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.
Algebra worksheet — GCSE Higher
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- 1.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
- 2.A school trip costs a £15 deposit plus £9 per student for the coach. The total cost for a class is £186. Work out how many students went on the trip.
- 3.The table shows the velocity, in m/s, of a train at various times, in seconds: 0 at t = 0, 20 at t = 4, 35 at t = 10, 27 at t = 14. Assuming the velocity changes at a constant rate between each pair of readings, estimate the distance the train travels between t = 0 and t = 12, using the trapezium rule.
- 4.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?
- 5.A line has equation y = −4x + 1. Work out the equation of the line perpendicular to it that passes through the point (0, 3).y = -4x + 1
- 6.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.
- 7.The nth term of a sequence is 4n − 3. Work out the 7th term of the sequence.
- 8.The equation x² − x − 6 = 0 has roots x = 3 and x = −2. It can be rearranged as xₙ₊₁ = xₙ² − 6. This formula is used with starting value x₀ = 2.9, close to the root x = 3. Work out what happens to the sequence of values as n increases.
- 9.Three vertices of a rectangle are (−4, −1), (2, −1) and (2, 3). The sides of the rectangle are parallel to the axes. Write down the coordinates of the fourth vertex.
- 10.A sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₃ = 11 and a₅ = 29, work out a₁.
- 11.A rule turns each input x into an output y. The inputs are x = 0, 1, 2, 3 and the outputs are y = 4, 7, 10, 13. Work out the output when x = 5.
- 12.Work out the gradient of a line that is perpendicular to the line with equation 2x + 3y = 6.
- 13.Solve the inequality 9 − 2x ≥ 1.
- 14.The formula for the area of a trapezium is A = (a + b)h / 2, where a and b are the parallel side lengths and h is the height. Make h the subject of the formula.
- 15.Show that the equation x³ − x − 3 = 0 has a solution between x = 1 and x = 2, by working out f(1) and f(2), where f(x) = x³ − x − 3.y = x
Answer key
- (d) 5n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5, so the terms increase by 5 each time and the nth term has the form 5n + c. Substituting n = 1: 5(1) + c = 9, so c = 4. Answer: the nth term is 5n + 4. The value 5n comes from leaving out the constant. The value 5n + 9 comes from using the first term as the constant directly, without subtracting the common difference first. The value 9n + 5 comes from swapping the roles of the first term and the common difference — using the first term, 9, as the coefficient of n and the difference, 5, as the constant.
- (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.
- (a) 271 m — The reading at t = 12 isn't in the table, so it must be interpolated between t = 10 (35 m/s) and t = 14 (27 m/s): the velocity falls by 8 m/s over the 4-second gap, so over 2 seconds it falls by 4 m/s, giving 35 − 4 = 31 m/s. The trapezium rule then uses three strips of unequal width, 4, 6 and 2 seconds, with areas of 40, 165 and 66: 40 + 165 + 66 = 271 m. Writing 267 m comes from skipping the interpolation and using the table's t = 14 reading, 27, directly as the height at t = 12. Writing 282 m comes from treating all three strips as though they were 4 seconds wide, instead of using the true gaps of 4, 6 and 2 seconds between the readings. Writing 275 m comes from using the t = 10 reading, 35, at both ends of the final strip, instead of interpolating a new value at t = 12.
- (a) They always charge the same, since 3(2n + 4) = 6n + 12. — Expand Advert A's formula by multiplying both terms inside the bracket by 3: 3 × 2n = 6n, and 3 × 4 = 12, giving 3(2n + 4) = 6n + 12, which is identical to Advert B's formula — so the two adverts always charge the same amount, whatever n is. Getting 6n + 4 comes from multiplying the 2n by 3 but leaving the 4 unmultiplied. Getting 2n + 7 comes from adding 3 to the bracket instead of multiplying by it. Saying it depends on n avoids expanding the bracket at all — once expanded, both formulas are identical for every value of n, so the cost can be compared directly.
- (d) y = (1/4)x + 3 — The given line has gradient −4, so the perpendicular gradient is the negative reciprocal, 1/4. Since (0, 3) is the y-intercept, the perpendicular line is y = (1/4)x + 3. Distractor routes: y = −4x + 3 uses the original line's own gradient instead of the perpendicular gradient, only changing the intercept for the new point. y = (−1/4)x + 3 takes the reciprocal of −4 but forgets to change its sign, so it is not the true negative reciprocal. y = 4x + 3 negates the gradient −4 but does not take its reciprocal, giving 4 instead of 1/4.
- (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) 25 — Substitute n = 7: 4 × 7 − 3 = 28 − 3 = 25. Forgetting to subtract 3 gives 4 × 7 = 28. Subtracting 3 from 7 before multiplying by 4, 4 × (7 − 3) = 16, applies the operations in the wrong order. Substituting n = 8 by miscounting the position gives 4 × 8 − 3 = 29.
- (c) The sequence diverges, moving away from x = 3 — Starting from x₀ = 2.9: x₁ = 2.9² − 6 = 2.41, x₂ = 2.41² − 6 = −0.19, x₃ = (−0.19)² − 6 = −5.96, x₄ = (−5.96)² − 6 = 29.56 — the values swing away from 3 and grow rapidly, so the sequence diverges rather than settling anywhere. Choosing 'settles towards x = 3' assumes that starting close to a root is enough for a rearrangement to converge to it, which is not always true — this rearrangement changes values too steeply near x = 3 to stay there. Choosing 'settles towards x = −2' assumes a diverging sequence must eventually land on the other root; instead it runs away to increasingly large values. Choosing 'stays constant at 2.9' ignores that applying the formula changes the value at every step.
- (b) (−4, 3) — Method: in a rectangle whose sides are parallel to the axes only two different x-coordinates and two different y-coordinates appear, and each of them is shared by a pair of vertices, so the missing vertex takes the x-coordinate and the y-coordinate that so far appear only once. Working: the x-coordinates given are −4, 2 and 2, so 2 is already used twice and −4 is used once; the y-coordinates given are −1, −1 and 3, so −1 is already used twice and 3 is used once; the fourth vertex therefore has x = −4 and y = 3. Answer: (−4, 3). The distractors: (3, −4) comes from picking the two unpaired coordinates correctly and then writing them in the wrong order; (−4, −5) comes from matching the 4-unit vertical side but measuring it downwards from (−4, −1) instead of upwards; (8, 3) comes from carrying on round the shape with the horizontal step used earlier, adding 6 to the x-coordinate of (2, 3) instead of closing the rectangle.
- (a) 4 — a₅ = a₄ + a₃, so a₄ = a₅ − a₃ = 29 − 11 = 18. a₄ = a₃ + a₂, so a₂ = a₄ − a₃ = 18 − 11 = 7. a₃ = a₂ + a₁, so a₁ = a₃ − a₂ = 11 − 7 = 4. Checking forwards: 4, 7, 11, 18, 29. Answering 7 stops one step early, reporting a₂ = 7 instead of continuing back one more step to a₁ — wrong, because the question asks for a₁, not a₂. Answering 18 reports a₄ = 18, an intermediate value found along the way, instead of a₁ — wrong, because a₄ is a term used to reach the answer, not the term the question asks for. Answering 3 takes one backward step too many, working out a further term a₀ = a₂ − a₁ = 7 − 4 = 3 — wrong, because the sequence starts at a₁, so a₁ = 4 is as far back as the question goes.
- (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
- (b) 3/2 — Rearrange 2x + 3y = 6 into y = mx + c: 3y = −2x + 6, so y = −(2/3)x + 2. The gradient of this line is −2/3. The perpendicular gradient is the negative reciprocal: 3/2. Distractor routes: −1/2 comes from reading the gradient straight off the x-coefficient, 2, without dividing by the y-coefficient, 3, first, then taking its negative reciprocal. −3/2 correctly finds the gradient −2/3 but only takes its reciprocal without also changing the sign, giving −3/2 instead of 3/2. 2/3 comes from negating the gradient −2/3 to 2/3, but forgetting to also take the reciprocal.
- (c) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
- (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.
- (b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
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