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 · Calculator · 15 questions

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1.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.
(a)130
(b)13
(c)65
(d)490
2.By completing the square, find the turning point of the curve y = x² + 8x − 3.
y = x² + 8x − 3
(a)x = −8, y = −67
(b)x = −4, y = 19
(c)x = 4, y = −19
(d)x = −4, y = −19
3.f(x) = x³ − 3x − 20, and the equation f(x) = 0 has exactly one solution. Work out the pair of consecutive integers between which that solution lies.
y = x
(a)2 and 3
(b)3 and 4
(c)4 and 5
(d)6 and 7
4.The iterative formula xₙ₊₁ = 5 − 3/xₙ is used with starting value x₀ = 2.5, so that x₁ is the value after the formula has been used once. Work out x₄ correct to 3 significant figures.
(a)3.80
(b)4.29
(c)4.30
(d)2.50
5.A cuboid has a square base of side x metres and a height that is 3 m more than x. Its volume is 150 m³. This gives the equation x³ + 3x² − 150 = 0, which can be solved using the iterative formula xₙ₊₁ = ∛(150 − 3xₙ²). Taking x₀ = 4, work out x₂ correct to 2 decimal places.
(a)84.51
(b)4.39
(c)4.67
(d)6.32
6.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
7.Two students each use the trapezium rule to estimate the area under the same curve between x = 0 and x = 8. Student A uses 4 strips of width 2. Student B uses 8 strips of width 1. Which estimate is more likely to be closer to the true area, and why?
(a)Student A's — fewer strips means fewer rounding errors.
(b)Student B's — narrower strips fit the curve more closely.
(c)They are always the same; the trapezium rule is exact.
(d)Student A's — wider strips give a better average height.
8.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.
y = x
(a)4 and 5
(b)2 and 3
(c)0 and 1
(d)3 and 4
9.Solve 2x² − 32 = 0.
(a)x = 8
(b)x = 4
(c)x = 16
(d)x = 4 or x = −4
10.Point R has coordinates (3, −5). Point R is reflected in the line y = 2 to point S. Write down the coordinates of S.
(a)(3, 5)
(b)(−3, 9)
(c)(3, 9)
(d)(3, 7)
11.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
(a)y stays the same however large x becomes
(b)y decreases towards zero but never reaches it
(c)y increases towards a limit but never reaches it
(d)y decreases at a steady rate and reaches zero
12.A rectangular photo has length (x + 3) cm and width x cm. Its area is 40 cm². Work out the value of x.
(a)x = −8
(b)x = 5
(c)x = 5 or x = −8
(d)x = 4
13.A regular hexagon has sides of length (x + 2) cm. Write down an expression, in terms of x, for the perimeter of the hexagon.
(a)6x + 12
(b)6x + 2
(c)x + 12
(d)8x
14.f(x) = x³ − 3x − 5. Given that f(2.2) = −0.952 and f(2.3) = 0.267, work out what this shows about the equation x³ − 3x − 5 = 0.
y = x
(a)It has a solution between x = −2.2 and x = −2.3
(b)It has exactly two solutions in this interval
(c)It has no solution between x = 2.2 and x = 2.3
(d)It has a solution between x = 2.2 and x = 2.3
15.A van's value depreciates by 18% each year. After 2 years it is worth £5,379.20. Work out its value when it was new, correct to the nearest pound.
(a)£6,560
(b)£3,617
(c)£8,000
(d)£8,405

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