Question1
1: E1.1
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: E1.6
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: E2.2
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: E5.4
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}
$
Question5
5: E8.1
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
$1 – 0.15 = 0.85$
Question6
6: E2.4
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)}
$
Question7
7: E1.11
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}
$
Question8
8: E1.4
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}
$
Question9
9: E6.2
The diagram shows a right-angled triangle.
Calculate $AB.$
▶️Answer/Explanation
$6.39$
the cosine rule
$\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}$
$\cos(37^\circ) = \frac{AB}{8}$
$AB = 8 \times \cos(37^\circ)$
$\cos(37^\circ) \approx 0.7986$
$AB = 8 \times 0.7986$
$AB \approx 6.39 \, \mathrm{cm}$
Question10
10: E3.2
Find the gradient of the line joining the points (-2,7) and (3,1).
▶️Answer/Explanation
$-\frac{6}{5} \, \text{oe}$
$
\text{Gradient (} m \text{)} = \frac{y_2 – y_1}{x_2 – x_1}
$
For the points (-2, 7) and (3, 1):
\((x_1, y_1) = (-2, 7)\)
\((x_2, y_2) = (3, 1)\)
$
m = \frac{1 – 7}{3 – (-2)}
$
$
= \frac{-6}{3 + 2}
$
$
= \frac{-6}{5}
$
Question11
11: E2.5
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
$
Question12
12: E2.4
Simplify.
$( \mathbf{a} )$ $\frac {32g^{16}}{16g^{8}}$
$( \mathbf{b} )$ $( 625k^8) ^{\frac {1}{4}}$
▶️Answer/Explanation
[(a)] $2g^8$ (final answer)
[(b)] $5k^2$ (final answer)
(a)
$
\frac{32g^{16}}{16g^8}
$
Dividing the coefficients
$
\frac{32}{16} = 2
$
dividing powers
$
\frac{g^{16}}{g^8} = g^{16-8} = g^8
$
$
2g^8
$
(b)
$(625k^8)^{\frac{1}{4}}$
$
= 625^{\frac{1}{4}} \times (k^8)^{\frac{1}{4}}
$
$
625^{\frac{1}{4}} = 5
$
$
(k^8)^{\frac{1}{4}} = k^{8 \times \frac{1}{4}} = k^2
$
$
5k^2
$
Question13
13(a): E4.5
13(b): E4.5
$(\mathbf{a})$
Shade the region $A\cup B^{\prime}.$
$(\mathbf{b})$
Use set notation to describe the shaded region
▶️Answer/Explanation
(a) 
(b) $R \cap (P \cup Q’) \, \text{or} \, R \cap P’ \cap Q’ \, \text{oe}$
(a)
\( A \cup B’ \) means everything in set A or not in set B (the complement of B) which include outside area.
(b)
\( P \cup Q \) is everything in P or Q.
\( (P \cup Q)’ \) is the complement, meaning everything outside P and Q.
Intersection with R (\( R \cap (P \cup Q)’ \)) gives the part of R that lies outside both P and Q.
$
R \cap (P \cup Q)’
$
\( P’ \): Everything not in P
\( Q’ \): Everything not in Q
Intersection with R means we take the part of R that’s outside both P and Q.
$
R \cap P’ \cap Q’
$
Question14
14(a): E4.7
14(b): E4.7
$P,Q,R$ and $T$ are points on the circle.
$AB$ is a tangent to the circle at $T.$
Angle $ATP=50^\circ$, angle $PTR=48^\circ$ and $PQ=QR$
(a) Find angle $PRT.$
$( \mathbf{b} )$ Find angle $QPR.$
▶️Answer/Explanation
(a) $50$
(b) $24$
(a)
Tangent-Secant Theorem:
The angle between the tangent and a chord through the point of contact is equal to the angle in the alternate segment.
$
\angle PRT = \angle ATP = 50^\circ
$
(b)
Triangle PQR is isosceles (\( PQ = QR \)),
$
\angle QPR = \angle QRP
$
Opposite angles in a cyclic quadrilateral add up to \( 180^\circ \).
$
\angle PTR + \angle PQR = 180^\circ
$
$
48^\circ + \angle PQR = 180^\circ
$
$
\angle PQR = 132^\circ
$
The angles in a triangle must sum to \( 180^\circ \):
$
\angle QPR + \angle QRP + \angle PQR = 180^\circ
$
\( \angle QPR = \angle QRP \), :
$
2\angle QPR + 132^\circ = 180^\circ
$
$
2\angle QPR = 180^\circ – 132^\circ
$
$
2\angle QPR = 48^\circ
$
$
\angle QPR = 24^\circ
$
Question15
15(a): E9.6
15(b): E9.6
The time taken for each of 200 students to complete a calculation is measured. The cumulative frequency diagram shows the results.
Use the diagram to find an estimate for
$(\mathbf{a})$ the interquartile range
$(\mathbf{b})$ the number of students taking more than 40 seconds to complete the calculation.
▶️Answer/Explanation
(a) $11$
(b) $6$
(a)
The interquartile range
$
\text{IQR} = Q_3 – Q_1
$
Total number of students: 200
Lower quartile (Q1): The 25th percentile → \( 25\% \times 200 = 50 \)
From the graph, when the cumulative frequency is 50, the corresponding time is approximately 16 seconds.
Upper quartile (Q3): The 75th percentile → \( 75\% \times 200 = 150 \)
From the graph, when the cumulative frequency is 150, the corresponding time is approximately 27 seconds.
$
\text{IQR} = 27 – 16 = 11 \text{ seconds}
$
(b)
When the time is 40 seconds, the cumulative frequency is approximately 194.
Total number of students: 200
Students taking more than 40 seconds:
$
200 – 194 = 6
$
Question16
16: E2.12
$$A=\pi r^2+\pi dh$$
Rearrange the formula to make $h$ the subject
▶️Answer/Explanation
$\frac{A – \pi r^2}{\pi d} \, \text{oe final answer}$
$
A = \pi r^2 + \pi d h
$
$
A – \pi r^2 = \pi d h
$
$
h = \frac{A – \pi r^2}{\pi d}
$
Question17
17(a): E1.8
17(b): E1.8
Work out, giving each answer in standard form.
$( \mathbf{a} )$ $\left ( 2. 1\times 10^{101}\right ) \times \left ( 8\times 10^{101}\right )$
$( \mathbf{b} )$ $\left ( 2. 1\times 10^{101}\right ) + \left ( 2. 1\times 10^{100}\right )$
▶️Answer/Explanation
(a) $1.68 \times 10^{203}$
(b) $2.31 \times 10^{101}$
(a)
$
(2.1 \times 10^{101}) \times (8 \times 10^{101})
$
coefficients
$
2.1 \times 8 = 16.8
$
exponents (laws of indices):
$
10^{101} \times 10^{101} = 10^{202}
$
$
= 16.8 \times 10^{202}
$
$
= 1.68 \times 10^{203}
$
(b)
$
(2.1 \times 10^{101}) + (2.1 \times 10^{100})
$
common coefficient
$
= 2.1 \times 10^{100}(10^1 + 1)
$
$
= 2.1 \times 10^{100}(10 + 1)
$
$
= 2.1 \times 10^{100} \times 11
$
$
= 23.1 \times 10^{100}
$
$
= 2.31 \times 10^{101}
$
Question18
18: E4.4
\(\begin{aligned}&\text{The diagram shows two sides, }VA\mathrm{~and~}VB,\text{ of a regular polygon.}\\&AVX\text{ is a straight line.}\\&\mathrm{Angle~}BVX=y^{\circ}\text{ and angle }AVB=11.5y^{\circ}.\\&\text{Find the number of sides of this polygon.}\end{aligned}\)
▶️Answer/Explanation
$25$
\( \angle AVB = 11.5y^\circ \)
\( \angle BVX = y^\circ \)
\( AVX \) is a straight line.
Since \( AVX \) is a straight line
$
\angle AVB + \angle BVX = 180^\circ
$
$
11.5y + y = 180
$
$
12.5y = 180
$
$
y = \frac{180}{12.5} = 14.4^\circ
$
number of sides \( n \) of the regular polygon
$
\frac{360^\circ}{y} = n
$
$
\frac{360}{14.4} = 25
$
Question19
19(a): E7.1
19(b): E7.1
(a) Describe fully the single transformation that maps triangle T onto triangle W.
(b) Draw the enlargement of triangle T with scale factor -2 and centre of enlargement (-1,1).}
▶️Answer/Explanation
(a) $\text{Rotation: 90° clockwise oe}$
$(0, -2)$
(b) $\text{Triangle at: } (-5, -1), (-5, -7), (-7, -7)$
(a) $\text{Rotation: 90° clockwise}$
(b)
Question20
20(a): E2.13
20(b): E2.13
$\mathrm{f}(x)=3^x+2$
$( \mathbf{a} )$ Find $x$ when f$(x)=245.$
$( \mathbf{b} )$ Find $x$ when $f^-1(x)=7.$
▶️Answer/Explanation
(a) $5$
(b) $2189$
(a)
$
f(x) = 3^x + 2
$
$
3^x + 2 = 245
$
$
3^x = 243
$
Since \( 243 = 3^5 \)
$
x = 5
$
(b)
$
y = 3^x + 2
$
$
y – 2 = 3^x
$
Taking the logarithm
$
x = \log_3(y – 2)
$
$
f^{-1}(x) = \log_3(x – 2)
$
$f^{-1}(x) = 7$
$
\log_3(x – 2) = 7
$
$
x – 2 = 3^7
$
\( 3^7 = 2187 \):
$
x = 2187 + 2
$
$
x = 2189
$
Question21
21: E1.4
Write the recurring decimal $0.41$ as a fraction in its simplest form
You must show all your working.
▶️Answer/Explanation
$41.11\ldots – 41.11\ldots \, \text{oe} \\
\frac{37}{90} \, \text{cao}$
$
x = 0.\overline{41}
$
This means the decimal repeats forever \( 0.41414141…\)
Multiply both sides by 100
$
100x = 41.41414141…
$
$
100x – x = (41.41414141…) – (0.41414141…)
$
$
99x = 41
$
$
x = \frac{41}{99}
$
Question22
22: E6.4
Solve the equation $tan~x+ \sqrt 3= 0$ for $0^{\circ}\leq x\leq360^{\circ}.$
▶️Answer/Explanation
$120, \, 300
$
$
\tan x + \sqrt{3} = 0
$
$
\tan x = -\sqrt{3}
$
$
\tan x = \pm \sqrt{3} \implies \text{Reference angle} = 60^\circ
$
Since the value is negative, the solutions will be in Quadrants II and IV .
In Quadrant II:
$
x = 180^\circ – 60^\circ = 120^\circ
$
In Quadrant IV:
$
x = 360^\circ – 60^\circ = 300^\circ
$
Question23
23: E2.2
Simplify.
$$\frac{2}{y+1}-\frac{3}{y}$$
Give your answer as a single fraction in its simplest form.
▶️Answer/Explanation
$\frac{-y – 3}{y(y + 1)} \, \text{or} \, \frac{-y – 3}{y^2 + y} \, \text{or} \, \frac{-y + 3}{y(y + 1)} \\
\text{or} \, \frac{-y + 3}{y^2 + y} \, \text{(final answer)}$
$
\frac{2}{y+1} – \frac{3}{y}
$
$
= \frac{2 \cdot y}{y(y+1)} – \frac{3(y+1)}{y(y+1)}
$
$
= \frac{2y – (3y + 3)}{y(y+1)}
$
$
= \frac{2y – 3y – 3}{y(y+1)}
$
$
= \frac{-y – 3}{y(y+1)}
$
Question24
24: E5.4
The diagram shows a triangular prism with cross-section triangle $BCV.$
Angle $BCV=90^{\circ},BC=5$cm,$CV=4$cm and $AB=15$cm.
Calculate the angle between $AV$ and the base $ABCD.$
▶️Answer/Explanation
$14.2 \, \text{or} \, 14.19 \, \text{to} \, 14.20
$
In triangle \( ABC \):
$
AC = \sqrt{AB^2 + BC^2}
$
$
AC = \sqrt{15^2 + 5^2}
$
$
= \sqrt{225 + 25}
$
$
= \sqrt{250}
$
$
AC = 5\sqrt{10} \approx 15.81 \, \text{cm}
$
$
\tan(x) = \frac{VC}{AC}
$
$
\tan(x) = \frac{4}{15.81}
$
$
\tan(x) \approx 0.253
$
$
x = \tan^{-1}(0.253)
$
$
x \approx 14.2^\circ
$
Question25
25: E7.4
\(\begin{array}{cc}\mathrm{Simplify.}&\\&\underline{pt-p-t+1}\\&1-t^2\end{array}\)
▶️Answer/Explanation
$\frac{1 – p}{1 + t} \, \text{(final answer)}
$
$
\frac{pt – p – t + 1}{1 – t^2}
$
Numerator:
$
pt – p – t + 1 = p(t – 1) – (t – 1)
$
$
= (p – 1)(t – 1)
$
Denominator:
$
1 – t^2 = (1 – t)(1 + t)
$
$
= -(t – 1)(t + 1)
$
$
= \frac{(p – 1)(t – 1)}{-(t – 1)(t + 1)}
$
$
= \frac{1 – p}{t + 1}
$
Question26
26: E7.4
In the diagram, $O$ is the origin.
$\overrightarrow{OP}=\mathbf{p}$ and $\overrightarrow {OQ}=\mathbf{q}.$
$R$ is the point of intersection of $PQ$ and $OS$,with $PR: RQ= 1: 2$ and $OR=RS.$
Find the position vector of $S$ in terms of p and q
Give your answer in its simplest form.
▶️Answer/Explanation
$\frac{4}{3}p + \frac{2}{3}q \, \text{oe}$
\( \mathbf{OP} = \mathbf{p} \)
\( \mathbf{OQ} = \mathbf{q} \)
\( PR : RQ = 1 : 2 \)
\( OR = RS \)
Since point \( R \) divides line segment \( PQ \) in the ratio \( 1:2 \)
$\mathbf{R} = \frac{1( \mathbf{q}) + 2( \mathbf{p})}{1 + 2}$
$\mathbf{R} = \frac{2\mathbf{p} + \mathbf{q}}{3}$
\( OR = RS \)
\( R \) is the midpoint of segment \( OS \).
$\mathbf{R} = \frac{\mathbf{O} + \mathbf{S}}{2}$
Since \( O \) is the origin, \( \mathbf{O} = \mathbf{0} \).
$\mathbf{R} = \frac{\mathbf{S}}{2}$
$\frac{\mathbf{S}}{2} = \frac{2\mathbf{p} + \mathbf{q}}{3}$
$\mathbf{S} = \frac{4\mathbf{p} + 2\mathbf{q}}{3}$