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 company's cost, in £, for producing x items is shown on a graph. The tangent to the curve at x = 50 passes through (30, 400) and (70, 800). Interpret the gradient of this tangent in the context of the company's costs.
- 2.A circle has centre (0, 0) and equation x² + y² = 64. Write down the radius of the circle.
- 3.The diagram shows the graph of a quadratic function. Which equation could it represent?
- 4.A quadratic graph has equation y = (x − 4)². A student says this graph crosses the x-axis at two different points. Explain why the student is wrong.
- 5.Work out the maximum value of y = sin x, and the smallest positive value of x, in degrees, at which it occurs.y = sin(x)
- 6.A company's weekly profit, P thousand pounds, when it makes x thousand items is modelled by P = x² − 8x + 12, for x ≥ 0. Work out the values of x for which the company makes a loss.
- 7.A candidate solves the inequality x² > 9 and writes: "x² > 9, so x > 3." Which of these is a correct comment on the candidate's work?
- 8.A plumber charges a £40 call-out fee plus £25 for each hour worked. The total charge, £C, for a job lasting h hours is given by C = 40 + 25h. Work out the charge for a job lasting 3 hours, and name what kind of statement C = 40 + 25h is.
- 9.The table shows the height, in metres, of a firework rocket at various times, in seconds, during its flight: 40 at t = 2, 54 at t = 3, and 60 at t = 4. Use the chord between t = 2 and t = 4 to estimate the gradient of the height-time graph at t = 3, stating the correct units.
- 10.A rule turns each input x into an output y. An input of 1 gives an output of 1, an input of 2 gives an output of 4 and an input of 3 gives an output of 9. Work out the rule.
- 11.A student says that (x + 4)² is equivalent to x² + 16. For which value of x do the two expressions give the SAME result, making it look (misleadingly) like the student could be right?
- 12.Solve the simultaneous equations y = 3 − x and x² + y² = 9, giving both pairs of solutions.
- 13.A phone company works out a monthly bill, £B, using the formula B = 18 + 0.05t, where £18 is the fixed monthly charge and t is the number of extra text messages sent beyond the free allowance, each charged at 5p. Farida's bill for one month is £24.50. Work out the number of extra text messages, t, she sent.
- 14.Write down the gradient of the line with equation y = −x + 3.y = −x + 3
- 15.Which of these values of x is a solution of x² + 2x − 15 = 0?
Answer key
- (d) The cost increases by about £10 per extra item — Gradient = change in cost ÷ change in items = (800 − 400) ÷ (70 − 30) = 400 ÷ 40 = 10. The units of the gradient are £ per item, so the cost is increasing by about £10 for every extra item produced, near x = 50. Subtracting in the wrong order, (400 − 800) ÷ (70 − 30) = −10, gives the right size but the wrong sign — check which point comes first each time. Leaving out the division by 40 gives £400 per item; dividing the wrong way round, 40 ÷ 400 = 0.1, gives £0.10 per item.
- (b) 8 — Method: a circle centred on the origin has equation x² + y² = r², where r is the radius, so the number on the right-hand side is the square of the radius and not the radius itself. Working: comparing x² + y² = 64 with x² + y² = r² gives r² = 64, so r = √64 = 8. Answer: the radius is 8. The distractors: 64 is r² read straight off the equation as though the right-hand side were the radius, which is the commonest error on this form; 32 comes from halving 64, treating the right-hand side as a diameter that has to be halved; 16 is the diameter, 2 × 8, quoted in place of the radius.
- (d) $y = x^2 - 2x - 3$ — Method: read the two x-intercepts (roots) from the graph, write the quadratic as the product of the corresponding factors, then expand. Working: the curve crosses the x-axis at x = −1 and x = 3, so the equation factorises as (x + 1)(x − 3), which expands to x² − 2x − 3. Answer: y = x² − 2x − 3. Distractor refutation: y = x² − x − 6 comes from misreading the left-hand crossing point as x = −2 instead of x = −1, giving factors (x + 2)(x − 3). y = x² − x − 2 comes from misreading the right-hand crossing point as x = 2 instead of x = 3, giving factors (x + 1)(x − 2). y = x² + 2x − 3 comes from writing the factors as (x − 1)(x + 3), swapping which root gets the plus sign and which gets the minus sign, giving the wrong sign on the x term.
- (a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
- (b) 1 at x = 90° — The graph of y = sin x rises from (0°, 0) to its maximum value of 1 at x = 90°, a quarter of the way through the period. Thinking the maximum occurs where the graph crosses the y-axis, at the start of the curve, gives 1 at x = 0°, but sin 0° = 0, not the maximum. Thinking the maximum occurs halfway to x = 360°, rather than a quarter of the way, gives 1 at x = 180°, but sin 180° = 0, not the maximum. Swapping the roles of the maximum value and the x co-ordinate at which it occurs gives 90 at x = 1, but the maximum value of sin x is 1, not 90.
- (b) 2 < x < 6 — Factorise: x² − 8x + 12 = (x − 2)(x − 6), giving roots at x = 2 and x = 6. Since the coefficient of x² is positive, the graph is a U-shape that is negative, below the axis, between its roots. So P < 0 for 2 < x < 6. Distractor routes: x < 2 or x > 6 takes the region outside the roots, which is where P is positive, a profit, the opposite of a loss. 2 ≤ x ≤ 6 includes the endpoints, where P = 0 exactly, break-even rather than a loss, since the inequality is strict. −6 < x < −2 comes from factorising as (x + 2)(x + 6), reversing the sign of both roots.
- (d) Wrong — the correct solution is x < −3 or x > 3. — Squaring or unsquaring an inequality is not a safe one-step move: the solution must be split into two branches, since a number less than −3 also squares to more than 9. So x > 3 finds only half the solution set. The full answer is x < −3 or x > 3. Distractor routes: "Right — square rooting both sides gives x > 3" treats the square root of an inequality the same as the square root of an equation and misses the negative branch entirely. "Wrong — the correct solution is −3 < x < 3" applies the between-the-roots pattern that belongs to the opposite inequality, x² < 9. "Right, but x = −3 and x = 3 should also work" treats the inequality as if it also allowed equality, when x² > 9 is strict and excludes both x = 3 and x = −3.
- (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.
- (c) 10 m/s — A symmetric chord gradient uses the two points either side of t = 3: (2, 40) and (4, 60). The gradient is the change in height divided by the change in time: (60 − 40) ÷ (4 − 2) = 20 ÷ 2 = 10 m/s. Using only the values either side of one gap, (2, 40) and (3, 54), instead of the full symmetric chord, gives (54 − 40) ÷ (3 − 2) = 14 m/s. Getting the correct number but dropping the time unit, leaving only metres, gives 10 m. Dividing time by height instead of height by time inverts the calculation to (4 − 2) ÷ (60 − 40) = 0.1 s/m.
- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (c) x = 0 — Expand (x + 4)² correctly: (x + 4)² = x² + 8x + 16. This equals x² + 16 only when 8x is zero, i.e. when x = 0 — at every other value of x the two expressions differ by 8x. Choosing x = 4 confuses the constant inside the bracket with the value of x that makes the expressions match. Choosing x = −4 makes the same confusion but with the sign flipped. Choosing x = 8 mistakes the coefficient of the middle term, 8x, for the value of x itself.
- (c) x = 0, y = 3 and x = 3, y = 0 — Substitute y = 3 − x into x² + y² = 9: x² + (3 − x)² = 9. Expanding (3 − x)² = 9 − 6x + x² gives x² + 9 − 6x + x² = 9, which simplifies to 2x² − 6x = 0, or 2x(x − 3) = 0, so x = 0 or x = 3. Using y = 3 − x: x = 0 gives y = 3; x = 3 gives y = 0. Distractor routes: x = 0, y = 3 alone stops after the factor 2x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = 3 and x = −3, y = 6 comes from factorising 2x² − 6x as 2x(x + 3), a sign error that gives a second root of −3 instead of 3. x = 0, y = 3 and x = 6, y = −3 comes from expanding (3 − x)² as 9 − 6x, dropping the x² term, which changes the quadratic to x² − 6x = 0 and its second root to 6.
- (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.
- (b) −1 — Method: in the form y = mx + c the gradient m is the number multiplying x, and an x term written with no number in front of it has a coefficient of 1. Working: y = −x + 3 is the same equation as y = (−1)x + 3, so comparing it with y = mx + c gives m = −1 and c = 3. Answer: −1. The distractors: 1 comes from taking the coefficient as 1 and leaving the minus sign out of the answer; 3 comes from reading the constant as the gradient, confusing m with c; −3 comes from moving the minus sign across to the constant term and then reading the gradient off that term instead of the x term.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
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