Question1
1: C1.1
Write the number two million two thousand and two in figures.
▶️Answer/Explanation
$2002002$
- Two million → This is 2,000,000
- Two thousand → This is 2,000
- Two → This is 2
write them as a single number
$2,002,002$
Question2
2: C1.6
Put one pair of brackets into this calculation to make it correct.
$$\begin{matrix}5&+&4&\times&3&+&9&=&53\end{matrix}$$
▶️Answer/Explanation
$5+4\times(3+9)=53$
$
5 + 4 \times (3 + 9)
$
First, calculate inside the brackets
$
3 + 9 = 12
$
$
4 \times 12 = 48
$
$
5 + 48 = 53
$
Question3
3: C2.2
Simplify.
$$7x-8y-x-y$$
▶️Answer/Explanation
$6x-9y$ or $3(2x-3y)$ final answer
$
7x – 8y – x – y
$
For the \(x\) terms
$
7x – x = 6x
$
For the \(y\) terms
$
-8y – y = -9y
$
$
6x – 9y
$
Question4
4(a): C1.9
4(b): C1.9
4(c): C1.10
(a) Write $164703$ correct to the nearest thousand.
(b) Write $16.983$ correct to l decimal place.
(c) Write $0.037665$ correct to 2 significant figures.
▶️Answer/Explanation
(a) $165 000$ cao
(b) $17.0$ cao
(c) $0.038$ cao
(a)
Look at the hundreds digit (7).
Since 7 is 5 or more, round up the thousands digit.
$164703 \approx 165000$
(b)
Look at the second decimal digit (8).
Since 8 is 5 or more, round up the first decimal place
$16.983 \approx 17.0$
(c)
The first significant figure is 3, and the second is 7.
Look at the next digit (6).
Since 6 is 5 or more, round up the second significant figure.
$0.037665 \approx 0.038$
Question5
5(a): C4.5
5(b): C4.5
(a)On the diagram, draw any lines of symmetry.
(b) Write down the order of rotational symmetry of this shape.
▶️Answer/Explanation
(a) Correct line of symmetry
(b) $3$
(a)
(b)
Question6
6: C1.5
Write these numbers in order, starting with the smallest.
$$0.45\quad \quad \quad 42\%\quad \quad \quad \frac4{11}\quad \quad \quad \frac25$$
▶️Answer/Explanation
$\frac 4{11}$ $\frac 25$ $42\% 0. 45$
\( 0.45 \) is already a decimal.
\( 42\% = 0.42 \) (since percent means “out of 100”)
\( \frac{4}{11} \approx 0.3636 \) (divide 4 by 11)
\( \frac{2}{5} = 0.4 \)
$
0.3636, \quad 0.4, \quad 0.42, \quad 0.45
$
From smallest to largest
$
\frac{4}{11}, \quad \frac{2}{5}, \quad 42\%, \quad 0.45
$
Question7
7: C5.4
The base of a cuboid measures l0cm by $7$cm.
The volume of the cuboid is 280 cm$^{3}$
Calculate the height of the cuboid.
▶️Answer/Explanation
$4$
- Base dimensions: 10 cm × 7 cm
- Volume: 280 cm³
$
\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}
$
$
280 = 10 \times 7 \times h
$
$
280 = 70h
$
$
h = \frac{280}{70}
$
$
h = 4 \, \text{cm}
$
Question8
8: C8.1
In a city, the probability that it will rain today is $0.15$.
Find the probability that it will not rain today in this city
▶️Answer/Explanation
$0.85$
We know the probability of rain is 0.15.
The total probability for any event is 1. So, the probability it won’t rain is
$1 – 0.15 = 0.85$
Question9
9(a): C1.9
9(b): C1.9
One day the temperature in Tokyo is $-5^\circ$C and the temperature in Manila is $18^\circ$C .
(a) Work out the difference between these two temperatures
(b) The temperature in Tokyo rises by $4^{\circ}$C.
Find the new temperature in Tokyo.
▶️Answer/Explanation
(a) $23$
(b) $-1$
(a)
Temperature in Tokyo \( -5^\circ \mathrm{C} \)
Temperature in Manila \( 18^\circ \mathrm{C} \)
The difference is
$
18 – (-5)
$
$
18 + 5 = 23^\circ \mathrm{C}
$
(b)
Starting temperature in Tokyo: \( -5^\circ \mathrm{C} \)
Rise in temperature: \( +4^\circ \mathrm{C} \)
$
-5 + 4 = -1^\circ \mathrm{C}
$
Question10
10(a)(i): C2.7
10(a)(ii): C2.7
10(b): C2.7
(a) These are the first four terms of a sequence.
$3$ $10$ $17$ $24$
(i) Write down the next term.
(ii) Write down the term to term rule for continuing the sequence.
(b) These are the first four terms of another sequence.
$16$ $14$ $11$ $7$
Write down the next two terms of this sequence.
▶️Answer/Explanation
(a)(i) $31$
(a)(ii) add $7$
(b) $2$ $-4$
(a)(i)
$
3, \, 10, \, 17, \, 24
$
the difference between consecutive terms
$
10 – 3 = 7, \quad 17 – 10 = 7, \quad 24 – 17 = 7
$
So, to find the next term
$
24 + 7 = 31
$
(a)(ii)
Since each term increases by 7, the term-to-term rule is
$\text{Add 7}$
(b)
$
16, \, 14, \, 11, \, 7
$
the difference between consecutive terms
$
14 – 16 = -2, \quad 11 – 14 = -3, \quad 7 – 11 = -4
$
The differences are -2, -3, -4, so it looks like the sequence is decreasing by increasing negative values
Subtract 5 from the last term
$
7 – 5 = 2
$
Subtract 6 from the next term
$
2 – 6 = -4
$
$
2, \, -4
$
Question11
11: C4.6
The diagram shows an isosceles triangle.
Find the value of x.
▶️Answer/Explanation
$108$
it’s an isosceles triangle, the two base angles are equal. One base angle is given as 36°, so the other base angle is also 36°.
$
\text{Angle sum in a triangle} = 180^\circ
$
$
36^\circ + 36^\circ + x^\circ = 180^\circ
$
$
72^\circ + x^\circ = 180^\circ
$
$
x^\circ = 180^\circ – 72^\circ
$
$
x^\circ = 108^\circ
$
Question12
12: C5.4
The diagram shows a cuboid.
on the $1 \rm {cm}^3$ grid, complete a net of this cuboid.F
One face has been drawn for you.
▶️Answer/Explanation
Fully correct net
Question13
13: C2.2
Factorise completely.
$4x^2y-5xy^2$
▶️Answer/Explanation
$xy(4x – 5y)$ final answer
Both terms have a common factor of \( xy \). So
$
= xy(4x – 5y)
$
The fully factorised expression is
$
\mathbf{xy(4x – 5y)}
$
Question14
14: C4.3
The scale of a map is $1:40 000$.
On the map the distance between two villages is 37cm. Calculate the actual distance between the two villages. Give your answer in kilometres.
▶️Answer/Explanation
$14.8$
The scale is 1 : 40,000, meaning 1 cm on the map represents 40,000 cm in real life.
To find the actual distance, multiply by the scale factor:
$
\text{Actual distance} = 37 \times 40,000
$
$
= 1,480,000 \, \mathrm{cm}
$
1 kilometre = 100,000 cm
$
1,480,000 \div 100,000 = 14.8 \, \mathrm{km}
$
Question15
15: C1.4
Without using a calculator, work out $\frac{3}{7}-\frac{1}{14}.$
You must show all your working and give your answer as a fraction in its simplest form
▶️Answer/Explanation
$\frac{6}{14}-\frac{1}{14}$
$\frac{5}{14}$ cao
The fractions
$
\frac{3}{7} – \frac{1}{14}
$
The LCM (Least Common Multiple) of 7 and 14 is 14.
Convert \(\frac{3}{7}\) to a denominator of 14:
$
\frac{3}{7} = \frac{3 \times 2}{7 \times 2} = \frac{6}{14}
$
$
\frac{6}{14} – \frac{1}{14} = \frac{6 – 1}{14} = \frac{5}{14}
$
Question16
16: C1.13
The price of a game increases from \(\$48\) to \(\$56.40\) .
Calculate the percentage increase in the price.
▶️Answer/Explanation
$17.5$
$
\text{Percentage increase} = \frac{\text{New price} – \text{Original price}}{\text{Original price}} \times 100
$
$
= \frac{56.40 – 48}{48} \times 100
$
$
= \frac{8.40}{48} \times 100
$
$
= 0.175 \times 100
$
$
= 17.5\%
$
Question17
17: C6.2
The diagram shows a right-angled triangle.
Calculate $AB.$
▶️Answer/Explanation
$6.39$
the cosine rule
$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
$\cos(37^\circ) = \frac{AB}{8}$
$AB = 8 \times \cos(37^\circ)$
$\cos(37^\circ) \approx 0.7986$
$AB = 8 \times 0.7986$
$AB \approx 6.39 \, \mathrm{cm}$
Question18
18: C1.10
The length, s metres, of a ship is 83 m, correct to the nearest metre.
Complete this statement about the value of s.
▶️Answer/Explanation
$82.5$
When a value is rounded to the nearest meter:
- It could be up to 0.5 meters above or below the rounded value.
- The lower bound is 83 – 0.5 → 82.5 meters
- The upper bound is 83 + 0.5 → 83.5 meters
\( 82.5 \leq s < 83.5 \)
Question19
19: C2.5
Solve the simultaneous equations.
$$\begin{array}{l}5t-2w=19\\3t+2w=5\end{array}$$
▶️Answer/Explanation
$[t = ] 3$
$[w = ] –2$
Since the coefficients of \( w \) are +2 and -2
adding the equations will eliminate \( w \).
$
(5t – 2w) + (3t + 2w) = 19 + 5
$
$
5t + 3t = 24
$
$
8t = 24
$
$
t = 3
$
Substitute \( t = 3 \) into one equation
$
3(3) + 2w = 5
$
$
9 + 2w = 5
$
$
2w = 5 – 9
$
$
2w = -4
$
$
w = -2
$
$
t = 3, \quad w = -2
$
Question20
20: C4.6
The diagram shows the positions of three towns $A,B$ and $C.$
Angle $ABC=103^{\circ}.$
The bearing of town $B$ from town $A$ is 48°.
Town $C$ is due east of town $A.$
Find the bearing of town $C$ from town $B.$
▶️Answer/Explanation
$125$
- Angle \( ABC = 103^\circ \)
- Bearing of B from A = 48°
- C is due east of A, so the line AC is horizontal.
bearing of C from B.
- Bearing of B from A = 48°, so the angle between the north line at A and line AB is 48°.
- Since C is due east of A, the angle NAC = 90°.
So, the bearing of C from A = 90°.
The external angle at B is:
$
180^\circ – 103^\circ = 77^\circ
$
the bearing of C from B is measured clockwise from north,
add this 77° to the 48° bearing of B from A:
$
\text{Bearing of C from B} = 48^\circ + 77^\circ = 125^\circ
$
Question21
21(a): C1.2
21(b): C1.2
(a)
$\mathcal{E}= \{ 1, 4, 5, 8, 9, 12, 16, 64\}$
$C=\{$cube numbers$\}$
$S=\{$square numbers$\}$
(i) Complete the Venn diagram
(ii) Find n$(C\cup S).$
(b) 
On this Venn diagram, shade the region $A\cap B.$
▶️Answer/Explanation
(a)(i)
(a) (ii) $6$
(b)
(a)
Cube Numbers (Set \( C \))
$
1 = 1^3, \quad 8 = 2^3, \quad 64 = 4^3
$
$
C = \{ 1, 8, 64 \}
$
Square Numbers (Set \( S \))
$
1 = 1^2, \quad 4 = 2^2, \quad 9 = 3^2, \quad 16 = 4^2, \quad 64 = 8^2
$
$
S = \{ 1, 4, 9, 16, 64 \}
$
Intersection \( C \cap S \): Numbers that are both squares and cubes
$
C \cap S = \{ 1, 64 \}
$
(ii)
\( C = \{ 1, 8, 64 \} \) (cube numbers)
\( S = \{ 1, 4, 9, 16, 64 \} \) (square numbers)
The union of the sets includes all unique elements
$
C \cup S = \{ 1, 4, 8, 9, 16, 64 \}
$
$
n(C \cup S) = 6
$
(b)
In a Venn diagram, the symbol \( A \cap B \) represents the intersection of sets \( A \) and \( B \).
It includes all the elements that are common to both sets \( A \) and \( B \).
The overlapping region (where the circles intersect) is the shaded part, representing \( A \cap B \).
Question22
22(a)(i): C1.8
22(a)(ii): C1.8
22(b): C1.8
(a) Write these numbers in standard form.
(i) $0.007$
(ii) $700000000$
(b) Calculate $\frac {3200\times 5. 4\times 10^{- 3}}{4. 8\times 10^{- 4}}.$
Give your answer in standard form.
▶️Answer/Explanation
(a) (i) $7\times10^{-3}$
(a) (ii) $7\times10^8$
(b) $3.6\times10^4$
(a)
$
a \times 10^n
$
Where \( 1 \leq a < 10 \) and \( n \) is an integer.
Move the decimal point 3 places to the right
$
0.007 = 7 \times 10^{-3}
$
(ii)
Move the decimal point 8 places to the left
$
700000000 = 7 \times 10^8
$
(b)
$
\frac{3200 \times 5.4 \times 10^{-3}}{4.8 \times 10^{-4}}
$
$
= \frac{(3200 \times 5.4)(10^{-3})}{4.8 \times 10^{-4}}
$
$
= \frac{3200 \times 5.4}{4.8} \times \frac{10^{-3}}{10^{-4}}
$
$
= 3.6 \times 10^3 \times 10^1
$
Question23
23: C5.4
The diagram shows a sphcrical tank with radius 0.5 m and a cylindrical jug with diameter $24$cm and height $32$cm. The tank is full of water.
Calculate how many times the jug can be completely filled with water from the tank
[The volume, $V$, of a sphere with radius $r$ is $\frac43\pi r^3.]$
▶️Answer/Explanation
$36$
$V_{\text{sphere}} = \frac{4}{3}\pi r^3$
The radius of the tank is \( r = 0.5 \, \mathrm{m} = 50 \, \mathrm{cm}\).
$V_{\text{sphere}} = \frac{4}{3}\pi (50)^3$
$= \frac{4}{3}\pi (125000)$
$= \frac{500000}{3}\pi \, \mathrm{cm^3}$
The volume of a cylinder $V_{\text{jug}} = \pi r^2 h$
The diameter of the jug is \( 24 \, \mathrm{cm} \), so the radius is
$r = \frac{24}{2} = 12 \, \mathrm{cm}$
The height is \( h = 32 \, \mathrm{cm} \).
$V_{\text{jug}} = \pi (12)^2 (32)$
$= \pi (144)(32)$
$= 4608\pi \, \mathrm{cm^3}$
$\text{Number of fills} = \frac{V_{\text{sphere}}}{V_{\text{jug}}}$
$= \frac{\frac{500000}{3}\pi}{4608\pi}$
$\approx 36.17$
The jug can be completely filled 36 times from the full spherical tank