Grade 8/9 Maths Quizzes

Question 1 8B Transforming graphs y=f(x)

[2 marks]

Here is the graph of y = f(x)

(a) On the grid below, draw the graph of y = f(x) - 4 [1 mark]

(b) On the grid below, draw the graph of y = f(-x) [1 mark]

Worked Solution & Answer

(a)

Translation by (0, –4) B1

Answer (a): Trapezium translated 4 units down with vertices (–3, –4), (–2, –3), (–1, –3), (0, –4)

(b)

Reflection in the y-axis B1

Answer (b): Trapezium reflected in the y-axis with vertices (0, 0), (1, 1), (2, 1), (3, 0)

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Question 2 8A Quad. sim. equations

[3 marks]

The diagram shows the graph of x² + y² = 30.25

Use the graph to find estimates for the solutions of the simultaneous equations
x² + y² = 30.25
y - 2x = 1

Worked Solution & Answer

Plot line y = 2x + 1 through (0, 1) and (2, 5) M2
Intersections with circle are at (2.1, 5.1) and (–2.9, –4.7). A1

Answer: x = 2.1, y = 5.1 and x = -2.9, y = -4.7

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Question 3 8E Quadratic inequalities AQA Jun 2019 2H Q24

[3 marks]

Here is a sketch of the curve y = x² + 4x - 12

Work out the values of x for which x² + 4x - 12 < 0
Give your answer as an inequality.

Answer

Worked Solution & Answer

Solve x² + 4x - 12 = 0: M1
(x + 6)(x - 2) = 0 ⇒ x = -6 or x = 2 M1
The curve is a parabola opening upwards, so y < 0 between the two roots:
-6 < x < 2 A1

Answer: Answer -6 < x < 2

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Question 4 8H Circles and tangents Specimen Set 1 2H Q23

[3 marks]

Here is a circle, centre O, and the tangent to the circle at the point P(4, 3) on the circle.

Find an equation of the tangent at the point P.

y =

Worked Solution & Answer

Gradient of radius OP = 3/4. Gradient of tangent = –4/3. Equation: y – 3 = –(4/3)(x – 4) M2
⇒ y = –(4/3)x + 25/3 (or 4x + 3y = 25). A1

Answer: y = y = -43x + 253

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Question 5 8K Vector proof AQA Nov 2022 2H Q25

[3 marks]

Two congruent parallelograms, PQRV and VRST, are joined.

QP→ = a PV→ = b
X is the midpoint of VT.
VW : WR = 1 : 2
Prove that Q, W and X lie on a straight line.

Worked Solution & Answer

VR = PQ = –a M1
QW = QP + PV + VW = a + b – 1/3 a = 2/3 a + b
QX = QP + PV + VX = a + b + 1/2 b = a + 3/2 b M1
QX = 3/2 (2/3 a + b) = 3/2 QW
Since QX is a multiple of QW and Q is a common point, Q, W and X are collinear. C1

Answer: QW = a + b – 1/3 a = 2/3 a + b.
QX = a + 3/2 b = 3/2 (2/3 a + b) = 3/2 QW.
Since QX is a scalar multiple of QW and they share point Q, Q, W and X lie on a straight line.

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Question 6 8G Velocity time graphs

[3 marks]

The graph shows the speed of a car, in metres per second, during the first 20 seconds of a journey.

(a) Work out an estimate for the distance the car travelled in the first 20 seconds.
Use 4 strips of equal width. [3 marks]

metres

Worked Solution & Answer

(a)

(Average speed = distance ÷ time)
Strip width = 5 s M1
Speeds at t = 0, 5, 10, 15, 20 are 0, 22, 28, 32, 35 m/s M1
Area = 1/2×5×22 + 1/2×5×(22+28) + 1/2×5×(28+32) + 1/2×5×(32+35) = 55 + 125 + 150 + 167.5 = 497.5 m A1

Answer (a): 497.5 metres

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Question 7 8C Algebraic proof IGCSE Jun 2019 2HR Q17

[4 marks]

The table gives information about the first six terms of a sequence of numbers.

Prove algebraically that the sum of any two consecutive terms of this sequence is always a square number.

Worked Solution & Answer

nth term + (n+1)th term = ½n(n+1) + ½(n+1)(n+2) = ½(n² + n + n² + 3n + 2) = ½(2n² + 4n + 2) = n² + 2n + 1 = (n + 1)² M3 C1

Answer: n(n+1)2 + (n+1)(n+2)2 = (n+1)(2n+2)2 = (n+1)², which is a square number

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Question 8 8F Quadratic sequences AQA Nov 2021 1H Q21

[4 marks]

The first two terms of a quadratic sequence are 10 and 17
Here is some information about the sequence.

Work out an expression for the nth term of the sequence.

Answer

Worked Solution & Answer

Second difference is +6, so a = 62 = 3, giving 3n² M1
Comparing terms with 3n²:
n = 1: 10 - 3(1)² = 7 M1
n = 2: 17 - 3(2)² = 5
n = 3: 30 - 3(3)² = 3 M1
The sequence of differences 7, 5, 3, … has first difference -2, giving -2n + 9
Therefore, the nth term is 3n² - 2n + 9. A1

Answer: Answer 3n² - 2n + 9

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Question 9 8J Circle theorem proof AQA Nov 2022 1H Q26

[4 marks]

P, Q and R are points on a circle.
SP is a tangent to the circle.
RQ = PQ

Prove that y = 90° - x

Worked Solution & Answer

By the alternate segment theorem, angle PQR = angle RPS = 2x M2
Since RQ = PQ, triangle PQR is isosceles, so angle QPR = angle PRQ = y M1
Angles in triangle PQR add to 180°: y + y + 2x = 180° ⇒ 2y = 180° – 2x ⇒ y = 90° – x. C1

Answer: Angle PQR = 2x (alternate segment theorem).
Angle QPR = y (isosceles triangle PQR as RQ = PQ).
2y + 2x = 180° (angles in a triangle sum to 180°), so y = 90° – x.

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Question 10 8D Completing the square

[5 marks]

Sketch the graph of
y = x² - 6px - 7\quadwhere p > 0
showing the coordinates of the turning point, in terms of p, and the coordinates of the
intercept with the y-axis.
You must show all your working.

Worked Solution & Answer

y = (x – 3p)² – 9p² – 7 ⇒ turning point at (3p, –9p² – 7) in 4th quadrant since p > 0 M4
y-intercept at (0, –7) A1

Answer: Parabola sketched with y-intercept at (0, –7) and minimum turning point at (3p, –9p² – 7) in the 4th quadrant

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Grade 8/9 Quizzes

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