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Question 1

Topic – C1.1

Write the number twenty-five million in figures.

▶️ Answer/Explanation

Solution

Ans: 25,000,000

Twenty-five million is written as 25 followed by six zeros.

In scientific notation: $ 25 \times 10^6 $.

Commas separate thousands for clarity: 25,000,000.

Question 2

Topic – C1.4

$( \mathbf{a} )$ Write $0.7$ as a fraction.

$( \mathbf{b} )$ Write $\frac{13}{20}$ as a percentage.

▶️ Answer/Explanation

Solution

(a) Ans: $\frac{7}{10}$

$0.7$ means $7$ tenths, so it directly converts to $\frac{7}{10}$. This fraction is already in its simplest form.

(b) Ans: $65\%$

To convert $\frac{13}{20}$ to a percentage, multiply by $100\%$:

$\frac{13}{20} \times 100\% = 65\%$.

Question 3

Topic – C1.5

$-7\quad \quad -12\quad \quad-3\quad \quad-28\quad \quad-6\quad \quad15\quad \quad-4\quad \quad-8$

From the list of numbers, find

(a) All the numbers which are less than $-5$

(b) The product of the largest number and the smallest number.

▶️ Answer/Explanation

Solution

(a) Ans: $-7, -12, -28, -6, -8$

Numbers less than $-5$ are those left of $-5$ on the number line. From the list, these are $-7, -12, -28, -6,$ and $-8$.

(b) Ans: $-420$

The smallest number is $-28$ (farthest left), and the largest is $15$ (farthest right). Their product is $15 \times (-28) = -420$.

Question 4

Topic – C1.15

An exam starts at 11:50 and lasts for $ 2\frac{1}{4} $ hours.
Work out the time that the exam finishes.

▶️ Answer/Explanation

Solution

Ans: 14:05 (2:05 PM)

$ 2\frac{1}{4} $ hours = 2 hours 15 minutes.

Start time: 11:50 + 2 hours = 13:50.

Add 15 minutes: 13:50 + 15 mins = 14:05.

Thus, the exam ends at 14:05.

Question 5

Topic – C1.4

Write $56.17345$ correct to $1$ decimal place.

▶️ Answer/Explanation

Solution

Ans: 56.2

To round $56.17345$ to 1 decimal place:

1. Identify the first decimal digit: 1 (in $56.17345$).

2. Look at the second decimal digit (7) to decide rounding.

3. Since $7 \geq 5$, we round the first decimal digit up from 1 to 2.

Thus, $56.17345$ rounded to 1 decimal place is 56.2.

Question 6

Topic – C1.15

Work out the number of seconds in $5$ hours.

▶️ Answer/Explanation

Solution

Ans: 18000

First, note that $1$ hour $= 60$ minutes and $1$ minute $= 60$ seconds.

So, for $5$ hours:

$5 \text{ hours} = 5 \times 60 \text{ minutes} = 300 \text{ minutes}$.

Now, convert minutes to seconds:

$300 \text{ minutes} = 300 \times 60 \text{ seconds} = 18000 \text{ seconds}$.

Thus, there are $18000$ seconds in $5$ hours.

Question 7

Topic – C1.1

From the list of numbers, write down

(a) a cube number

(b) a prime number.

▶️ Answer/Explanation

Solution

(a) Ans: 27

A cube number is obtained by multiplying a number by itself three times (n × n × n). In the given list, 27 is the only number that satisfies this (3 × 3 × 3 = 27).

(b) Ans: 29

A prime number has exactly two distinct factors: 1 and itself. Checking the list, 29 is the only number that meets this condition, as it is divisible only by 1 and 29.

Question 8

Topic – C7.1

$\textbf{v}= \begin{pmatrix} -1\\ 3\end{pmatrix}$, $\textbf{y}= \begin{pmatrix} 2\\ 5\end{pmatrix}$

Find:

$(a)\; \textbf{v} – \textbf{y}$

$(b)\; 2\textbf{v}$

▶️ Answer/Explanation

Solution

Ans:

(a) $ \begin{pmatrix} -3 \\ -2 \end{pmatrix} $

(b) $ \begin{pmatrix} -2 \\ 6 \end{pmatrix} $

Explanation:

(a) Subtract corresponding components of vectors $\textbf{v}$ and $\textbf{y}$:

\[ \textbf{v} – \textbf{y} = \begin{pmatrix} -1 – 2 \\ 3 – 5 \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \end{pmatrix} \]

(b) Multiply each component of $\textbf{v}$ by 2:

\[ 2\textbf{v} = \begin{pmatrix} -2 \\ 6 \end{pmatrix} \]

Question 9

Topic – C1.16

A suit costs 6500 rupees.

Calculate the cost of the suit in dollars when the exchange rate is 1 rupee = $0.013.

▶️ Answer/Explanation

Solution

Ans: \$84.5

Given:

Cost of suit = 6500 rupees
Exchange rate = 1 rupee = \$0.013

To find the cost in dollars:

\[ 6500 \times 0.013 = 84.5 \]

Thus, the suit costs \$84.5.

Question 10

Topic – C5.4

The diagram shows one face of a cuboid on a 1 cm² grid.

The cuboid has a volume of 24 cm³.
Complete a net of this cuboid.

▶️ Answer/Explanation

Solution

Ans: The cuboid has dimensions 4 cm × 3 cm × 2 cm.

1. From the grid, we observe one face measures 4 cm × 3 cm.

2. Let the unknown dimension be x cm.

3. Volume = 4 × 3 × x = 24 cm³.

4. Solving: 12x = 24 → x = 2 cm.

5. Thus, the cuboid’s net includes six faces: two 4×3, two 4×2, and two 3×2 rectangles.

Question 11

Topic – C9.3

The median of six numbers is $61$.
Five of the numbers are $24, 43, 58, 71$ and $85$.
Work out the sixth number.

▶️ Answer/Explanation

Solution

Ans: 64

For six numbers, the median is the average of the 3rd and 4th numbers when arranged in order.

The possible positions for the sixth number (x) give three cases:

1) If x is between 58 and 71: (58 + x)/2 = 61 → x = 64

2) If x > 71: (71 + x)/2 = 61 → x = 51 (invalid as x must be >71)

3) If x < 58: (x + 58)/2 = 61 → x = 64 (invalid as x must be <58)

Only x = 64 satisfies all conditions.

Question 12

Topic – C4.6

Work out the size of one interior angle of a regular 9-sided polygon.

▶️ Answer/Explanation

Solution

Ans: 140°

1. Sum of interior angles = $(n-2) \times 180°$, where $n = 9$.

2. Substitute: $(9-2) \times 180° = 1260°$.

3. Divide by 9 (equal angles): $1260° \div 9 = 140°$.

Each interior angle is $140°$.

Question 13

Topic – C1.2

On the Venn diagram, shade the region $A \cap B$.

▶️ Answer/Explanation

Solution

Ans:

The intersection $A \cap B$ represents all elements that are common to both sets $A$ and $B$. On a Venn diagram, this is the overlapping area between the two circles labeled $A$ and $B$. The correct shading covers only this shared region.

Question 14

Topic – C2.2

Factorise completely. $8g – 2g^2$

▶️ Answer/Explanation

Solution

Ans: $2g(4 – g)$

1. Identify the common factor: Both terms ($8g$ and $-2g^2$) share a factor of $2g$.

2. Factor out $2g$: $$ 8g – 2g^2 = 2g(4) – 2g(g) = 2g(4 – g). $$

3. Verify completeness: The expression $(4 – g)$ cannot be factored further, so $2g(4 – g)$ is the complete factorisation.

Question 15

Topic – C4.7

Circle diagram

The diagram shows a circle, centre O, with diameter AC.
A, B, C, D, and E lie on the circumference.
(a) Find the value of $ x $.
Give a reason for your answer.

(b) Find the value of $ y $.
Give a reason for your answer.

▶️ Answer/Explanation

Solution

Ans:

(a) $ x = 25^\circ $

(b) $ y = 46^\circ $

Explanation:

(a) Since AC is the diameter, $ \angle ABC = 90^\circ $ (angle in a semicircle).
In $ \triangle ABC $:
$ 90^\circ + 65^\circ + x = 180^\circ $ (sum of angles in a triangle).
Solving gives $ x = 25^\circ $.

(b) $ \triangle DOE $ is isosceles (OD = OE as radii).
Thus, base angles are equal: $ y = 46^\circ $.

Question 16

Topic – C1.4

Without using a calculator, work out $\frac{4}{7}\div 8$.

You must show all your working and give your answer as a fraction in its simplest form.

▶️ Answer/Explanation

Solution

Ans: $\frac{1}{14}$

$\frac{4}{7} \div 8 = \frac{4}{7} \times \frac{1}{8}$

Multiply numerators and denominators: $\frac{4 \times 1}{7 \times 8} = \frac{4}{56}$

Simplify by dividing numerator and denominator by 4: $\frac{1}{14}$

Question 17

Topic – C9.3

A school records how many calculators it sells each week for 40 weeks. The results are shown in the table.

Work out the mean number of calculators the school sells each week.

▶️ Answer/Explanation

Solution

Ans: 1.375

First, multiply each number of calculators by its frequency:

$ (0 \times 14) + (1 \times 12) + (2 \times 6) + (3 \times 5) + (4 \times 0) + (5 \times 2) + (6 \times 1) = 55 $

Total weeks = $ 14 + 12 + 6 + 5 + 2 + 1 = 40 $.

Mean = $ \frac{55}{40} = 1.375 $.

Question 18

Topic – C1.10

The mass, $m$ kg, of a bag of sand is 12 kg, correct to the nearest kilogram.

Complete the statement about the value of $m$.

▶️ Answer/Explanation

Solution

Answer: $11.5 \leq m < 12.5$

1. The mass is rounded to the nearest kg, so the true value is within ±0.5 kg of 12 kg.

2. Lower bound: $12 – 0.5 = 11.5$ kg.

3. Upper bound: $12 + 0.5 = 12.5$ kg (but $m$ must be less than 12.5 kg to round down to 12 kg).

Thus, $m$ is in the range $11.5 \leq m < 12.5$.

Question 19

Topic – C1.16

Qianna invests $\$3000$ at a rate of $4\%$ per year compound interest.
Calculate the value of her investment at the end of $6$ years.

▶️ Answer/Explanation

Solution

Answer: \$3795.96

Using the compound interest formula:

\[ A = P \left(1 + \frac{r}{100}\right)^n \]

Substitute $ P = 3000 $, $ r = 4 $, and $ n = 6 $:

\[ A = 3000 \times (1.04)^6 \]

Calculate $ (1.04)^6 \approx 1.2653 $:

\[ A \approx 3000 \times 1.2653 = 3795.96 \]

Question 20

Topic – C2.5

Solve.

$\frac{25-2u}{3}=2$

▶️ Answer/Explanation

Solution

Ans: B

$ \frac{25 – 2u}{3} = 2 $

$ 25 – 2u = 6 $

$ -2u = -19 $

$ u = 9.5 $

Question 21

Topic – C1.8

Calculate $0.3^2.$
Give your answer in standard form.

▶️ Answer/Explanation

Solution

Ans: $9 \times 10^{-2}$

$0.3^2 = 0.09$

In standard form, we write this as:

$9 \times 10^{-2}$

(The decimal moves 2 places right to become 9, hence the $-2$ exponent)

Question 22

Topic – C8.1

The probability of passing a driving test is $0.36$

$600$ people take this driving test.

Work out the expected number of these people that will pass

▶️ Answer/Explanation

Solution

Answer: 216

The expected number of people passing is calculated by multiplying the probability of passing by the number of test-takers.

\[ \text{Expected passes} = 0.36 \times 600 \]

Breaking it down:

\[ 0.36 \times 600 = (0.3 \times 600) + (0.06 \times 600) = 180 + 36 = 216 \]

Thus, the expected number of people who pass is 216.

Question 23

Topic – C2.5

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

$$\begin{matrix}3x-2y=19\\x+y=3\end{matrix}$$

▶️ Answer/Explanation

Solution

Ans: x = 5, y = -2

From $ x + y = 3 $, express $ x = 3 – y $.

Substitute into $ 3x – 2y = 19 $:

$ 3(3 – y) – 2y = 19 $

Simplify: $ 9 – 5y = 19 $ → $ y = -2 $.

Then, $ x = 3 – (-2) = 5 $.

Question 24

Topic – C6.2

The diagram shows a right-angled triangle.

Show that angle y is 31.9°, correct to 1 decimal place.

▶️ Answer/Explanation

Solution: 31.86 to 31.87

Using tangent ratio: $ \tan(y) = \frac{\text{opposite}}{\text{adjacent}} = \frac{46}{74} $

Calculate angle: $ y = \tan^{-1}\left(\frac{46}{74}\right) $

$ y = 31.86° $ (unrounded)

Rounded to 1 decimal place: $ y = 31.9° $

Question 25

Topic – C6.1

The diagram shows two right-angled triangles, ABC and ACD, sharing side AC.

Find the length of AD.

▶️ Answer/Explanation

Answer: 60

Detailed Solution:

First, find AC using triangle ABC:
Using Pythagoras’ theorem:
$ AC = \sqrt{AB^2 + BC^2} = \sqrt{39^2 + 52^2} = \sqrt{1521 + 2704} = \sqrt{4225} = 65 \text{ cm} $
Now find AD using triangle ACD:
Using Pythagoras’ theorem:
$ AD = \sqrt{AC^2 – CD^2} = \sqrt{65^2 – 25^2} = \sqrt{4225 – 625} = \sqrt{3600} = 60 \text{ cm} $

Question 26

Topic – C5.3

A circle has an area of $25\pi\mathrm{cm}^{2}.$

$\mathbf{(a)}$ Work out the circumference of the circle.
Give your answer in terms of $\pi.$

$\mathbf{(b)}$ Two of the circles are used as the ends of a cylinder, with height $h$ cm.

The total surface area of the cylinder is $170\pi\mathrm{cm}^{2}.$

Work out the value of $h.$

▶️ Answer/Explanation

Solution

Ans: (a) $10\pi$ cm (b) $12$ cm

Part (a):

Given area $A = 25\pi$ cm². Using $A = \pi r^2$, solve for $r$: $\pi r^2 = 25\pi$ ⇒ $r = 5$ cm.

Circumference $C = 2\pi r = 2\pi \times 5 = 10\pi$ cm.

Part (b):

Total surface area = $170\pi$ cm². Circular ends contribute $2 \times 25\pi = 50\pi$ cm².

Curved surface area: $2\pi r h = 10\pi h$ cm². Solve $50\pi + 10\pi h = 170\pi$ ⇒ $h = 12$ cm.

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