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Question 1 7G Trigonometric & exp. Graphs
[1 mark]
Sana needs to draw the graph of y = 3x for 0 ≤ x ≤ 4
She draws the graph shown on the grid.
yx12345102030405060708090O
Write down one thing Sana has done wrong.
Worked Solution & Answer
When x = 0, y = 3⁰ = 1, so the y-intercept should be (0, 1) C1
yx12345102030405060708090O
Answer: The graph should cross the y-axis at (0, 1), not (0, 0), because 3⁰ = 1
Question Paper (PDF) • Mark Scheme (PDF)
Question 2 7D Algebraic fractions AQA Nov 2023 3H Q22
[2 marks]
Write 72a2 - 35a as a single fraction in its simplest form.
Answer
Worked Solution & Answer
Common denominator is 10a²: M1
(7 × 5) / (10a²) – (3 × 2a) / (10a²) = (35 – 6a) / (10a²). A1
Answer: Answer 35 - 6a10a2
Question Paper (PDF) • Mark Scheme (PDF)
Question 3 7E The quadratic formula AQA Jun 2025 2H Q15
[2 marks]
Solve 3x2 + 5x - 9 = 0
Give your solutions as decimals.
Worked Solution & Answer
x = [–5 ± √(5² – 4(3)(–9))] / (2 × 3) = [–5 ± √133] / 6. M1
x ≈ 1.09 and x ≈ –2.76. A1
Answer: 1.09 and –2.76
Question Paper (PDF) • Mark Scheme (PDF)
Question 6 7B Bounds IGCSE Jan 2022 2F Q19 / 2H Q4
[2 marks]
Each side of a regular octagon has a length of 18 mm, correct to the nearest 0.5 mm
Diagram NOTaccurately drawn
(a) Write down the lower bound of the length of each side of the octagon. [1 mark]
mm
(b) Write down the upper bound of the length of each side of the octagon. [1 mark]
mm
Worked Solution & Answer
(a)
18 – 0.25 = 17.75 mm B1
Answer (a): 17.75 mm
(b)
18 + 0.25 = 18.25 mm B1
Answer (b): 18.25 mm
Question Paper (PDF) • Mark Scheme (PDF)
Question 7 7C Proportion (harder)
[2 marks]
Work out
(a) Sketch graph for y is directly proportional to x (1) (b) Sketch graph for y is inversely proportional to x (1)
(a) Using the axes below, sketch a graph to represent the statement
y is directly proportional to x [1 mark]
xyO
(b) Using the axes below, sketch a graph to represent the statement
y is inversely proportional to x [1 mark]
xyO
Worked Solution & Answer
(a)
y = kx is a straight line through (0, 0) B1
xyO
Answer (a): Straight line with positive gradient passing through the origin O
(b)
y = k/x is a decreasing hyperbola in quadrant 1 B1
xyO
Answer (b): Reciprocal curve y = k/x in the first quadrant asymptotic to both axes
Question Paper (PDF) • Mark Scheme (PDF)
Question 8 7A Surds AQA Jun 2023 1H Q22
[3 marks]
a = √(2) and b = √(18)
Match each expression to its value.
One has been done for you.
a2a + babb / a236364√210√20
a = √2 and b = √18. Match each expression (a², a + b, ab, b/a) to its value. [3 marks]
Worked Solution & Answer
b = √18 = 3√2. M1
a + b = √2 + 3√2 = 4√2. M1
ab = √2 × √18 = √36 = 6.
b/a = √18 / √2 = √9 = 3. A1
a2a + babb / a236364√210√20
Answer: a + b → 4√2, ab → 6, b/a → 3
Question Paper (PDF) • Mark Scheme (PDF)
Question 9 7K Iteration AQA Jun 2024 2H Q22
[3 marks]
@@TABULAR0@@
Use x1 = 1 to work out an approximate solution to x = 5 - 1x
Give your answer to 4 significant figures.
x =
Worked Solution & Answer
x₂ = 5 – 1/1 = 4. M1
x₃ = 5 – 1/4 = 4.75.
x₄ = 5 – 1/4.75 = 4.78947... M1
x₅ = 5 – 1/4.78947... = 4.79120...
x₆ = 4.79128... → 4.791 (to 4 s.f.). A1
Answer: x = 4.791
Question Paper (PDF) • Mark Scheme (PDF)
Question 10 7J Functions (harder) AQA Jun 2017 2H Q27
[4 marks]
(a) h(x) = 3√(x) for all values of x
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2
27 (b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360°
(a) h(x) = 3√(x) for all values of x
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2 [2 marks]
–2–112–8–6–4–22468xyO
(b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360° [2 marks]
0°90°180°270°360°–101xy
Worked Solution & Answer
(a)
The inverse function is h-1(x) = x3 M1
Values: (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
Plot these points and draw a smooth cubic curve. A1
–2–112–8–6–4–22468xyO
Answer (a): Cubic graph y = x³ drawn through (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
(b)
fg(x) = f(g(x)) = \sin(x + 90°) = \cos x M1
Plot the cosine wave: 0° → 1, 90° → 0, 180° → -1, 270° → 0, 360° → 1. A1
0°90°180°270°360°–101xy
Answer (b): Cosine curve drawn from (0°, 1) to (90°, 0), (180°, -1), (270°, 0), (360°, 1)
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 23 marks)

Grade 7 Quizzes by Topic

19 Topics
7A - Surds7B - Bounds
7C - Proportion (harder)7D - Algebraic fractions
7E - The quadratic formula7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs7H - Rearranging (harder)
7J - Functions (harder)7K - Iteration
7L - Area of any triangle7M - The sine rule
7N - The cosine rule7P - Congruent triangles
7Q - 3D pythagoras / trig.7R - Histograms
7S - Conditional probability7T - Exact trig. values (harder)
7U - Area of a segment