 4 or x = −4 — Method: divide both sides by 2 to get x² = 16, then take the square root of both sides: x = 4 or x = −4. Distractor origins: x = 4 forgets the negative root; x = 8 comes from halving 16 instead of taking its square root; x = 16 stops at x² = 16 without ever taking the square root.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (b) n² + 4n — First differences of 5, 12, 21, 32, 45 are 7, 9, 11, 13. Second differences are 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the totals (5, 12, 21, 32, 45) leaves 4, 8, 12, 16, 20, the linear expression 4n. So the number of oranges is n² + 4n. Using the second difference itself as the coefficient of n², without halving it, gives 2n² + 4n. Finding a = 1 correctly but dropping the linear remainder 4n entirely leaves n². Treating the first first difference (7) as a constant common difference and building a linear formula 5 + 7(n − 1) = 7n − 2 fits only the first two totals, and gives 19 for n = 3 instead of 21.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (c) h = V / (lw) — Method: undo the multiplication by lw by dividing both sides by lw. Working: V = lwh, so dividing both sides by lw gives h = V / (lw). The value h = V − lw comes from subtracting lw instead of dividing by it. The value h = Vlw comes from multiplying by lw instead of dividing. The value h = lw / V comes from inverting the fraction, dividing lw by V instead of V by lw.
- (a) 842.0 — The radius to (20, 21) has gradient 21/20, so the tangent's gradient is −20/21. The tangent line is y − 21 = −20/21(x − 20), i.e. y = −20/21x + 841/21. Setting y = 0 gives the x-intercept x = 841/20 = 42.05; setting x = 0 gives the y-intercept y = 841/21 ≈ 40.048. The area of triangle OPQ is 1/2 × 42.05 × 40.048 ≈ 842.0. 1684.0 comes from multiplying the two intercepts without the 1/2 that a triangle's area needs — twice the correct area. 580.7 comes from using the circle's radius, 29, as a side of the triangle instead of the x-intercept, 42.05: 1/2 × 29 × 40.048 ≈ 580.7. 2.0 comes from a sign error in the tangent's gradient — using 20/21 instead of −20/21 — which gives a different line, with intercepts x ≈ −2.05 and y ≈ 1.952, and area 1/2 × 2.05 × 1.952 ≈ 2.0.
- (d) n² + 3 — First differences: 3, 5, 7, 9. Second differences: 2, 2, 2, so the sequence is quadratic and the coefficient of n² is half the second difference: a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (4, 7, 12, 19, 28) leaves 3, 3, 3, 3, 3, a constant, so the nth term is n² + 3. Using the second difference itself as a, without halving it, gives 2n² + 3. Finding a = 1 correctly but then dropping the constant remainder gives n². Treating the first first difference (3) as a common difference and building a linear formula a + (n − 1)d = 4 + 3(n − 1) gives 3n + 1, which fits only the first term.
- (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) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (d) 11 — f⁻¹(x) = 5 means x = f(5), since applying f to both sides undoes the inverse. f(5) = 2 × 5 + 1 = 11. Writing 2 comes from confusing f⁻¹(x) = 5 with f(x) = 5, and solving 2x + 1 = 5 instead: 2x = 4, x = 2. Writing 9 comes from finding f⁻¹(x) with a sign error, f⁻¹(x) = (x + 1)/2 instead of (x − 1)/2, then setting this equal to 5: x + 1 = 10, x = 9. Writing 6 comes from finding f⁻¹(x) without dividing by 2 at all, f⁻¹(x) = x − 1, then setting this equal to 5: x = 6.
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
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