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.The function f(x) = x² for all real values of x has no inverse function, but g(x) = x² for x ≥ 0 does have one. Which statement correctly explains this?
y = x²
(a)Restricting any domain always creates an inverse
(b)Squares can never be reversed, under any conditions
(c)g's outputs are positive; f's could be negative
(d)g is one-to-one: f(3) = f(−3), removed by x ≥ 0
2.Which of these values of x is a solution of x² + 2x − 15 = 0?
(a)x = 3
(b)x = 5
(c)x = 15
(d)x = −3
3.The area under a speed-time graph between t = 0 and t = 6 seconds is estimated using three strips of equal width, using the speeds, in m/s, at t = 0, 2, 4 and 6: 0, 5, 9 and 12. Using the trapezium rule with these three trapezia, estimate the distance travelled.
(a)28 m
(b)80 m
(c)52 m
(d)40 m
4.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
(a)s = t / d
(b)s = d / t
(c)s = d − t
(d)s = dt
5.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
6.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
(a)t = 4
(b)t = 8
(c)t = 10
(d)t = 5
7.A plumber charges a call-out fee plus an hourly rate for a job. The total cost, £C, for a job lasting h hours is given by the formula C = 40 + 25h. What does the number 40 represent in this formula?
(a)The total cost for a job lasting 1 hour.
(b)The hourly rate the plumber charges.
(c)The number of hours worked before charging starts.
(d)The fixed call-out fee, charged before the hourly rate.
8.A geometric sequence has first term 5, and each term after that is 3 times the term before it. Work out the first term of the sequence that is greater than 1000.
(a)405
(b)1215
(c)2187
(d)3645
9.Write down the expression that means the same as (x + 7) ÷ 4.
(a)4(x + 7)
(b)4/(x + 7)
(c)(x + 7)/4
(d)(x + 7) − 4
10.The point (3, 6) lies on the circle x² + y² = 45. The tangent to the circle at (3, 6) crosses the x-axis at the point P. Work out the coordinates of P.
(a)(15, 0)
(b)(6, 0)
(c)(0, 0)
(d)(0, 7.5)
11.Expand and simplify (x + 1)(x + 2)(x + 3).
(a)x³ + 6x² + 11x + 5
(b)x³ + 6x² + 6x + 6
(c)x³ + 5x² + 11x + 6
(d)x³ + 6x² + 11x + 6
12.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.
(a)485
(b)1.1
(c)297.5
(d)110
13.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.
(a)14 m/s
(b)10 m
(c)10 m/s
(d)0.1 s/m
14.The nth term of a sequence is 2n² + 1. Work out the 4th term of the sequence.
(a)33
(b)65
(c)32
(d)17
15.The point (−4, 7) is moved 4 units to the right and 9 units down. Write down the coordinates of the point it reaches.
(a)(0, 16)
(b)(0, −2)
(c)(−8, −2)
(d)(5, 3)

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