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

Maths · GCSE Higher · Algebra · 15 questions

MathsUKwww.geekhero.co.uk

1.Which of these is a factor of 15x + 20?
(a)5
(b)3
(c)4
(d)10
2.The diagram shows the graph of the cost, in pounds, of a taxi journey plotted against the distance travelled, in miles, for journeys of up to 4 miles. The same fixed charge and the same cost per mile apply to longer journeys. Work out the cost of a 6-mile journey.
(a)£24
(b)£14
(c)£12
(d)£8
3.A cycle route is 84 km long. Freya sets off along it at a steady 14 km/h. Write down the function for the distance y, in kilometres, that is still to be cycled after x hours.
(a)y = 84 + 14x
(b)y = 84 − 14x
(c)y = 14 − 84x
(d)y = 14x − 84
4.A proof that the product of two consecutive even numbers is always a multiple of 8 begins: Let the two consecutive even numbers be 2n and 2n + 2, so their product is 2n(2n + 2) = 4n(n + 1). Which line correctly completes the proof?
(a)4n(n + 1) is a multiple of 4, and because n and n + 1 are consecutive integers, it must be a multiple of 8.
(b)Expanding gives 4n(n + 1) = 4n² + 4n, which is clearly a multiple of 8.
(c)4n is always a multiple of 4 for any integer n, so 4n(n + 1) is a multiple of 4 as well, which means it is a multiple of 8.
(d)n and n + 1 are consecutive integers, so one of them must be even; this makes n(n + 1) even, so 4n(n + 1) is 4 × an even number, which is a multiple of 8.
5.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
(a)£14.50
(b)£13.50
(c)£12.50
(d)£24.50
6.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
(a)9 m
(b)18 m
(c)37 m
(d)16 m
7.The diagram shows two curves drawn on the same grid. Curve A is the graph of $y = x^{2}$ . Which equation could represent curve B?
(a) $y = (x - 3)^{2}$
(b) $y = x^{2} - 3$
(c) $y = x^{2} + 3 x$
(d) $y = x^{2} + 3$
8.The discriminant of a quadratic equation is greater than zero. Which statement about the solutions of that equation is correct?
(a)It has two whole-number solutions
(b)It has one repeated real solution
(c)It has two different real solutions
(d)It has no real solutions at all
9.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
(a)9.78
(b)8.39
(c)5.32
(d)7
10.The simultaneous equations 2x + 3y = 12 and x − y = 1 are given. Work out the value of y.
(a)y = 2
(b)y = 2.2
(c)y = 3
(d)y = 10
11.The equation x³ − 3x − 4 = 0 has a root near x = 2. Four students each try a different iterative formula, all starting from x₀ = 2: xₙ₊₁ = ∛(3xₙ + 4); xₙ₊₁ = (xₙ³ − 4) ÷ 3; xₙ₊₁ = 4 ÷ (xₙ² − 3); xₙ₊₁ = xₙ³ − 2xₙ − 4. Only one of these formulas keeps producing values that settle near the root when it is repeated. Work out x₁, correct to 3 decimal places, for the formula that does this.
(a)2.154
(b)1.333
(c)0.000
(d)4.000
12.A car accelerates uniformly from rest at 2.5 m/s² until it reaches a speed of 20 m/s, then travels at this constant speed for a further 30 seconds. Work out the total distance travelled.
(a)900 m
(b)760 m
(c)240 m
(d)680 m
13.The diagram shows part of the graph of a reciprocal function of the form $y = \frac{k}{x}$ , passing through the labelled point. Work out the value of k.
(a)5
(b)6
(c)9
(d)1.5
14.By completing the square, find the turning point of the curve y = 3x² + 12x + 7.
y = 3x² + 12x + 7
(a)x = −2, y = 5
(b)x = −4, y = −41
(c)x = 2, y = −5
(d)x = −2, y = −5
15.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
(a)t = 1 second and t = 4 seconds
(b)t = 1 second and t = −5 seconds
(c)t = 1 second and t = 5 seconds
(d)t = −1 second and t = 5 seconds

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