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Question1

Q1:
(a)
(i) C9.4
(ii) C9.3
(b)
(i) C9.4
(ii) C1.13
 
(a) $40$ football players vote on the colour of new shirts.

The results for red, yellow and green are shown in the bar chart.

(i)  Twice as many football players vote blue than vote orange.
Complete the bar chart.

(ii) Write down the mode.

(b) $40$ hockey players vote on the colour of new shirts.
 
The table shows the results.
 
 
(i) Complete the pie chart.
 
 
(ii) Work out the percentage of hockey players who vote grey
▶️Answer/Explanation

(a)(i): Correct bars at [blue] $10$ and [orange] $5$
(a)(ii): Red
(b)(i): Correct pie chart
(b)(ii): $30$

Detailed Answer

(a)(i)
Votes for Red: 11
Votes for Yellow: 6
Votes for Green: 8
$
11 + 6 + 8 + \text{Blue votes} + \text{Orange votes} = 40
$
$
25 + \text{Blue votes} + \text{Orange votes} = 40
$
$
\text{Blue votes} + \text{Orange votes} = 15
$

Blue votes = 2 × Orange votes. Let’s the Orange votes x
$
\text{Blue votes} = 2x
$
$
2x + x = 15
$
$
3x = 15
$
$
x = 5
$

Orange votes = 5
Blue votes = 10

(ii)

The mode is the most frequent value – the color with the highest number of votes.
The highest frequency is 11 votes for Red.

The mode is: Red
Let’s solve this step by step!

(c)(i)

The total number of players is 40, and a full circle is 360°.

White:
$
\frac{21}{40} \times 360 = 189^\circ
$

Grey:
$
\frac{12}{40} \times 360 = 108^\circ
$

Pink:
$
\frac{7}{40} \times 360 = 63^\circ
$

(c)(ii)
Frequency of grey =21
$
\text{Percentage for Grey} = \frac{12}{40} \times 100 = 30\%
$

Question2

Q2:
(a)
(i) C1.15
(ii) C1.15
(iii) C1.11
(b) C1.16
(c)
(i) C1.11
(ii) C1.13
(iii) C1.13

(a) Here is part of the timetable for buses from the station to the city centre.
All buses take the same time to travel from the station to the city centre.

(i) Complete the time table.

(ii) Beth walks 4 km from her home to the station at a speed of 6 km/h.
She wants to travel on the 09 24 bus.
Work out the latest time she can leave her home.

(iii) 45 seats on the bus are occupied.
This is $\frac{3}{5}$ of the total number of seats on the bus.
Work out the total number of seats on the bus.

(b) Beth buys $2.4$kg of onions costing \(\$1.25\) per kilogram and $4.5$kg of potatoes.
The total cost is \(\$11.64\) .
Find the cost of $1$kg of potatoes.

$( \mathbf{c} )$ $( \mathbf{i} )$ One day 140 people enter a shop
The ratio \(\textbf{adults:children=3:2}\).
Find the number of adults who enter the shop

(ii) The price of a television in this shop is $\$624.$

$37.5\%$ of this price is profit.
Calculate the profit on this television.

(iii) The price of a phone in this shop is \(\$420\).

This price increases by $12\%$.

Calculate the new price.

▶️Answer/Explanation

(a)(i): $1145$
(a)(ii): $0844$
(a)(iii): $75$
(b): $1.92$
(c)(i): $84$
(c)(ii): $234$
(c)(iii): $470.40$
Detailed Answer

(a)
Departure from station: 09:24
Arrival at city center: 10:03

Travel time
$
10:03 – 09:24 = 39 \text{ minutes}
$

For the 11:06 bus
$
11:06 + 00:39 = 11:45
$

(ii)
Distance = 4 km, Speed = 6 km/h
Time = Distance ÷ Speed:
$
\frac{4}{6} = 0.6667 \text{ hours} = 40 \text{ minutes}
$
She needs to catch the 09:24 bus
Time to leave home:
$
09:24 – 00:40 = 08:44
$

(iii)

If 45 seats are occupied and that’s \(\frac{3}{5}\) of the total seats
Let total seats = x
$
\frac{3}{5}x = 45
$
$
x = 45 \times \frac{5}{3}
$
$
x = 75$

(b)

Onions: $2.4 kg × \$1.25/kg = \$3.00$
Total cost: $\$11.64$
Cost of potatoes: $\$11.64 – \$3.00 = \$8.64$

Weight of potatoes: 4.5 kg

Price per kg
$
\frac{8.64}{4.5} = 1.92
$

(c)(i)
Total people: 140, Ratio: 3:2
Total parts: 3 + 2 = 5

Adults
$
\frac{3}{5} \times 140 = 84
$

Number of adults: 84

(ii)

Price: $\$624$, Profit = 37.5%

Profit amount:
$
0.375 \times 624 = 234
$

Profit: $\$234$

(iii)
Price: $420, Increase: 12%

Price increase
$
0.12 \times 420 = 50.40
$

New price:
$
420 + 50.40 = \$470.40
$

Question3

Q3:
(a) C7.1
(b)
(i) C3.1
(ii) C7.1
(iii) C7.1
(iv) C7.1

(a) The grid shows a trapezium.

 

On the grid, draw an enlargement of the trapezium with scale factor $3. $

$\mathbf{( b) }$ The diagram shows four triangles, $A,B,C$ and $T$,and a point $P$ on a grid

(i) Write down the coordinates of point $P.$

(ii) Describe fully the single transformation that maps

 (a) triangle $T$ onto triangle $A$

 (b) triangle $T$ onto triangle $B$

(c) triangle $T$ onto triangle $C.$

▶️Answer/Explanation

(a): Correct enlargement
(b)(i): $4,3$
(b)(ii)(a): Reflection, $x = -1$
(b)(ii)(b): Translation $\begin{pmatrix} 3 \\ -5 \end{pmatrix}$
(b)(ii)(c): Rotation, $(0, 0)$, $180^\circ$
Detailed Answer

(a)

(b)(i)

(b)(ii)

(a)
 When reflecting across the vertical line \( x = -1 \), every point on triangle \( T \) is flipped horizontally to the left.
The shape stays the same, and the orientation is reversed, perfectly matching triangle \( A \)’s position and size.

(b)
Translation means shifting the shape without rotating or flipping it.
3 units right (positive x-direction)
5 units down (negative y-direction)
After this movement, triangle \( T \)  exactly on triangle \( B \).

(c)
A 180° rotation around the origin flips every point across both the x-axis and y-axis.
In this case, triangle \( T \) rotates to the bottom-left quadrant, exactly aligning with triangle \( C \).

Question4

Q4:
(a) C1.1
(b) C1.1
(c) C1.1
(d) C1.7
(e) C1.3
(f) C1.4
(g) C1.4
(h) C1.1
(i) C1.3

(a) Write down the value of the $8$ in the number $39829$

(b) Write down all the factors of $18$

(c) Show that 57 is not a prime number.

$(\mathbf{d})$

$$\sqrt{x}=64$$

Find the value of $x.$

$( \mathbf{e} )$ Find the first multiple of $40$ that is greater than 620

$( \mathbf{f} )$ Find the reciprocal of $\frac{2}{3}.$

\(\begin{aligned}&(\mathbf{g})\quad\text{Find a fraction between }\frac{1}{5}\mathrm{~and~}\frac{1}{4}.\\&(\mathbf{h})\quad\text{Write down an irrational number with a value between 9 and 10.}\\&&&\\&(\mathbf{i})\quad\mathrm{Find~the~highest~common~factor~(HCF)~of~72~and~180.}\end{aligned}\)

▶️Answer/Explanation

(a): $800$
(b): $1, 2, 3, 6, 9, 18$
(c): $57 = 3 \times 19$
(d): $4096$
(e): $640$
(f): $1\frac{1}{2}$ or $1.5$
(g): Any correct fraction
(h): Any correct irrational number
(i): $36$
Detailed Answer

(a)
The 8 is in the hundreds place
$
8 \times 100 = 800
$

(b)
The factors of 18 are the numbers that divide 18 exactly
$
1, 2, 3, 6, 9, 18
$

(c)
A prime number has exactly 2 factors (1 and itself).
$
57 = 3 \times 19
$
It has factors: \( 1, 3, 19, 57 \), so it’s not prime.

(d)
\( \sqrt{x} = 64 \)
$
x = 64^2
$
$
x = 4096
$

(e)
Divide 620 by 40
$
620 \div 40 = 15.5
$
The next integer is 16, so the next multiple is:
$
40 \times 16 = 640
$

(f)
Reciprocal of \( \frac{2}{3} \)
The reciprocal is just flipping the fraction:
$
\frac{3}{2}
$

(g)
\( \frac{1}{5} = \frac{4}{20} \)
\( \frac{1}{4} = \frac{5}{20} \)

A fraction between them
$
\frac{9}{40} \text{ or } \frac{1}{4.5}
$
$
\frac{11}{50}
$

(h)

An irrational number cannot be expressed as a simple fraction and goes on forever without repeating.

Example:
$
\sqrt{90} \approx 9.49
$
Or:
$
\pi \approx 9.14
$

(i)
\( 72 = 2^3 \times 3^2 \)
\( 180 = 2^2 \times 3^2 \times 5 \)

$
\text{HCF} = 2^2 \times 3^2 = 4 \times 9 = 36
$

Question5

Q5:
(a)
(i) C4.6
(ii) C4.6
(b) C4.6
(c) C4.6
(d) C4.7
(e) C4.6
(f) C6.2

(a)   

(i) Measure angle $k.$

(ii) Write down the mathematical name for this type of angle

$\mathbf{(b)}$ The diagram shows a pair of parallel lines and a straight line.

Angles $a,b,c,d$ and $x$ are labelled.

Complete the statements.

Angle ……………….. is alternate to angle x.
Angle ……………….. is corresponding to angle x.

(c) The diagram shows a parallelogram.

Find the value of y.

(d)  

$A,B$ and $C$ lie on a circle, centre $O.$

Find angle $ACB.$

$( \mathbf{e} )$ The interior angle of a regular polygon is 171°.
Work out the number of sides of this polygon.

$(\mathbf{f})$

Calculate the value of $x.$

▶️Answer/Explanation

(a)(i): $326$
(a)(ii): reflex
(b): $c$, $d$
(c): $138$
(d): $73^\circ$
(e): $40$
(f): $23.6$ or $23.57$ to $23.58$

Detailed Answer

(a) (i)

(ii)
The angle \( k \) is reflex angle.

(b)
Alternate angle to \( x \): Angle \( c \)
Alternate angles (also called Z-angles) are equal and form a “Z” shape.

Corresponding angle to \( x \): Angle \( d \)
Corresponding angles are in the same relative position at each intersection (F-angles).

(c)
In a parallelogram, adjacent angles are supplementary (add up to \( 180^\circ \)).
$
y + 42^\circ = 180^\circ
$
$
y = 180^\circ – 42^\circ = 138^\circ
$
So, \( y = 138^\circ \).

(d)
According to the Central Angle Theorem, the measure of a central angle is twice the measure of an inscribed angle that subtends the same arc.
$\angle BOC = 2 \times \angle BAC = 2 \times 17^\circ = 34^\circ$.

In isosceles triangle BOC, $\angle OBC = \angle OCB$.
The sum of angles in triangle BOC is 180 degrees.
So, $\angle BOC + \angle OBC + \angle OCB = 180^\circ$.
 $34^\circ + \angle OBC + \angle OCB = 180^\circ$. 

$\angle OBC = \angle OCB$

$34^\circ + 2 \angle OCB = 180^\circ$.
$2 \angle OCB = 180^\circ – 34^\circ = 146^\circ$.
$\angle OCB = \frac{146^\circ}{2} = 73^\circ$.
$\angle ACB = 73^\circ$.

(e)

$
\text{Interior Angle} = \frac{(n-2) \times 180^\circ}{n}
$
$
171 = \frac{(n-2) \times 180}{n}
$
$
171n = 180(n – 2)
$
$
171n = 180n – 360
$
$
180n – 171n = 360
$
$
9n = 360
$
$
n = 40
$

(f)

$
\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}}
$

Opposite side = 7.4 cm
Hypotenuse = 18.5 cm
$
\sin(x) = \frac{7.4}{18.5}
$
$
\sin(x) = 0.4
$

Take the inverse sine
$
x = \sin^{-1}(0.4) \approx 23.58^\circ
$

Question6

Q6:
(a) C1.6
(b) C2.5
(c) C2.5
(d) C2.2
(e) C2.5
(f) C2.5

(a) In a sport, teams are given points using the formula

number of points=number of wins$\times4+$number of draws$\times2+$bonus points
One team has 15 wins, 7 draws and 6 bonus points Calculate the total number of points for this team.

$(\mathbf{b})$ Solve. $\frac{ x}{2}= 18$

$(\mathbf{c})$ Solve.

$4x+12=18$

$( \mathbf{d} )$ Expand and simplify

$$6(3x-4)+5(x-2)$$

$(\mathbf{e})$

$$T=5r-6$$

Make $r$ the subject of this formula.

(f)Bo has a green bag and a blue bag. Each bag contains some marbles. The green bag has x marbles. There are 5 times as many marbles in the blue bag than in the green bag.

Bo now adds 6 marbles to each bag. There are now 4 times as many marbles in the blue bag than in the green bag.

Use this information to write down an equation and solve it to find the value of x.

▶️Answer/Explanation

(a): $80$
(b): $36$
(c): $1\frac{1}{2}$ or $1.5$
(d): $23x – 34$
(e): $r = \frac{T + 6}{5}$
(f): $5x + 6 = 4(x + 6)\Rightarrow x=18$

Detailed Answer

(a)
$
\text{Points} = (\text{Number of wins} \times 4) + (\text{Number of draws} \times 2) + (\text{Bonus points})
$
$
= (15 \times 4) + (7 \times 2) + 6
$
$
= 60 + 14 + 6
$
$
= 80
$

(b)
\(\frac{x}{2} = 18\)
Multiply both sides by 2
$
x = 18 \times 2
$
$
x = 36
$

(c)
\(4x + 12 = 18\)
$
4x = 6
$
$
x = \frac{6}{4} = 1.5
$

(d)
$
6(3x – 4) + 5(x – 2)
$
$
= 18x – 24 + 5x – 10
$
$
= 23x – 34
$

(e)
$
T = 5r – 6
$
$
T + 6 = 5r
$
$
r = \frac{T + 6}{5}
$

(f)
Green bag: \( x \) marbles
Blue bag: \( 5x \) marbles

After adding 6 marbles to each bag:
Green bag: \( x + 6 \)
Blue bag: \( 5x + 6 \)

The blue bag has 4 times as many marbles as the green bag:

$
5x + 6 = 4(x + 6)
$
$
5x + 6 = 4x + 24
$
$
5x – 4x = 24 – 6
$
$
x = 18
$

Question7

Q7:
(a)
(i) C8.1
(ii) C8.1
(iii) C8.1
(b) C8.1
(c)
(i) C8.3
(ii) C8.3
(iii) C8.3
(d)
(i) C8.3
(ii) C8.3

(a) Li spins a fair 6-sided spinner numbered $1$ to $6$.
On the probability scale, draw an arrow $(\downarrow )$ to show the probability that the spinner lands on he number 2.

 

$( \mathbf{ii})$ Find the probability that the spinner lands on a prime number

$( \textbf{iii})$ Find the probability that the spinner lands on the number 7.

(b) A bag contains 3 red balls and 12 green balls
Li picks a ball at random.

Find the probability that it is a green ball.
Give your answer as a fraction in its simplest form

$(\mathbf{c})$ Li spins two fair 4-sided spinners, each numbered l to 4.
The two numbers are multiplied to give the score

Find the probability that the score is

(i) an even number

(ii) an integer

(iii) at least 10.

$( \mathbf{d} )$ A bag contains red discs and blue discs.
The probability that a disc picked at random is red is $\frac{1}{5}.$
Li picks a disc at random, notes its colour and then replaces it in the bag
She then picks another disc at random.
(i) Complete the tree diagram.

$( \textbf{ii) }$ Work out the probability that both of the discs she picks are blue.

▶️Answer/Explanation

(a)(i): Arrow at $\frac{1}{6}$
(a)(ii): $\frac{1}{2}$
(a)(iii): $0$
(b): $\frac{4}{5}$
(c)(i): $\frac{3}{4}$
(c)(ii): $1$
(c)(iii): $\frac{3}{16}$
(d)(i):
(d)(ii): $\frac{16}{25}$

Detailed Answer

(a)

The spinner has 6 sides, so the probability of landing on 2
$
\frac{1}{6}
$

(ii)
Prime numbers between 1 and 6: 2, 3, 5
There are 3 prime numbers out of 6
So the probability is:
$
\frac{3}{6} = \frac{1}{2}
$

(iii)
Since the spinner is numbered 1 to 6, landing on 7 is impossible:
$
0
$

(b)
Total balls: \( 3 \text{ red} + 12 \text{ green} = 15 \)
Green balls: 12

The probability is
$
\frac{12}{15} = \frac{4}{5}
$

(c)
(i)
Even results: 2, 4, 6, 8, 12, 16
favoured outcomes: 12
Total outcomes= 16
Probability
$
\frac{12}{16} = \frac{3}{4}
$

(ii)
Every outcome is an integer, so the probability is:
$
1
$

(iii)
Scores ≥ 10: 12, 16
favoured outcomes: 3
Total outcomes = 16
Probability:
$
\frac{3}{16}
$

(i)
The probability of picking a red disc is given as \( \frac{1}{5} \).
Probability of picking a blue disc
$
1 – \frac{1}{5} = \frac{4}{5}
$

Since the disc is replaced, the probabilities stay the same for the second pick
First disc: Red (\( \frac{1}{5} \))
Second disc: Red → \( \frac{1}{5} \)
Second disc: Blue → \( \frac{4}{5} \)

First disc: Blue (\( \frac{4}{5} \))
Second disc: Red → \( \frac{1}{5} \)
Second disc: Blue → \( \frac{4}{5} \)

(ii)

(First Blue AND Second Blue)
Multiply the probabilities along this path
$
\frac{4}{5} \times \frac{4}{5} = \frac{16}{25}
$

Question8

Q8:
(a) C5.2
(b) C5.2
(c) C5.3
(d) C5.5

(a) Fine area of this trapezium

 

(b)

The area of this triangle is $15.3$ cm$^{2}$.

Find the value of $b.$

$( \mathbf{c} )$ A circle has a circumference of 58.6cm.

Find the radius of this circle.

(d)

The diagram shows a rectangle with two semicircles removed.

Calculate the shaded area.

▶️Answer/Explanation

(a): $56.2$ or $56.16$
(b): $6.8$
(c): $9.33$ or $9.325$ to $9.327$
(d): $223$ or $222.8$ to $222.91$

Detailed Answer

(a)
Area of a trapezium is
$
\text{Area} = \frac{1}{2}(a + b)h
$

\( a = 8.2 \, \text{cm} \) (shorter parallel side)
\( b = 12.6 \, \text{cm} \) (longer parallel side)
\( h = 5.4 \, \text{cm} \) (height)
$
\text{Area} = \frac{1}{2}(8.2 + 12.6)(5.4)
$
$
= \frac{1}{2}(20.8)(5.4)
$
$
= \frac{112.32}{2}
$
$
= 56.16 \, \text{cm}^2
$

(b)
$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
$
Area = \( 15.3 \, \text{cm}^2 \)
Height = \( 4.5 \, \text{cm} \)
Base = \( b \, \text{cm} \)

$
15.3 = \frac{1}{2} \times b \times 4.5
$
$
30.6 = b \times 4.5
$
$
b = \frac{30.6}{4.5}
$
$
b = 6.8 \, \text{cm}
$

(c)
$
C = 2 \pi r
$
Circumference \( C = 58.6 \, \text{cm} \)
$
58.6 = 2 \pi r
$
$
r = \frac{58.6}{2 \pi}
$
$
= \frac{58.6}{6.2832}
$
$
\approx 9.33 \, \text{cm}
$

(d)
$
\text{Area of rectangle} = \text{Length} \times \text{Width}
$
Length = \( 28 \, \text{cm} \)
Width = \( 12 \, \text{cm} \)

$
= 28 \times 12 = 336 \, \text{cm}^2
$

$
\text{Radius} = \frac{12}{2} = 6 \, \text{cm}
$
Since the figure has **2 semicircles** their combined area is the same as **1 full circle**:
$
\text{Area of circle} = \pi r^2 = \pi (6)^2 = 36\pi \approx 113.1 \, \text{cm}^2
$
$
\text{Total area of semicircles} = 113.1 \, \text{cm}^2
$
$
\text{Shaded area} = \text{Area of rectangle} – \text{Area of semicircles}
$
$
= 336 – 113.1
$
$
\approx 222.9 \, \text{cm}^2
$

Question9

Q9:
(a) C3.5
(b) C3.5
(c)
(i) C2.10
(ii) C2.10
(iii) C2.11
(iv) C2.10

$( \mathbf{a} )$ Line$L$ has a gradient of 4 and passes through the point (0,3).
Write down the equation of line $L$ in the form $y=mx+c.$

$\begin{array}{ll}{\mathbf{(b)}}&{{\mathrm{Line~}G\text{ has the equation }y=2-6x.}}\\&{{\mathrm{Line~}G\text{ passes through the point}(a,5).}}\end{array}$

Find the value of $a.$

$( \mathbf{c} )$ $( \mathbf{i} )$ Complete the table of values for $y= x^2- 6.$

(ii) One the grid , draw the graph of $y=x^2-6$ for $-4\leqslant x\leqslant 4$

\((\mathbf{iii})\quad\text{Write down the equation of the line of symmetry of the graph.}\)

\((\mathbf{iv})\quad\text{Use your graph to solve the equation}\quad x^2-6=0\quad\mathrm{for~}x>0.\)

▶️Answer/Explanation

(a): $y = 4x + 3$
(b): $-\frac{1}{2}$
(c)(i): $3 \quad -6 \quad 3$
(c)(ii): Correct curve
(c)(iii): $x = 0$
(c)(iv): $2.3$ to $2.6$

Detailed Answer

(a)

$
y = mx + c
$

\( m \) is the gradient (slope) of the line.
\( c \) is the y-intercept (the point where the line crosses the y-axis).

Gradient \( m = 4 \)
The line passes through the point \( (0, 3) \), so the y-intercept \( c = 3 \).
$
y = 4x + 3
$

(b)
$
y = 2 – 6x
$
It passes through the point \( (a, 5) \).
$
5 = 2 – 6a
$
$
6a = 2 – 5
$
$
6a = -3
$
$
a = -\frac{3}{6} = -\frac{1}{2}
$

(c)(i)

$
y = x^2 – 6
$

x= -3 $\Rightarrow$ \( y =9 – 6 = 3 \)
x= 0 $\Rightarrow$ \(y = 0 – 6 = -6 \)
x= 3 $\Rightarrow$ \( y =9 – 6 = 3 \)

(c)(ii)


(c)(iii)
$
x = \text{the x-coordinate of the vertex}
$
Since the vertex is at \( (0, -6) \):
$
\text{Line of symmetry: } x = 0
$

(c)(iv)
$
x^2 – 6 = 0
$
$
x^2 = 6
$
for \( x > 0 \)
$
x = \sqrt{6} \approx 2.45
$

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