Question1
1: E3.1
The diagram shows two sides of a parallelogram ABCD.
Find the coordinates of point D.
▶️Answer/Explanation
$(-3,7)$
In a parallelogram, the diagonals bisect each other
Midpoint formula
$
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$
$
M = \left( \frac{-7 + 3}{2}, \frac{5 + 3}{2} \right)
$
$
= (-2, 4)
$
Since diagonals bisect, point D should keep the midpoint the same
$
(-2, 4) = \left( \frac{-1 + x_D}{2}, \frac{1 + y_D}{2} \right)
$
$
-2 = \frac{-1 + x_D}{2}
$
$
x_D = -3
$
$
4 = \frac{1 + y_D}{2}
$
$
y_D = 7
$
$
D(-3, 7)
$
Question2
2(a): E8.1
2(b): E8.1
Geetha has a box of toys.
She picks a toy at random from the box.
The probability that she picks a wooden toy is $0.6$ .
(a) Work out the probability that she does not pick a wooden toy.
(b) The box contains three types of toys, wooden, plastic or metal.
Complete the table.
▶️Answer/Explanation
(a) $0.4$
(b) 
(a) The total probability for any event is 1.
If the probability of picking a wooden toy is 0.6, the probability of not picking a wooden toy is:
$
1 – 0.6 = 0.4
$
(b)
$
\frac{\text{Number of wooden toys}}{\text{Total number of toys}} = 0.6
$
Let the number of wooden toys be \( W \).
The total number of toys is:
$
W + 14 + 14 = W + 28
$
$
\frac{W}{W + 28} = 0.6
$
$
W = 0.6W + 16.8
$
$
W = \frac{16.8}{0.4}
$
$
W = 42
$
The total number of toys is
$
42 + 14 + 14 = 70
$
Plastic toy probability
$
\frac{14}{70} = 0.2
$
Metal toy probability
$\frac{14}{70} = 0.2$
Question3
3(a): E2.7
3(b): E2.7
The table shows some information about two sequences.
(a) Complete the table.
(b) Find the smallest positive number in sequence B.
▶️Answer/Explanation
(a)40, –275
(b) $24$
(a) To get the 5th term put $n=5$ in both sequence.
sequence A = $60 -4n=60-20=40$
sequence B= $n^2 -300=(5)^2-300=-275$
(b)
$
B = n^2 – 300
$
For the smallest positive number
$
n^2 – 300 > 0 \implies n^2 > 300
$
$
n > \sqrt{300} \approx 17.32
$
Since n must be an integer, the smallest value of n is 18.
$
B = 18^2 – 300 = 324 – 300 = 24
$
Question4
4: E1.1
Find the greatest odd number that is a factor of $140$ and a factor of $210$.
▶️Answer/Explanation
$35$
Prime factorization of 140:
$
140 = 2^2 \times 5 \times 7
$
Prime factorization of 210
$
210 = 2 \times 3 \times 5 \times 7
$
The odd common factors powers of 2.
Common odd factors: 5 and 7.
The greatest odd factor is:
$35$
Question5
5(a): E1.6
5(b): E1.6
Calculate.
$(\mathbf{a})\sqrt[3]{343}-\sqrt{40.96}$
$( \mathbf{b} )$ $( 192+ 4\times 16) ^{1. 25}$
▶️Answer/Explanation
(a) $0.6$
(b) $1024$
(a)
$
\sqrt[3]{343} – \sqrt{40.96}
$
\( \sqrt[3]{343} = 7 \) (because \( 7^3 = 343 \))
\( \sqrt{40.96} = 6.4 \)
$
7 – 6.4 = 0.6
$
(b)
$
(192 + 4 \times 16)^{1.25}
$
$
192 + 64 = 256
$
$
\sqrt[4]{256} = 4 \quad \text{(because \( 4^4 = 256 \))}
$
$
256^{1.25} = 256^{1 + \frac{1}{4}} = 256 \times 4 = 1024
$
Question6
6: E4.6
The diagram shows 5 kites that are congruent to kite $ABCD$
Each kite is joined to the next kite along one edge.
Angle $DAB=40^{\circ}$ and $DCE$ is a straight line.
Find the value of $x.$
▶️Answer/Explanation
$145$
Properties of a kite:
1. Two pair of adjacent sides are equal.
$
\begin{aligned}
& A C=A D \\
& B C=B D
\end{aligned}
$
2. One pair of opposite angles (obtuse) are equal.
$
\angle C=\angle D
$
3. Sum of interior angles of a kite is 360 degrees.
$DCE$ is straight line
It means at point $C$ Sum of all angles will be $180^{\circ}$ and it divided by 6 equal angles.
$\begin{aligned} & \angle A+\angle B+\angle C+\angle D=360^{\circ} \\ & 40^{\circ}+x+30+x=360 \\ & 2 x=360-40-30 \\ & x=\frac{290}{ 2}=145^{\circ}\end{aligned}$
Question7
7(a): E5.2
7(b): E5.2
The diagram shows a shape made from a triangle $JKL$ and a semicircle with diameter $JL$
$JKL$ is an isosceles right-angled triangle with $\tilde{J}K=JL=12.8$cm
(a) Calculate the area of this shape.
(b) Calculate the perimeter of this shape.
▶️Answer/Explanation
(a) 146
(b) 51
(a)
Since JKL is an isosceles right-angled triangle,
$
\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}
$
both the base and height are 12.8 cm
$
= \frac{1}{2} \times 12.8 \times 12.8
$
$
= \frac{1}{2} \times 163.84
$
$
= 81.92 \, \text{cm}^2
$
The diameter of the semicircle is JL = 12.8 cm
so the radius is:
$
r = \frac{12.8}{2} = 6.4 \, \text{cm}
$
The area of a semicircle is half the area of a full circle
$
\text{Area of semicircle} = \frac{1}{2} \times \pi r^2
$
$
= \frac{1}{2} \times \pi (6.4)^2
$
$
= \frac{1}{2} \times \pi \times 40.96
$
$
= \frac{1}{2} \times 128.77
$
$
= 64.39 \, \text{cm}^2
$
$
\text{Total area} = 81.92 + 64.39
$
$
= 146.31 \, \text{cm}^2
$
(b)
First Find KL
In right-angled triangle JKL, we know:
$
KL^2 = JK^2 + JL^2
$
$
KL^2 = (12.8)^2 + (12.8)^2
$
$
KL^2 = 163.84 + 163.84
$
$
KL^2 = 327.68
$
$
KL \approx 18.1 \, \text{cm}
$
$
\text{Perimeter} = JK + KL + \text{Arc length of semicircle}
$
$
r = \frac{12.8}{2} = 6.4 \, \text{cm}
$
The arc length of a semicircle is half the circumference of a full circle
$
\text{Arc length} = \frac{1}{2} \times 2\pi r = \pi r
$
$
\approx 20.11 \, \text{cm}
$
$
\text{Perimeter} = 12.8 + 18.1 + 20.11
$
$
= 51.01 \, \text{cm}
$
Question8
8: E2.7
These are the first five terms of a sequence
$11 \quad \quad 18 \quad \quad 25 \quad \quad 32 \quad \quad 39$
Find an expression for the $n$th term of the sequence.
▶️Answer/Explanation
$7n+4$
$
11, 18, 25, 32, 39
$
the difference between consecutive terms
\( 18 – 11 = 7 \)
\( 25 – 18 = 7 \)
\( 32 – 25 = 7 \)
\( 39 – 32 = 7 \)
The common difference is 7.
The formula for the \( n \)th term of an arithmetic sequence is
$
a_n = a_1 + (n – 1)d
$
Where
\( a_1 \) is the first term (\( 11 \))
\( d \) is the common difference (\( 7 \))
$
a_n = 11 + (n – 1)(7)
$
$
a_n = 11 + 7n – 7
$
$
a_n = 7n + 4
$
Question9
9: E1.17
The value of a car is \( \$8000\).
Each year the value of the car decreases exponentially by $25\%$
Calculate the value of this car after $3$ years
▶️Answer/Explanation
$3375$
The formula for exponential depreciation is
$
V = P(1 – r)^t
$
\( V \) = value of the car after time
\( P = 8000 \) (initial value of the car)
\( r = 0.25 \) (the percentage decrease as a decimal)
\( t = 3 \) (time in years)
$
V = 8000(1 – 0.25)^3
$
$
= 8000(0.75)^3
$
$
= 8000(0.421875)
$
$
= 3375
$
Question10
10: E1.13
Amir invests \(\$1500\) in an account.
The account pays compound interest at a rate of $r\%$ per year At the end of 8 years the value of his investment is \( \$1656.73\) . Find the value of $r.$
▶️Answer/Explanation
$1.25$
Compound interest formula
$
A = P(1 + r)^t
$
\( A = 1656.73 \) (final amount)
\( P = 1500 \) (initial investment)
\( t = 8 \) years
\( r \) is the annual interest rate
$
1656.73 = 1500(1 + r)^8
$
$
\frac{1656.73}{1500} = (1 + r)^8
$
$
1.10449 = (1 + r)^8
$
$
1 + r = (1.10449)^{\frac{1}{8}}
$
$
1 + r = 1.0125
$
$
r = 0.0125
$
$
r = 1.25\%
$
Question11
11: E2.6
Find the inequalities that define the unshaded region, R.
▶️Answer/Explanation
\(\begin{array}{l}{y<x}\\{x<6}\\{1\leqslant y\leqslant5\mathrm{oe}}\end{array}\)
The vertical boundary on the right is at \( x = 6 \).
Since the region is to the left of this line
$
x \le 6
$
The horizontal boundary at the top is at \( y = 5 \).
Since the region is below this line:
$
y \le 5
$
Since the region is below this line
$
y \le x
$
Question12
12: E2.5
Solve the simultaneous equations. You must show all your working.
$$\frac{3x}{2}+5y=5\\4x-3y=46$$
▶️Answer/Explanation
Correctly equating one set of coefficients
Correct method to eliminate one variable
$x = 10, y = –2$
$
\frac{3x}{2} + 5y = 5 \quad \text{(Equation 1)}
$
$
4x – 3y = 46 \quad \text{(Equation 2)}
$
$
2 \times \left( \frac{3x}{2} + 5y = 5 \right)
$
$
3x + 10y = 10 \quad \text{(Equation 3)}
$
From Equation 3
$
3x = 10 – 10y
$
$
x = \frac{10 – 10y}{3}
$
Substitute this value of \( x \) into Equation 2
$
4x – 3y = 46
$
$
4\left( \frac{10 – 10y}{3} \right) – 3y = 46
$
$
\frac{4(10) – 4(10y)}{3} – 3y = 46
$
$
40 – 40y – 9y = 138
$
$
40 – 49y = 138
$
$
-49y = 98
$
$
y = -2
$
Substitute \( y = -2 \)
$
x = \frac{10 – 10(-2)}{3}
$
$
x = \frac{10 + 20}{3}
$
$
x = 10
$
Question13
13: E4.7
The diagram shows a cyclic quadrilateral.
Find the value of p.
▶️Answer/Explanation
$62$
The opposite angles in a cyclic quadrilateral add up to \( 180^\circ \).
\( 4m^\circ \) and \( 5m^\circ \) are opposite angles
$
4m + 5m = 180^\circ
$
\( (4m + 38)^\circ \) and \( p^\circ \) are opposite angles
$
(4m + 38) + p = 180^\circ
$
$
4m + 5m = 180
$
$
9m = 180
$
$
m = 20
$
\( 4m + 38 = 80 + 38 = 118^\circ \))
Use the equation for the opposite angles
$
4m + 38 + p = 180
$
$
p = 180 – 118
$
$
p = 62^\circ
$
Question14
14: E5.3
The diagram shows a circle with radius 9cm.
Calculate the area of the shaded major sector.
▶️Answer/Explanation
$221$
$
\text{Area of circle} = \pi r^2
$
Since the radius is 9 cm
$
= \pi (9)^2 = 81\pi \approx 254.47 \, \text{cm}^2
$
The angle of the minor sector is 48°
$
\text{Area of minor sector} = \frac{48^\circ}{360^\circ} \times 81\pi
$
$
= \frac{48}{360} \times 81\pi = \frac{2}{15} \times 81\pi = \frac{162\pi}{15}
$
$
= \frac{162 \times 3.1416}{15} \approx \frac{508.94}{15} \approx 33.93 \, \text{cm}^2
$
Subtract the minor sector’s area from the total circle area:
$
\text{Area of major sector} = 81\pi – 33.93
$
$
= 254.47 – 33.93 \approx 221 \, \text{cm}^2
$
Question15
15: E1.4
Write $0.146$ as a fraction in its simplest form.
You must show all your working.
▶️Answer/Explanation
$\frac{29}{198}$
Let
$
x = 0.1464646\ldots
$
The repeating part is “4646”.
Multiply both sides by 1000 to shift the decimal point three places right
$
1000x = 146.464646\ldots
$
Multiply both sides by 10 to shift the decimal point just before the repeating part
$
10x = 1.464646\ldots
$
$
1000x – 10x = (146.4646\ldots) – (1.4646\ldots)
$
$
990x = 145
$
$
x = \frac{145}{990}
$
Divide the fraction with their GCD(5)
$
= \frac{29}{198}
$
Question16
16(a): E1.2
16(b): E1.2
(a) In the Venn diagram, shade the region $M^{\prime}\cap N^{\prime}.$
(b) Find n$(B\cap(A^{\prime}\cup C)).$
▶️Answer/Explanation
(a) 
(b) $17$
(a)
\( M’ \) is the complement of set M, meaning everything outside of set M.
\( N’ \) is the complement of set N, meaning everything outside of set N.
The intersection of these complements, \( M’ \cap N’ \), represents the region that is outside both sets M and N .
(b)
\( B \cap (A’ \cup C) \) means the elements in set B that are also in the union of not A (A’) and set C.
Set A elements: 33, 16, 10, 18
Set B elements: 3, 16, 10, 4
Set C elements: 9, 18, 10, 4
Outside A elements (A’): 3, 4, 9, 20
\( A’ \cup C \):
This includes all elements outside A or in C:
Elements in A’: 3, 4, 9, 20
Elements in C: 9, 18, 10, 4
$
A’ \cup C = \{3, 4, 9, 10, 18, 20\}
$
Set B elements: 3, 16, 10, 4
Elements in both B and \( A’ \cup C \):
$
\{3, 10, 4\}
$
3 elements in B only → count: 3
10 elements in A ∩ B ∩ C → count: 10
4 elements in B ∩ C (not in A) → count: 4
Total count
$
3 + 10 + 4 = 17
$
Question17
17: E5.2
Calculate the area of triangle ABC.
▶️Answer/Explanation
$19.5$
$
\text{Area} = \frac{1}{2}ab \sin(C)
$
\( a = 6.7 \, \mathrm{cm} \)
\( b = 5.9 \, \mathrm{cm} \)
\( \angle B = 81^\circ \)
$
\text{Area} = \frac{1}{2} \times 6.7 \times 5.9 \times \sin(81^\circ)
$
$
\sin(81^\circ) \approx 0.9877
$
$
\text{Area} = \frac{1}{2} \times 6.7 \times 5.9 \times 0.9877
$
$
\text{Area} = \frac{39.05}{2} \approx 19.52 \, \mathrm{cm}^2
$
Question18
18: E2.10
The diagram shows the graph of $y=x^3+4x^2-2$ for $-3\leqslant x\leqslant1.5.$
By drawing a suitable straight line, solve the equation $x^3+4x^2-2=2x$ for $-3\leqslant x\leqslant1.5.$
▶️Answer/Explanation
$y=2x$ ruled
$x=-0.5$ to $-0.55$
$x=0.85$ to $0.9$
$
x^3 + 4x^2 – 2 = 2x
$
Split the equation into two parts to draw a straight line
$
x^3 + 4x^2 – 2 – 2x = 0
$
$
x^3 + 4x^2 – 2 =2x
$
$
y = x^3 + 4x^2 – 2 \quad \text{and} \quad y = 2x
$
The curve \( y = x^3 + 4x^2 – 2 \) is already given in the diagram.
\( y = 2x \), This is a straight line with a slope of 2, passing through the origin.
Question19
19(a): E2.2
19(b): E2.2
Factorise completely.
$( \mathbf{a} )$ $12m^2- 75t^2$
$( \mathbf{b} )$ $xy+ 15+ 3y+ 5x$
▶️Answer/Explanation
(a) $3(2m+5t)(2m-5t)$ final answer (b) $(x+3)(y+5)$ final answer
(a)
$
12m^2 – 75t^2
$
The coefficients: 12 and 75.
The greatest common factor (GCF) is 3.
$
12m^2 – 75t^2 = 3(4m^2 – 25t^2)
$
The expression inside the brackets is a difference of squares.
$
a^2 – b^2 = (a – b)(a + b)
$
\( 4m^2 = (2m)^2 \)
\( 25t^2 = (5t)^2 \)
$
= 3(2m – 5t)(2m + 5t)
$
(b)
$
xy + 15 + 3y + 5x
$
$
= (xy + 3y) + (5x + 15)
$
$
= y(x + 3)
$
$
= 5(x + 3)
$
(x + 3) is common in both terms
$
= (x + 3)(y + 5)
$
Question20
20: E6.4
Solve the equation $8\sin x+ 6= 1$ for $0^{\circ}\leqslant x\leqslant360^{\circ}.$
▶️Answer/Explanation
$218.7,321.3$
$
8 \sin x + 6 = 1
$
$
8 \sin x = 1 – 6
$
$
8 \sin x = -5
$
$
\sin x = -\frac{5}{8}
$
$
\theta = \sin^{-1}\left(\frac{5}{8}\right)
$
$
\theta \approx 38.68^\circ
$
Since the sine value is negative, the solutions are in the third and fourth quadrants
Third quadrant
$
x = 180^\circ + 38.68^\circ = 218.68^\circ
$
Fourth quadrant
$
x = 360^\circ – 38.68^\circ = 321.32^\circ
$
Question21
21: E6.6
$HD=4$cm$,EH=6.5$cm and $EF=9.1$ cm.
Calculate the angle between $CE$ and the base $CDHG.$
▶️Answer/Explanation
$33.2$ or $33.18…$
\( HD = 9.1 \text{ cm} \) (length of the base)
\( DC = 4 \text{ cm} \) (height)
Using the Pythagorean theorem
$
HC = \sqrt{HD^2 + DC^2}
$
$
HC = \sqrt{9.1^2 + 4^2}
$
$
= \sqrt{82.81 + 16}
$
$
= \sqrt{98.81}
$
$
HC \approx 9.94 \, \text{cm}
$
In triangle \( HGC \)
$
\tan(x) = \frac{ \text{height of the cuboid} }{ HC }
$
$
\tan(x) = \frac{6.5}{9.94}
$
$
\tan(x) \approx 0.654
$
$
x = \tan^{-1}(0.654)
$
$
x \approx 33.01^\circ
$
Question22
22: E8.3
Bag $A$ and bag $B$ each contain red counters and blue counters only
Stephan picks a counter at random from bag $A$ and Jen picks a counter at random from bag $B.$
The probability that Stephan picks a red counter is $0.4$.
The probability that Stephan and Jen both pick a red counter is $0.25$.
Find the probability that Stephan and Jen both pick a blue counter.
▶️Answer/Explanation
$0.225$
$
P(\text{Red from A}) = 0.4
$
$
P(\text{Red from A and Red from B}) = 0.25
$
Formula for joint probability
$
P(\text{Red from A and Red from B}) = P(\text{Red from A}) \times P(\text{Red from B})
$
$
0.25 = 0.4 \times P(\text{Red from B})
$
$
P(\text{Red from B}) = \frac{0.25}{0.4} = 0.625
$
Probability Stephan picks blue:
$
P(\text{Blue from A}) = 1 – P(\text{Red from A}) = 1 – 0.4 = 0.6
$
$
P(\text{Blue from B}) = 1 – P(\text{Red from B}) = 1 – 0.625 = 0.375
$
Events are independent:
$
P(\text{Blue from A and Blue from B}) = P(\text{Blue from A}) \times P(\text{Blue from B})
$
$
= 0.6 \times 0.375 = 0.225
$