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) (3, 0) — Method: every point on the x-axis has y-coordinate 0, so substituting y = 0 into the equation and solving gives the x-coordinate of the crossing point. Working: 0 = 2x − 6 gives 2x = 6, so x = 6 ÷ 2 = 3 and the graph crosses the x-axis at (3, 0). Answer: (3, 0). The distractors: (0, −6) is where the graph crosses the y-axis, found by substituting x = 0 rather than y = 0; (−3, 0) comes from moving the 6 across the equals sign without changing its sign, giving 2x = −6; (6, 0) comes from reading the constant straight off as the crossing point and never dividing by the gradient 2.
- (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.
- (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (a) (−6, −3) — Reflecting in the y-axis changes the sign of the x-coordinate and keeps the y-coordinate the same: (−6, −3). (6, 3) comes from reflecting in the x-axis instead, which changes the sign of the y-coordinate. (−6, 3) comes from reflecting in both axes. (6, −3) comes from not applying the reflection at all.
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (d) 5 — Each term is found by multiplying the previous term by the common ratio, 0.5: 80, 40, 20, 10, and the next term is 10 × 0.5 = 5. A candidate who instead subtracts the same amount each time (repeating the last difference of 10) would reach 10 − 10 = 0. A candidate who divides by 4 instead of by 2 would reach 10 ÷ 4 = 2.5. A candidate who multiplies by 2 instead of dividing (reversing the direction of the sequence) would reach 10 × 2 = 20.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (d) 44 m — The trapezium rule between t = 2 and t = 5 uses only the speeds at t = 2, 3, 4 and 5 — 12, 15, 16 and 14 — with three strips of width 1: adding the first and last readings and twice the sum of the middle readings gives 12 + 14 + 2 × (15 + 16) = 88, and half of that is 44 m. Writing 68.5 m comes from applying the trapezium rule to the whole table, from t = 0 to t = 6, instead of restricting it to the interval t = 2 to t = 5 that the question asks for. Writing 28.5 m comes from forgetting to double the two middle readings, 15 and 16, in the trapezium rule formula. Writing 39 m comes from averaging only the first and last speeds in the interval, 12 and 14, and multiplying by the 3-second interval, ignoring the readings at t = 3 and t = 4 in between.
- (a) n² + n — n² + n = n(n + 1), the product of two consecutive integers. One of any two consecutive integers is always even, so their product is always even, whatever whole number n is. n² − n + 1 = n(n − 1) + 1 is always ODD, since n(n − 1) is even and adding 1 makes it odd — the opposite of what's asked. 2n + 1 is always odd by definition, not even. n² + 1 is not always even at all: it depends on whether n is odd or even, and testing n = 2 gives 5, which is odd.
- (a) 2 m/s² — Method: acceleration = change in speed ÷ time = (17 − 5) ÷ 6 = 12 ÷ 6 = 2 m/s². Distractor origins: 12 m/s² stops after finding the change in speed and forgets to divide by the time; 22 m/s² adds the two speeds instead of subtracting them, and also forgets to divide by time (5 + 17 = 22); 72 m/s² multiplies the change in speed by the time instead of dividing (12 × 6 = 72).
- (b) 20 — The rate is the gradient: the change in litres divided by the change in time. From 2 to 7 minutes, the tank gains 150 − 50 = 100 litres over 7 − 2 = 5 minutes, so the rate is 100 ÷ 5 = 20 litres per minute. 100 comes from working out the change in litres but forgetting to divide by the change in time. 25 comes from using only the first reading, 50 ÷ 2, and ignoring the second reading entirely. 0.05 comes from dividing the change in time by the change in litres instead of the other way round, 5 ÷ 100.
- (b) x = 2, y = −5 — 2x² − 8x + 3 rewrites as 2(x² − 4x) + 3, then as 2[(x − 2)² − 4] + 3, which simplifies to 2(x − 2)² − 5, since −2 × 4 + 3 = −5. Substituting x = 2: 2 × 2² = 8, 8 × 2 = 16, so 8 − 16 + 3 = −5, confirming the minimum value −5 at x = 2: turning point x = 2, y = −5. Halving b instead of halving b/a — using 4 as the shift instead of 2 — lands on turning point x = 4, y = −29, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = −2, y = −5 — wrong, because (x − 2)² is zero at x = 2, not x = −2. Computing 8 − 3 = 5 instead of 3 − 8 = −5 flips the sign of the constant, giving x = 2, y = 5 — wrong, since the completed square's constant must be evaluated as 3 minus 8, not 8 minus 3. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (b) £115, and C = 40 + 25h is a formula — Substitute h = 3 into the rule, multiplying before adding. The hours cost 25 × 3 = 75, and adding the call-out fee gives 40 + 75 = 115, so the charge is £115. The rule itself links two different quantities, C and h, and lets one be worked out from the other, so it is a formula; an expression would have no equals sign in it. Adding the fee before multiplying gives 65 × 3 = 195, which charges the call-out fee three times over, and stopping at 40 + 25 leaves £65, the charge for a single hour.
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (b) (3x + 1)(2x − 3) — To factorise 6x² − 7x − 3, look for two numbers that multiply to 6 × (−3) = −18 and add to −7: these are −9 and 2. Splitting the middle term gives 6x² − 9x + 2x − 3, which groups to 3x(2x − 3) + 1(2x − 3) = (3x + 1)(2x − 3). Writing (3x − 1)(2x + 3) comes from using the pair 9 and −2 instead — the right product but the wrong signs, giving +7x instead of −7x. Writing (6x − 1)(x + 3) comes from picking the factor pair 18 and −1: it multiplies to −18 correctly, but it adds to 17, not −7, so checking only the product splits the middle term as 6x² + 18x − x − 3 and pairs the wrong factors of 6 and 3 together. Writing (x − 3)(6x + 1) comes from that same unchecked pair used the other way round, splitting the middle term as 6x² + x − 18x − 3.
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