1. [Maximum mark: 3]
Write 1400 as a product of powers of its prime factors. Show your working clearly.
▶️Answer/Explanation
Solution:
To express 1400 as a product of powers of its prime factors, we perform prime factorization:
1400 = 2 × 700
700 = 2 × 350
350 = 2 × 175
175 = 5 × 35
35 = 5 × 7
Therefore, 1400 = 2³ × 5² × 7¹.
2. [Maximum mark: 4]
(a) Describe fully the single transformation that maps shape A onto shape B.
(b) On the grid above, rotate shape A 180° about (−1, 0). Label your shape C.
▶️Answer/Explanation
(a) Solution:
The transformation is a translation by the vector \(\begin{pmatrix} 3 \\ -5 \end{pmatrix}\).
(b) Solution:
To rotate shape A 180° about (−1, 0), each point (x, y) of shape A is mapped to (2(−1) − x, 2(0) − y) = (−2 − x, −y). The rotated shape is labeled C.
3. [Maximum mark: 2]
Here is a list of four numbers written in ascending order of size: \( x, x, y, 15 \), where \( x \) and \( y \) are integers. The numbers have a median of 12.5 and a range of 4.
Find the value of \( x \) and the value of \( y \).
▶️Answer/Explanation
Solution:
(x=) 11 (and) (y=) 14
4. [Maximum mark: 5]
Given the universal set \( \mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \), set \( A = \{factors of 6\} \), and set \( B = \{prime numbers\} \):
(a)(i) List the members of \( A \cup B \).
(a)(ii) List the members of \( A’ \).
(b) Explain why Harpreet’s statement \( A \cap B = \emptyset \) is incorrect.
(c) List the 4 members of set \( C \) such that \( A \cap C \) has 2 members, \( B \cap C \) has 2 members, and \( A \cap C \) and \( B \cap C \) have no common members.
▶️Answer/Explanation
(a)(i) Solution:
\( A \cup B = \{1, 2, 3, 5, 6, 7\} \).
(a)(ii) Solution:
\( A’ = \{4, 5, 7, 8, 9, 10\} \).
(b) Solution:
\( A \cap B = \{2, 3\} \), which is not empty. Thus, Harpreet is incorrect.
(c) Solution:
Set \( C \) must include 2 members from \( A \) (1, 2, 3, 6) and 2 members from \( B \) (2, 3, 5, 7), with no overlap. One possible set is \( C = \{1, 5, 6, 7\} \).
5. [Maximum mark: 4]
The diagram shows the design for a badge, which will be made using wire. The design is a circle inside a square ABCD.
The circle touches the square at points E, F, G, and H. The area of the square is 81 cm². Calculate the total length of wire needed to make the square and the circle, correct to 3 significant figures.
▶️Answer/Explanation
Solution:
The side length of the square is \( \sqrt{81} = 9 \) cm. The perimeter of the square is \( 4 \times 9 = 36 \) cm.
The diameter of the circle is equal to the side length of the square, so the radius is \( \frac{9}{2} = 4.5 \) cm. The circumference of the circle is \( 2\pi \times 4.5 = 9\pi \) cm ≈ 28.274 cm.
The total length of wire needed is \( 36 + 28.274 = 64.274 \) cm, which rounds to 64.3 cm (3 significant figures).
6. [8 marks]
(a) Solve \(\frac{2f}{3}=4f-17\)
(b) Simplify \((e+12)^0\)
(c) Simplify \(\frac{12a^4h^6}{4ah^2}\)
(d) Factorize \(20x^5y+12x^3y^4\)
▶️Answer/Explanation
(a) f=5.1
(b) 1
(c) 3a3h4
(d) 4x3y(5x2+3y3)
7. [2 marks]
Solve \(\frac{3^2 × 3^5}{3^{10}} = 3^n\) for n.
▶️Answer/Explanation
\(\frac{3^7}{3^{10}} = 3^{-3}\)
∴ n=-3
8. [Maximum mark: 3]
In a sale, all normal prices are reduced by 17%. The sale price of a fridge is 6225 rupees. Work out the normal price.
▶️Answer/Explanation
Solution:
Sale price = 83% of normal price
6225 = 0.83 × normal price
Normal price = 6225 ÷ 0.83 = 7500 rupees
9. [Maximum mark: 4]
(a) Write \( 6.04 \times 10^{5} \) as an ordinary number.
(b) Write \( 0.00007 \) in standard form.
(c) Work out \( \frac{7.6 \times 10^{10}}{4 \times 10^{5} – 2 \times 10^{4}} \) and give your answer in standard form.
▶️Answer/Explanation
Solution:
(a) 604 000 1
(b) $7 \times 10^{-5}$ 1
(c) 380 000 or $3.8 \times 10^5$ or $38 \times 10^4$ oe 2
Working not required, so correct answer scores full marks (unless from obvious incorrect working) $2 \times 10^5$
10. [Maximum mark: 5]
The diagram shows a hexagon ABCDEF
with AB = 11cm, angle BCF = 30°, and other given measurements. Calculate the area of ABCDEF.
▶️Answer/Explanation
Solution:
Using trigonometry to find height of trapezium:
Height = 5 × sin30° = 2.5 cm
Area of trapezium ABED = ½ × (11 + 23) × 2.5 = 42.5 cm²
Area of triangle BCF = ½ × 4.7 × 2.5 = 5.875 cm²
Area of triangle CDE = ½ × 6.5 × 3 × sin(108°) ≈ 9.27 cm²
Total area ≈ 42.5 + 5.875 + 9.27 ≈ 57.645 cm²
11. [Maximum mark: 7]
The cumulative frequency table gives information about the time, in hours, that each of 60 workers spent working from home in one week.
(a) On the grid below, draw a cumulative frequency graph for the information in the table.
(b) Use your graph to find an estimate for the interquartile range of the times.
25 workers spent more than $W$ hours working from home.
(c) Use your graph to find an estimate for the value of $W$.
One of the 60 workers is chosen at random. This worker spent $H$ hours working from home.
(d) Find the probability that $5 < H \le 10$
▶️Answer/Explanation
(a) Plot points at (5,6), (10,17), (15,27), (20,42), (25,53), (30,60) and join with smooth curve
(b) Interquartile range ≈ 13 hours (using Q3 ≈ 21.5 and Q1 ≈ 8.5)
(c) W ≈ 17 hours (from CF = 35)
(d) P(5 < H ≤ 10) = (17-6)/60 = 11/60 ≈ 0.183
12. [Maximum mark: 4]
In the diagram, ABC and AED are straight lines. BE is parallel to CD. AE = 10 cm and CD = 1.5 × BE.
(a) Work out the length of ED
Given AB = (2x + 5) cm and BC = (3x – 5) cm,
(b) work out the value of x.
▶️Answer/Explanation
(a) Using similar triangles: ED = 5 cm
(b) AB + BC = 1.5 × AB ⇒ 5x = 1.5(2x + 5) ⇒ x = 3.75
13. [Maximum mark: 3]
OAB is a sector of a circle with center O and radius r cm.
Angle AOB = 60°. The perimeter of the sector is P cm. Find a formula for P in terms of r.
Give your answer in the form $P = r(c\pi + k)$ where $c$ and $k$ are values to be found.
▶️Answer/Explanation
Solution:
Arc length = (60/360) × 2πr = πr/3
Perimeter P = 2r + πr/3 = r(2 + π/3)
Answer: P = r(π/3 + 2)
14. [Maximum mark: 3]
Adriana is going to roll a biased dice and spin a biased coin. The probability that the coin will land on Heads is 0.8. The probability that the dice will land on 6 and the coin will land on Heads is 0.24. Work out the probability that the dice will land on 6 and the coin will land on Tails.
▶️Answer/Explanation
Solution:
P(6 and Heads) = P(6) × P(Heads) ⇒ 0.24 = P(6) × 0.8 ⇒ P(6) = 0.3
P(6 and Tails) = P(6) × P(Tails) = 0.3 × 0.2 = 0.06
15. [Maximum mark: 3]
AB, BC and CD are three sides of a regular pentagon and CDE is a triangle. BCE is a straight line. CD = 6.5 cm, CE = 3 cm. Work out the area of triangle CDE (to 3 significant figures).
▶️Answer/Explanation
Solution:
Interior angle of pentagon = 108° ⇒ Angle DCE = 180° – 108° = 72°
Area = ½ × 6.5 × 3 × sin72° ≈ 9.27 cm²
16. [Maximum mark: 2]
Write down the letter of the graph that could have the equation:
(i) \( y = -\frac{1}{x} \)
(ii) \( y = \sin x^\circ \).
▶️Answer/Explanation
Solution:
(i) Graph E.
(ii) Graph A.
17. [Maximum mark: 3]
Given f(x) = x/(2x-4) and g(x) = 3x + 1, and that fg(k) = 2, work out the value of k.
▶️Answer/Explanation
Solution:
fg(k) = f(3k+1) = (3k+1)/(2(3k+1)-4) = 2
(3k+1)/(6k-2) = 2 ⇒ 3k+1 = 12k-4 ⇒ 5 = 9k ⇒ k = 5/9
18. [Maximum mark: 2]
Use algebra to show that 0.306 (recurring) = 34/111.
▶️Answer/Explanation
Solution:
Let x = 0.306306…
1000x = 306.306306…
Subtract: 999x = 306 ⇒ x = 306/999 = 34/111
19. [Maximum mark: 3]
Aviv’s cycle journey has average speed = 19 km/h (to nearest whole number) and time = 1.5 hours (to 1 decimal place). Work out the upper bound for the distance traveled.
▶️Answer/Explanation
Solution:
Upper bound speed = 19.5 km/h
Upper bound time = 1.55 hours
Upper bound distance = 19.5 × 1.55 ≈ 30.2 km
20. [Maximum mark: 4]
Solve the inequality 6x² – 7x – 20 > 0.
▶️Answer/Explanation
Solution:
First solve 6x² – 7x – 20 = 0 ⇒ (3x+4)(2x-5) = 0 ⇒ x = -4/3 or 5/2
The parabola opens upwards (a=6>0), so inequality holds when x < -4/3 or x > 5/2
21. [Maximum mark: 4]
ABCD is a square. Point A has coordinates (-5,2) and B has (3,5). Find an equation of the line that passes through B and C in the form ax + by + c = 0.
▶️Answer/Explanation
Solution:
Gradient of AB = (5-2)/(3-(-5)) = 3/8
Gradient of BC (perpendicular) = -8/3
Equation: y – 5 = (-8/3)(x – 3)
Multiply by 3: 3y – 15 = -8x + 24 ⇒ 8x + 3y – 39 = 0
22. [Maximum mark: 5]
Solve the simultaneous equations:
x² + y² = y + 11
y = 3x – 1
▶️Answer/Explanation
Solution:
Substitute y: x² + (3x-1)² = (3x-1) + 11
x² + 9x² – 6x + 1 = 3x + 10 ⇒ 10x² – 9x – 9 = 0
Solve: (5x+3)(2x-3) = 0 ⇒ x = -0.6 or 1.5
Corresponding y values: y = -2.8 or 3.5
Solutions: (-0.6, -2.8) and (1.5, 3.5)
23. [Maximum mark: 2]
A curve has equation \( y = f(x) \). The coordinates of the minimum point on this curve are \( (6, -3) \). Write down the coordinates of the minimum point on the curve with equation:
(i) \( y = f(x) + 10 \)
(ii) \( y = f(3x) \).
▶️Answer/Explanation
Solution:
(i) The minimum point is \( (6, 7) \).
(ii) The minimum point is \( (2, -3) \).
24. [Maximum mark: 5]
A solid S is made from a cone and a hemisphere
with equal radii x cm. The total height of S is 4x cm. A separate sphere has radius kx cm and volume 12.5 × volume of S.
(a) Find k
(b) For a similar solid T with volume 512 × volume of S, find the scale factor d for surface area.
▶️Answer/Explanation
(a) Volume of S = (2/3)πx³ + (1/3)πx²(3x) = (5/3)πx³
Sphere volume = (4/3)π(kx)³ = 12.5 × (5/3)πx³ ⇒ k³ = 125/8 ⇒ k = 2.5
(b) Volume scale factor 512 = 8³ ⇒ linear scale factor d = 8 ⇒ surface area scale factor = 8² = 64
25. [Maximum mark: 6]
OPQR is a parallelogram
with vectors OP = 2a and OR = 3b. Points M and N divide PQ and RQ in given ratios. Find vectors ON and MR, then find k where OY = k × ON.
▶️Answer/Explanation
(a)(i) ON = (8/5)a + (3/5)b
(a)(ii) MR = (9/4)b – 2a
(b) Setting OY = kON = (1-λ)OR + λOM gives k = 5/8
26. [Maximum mark: 4]
Write \( 4 – \left[ (3x – 5) + \frac{3x^2 + x – 10}{4x – 1} \right] \) as a single fraction in its simplest form.
▶️Answer/Explanation
Solution:
\( (3x-5)(x+2) \) \[ \frac{4(3x^2+x-10)-(3x-5)(4x-1)}{3x^2+x-10} \quad \text{or} \quad \frac{27x-45}{3x^2+x-10} \] \[ \frac{4x-1}{x+2} \quad \frac{4(3x^2+x-10)-(3x-5)(4x-1)}{(3x-5)(x+2)} \] \[ \text{implies first } M1 \quad \text{or} \quad \frac{27x-45}{(3x-5)(x+2)} \] \[ \frac{4(x+2)-(4x-1)}{x+2} \quad \frac{9(3x-5)}{(3x-5)(x+2)} \] \[ \text{or} \quad \frac{4x+8-4x+1}{x+2} \] \[ \text{or} \quad \frac{4(x+2)}{x+2} – \frac{4x-1}{x+2} \] \[ \text{or} \quad \frac{4x+8}{x+2} – \frac{4x-1}{x+2} \]
Working not required, so correct answer scores full marks (unless from obvious incorrect working).