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.n = 3. Work out the value of (2n)².
- 2.The trapezium rule is used to estimate the area under a curve between x = 2 and x = 6. The curve is concave up (it curves upwards, like the inside of a bowl) throughout this interval. Which statement about the estimate is correct?
- 3.To solve 6x − 4 = 2x + 20, Yusuf's first step is to subtract 2x from both sides. Work out what equation this gives.
- 4.Solve 3x + 14 = 8x − 6
- 5.Solve (2x + 1)/3 = 5
- 6.Solve 2(x + 3) = 10
- 7.The nth term of a sequence is n² − 2n + 5. Work out the 10th term.
- 8.Four sequences are shown below. Sequence P: 4, 8, 16, 32, ... Sequence Q: 4, 8, 12, 16, ... Sequence R: 4, 7, 12, 19, ... Sequence S: 4, 9, 16, 25, ... Work out which of the sequences P, Q, R and S is geometric.
- 9.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.
- 10.Factorise x² − 3x − 10.
- 11.Describe a sequence of two transformations that maps the graph of y = cos x onto the graph of y = cos(x + 90°) − 2.y = cos(x)
- 12.A straight line passes through the points (−1, 2) and (3, 14). Work out the equation of the line.
- 13.The first five terms of a quadratic sequence are 2, 5, 10, 17, 26. Work out the next term in the sequence.
- 14.Rosie is asked whether 4(x + 3) = 4x + 12 is an equation or an identity. Which statement is correct? Give a reason for your answer.
- 15.A phone plan costs a fixed £15 per month plus 8p per minute of calls. In one month, Jamal's bill was £23.80. Form an equation using m for the number of minutes, and solve it to find how many minutes Jamal used that month.
Answer key
- (d) 36 — (2n)² means the whole of 2n is squared, so with n = 3: (2n)² = (2 × 3)² = 6² = 36. Answering 18 instead works out 2n² — squaring only the n and then multiplying by 2 — which is a different expression because the brackets around 2n are missing. Answering 12 squares only the coefficient, treating (2n)² as 2² × n = 4 × 3 = 12, and forgets to square the n as well. Answering 9 ignores the coefficient of 2 altogether and works out n² on its own. The value of (2n)² when n = 3 is 36.
- (a) Overestimates — trapezium edges lie above the curve. — For a concave-up curve, each chord connecting two points on the curve lies ABOVE the curve between those points, because the curve bends away from the chord underneath it. The trapezium rule uses these chords as the top edges of each trapezium, so the trapeziums cover a larger region than the true area under the curve — the estimate is an overestimate. Saying it underestimates because the edges lie below the curve gets the direction backwards: for a concave-up curve the edges lie above, not below. Saying it overestimates 'because more strips always increase it' gives the right verdict for the wrong reason — using MORE, narrower strips makes a trapezium estimate more accurate and brings it closer to the true area, it does not simply increase the total. Saying the estimate is exact ignores that straight edges cannot follow a curved line perfectly; there is always a gap between the chord and the curve on a bend, which is exactly what makes the rule an estimate rather than an exact calculation.
- (b) 4x − 4 = 20 — Method: subtract 2x from both sides of the equation, and simplify each side separately. Working: left side: 6x − 4 − 2x = 4x − 4. Right side: 2x + 20 − 2x = 20. Answer: 4x − 4 = 20. 4x = 20 drops the −4 from the left side, as though subtracting 2x also removes the constant term. 8x − 4 = 20 comes from moving the 2x across to the left without changing its sign: it is taken off the right side correctly, leaving 20, but added to the left side instead of subtracted, giving 6x + 2x = 8x. 4x − 4 = 2x + 20 comes from subtracting 2x from the left-hand side only and leaving the right-hand side unchanged; whatever is done to one side must be done to the other.
- (c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
- (b) 85 — Substitute n = 10: 10² − 2 × 10 + 5 = 100 − 20 + 5 = 85. A sign error on the −2n term, treating it as +2n, gives 100 + 20 + 5 = 125. Using n = 9 instead of n = 10 gives 81 − 18 + 5 = 68. Working out 10² − 2 × 10 but forgetting to add the final +5 gives 80.
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (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.
- (b) (x − 5)(x + 2) — We need two numbers that multiply to −10 and add to −3: these are −5 and 2, since −5 × 2 = −10 and −5 + 2 = −3. So x² − 3x − 10 = (x − 5)(x + 2). A candidate who swaps the signs, using +5 and −2, gets (x + 5)(x − 2), which expands to x² + 3x − 10 — the wrong middle term. A candidate who picks the factor pair 1 and 10 instead of 2 and 5 gets (x − 10)(x + 1), which expands to x² − 9x − 10. A candidate who makes both factors negative gets (x − 5)(x − 2), which expands to x² − 7x + 10 — the wrong sign on the constant term.
- (b) Translate −90° in x, then translate −2 in y. — cos(x + 90°) translates the graph 90° in the NEGATIVE x-direction, since a positive shift inside the bracket moves a graph left, not right, and subtracting 2 afterwards translates it 2 units in the negative y-direction (down). So the sequence is: translate −90° in x, then translate −2 in y. Using +90° in x reverses the direction of the horizontal shift — the sign inside the bracket moves the graph the opposite way to what it looks like. Using +2 in y reverses the direction of the vertical shift; subtracting 2 outside the function moves the graph down, not up. Describing the −2 as a reflection in the x-axis is wrong because a reflection turns positive y-values negative and vice versa, whereas here every y-value is simply reduced by the fixed amount 2, which is what a translation does, not a reflection.
- (c) y = 3x + 5 — Method: find the gradient from the two points, then substitute one point into y = mx + c to find c. Working: gradient = (14 − 2) ÷ (3 − (−1)) = 12 ÷ 4 = 3. Using the point (3, 14): 14 = 3(3) + c, so 14 = 9 + c, giving c = 5. Answer: y = 3x + 5. y = 6x − 4 comes from mishandling the negative x-coordinate, treating 3 − (−1) as 3 − 1 = 2, so the gradient becomes 12 ÷ 2 = 6, and then c = 14 − 18 = −4. y = 3x + 23 comes from a sign error isolating c, adding 9 to 14 instead of subtracting it: c = 14 + 9 = 23. y = x/3 + 13 comes from dividing the change in x by the change in y instead of the other way round, giving a gradient of 4 ÷ 12 = 1/3, and then c = 14 − 1 = 13.
- (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.
- (a) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.
- (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.
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