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Question 1

Topic – C4.5

Write down the letter of the shape that is congruent to the shaded shape.

▶️ Answer/Explanation

Solution

Ans: D

Congruent shapes are identical in shape and size. The shaded shape matches exactly with shape D when rotated.

Question 2

Topic – C1.1

Write down:

(a) all the factors of 32

(b) the reciprocal of $\frac{1}{8}$

▶️ Answer/Explanation

Solution

(a) Ans: 1, 2, 4, 8, 16, 32

(b) Ans: 8 (Reciprocal means flipping the fraction)

(c) Ans: 70,000 (The 7 is in the ten-thousands place)

Question 3

Topic – C4.5

Draw the lines of symmetry on this rhombus.

▶️ Answer/Explanation

Solution

Ans: A rhombus has two lines of symmetry – one vertical and one horizontal diagonal.

Draw one line connecting opposite vertices, and another line connecting the other pair of opposite vertices.

Question 4

Topic – C1.3

From the list, write down:

(a) a cube number

(b) a prime number

▶️ Answer/Explanation

Solution

(a) Ans: 64 (4×4×4)

(b) Ans: 61 (Only divisible by 1 and itself)

Note: 63, 66, 68, 69 are all composite numbers.

Question 5

Topic – C1.15

Tara goes on a journey by train.
The train leaves at 0648.
The journey takes 12 hours and 35 minutes.

Find the time when Tara arrives.

▶️ Answer/Explanation

Solution

Ans: 19:23 or 7:23 pm

Start at 06:48, add 12 hours to reach 18:48 (6:48 pm).

Then add 35 minutes: 18:48 + 35 minutes = 19:23.

Question 6

Topic – C9.3

Jamie records the masses of two samples of oranges, type A and type B. The stem-and-leaf diagram shows the mass, in grams, of each of 30 oranges of type A.

(a) Complete the table to show the range for type A oranges.

(b) Use the information in the table to write down two comments comparing the masses of type A and type B oranges.

▶️ Answer/Explanation

Solution

(a) 42 (calculated from stem-and-leaf diagram)

(b) 1. Type A oranges are lighter on average

2. Type A masses are more varied/less consistent

For (a), subtract smallest mass (176g) from largest (218g). For (b), compare means and ranges.

Question 7

Topic – C4.2

In triangle LMN, LN = 7.5 cm and MN = 8 cm.

(a) Using a ruler and compasses only, construct triangle LMN. Leave in your construction arcs.
The line LM has been drawn for you.

(b) Write down the mathematical name for this triangle.

▶️ Answer/Explanation

Solution

(a) Construction with arcs

(b) Scalene

For (a), draw arcs from L (radius 8cm) and M (radius 7.5cm) to find N. For (b), all sides are different lengths.

Question 8

Topic – C5.4

The surface area of a cube is 73.5 cm². Find the length of one side of the cube.

▶️ Answer/Explanation

Solution

3.5 cm

Surface area = 6 × side² → 73.5 ÷ 6 = 12.25 → √12.25 = 3.5 cm.

Question 9

Topic – C4.6

The diagram shows an equilateral triangle. Find the value of x.

▶️ Answer/Explanation

Solution

150

External angle = 360° – internal angle (60° for equilateral) → x = 360 – 60 – 60 – 60 = 150°.

Question 10

Topic – C7.1

Given vectors: a = $\begin{pmatrix}4\\9\end{pmatrix}$, b = $\begin{pmatrix}-6\\1\end{pmatrix}$, c = $\begin{pmatrix}13\\-2\end{pmatrix}$

(a) Work out a + b

(b) Work out 3c

▶️ Answer/Explanation

Solution

(a) $\begin{pmatrix}-2\\10\end{pmatrix}$

(b) $\begin{pmatrix}39\\-6\end{pmatrix}$

For (a), add corresponding components. For (b), multiply each component by 3.

Question 11

Topic – C2.2

Factorise completely.

$15v² – 3v$

▶️ Answer/Explanation

Solution

Ans: $3v(5v – 1)$

1. Identify common factor: both terms have 3 and v

2. Factor out 3v: $3v(5v) – 3v(1)$

3. Write final factorised form: $3v(5v – 1)$

Question 12

Topic – C8.1

Rama asks a group of students how they travel to school. The table shows the probability of how a student, chosen at random, travels to school.

(a) Complete the table.

(b) There are 1800 students at the school. Find the expected number of students that walk to school.

▶️ Answer/Explanation

Solution

Ans:

(a) Other probability = 0.11 (1 – 0.4 – 0.32 – 0.17)

(b) Expected walkers = 1800 × 0.32 = 576

1. For (a), subtract given probabilities from 1

2. For (b), multiply total students by walk probability

Question 13

Topic – C1.4

Without using a calculator, work out $1 \frac{5}{6} ÷ \frac{11}{15}$.

Give your answer as a mixed number in its simplest form.

▶️ Answer/Explanation

Solution

Ans: $2 \frac{1}{2}$

1. Convert mixed number to improper fraction: $\frac{11}{6}$

2. Change division to multiplication by reciprocal: $\frac{11}{6} × \frac{15}{11}$

3. Simplify: $\frac{15}{6} = \frac{5}{2} = 2 \frac{1}{2}$

Question 14

Topic – C4.1

The diagram shows triangle ABC. M is the midpoint of AC.

Triangle ABC is rotated 180° about centre M.
The image and the original triangle together form a quadrilateral ABCD.

(a) Write down the name of quadrilateral ABCD formed.

(b) Find angle BAD.

▶️ Answer/Explanation

Solution

Ans:

(a) Parallelogram

(b) 68°

1. 180° rotation creates parallelogram

2. Angle BAD = 180° – 112° = 68° (angles on straight line)

Question 15

Topic – C1.16

Shubhu invests $750 in a savings account for 5 years.
The account pays simple interest at 1.8% per year.

Calculate the total interest earned.

▶️ Answer/Explanation

Solution

Ans: $67.50

1. Calculate annual interest: $750 × 1.8% = $13.50

2. Multiply by 5 years: $13.50 × 5 = $67.50

Or in one step: $750 × 1.8% × 5

Question 16

Topic – C2.5

Solve the equation.

5x + 7 = 9x – 3

▶️ Answer/Explanation

Solution

Ans: 2.5

5x + 7 = 9x – 3

7 + 3 = 9x – 5x

10 = 4x

x = 2.5

Question 17

Topic – C3.5

(a) Find the equation of line L in the form y = mx + c

(b) On the grid, draw a line that is perpendicular to line L

▶️ Answer/Explanation

Solution

Ans:

(a) y = (1/3)x + 1 (from the graph, slope is 1/3 and y-intercept is 1)

(b) Any line with slope -3 (negative reciprocal of 1/3) would be perpendicular

Question 18

Topic – C2.1

A bar of chocolate costs $3 and a bag of sweets costs $5. Write down an expression for the total cost, in dollars, of x bars of chocolate and y bags of sweets.

▶️ Answer/Explanation

Solution

Ans: 3x + 5y

Cost of x chocolate bars = 3 × x = 3x

Cost of y sweet bags = 5 × y = 5y

Total cost = 3x + 5y

Question 19

Topic – C8.1

(a) A bag contains these cards.

One of these cards is picked at random. Find the probability that the number on the card is greater than 3.

(b) A box contains 3 blue and 7 red cards. Kim picks one card at random, notes its colour and then replaces it in the box. She then picks another card at random.

(i) Complete the tree diagram

(ii) Work out the probability that both of the cards Kim picks are blue.

▶️ Answer/Explanation

Solution

Ans:

(a) 4/7 (Numbers >3: 7,9,4,5 out of 7 total cards)

(b)(i) Tree diagram should show:

First branch: Blue (3/10) → Second branches: Blue (3/10) and Red (7/10)

First branch: Red (7/10) → Second branches: Blue (3/10) and Red (7/10)

(b)(ii) 9/100 (3/10 × 3/10 = 9/100)

Question 20

Topic – C5.3

The diagram shows a sector of a circle with radius r cm and sector angle 72°.
The arc length is 9.35 cm.

Calculate the value of r.

▶️ Answer/Explanation

Solution

Ans: 7.44

Arc length = (θ/360) × 2πr

9.35 = (72/360) × 2 × 3.142 × r

9.35 = 0.2 × 6.284 × r

r = 9.35 ÷ 1.2568 ≈ 7.44

Question 21

Topic – C1.2

ξ = {2, 4, 8, 9, 10, 12}
Q = {square numbers}
R = {multiples of 4}

(a) Use this information to complete the Venn diagram.

(b) Write down n(Q ∩ R).

▶️ Answer/Explanation

Solution

Ans:

(a) Venn diagram should show: 4, 8 in Q∩R; 9 in Q only; 12 in R only; 2,10 outside both circles

(b) 1 (only 4 is both a square number and multiple of 4 in the given set)

Square numbers in set: 4, 9

Multiples of 4 in set: 4, 8, 12

Intersection Q∩R contains only 4

Question 22

Topic – C1.1

Find the highest common factor (HCF) of 48 and 80.

▶️ Answer/Explanation

Solution

Ans: 16

Prime factors of 48: 2 × 2 × 2 × 2 × 3

Prime factors of 80: 2 × 2 × 2 × 2 × 5

Common prime factors: 2 × 2 × 2 × 2 = 16

Therefore, HCF is 16

Question 23

Topic – C2.5

Solve the simultaneous equations. You must show all your working.

3x + 5y = 23
6x – 4y = 11

▶️ Answer/Explanation

Solution

Ans: x = 3.5, y = 2.5

Multiply first equation by 2: 6x + 10y = 46

Subtract second equation: (6x+10y)-(6x-4y)=46-11 → 14y=35 → y=2.5

Substitute y=2.5 into first equation: 3x+12.5=23 → 3x=10.5 → x=3.5

Check in second equation: 6(3.5)-4(2.5)=21-10=11 ✓

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