Question1
1: E1.1
The temperature at midnight is $-4^{\circ}C$.
The temperature at noon is $25^{\circ}C$
Work out the difference between these two temperatures.
▶️Answer/Explanation
$29$
Midnight temperature: \(-4^\circ C\)
Noon temperature: \( 25^\circ C \)
The difference between the two temperatures is the distance between them on the number line.
$
25 – (-4) = 25 + 4 = 29^\circ C
$
Question2
2: E1.16
A gardener charges \( \$6.55\) for each hour he works plus a fixed charge of \( \$15.50\) .
Calculate the total amount he charges when he works for 4 hours.
▶️Answer/Explanation
$41.7$
Hourly rate: $\$6.55$
Fixed charge: $\$15.50$
Hours worked: 4
the cost for 4 hours of work
$
6.55 \times 4 = 26.20
$
Adding the fixed charge
$
26.20 + 15.50 = 41.70
$
Question3
3(a): E9.4
3(b): E9.3
A delivery driver records the number of pizzas she delivers each month for one year.
(a) Complete the stem-and-leaf diagram.
(b) Find the median.
▶️Answer/Explanation
>(a)
(b) 46
split the numbers into stems (tens digits) and leaves (ones digits).
- 20s: 22, 28
- 30s: 36, 39
- 40s: 41, 44, 48, 49
- 50s: 52, 54, 57, 57
(b)
To find the median, order the data
$
22, 28, 36, 39, 41, 44, 48, 49, 52, 54, 57, 57
$
There are 12 data points, so the median is the average of the 6th and 7th values
$
\frac{44 + 48}{2} = \frac{92}{2} = 46
$
Question4
4: E1.4
Jonah has \(\$750\).
He spends $\frac{1}{4}$ of this money on travel and some of this money on food
He now has \(\$437.50\).
Work out the fraction of the \(\$750\) he spends on food
▶️Answer/Explanation
$\frac{1}{6}$
Jonah spends \(\frac{1}{4}\) of his money on travel.
Since he starts with \$750
$
\frac{1}{4} \times 750 = 187.50
$
He now has $\$437.50$ left, so the amount he spent on food is the remaining part of the spent money
$
750 – 437.50 = 312.50
$
The amount spent on travel is already $\$187.50$
$
312.50 – 187.50 = 125
$
He spent $\$125$ on food out of $\$750$
$
\frac{125}{750}
$
$
= \frac{1}{6}
$
Question5
5: E1.15
The table shows part of a tram timetable.
All the trams take the same number of minutes to complete the journey from Newpoint to Westhill.
Complete the table.
▶️Answer/Explanation
$13.05$
The tram leaves Newpoint at 10:30 and arrives at Westhill at 11:17.
$
11:17 – 10:30 = 47 \text{ minutes}
$
So, the journey takes 47 minutes.
Since the journey takes 47 minutes
$12:18 + 47\text{ minutes}= 13:05$
$13:30 + 47\text{ minutes}= 14:17$
Question6
7: E1.2
Write $0.04628$ correct to 2 significant figures.
▶️Answer/Explanation
0.046 cao
The first 2 significant figures are 4 and 6.
the third digit (2). Since it’s less than 5, don’t round up.
$0.046$
Question7
On the Venn diagram, shade the region $A \cap B$, .
▶️Answer/Explanation
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 \( A \cap B \).
Question8
8: E1.13
Kai invests \(\$5000\) in an account paying simple interest at a rate of $r\%$ per year.
At the end of 8 years, the value of his investment is \(\$5700\).
Find the value of r.
▶️Answer/Explanation
$1.75$
Simple Interest Formula:
$
A = P(1 + rt)
$
\( A \) is the final amount (after interest) $= \$5700$
\( P \) is the principal amount (initial investment) $= \$5000$
\( r \) is the interest rate
\( t \) is the time in years $= 8$ years
$
5700 = 5000(1 + r(8))
$
$
\frac{5700}{5000} = 1 + 8r
$
$
1.14 = 1 + 8r
$
$
1.14 – 1 = 8r
$
$
0.14 = 8r
$
$
r = 0.0175
$
$
r = 1.75\%
$
Question9
9(a): E7.1
9(b): E7.1
(a) Describe fully the single transformation that maps triangle A onto triangle B.
(b) On the grid, draw the image of triangle A after a translation by the vector $\binom {-4}{3}$
▶️Answer/Explanation
(a) Enlargement
[s f] 2
[centre] $(1,-1)$
(b) image at $(-1,4)(-1,5)(1,4)$
(a)
Triangle B is a larger, identical version of triangle A
Both triangles face the same direction, so it’s a direct enlargement (not a reflection or rotation).
$
\text{Scale factor} = \frac{2}{1} = 2
$
(b)
The translation vector (-4, 3)
Move 4 units left (because -4).
Move 3 units up (because +3).
$
(x, y) \rightarrow (x – 4, y + 3)
$
Triangle A has vertices
\( (3, 1) \rightarrow (-1, 4) \)
\( (5, 1) \rightarrow (1, 4) \)
\( (3, 2) \rightarrow (-1, 5) \)
Question10
10: E1.8
Write $174000$ in standard form.
▶️Answer/Explanation
$1.74\times 10^5$
Standard form is written as
$
a \times 10^n \quad \text{where} \, 1 \leq a < 10
$
$
174000 = 1.74 \times 10^5
$
Question11
11: E8.2
A company surveys $40$ of its employees.
In the survey, $3$ employees say they walk to work.
The company has a total of $1240$ employees.
Find the expected number of employees in the company who walk to work.
▶️Answer/Explanation
$93$
3 out of 40 employees walk to work.
The company has a total of 1240 employees.
$
\frac{3}{40} = \frac{x}{1240}
$
\( x \) is the expected number of employees who walk to work.
$
3 \times 1240 = 40 \times x
$
$
3720 = 40x
$
$
x = \frac{3720}{40}
$
$
x = 93
$
Question12
12: E6.2
The diagram shows a right-angled triangle.
Calculate the value of x.
▶️Answer/Explanation
$52.6$ or $52.61$ to $52.62$
$
\cos(x) = \frac{\text{adjacent}}{\text{hypotenuse}}
$
$
\cos(x) = \frac{8.5}{14}
$
$
\cos(x) = 0.6071
$
$
x = \cos^{-1}(0.6071)
$
$
x \approx 52.6^\circ
$
Question13
13: E1.4
Without using a calculator, work out $2\frac{1}{4}\div 1\frac{7}{8]$
You must show all your working and give your answer as a mixed number in its simplest form.
▶️Answer/Explanation
$\frac{9}{4}\times\frac{8}{15}$oe
or $\frac{18}{8}\div\frac{15}{8}$oe with common denominator
$1\frac{1}{5}$ cao
Convert \( 2 \frac{1}{4} \) to an improper fraction
$2 \frac{1}{4} = \frac{2 \times 4 + 1}{4} = \frac{9}{4}$
Convert \( 1 \frac{7}{8} \) to an improper fraction
$1 \frac{7}{8} = \frac{1 \times 8 + 7}{8} = \frac{15}{8}$
Dividing by a fraction is the same as multiplying by its reciprocal
$\frac{9}{4} \div \frac{15}{8} = \frac{9}{4} \times \frac{8}{15}$
$= \frac{9 \times 8}{4 \times 15} = \frac{72}{60}$
Divide the numerator and denominator by GCD( 12)
$= \frac{72 \div 12}{60 \div 12} = \frac{6}{5}$
$= 1 \frac{1}{5}$
Question14
14: E3.5
\(\begin{aligned}&A\text{ is the point }(0,2)\mathrm{~and~}B\text{ is the point }(8,6).\\&\text{Find the equation of line }AB.\\&\text{Give your answer in the form }y=mx+c.\end{aligned}\)
▶️Answer/Explanation
$y=\frac{1}{2}x+2$
The slope formula is
$
m = \frac{y_2 – y_1}{x_2 – x_1}
$
The coordinates of points \( A(0, 2) \) and \( B(8, 6) \):
$
m = \frac{6 – 2}{8 – 0} = \frac{4}{8} = \frac{1}{2}
$
The equation of a line is
$
y = mx + c
$
\( m = \frac{1}{2} \) coordinates of point \( A(0, 2) \) , To find \( y \)-intercept (\( c \)):
$
2 = \frac{1}{2}(0) + c
$
$
c = 2
$
$
y = \frac{1}{2}x + 2
$
Question15
15: E4.6
Three towns, $A,B$ and $C$, are equidistant from each other.
The bearing of $C$ from $A$ is 104$^\circ.$
Calculate the bearing of $B$ from $C.$
▶️Answer/Explanation
$224$
Towns \( A \), \( B \), and \( C \) form an equilateral triangle , all angles are \( 60^\circ \) (since they’re equidistant).
The bearing of \( C \) from \( A \) is \( 104^\circ \).
$104^\circ + 60^\circ = 164^\circ$
Bearings are measured clockwise from north.
Since we know \( A \)’s bearing from \( C \) is \( 164^\circ \) and the triangle is equilateral
The bearing of \( B \) from \( C \) is \( 60^\circ \) counterclockwise from \( 164^\circ \)
$164^\circ + 60^\circ = 224^\circ$
Question16
16(a): E9.6
16(b): E9.6
The speed–time graph shows information about a car journey.
(a) Find the deceleration of the car between $240$ and $320$ seconds.
(b) Calculate the total distance the car travels during the $320$ seconds.
▶️Answer/Explanation
(a) $0.2$
(b) $4240$
(a)
Deceleration is the rate of change of velocity
$\text{Deceleration} = \frac{\text{Change in speed}}{\text{Time taken}}$
Speed at \( t = 240 \, s \) is \( 16 \, m/s \).
Speed at \( t = 320 \, s \) is \( 0 \, m/s \).
$320 – 240 = 80 \, s$
$\text{Deceleration} = \frac{16 – 0}{80} = \frac{16}{80} = 0.2 \, m/s^2$
(b)
Distance traveled is the area under the speed-time graph!
1. Acceleration phase (0 to 30 s) Triangle
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 16 = 240 \, m$
2. Constant speed phase (30 to 240 s) Rectangle
$\text{Area} = \text{base} \times \text{height} = (240 – 30) \times 16 = 210 \times 16 = 3360 \, m$
3. Deceleration phase (240 to 320 s) Triangle
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 80 \times 16 = 640 \, m$
Total distance
$240 + 3360 + 640 = 4240 \, m$
Question17
17: E8.1
$W=\{$students who walk to school$\}$
$G=\{$students who wear glasses$\}$
There are 20 students in a class.
- 8 walk to school
- 3 wear glasses and walk to school
- 2 do not wear glasses and do not walk to school.
Complete the Venn diagram.
▶️Answer/Explanation
The total number of students is 20.
- 5 who only walk
- 3 who walk and wear glasses
- 2 who do neither
the number of students who only wear glasses
$20-(5+3+2)\Rightarrow 10$
Question18
18: E2.10
The graph of $y=$f$(x)$ is drawn on the grid
(a) Draw the tangent to the graph at the point $x=3.$
(b) Use your tangent to find an estimate for the gradient of the curve at the point $x=3.$
▶️Answer/Explanation
(a) tangent ruled at $x = 3$
(b) $4.8$ to $5.8$
(a)
(b)
$
\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 – y_1}{x_2 – x_1}
$
From the diagram
Point 1: \((3, 5)\)
Point 2: \((2, 0)\)
$
\text{Gradient} = \frac{5 – 0}{3 – 2} = \frac{1}{1} = 1
$
So, the gradient of the line is 1
Question19
19(a): E2.8
19(b): E2.8
$\begin{array}{ll}(\mathbf{a})&y~\text{is directly proportional to}\left(x-1\right)^2.\\&\text{When }x=4,\:y=3.\end{array}$
Find $y$ when $x=7.$
(b) m is inversely proportional to the square root of $p.$
Explain what happens to the value of $m$ when the value of $p$ is multiplied by 9
▶️Answer/Explanation
(a) $12$
(b) divided by $3$
(a)
$
y \propto (x – 1)^2
$
$
y = k(x – 1)^2
$
\( x = 4 \), \( y = 3 \)
$
3 = k(4 – 1)^2
$
$
3 = k(3)^2
$
$
3 = 9k
$
$
k = \frac{3}{9} = \frac{1}{3}
$
Now, we want to find the value of \( y \) when \( x = 7 \).
$
y = \frac{1}{3}(7 – 1)^2
$
$
= \frac{1}{3}(6)^2
$
$
= \frac{1}{3}(36)
$
$
= 12
$
(b)
$
m \propto \frac{1}{\sqrt{p}}
$
$
m = \frac{k}{\sqrt{p}}
$
If the value of \( p \) is multiplied by 9, the square root of \( p \) becomes 3 times larger:
$
\sqrt{9p} = 3\sqrt{p}
$
Since \( m \) is inversely proportional to the square root of \( p \), multiplying the square root by 3 means that the value of \( m \) will be divided by 3
$
m_{\text{new}} = \frac{m}{3}
$
Question20
20: E4.4
Two parcels are mathematically similar.
The larger parcel has volume 80cm$^{3}$ and height 5.2cm.
The smaller parcel has volume 33.75 cm$^{3}.$ Calculate the height of the smaller parcel.
▶️Answer/Explanation
$3.9$
For similar shapes, the ratio of volumes is the **cube** of the ratio of their corresponding lengths
$
\frac{\text{Volume of large parcel}}{\text{Volume of small parcel}} = \left( \frac{\text{Height of large parcel}}{\text{Height of small parcel}} \right)^3
$
$
\frac{80}{33.75} = \left( \frac{5.2}{h} \right)^3
$
$
\frac{80}{33.75} = 2.3704
$
$
\frac{5.2}{h} = \sqrt[3]{2.3704}
$
$
\sqrt[3]{2.3704} \approx 1.33
$
$
h = \frac{5.2}{1.33}
$
$
h \approx 3.91 \, \mathrm{cm}
$
Question21
21: E2.5
Solve the simultaneous equations. You must show all your working.
$4y+3x=13$
$y=x^2-18$
▶️Answer/Explanation
$4x^{2}+3x-85[=0]$
or 16$y^2-113y+7[=0]$
oe simplified
correct method to solve their quadratic equation e.g. factors, quadratic formula, completing the square
$$\begin{aligned}&x=-5\:y=7\\&x=\frac{17}{4}\:\mathrm{oe}\:y=\frac{1}{16}\:\mathrm{oe}\end{aligned}$$
1. \( 4y + 3x = 13 \)
2. \( y = x^2 – 18 \)
substitute \( y = x^2 – 18 \) into the first equation
$
4(x^2 – 18) + 3x = 13
$
$
4x^2 – 72 + 3x = 13
$
$
4x^2 + 3x – 72 – 13 = 0
$
$
4x^2 + 3x – 85 = 0
$
using the quadratic formula
$
x = \frac{-3 \pm \sqrt{3^2 – 4(4)(-85)}}{2(4)}
$
$
= \frac{-3 \pm \sqrt{9 + 1360}}{8}
$
$
= \frac{-3 \pm \sqrt{1369}}{8}
$
Since \( \sqrt{1369} = 37 \):
$
x = \frac{-3 \pm 37}{8}
$
1. \( x = \frac{-3 + 37}{8} = \frac{34}{8} = 4.25 \)
2. \( x = \frac{-3 – 37}{8} = \frac{-40}{8} = -5 \)
For \( x = 4.25 \):
$
y = (4.25)^2 – 18
$
$
= 0.0625
$
For \( x = -5 \):
$
y = (-5)^2 – 18
$
$
= 7
$
$
(x, y) = (4.25, 0.0625) \quad \text{or} \quad (-5, 7)
$
Question22
22: E2.12
(a) For each sketch, put a ring around the correct type of function shown.
$\mathbf{( i) }$ On the grid, sketch the curve $y= \sin x$ for $0^\circ\leqslant x\leqslant360^\circ.$
(ii) Solve the equation $\sin x+ 0. 4= 0$ for $0^{\circ}\leqslant x\leqslant360^{\circ}$
▶️Answer/Explanation
(a)(i) cubic
(a)(ii) reciprocal
(b)(i) 
(b)(ii) $203.6$ and $336.4$
(a)
Graph (i):
The curve has multiple turning points, which is a cubic function.
Graph (ii):
The curve has asymptotes (it gets close to the axes but never touches them), characteristic of a reciprocal function (e.g., \( y = \frac{1}{x} \)).
(b) (i)
$
\sin(x) + 0.4 = 0
$
$
\sin(x) = -0.4
$
$
x_{\text{ref}} = \sin^{-1}(0.4)
$
$
x_{\text{ref}} = 23.58^\circ
$
Sine value is negative, the angle will be in the 3rd and 4th quadrants
In the 3rd quadrant
$
x = 180^\circ + 23.58^\circ = 203.58^\circ
$
In the 4th quadrant
$
x = 360^\circ – 23.58^\circ = 336.42^\circ
$
Question23
23(a): E8.3
23(b): E8.3
The Venn diagram shows information about the number of students in a class.
Some study English $(E)$, some study French $(F)$, some study Spanish $(S)$ and some do not study any of
these languages.
(a) Find n$((E\cup F)^\prime\cup S)$.
(b) One student is picked at random from those who study Spanish.
Find the probability that this student studies exactly two languages.
▶️Answer/Explanation
(a) $15$
(b) $\frac{1}{2}$
Only English → 8
Only French → 6
Both English and French → 1
English and Spanish → 2
French and Spanish → 3
All three languages → 1
(a)
Students not studying English or French:
$
(E \cup F)’ = 4 + 5 = 9
$
\( (E \cup F)’ \cup S \) – students not in English or French OR those in Spanish.
Students in Spanis
$
2 + 3 + 1 + 4 = 10
$
Combine with those outside \( E \cup F \)
$
n((E \cup F)’ \cup S) = 4 + 5 + 2 + 3 + 1 = 15
$
(b)
Total students studying Spanish:
$
2 + 3 + 1 + 4 = 10
$
English and Spanish (but not French) → 2
French and Spanish (but not English) → 3
Total studying exactly two languages
$
2 + 3 = 5
$
$
\text{Probability} = \frac{5}{10} = \frac{1}{2}
$
Question24
24(a): E7.4
24(b): E7.4
$O$ is the origin and $OPQR$ is a parallelogram
$M$ is the midpoint of $PQ$ and $\hat{N}$ divides $QR$ in the ratio 2:1.
$\overrightarrow{OP}=$a and $\overrightarrow OR=\mathbf{b}.$
$( \mathbf{a} )$ Find $\overrightarrow{MN}.$
Give your answer in terms of a and/or b and in its simplest form.
$(\mathbf{b})$ The lines $MN$ and $OR$ are extended to meet at S.
Find the position vector of $S.$
Give your answer in terms of a and/or b and in its simplest form.
▶️Answer/Explanation
(a) $\frac{1}{2}\mathbf{b}-\frac{2}{3}\mathbf{a}$
(b) $\frac{5}{4}\mathbf{b}$
(a)
\( \overrightarrow{OP} = \mathbf{a} \)
\( \overrightarrow{OR} = \mathbf{b} \)
Since \( OPQR \) is a parallelogram
\( \overrightarrow{OQ} = \mathbf{a} + \mathbf{b} \)
$
\overrightarrow{M} = \frac{1}{2}(\overrightarrow{P} + \overrightarrow{Q})
$
$
= \frac{1}{2}(\mathbf{a} + (\mathbf{a} + \mathbf{b}))
$
$
= \frac{1}{2}(2\mathbf{a} + \mathbf{b}) = \mathbf{a} + \frac{1}{2}\mathbf{b}
$
N divide QR in 2:1.
$
\overrightarrow{N} = \frac{1(\overrightarrow{Q}) + 2(\overrightarrow{R})}{1 + 2}
$
$
= \frac{1(\mathbf{a} + \mathbf{b}) + 2(\mathbf{b})}{3}
$
$
= \frac{\mathbf{a} + 3\mathbf{b}}{3}
$
$
\overrightarrow{MN} = \overrightarrow{N} – \overrightarrow{M}
$
$
= \frac{\mathbf{a} + 3\mathbf{b}}{3} – \left(\mathbf{a} + \frac{1}{2}\mathbf{b}\right)
$
$
= -\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}
$
(b)
The equation for line \( MN \) is:
$
\overrightarrow{S} = \overrightarrow{M} + k(\overrightarrow{MN})
$
$
\overrightarrow{S} = \left( \mathbf{a} + \frac{1}{2}\mathbf{b} \right) + k\left( -\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b} \right)
$
$
= \mathbf{a} + \frac{1}{2}\mathbf{b} – \frac{2k}{3}\mathbf{a} + \frac{k}{2}\mathbf{b}
$
this equal to the equation of line \( OR \) also
$
\overrightarrow{S} = t\mathbf{b}
$
Match coefficients of \( \mathbf{a} \) and \( \mathbf{b} \)
\( \mathbf{a} \)
$
1 – \frac{2k}{3} = 0 \implies k = \frac{3}{2}
$
\( \mathbf{b} \)
\( k = \frac{3}{2} \)
$
\frac{1}{2} + \frac{3}{4} = t \implies t = \frac{5}{4}
$
$
\overrightarrow{S} = \frac{5}{4}\mathbf{b}
$