Question1
1(a): C1.15
1(b)(i): C1.16
1(b)(ii): C1.13
1(c): C1.16
1(d)(i): C9.4
1(d)(ii): C9.4
(a) The table shows some information about the opening hours of a café. The café opens $4$ days a week.
Complete the table.
(b)$( \mathbf{i} )$ A waiter works $29$ hours a week in the café.
He is paid $\$9.50$ per hour.
He is paid for $52$ weeks of the year.
Work out his total pay for the year.
(ii) The chef is paid $32\%$ more than the waiter per hour
Work out how much the chef is paid per hour
(c) Here is part of the café’s menu.
Raj buys $2$ cups of coffee, $1$ cup of tea and $3$ slices of pizza
Calculate the change he receives from $20.
(d) The chef records the types of baguettes the café sells in one day.
(i) Complete the frequency table to show this information. You may use the tally column to help you.
(ii) On the grid, draw a bar chart to show this information.
▶️Answer/Explanation
<p1(a) 10.30 am, 4 [ .00 ] pm, 8 1/2
1(b)(i) 14326
1(b)(ii) 12.54
1(c) 1.6 [ 0 ]
1(d)(i) 7, 4, 11
1(d)(ii) Correct bar chart
Detailed Solution
(a)
Thursday Opening Time: 8:00 AM Closing Time: 4:30 PM Hours Open Calculation: From 8:00 AM to 4:00 PM = 8 hours From 4:00 PM to 4:30 PM = 0.5 hours Total Hours Open: 8.5 hours | Friday Opening Time: 8:30 AM Total Hours Open: 7.5 hours Closing Time Calculation: 8:30 AM + 7 hours = 3:30 PM 3:30 PM + 0.5 hours = 4:00 PM Closing Time: 4:00 PM |
Saturday Opening Time: 9:30 AM Closing Time: 5:30 PM Hours Open: 8 hours (Given) | Sunday Closing Time: 3:30 PM Total Hours Open: 5 hours Opening Time Calculation: 3:30 PM – 5 hours = 10:30 AM Opening Time: 10:30 AM |
(b) (i):
The waiter works 29 hours per week, earning \$9.50 per hour.
He is paid for 52 weeks in a year.
$
\text{Weekly Pay} = 29 \times 9.50
$
$
= 275.50
$
$
\text{Yearly Pay} = 275.50 \times 52
$
$
= 14,326
$
waiter’s total pay for the year is $\$14,326.$
(b)(ii):
The chef earns 32% more than the waiter per hour.
$
\text{Increase} = 9.50 \times \frac{32}{100}
$
$
= 9.50 \times 0.32
$
$
= 3.04
$
$
\text{Chef’s Pay Per Hour} = 9.50 + 3.04
$
$
= 12.54
$
chef is paid $\$12.54$ per hour.
(c)
Cost of 2 cups of coffee
$
2 \times 2.50 = 5.00
$
Cost of 1 cup of tea
$
1 \times 2.30 = 2.30
$
Cost of 3 slices of pizza
$
3 \times 3.70 = 11.10
$
$
5.00 + 2.30 + 11.10 = 18.40
$
$
\text{Change} = 20.00 – 18.40 = 1.60
$
Raj will receive $\$1.60$ as change.
(d)(i)
(d)(ii)
Question2
2(a)(i): C9.2
2(a)(ii): C9.4
2(a)(iii): C8.1
2(b)(i): C9.3
2(b)(ii): C9.3
2(b)(iii): C9.3
2(b)(iv): C9.3
(a) Manjit asks $30$ students whether they prefer joke books, puzzle books or poetry books.
The results are shown in the table.
(i) Complete the table.
(ii) Complete the pie chart.
(iii) One of the students is chosen at random.
Find the probability that this student prefers puzzle books.
The stem-and-leaf diagram shows the test scores for $24$ students.
$( \mathbf{i} )$ Write down the mode.
$( \ddot{\mathbf{i} } )$ $75\%$ of the $24$ students pass the test.
Work out the lowest score needed to pass the test.
(iii) Work out the range.
(iv) Frankie was absent on the day of the test.
His score is not on the stem-and-leaf diagram.
When he takes the test, his score increases the range by $3$ marks.
Write down the two possible values of Frankie’s score.
▶️Answer/Explanation
<p2(a)(i) 96, 216, 48
2(a)(ii) Correct pie chart drawn
2(a)(iii) 18/30 oe
2(b)(i) 51
2(b)(ii) 38
2(b)(iii) 45
2(b)(iv) 19, 70
Detailed Solution
(a)(i)
total number of students
$
8 + 18 + 4 = 30
$
The total angle in a pie chart is 360°.
$
\text{Sector angle} = \frac{\text{Number of students}}{\text{Total students}} \times 360
$
Joke books
$
\frac{8}{30} \times 360 = \frac{2880}{30} = 96^\circ
$
Puzzle books
$
\frac{18}{30} \times 360 = \frac{6480}{30} = 216^\circ
$
Poetry books
$
\frac{4}{30} \times 360 = \frac{1440}{30} = 48^\circ
$
(a)(ii)
(a)(iii)
Probability formula
$
P(\text{Puzzle}) = \frac{\text{Students who prefer puzzle books}}{\text{Total students}}
$
$
P(\text{Puzzle}) = \frac{18}{30}
$
$
= \frac{3}{5}
$
(b)(i)
The mode is the number that appears the most frequently in the dataset.
From the stem-and-leaf diagram,
$
22, 25, 26, 29, 33, 37, 38, 42, 43, 45, 45, 47, 48, 51, 51, 51, 56, 58, 59, 60, 62, 65, 67
$
The number 51 appears 3 times, which is more than any other number.
(b)(ii)
75% of the 24 students passed the test.
$
75\% \times 24 = \frac{75}{100} \times 24 = 18
$
So, the lowest passing score is the 18th highest score.
the scores in ascending order
$
22, 25, 26, 29, 33, 37, 38, 42, 43, 45, 45, 47, 48, 51, 51, 51, 56, \mathbf{58}, 59, 60, 62, 65, 67
$
The 18th score in this ordered list is 58.
(b)(iii)
$
\text{Range} = \text{Maximum score} – \text{Minimum score}
$
From the data:
Minimum score = 22
Maximum score = 67
$
\text{Range} = 67 – 22 = 45
$
(b)(iv)
Frankie’s score increases the range by 3 marks.
Current range = 45
New range = 45 + 3 = 48
The minimum score is 22.
To extend the range by 3,
$
22 – 3 = 19
$
The maximum score is 67.
To extend the range by 3,
$
67 + 3 = 70
$
Question3
3(a)(i): C1.11
3(a)(ii): C1.11
3(b)(i): C1.6
3(b)(ii): C1.13
$(\mathbf{a})$ A recipe for making $20$ biscuits uses $150$ g flour, $125$ g butter and $50$g sugar
(i) Write the ratio $\textbf{flour:butter: sugar}$ in its simplest form
(ii) Work out the amount of flour, butter and sugar needed to make 50 biscuits
$( \mathbf{b} )$ $( \mathbf{i} )$ A recipe for making one loaf of bread uses $600$g of flour.
A sack of flour contains $16$kg of flour.
Complete the statements.
One sack of flour makes a maximum of ………………………. loaves of bread.
The amount of flour left over is ………………………. g.
(ii) The amount of flour in a sack decreases from $16$kg to $15$kg
Work out the percentage decrease of flour in the sack
▶️Answer/Explanation
<p3(a)(i) 6 : 5 : 2
3(a)(ii) 375, 312.5, 125
3(b)(i) 26, 400
3(b)(ii) 6.25
Detailed Solution
(a)(i)
Flour = 150 g
Butter = 125 g
Sugar = 50 g
The ratio
$
150:125:50
$
greatest common divisor (GCD) of 150, 125, and 50.
GCD = 25
$
\frac{150}{25} : \frac{125}{25} : \frac{50}{25}
$
$
6:5:2
$
(a)(ii)
Since 20 biscuits require:
Flour = 150 g
Butter = 125 g
Sugar = 50 g
scaling factor
$
\frac{50}{20} = 2.5
$
Flour: \( 150 \times 2.5 = 375 \) g
Butter: \( 125 \times 2.5 = 312.5 \) g
Sugar: \( 50 \times 2.5 = 125 \) g
(b)(i)
1 loaf requires 600 g of flour.
$
16 \times 1000 = 16000 \text{ g}
$
$
\frac{16000}{600} = 26.67
$
the maximum number is 26 loaves.
Flour left over:
$
16000 – (26 \times 600) = 16000 – 15600 = 400 \text{ g}
$
(b)(ii)
The initial amount of flour was 16 kg and it decreased to 15 kg.
$
16 – 15 = 1 \text{ kg}
$
$
1 \times 1000 = 1000 \text{ g}
$
$
\frac{\text{Decrease}}{\text{Original amount}} \times 100
$
$
\frac{1000}{16000} \times 100
$
$
= \frac{1000 \times 100}{16000} = \frac{100000}{16000} = 6.25\%
$
Question4
4(a): C1.1
4(b): C1.1
4(c): C1.4
4(d): C1.6
4(e): C1.1
4(f): C1.9
(a) Write $6479$ correct to the nearest $100$.
$(\mathbf{b})$ Write down the multiple of $13$ that is between $100$ and $110$.
$(\mathfrak{c})$ Find the reciprocal of $0.6$.
$( d)$ Work out.
$3+4\times2$
(e) Write down an irrational number with a value between $15$ and $20$.
$(\mathbf{f})$ By writing each number in the calculation correct to $1$ significant figure, find an estimate for the
value of
$$\frac{423.8-78.4}{23.5}.$$
You must show all your working.
▶️Answer/Explanation
<p4(a) 6500
4(b) 104
4(c) 5/3 oe
4(d) 11
4(e) Any irrational number between 15 and 20
4(f) 400 – 80 / 20 = 16 nfww
Detailed Solution
(a)
at the tens digit:
The number is 6479
The tens digit is 7 (greater than 5),
$
6479 \approx 6500
$
(b)
multiples of 13:
$
13 \times 7 = 91
$
$
13 \times 8 = 104
$
$
13 \times 9 = 117
$
The only multiple between 100 and 110 is 104.
(c)
The reciprocal of a number \( x \) is
$
\frac{1}{x}
$
For \( x = 0.6 \):
$
\frac{1}{0.6} = \frac{10}{6} = \frac{5}{3}
$
(d)
Using BIDMAS/BODMAS (Order of operations: Brackets, Indices, Division/Multiplication, Addition/Subtraction):
First, multiply
$
4 \times 2 = 8
$
Then, add
$
3 + 8 = 11
$
(e)
An irrational number is a number that cannot be expressed as a fraction, and its decimal representation does not terminate or repeat.
Examples:
\( \sqrt{256} = 16 \) (not irrational)
\( \sqrt{250} \approx 15.81 \) (irrational)
(f)
$
\frac{423.8 – 78.4}{23.5}
$
Round each number to 1 significant figure
\( 423.8 \approx 400 \)
\( 78.4 \approx 80 \)
\( 23.5 \approx 20 \)
$
\frac{400 – 80}{20}
$
$
\frac{320}{20}
$
$
= 16
$
Question5
5(a): C4.6
5(b)(i): C4.6
5(b)(ii): C4.4
5(c): C5.2
5(d)(i): C7.1
5(d)(ii): C7.1
5(e)(i): C7.1
5(e)(ii): C7.1
The diagram shows three triangles, $A, B$ and $C,$ on a $1$ cm2 grid.
acute congruent obtuse
trigonometry cosine reflex
(ii) triangle $A$ after a reflection in the line $x=4.$
▶️Answer/Explanation
<p5(a) 18
5(b)(i) Acute
5(b)(ii) Congruent
5(c) 9 cm²
5(d)(i) Enlargement [centre] (-4, 0) [scale factor] 1/3
5(d)(ii) Rotation [centre] (2, 0) 180°
5(e)(i) Correct translation vertices at (8, -1), (5, -4), (5, -10)
5(e)(ii) Correct reflection vertices at (3, 9), (6, 6), (6, 0)
Detailed Solution
(a)
$
AB = \sqrt{(2 – 2)^2 + (0 – (-6))^2} = \sqrt{0 + 36} = 6 \Rightarrow AB^2 = 36
$
$
AC = \sqrt{(2 – (-1))^2 + (0 – (-9))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \Rightarrow AC^2 = 90
$
$
BC = \sqrt{(2 – (-1))^2 + (-6 – (-9))^2} = \sqrt{3^2 + 3^2} = \sqrt{18} \Rightarrow BC^2 = 18
$
$
\cos(C) = \frac{AB^2 + AC^2 – BC^2}{2 \cdot AB \cdot AC} = \frac{36 + 90 – 18}{2 \cdot 6 \cdot \sqrt{90}}
= \frac{108}{12 \cdot \sqrt{90}}
$
$
\cos(C) = \frac{108}{113.88} \approx 0.9483
$
$
C = \cos^{-1}(0.9483) \approx \rm{18.5^\circ}
$
(b) (i)
Angle \( c \)
$
\rm{\text{acute}}
$
(An acute angle is less than 90°.)
(b)(ii)
Triangles \( A \) and \( C \) are
$
\rm{\text{congruent}}
$
(They are the same shape and size-same dimensions and angles.)
(c)
Triangle \( A \) has a base of 3 units (horizontal from \( (2,0) \) to \( (5,9) \)) and height of 6 units (vertical from \( (2,0) \) to \( (2,6) \)).
$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 6 = \rm{9\ \text{cm}^2}
$
(d) (i)
Enlargement [centre] (-4, 0) [scale factor] 1/3
(d) (ii)
Rotation [centre] (2, 0) 180°
(e) (i)
Triangle \( A \) after a translation by vector \( \begin{pmatrix} 3 \\ -10 \end{pmatrix} \)
Move each point of triangle \( A \) 3 units right and 10 units down.
\( (2,0) \to (5,-10) \)
\( (2,6) \to (5,-4) \)
\( (5,9) \to (8,-1) \)
(e)(ii)
Triangle \( A \) after a reflection in the line \( x = 4 \)
Reflect each point of triangle \( A \) across the vertical line \( x = 4 \)
\( (2,0) \to (6,0) \)
\( (2,6) \to (6,6) \)
\( (5,9) \to (3,9) \)
Question6
6(a): C3.5
6(b): C3.1
6(c)(i): C2.10
6(c)(ii): C2.10
6(d)(i): C3.2
6(d)(ii): C2.5
$(\mathbf{a})$ Find the equation of line $L$ in the form $y=mx+c.$
$\mathbf{(b)}$ Write down the coordinates of the point where line $L$ crosses the x-axis
(c) (i) Complete the table of values for $y=x^2+5x+3.$
$( \mathbf{ii})$ On the grid, draw the graph of $y= x^2+ 5x+ 3$ for $-6\leqslant x\leqslant1.$
$( \mathbf{d} )$ $( \mathbf{i} )$ On the grid, draw the line $y=6.$
(ii) Use your graphs to solve the equation $x^2+5x+3=6.$
▶️Answer/Explanation
<p6(a) [y =] 2x + 7
6(b) (-3.5, 0)
6(c)(i) 3, -3, -3, 3, 9
6(c)(ii) Completely correct curve
6(d)(i) Correct ruled line drawn
6(d)(ii) 0.4 to 0.7, -5.7 to -5.4
Detailed Solution
(a)
The equation of a straight line
$
y = mx + c
$
The gradient formula
$
m = \frac{y_2 – y_1}{x_2 – x_1}
$
\( (3.5, 0) \) and \( (0,7) \)
$
m = \frac{7 – 0}{0 – 3.5}
$
$
m = \frac{7}{-3.5} = -2
$
Thus, the gradient is \( m = -2 \).
Line crosses the y-axis at \( (0, 7) \).
$
c = 7
$
$
y = -2x + 7
$
(b)
The x-intercept occurs where \( y = 0 \),
$
0 = -2x + 7
$
$
2x = 7
$
$
x = \frac{7}{2} = 3.5
$
(c)(i)
$
y = x^2 + 5x + 3
$
for \( x = -5 \)
$
y = (-5)^2 + 5(-5) + 3
$
$
= 25 – 25 + 3
$
$
= 3
$
for \( x = -3 \)
$
y = (-3)^2 + 5(-3) + 3
$
$
= 9 – 15 + 3
$
$
= -3
$
for \( x = -2 \)
$
y = (-2)^2 + 5(-2) + 3
$
$
= 4 – 10 + 3
$
$
= -3
$
for \( x = 1 \)
$
y = (1)^2 + 5(1) + 3
$
$
= 1 + 5 + 3
$
$
= 9
$
(c)(ii)
(d)(i)
(d)(ii)
Question7
7(a): C3.1
7(b): C3.1
7(c): C1.12
7(d): C3.1
7(e): C1.8
The scale drawing shows the positions of three towns, $R, S$ and $T,$ on a map.
$RS$ and $ST$ are straight roads between the towns.
The scale is $1$ centimetre represents $8$ kilometres.
$(\mathbf{a})$ Work out the actual distance between $R$ and $S.$
$\mathbf{( b) }$ Another town,$V$,is on a bearing of 163° from $R$ and on a bearing of 215° from $T$
Mark the position of $V$ on the map.
$(\mathbf{c})$ A man cycles at a constant speed of 24 km/h along the straight road from $S$ to $T.$
After l hour and 50 minutes he stops at a café, $C.$
Mark the position of $C$ on the map.
You must show all your working.
$(\mathbf{d})$ A hotel, $H$, is on a bearing of $321°$ from $R.$
Work out the bearing of $R$ from $H.$
(e) Write the scale lcm to $8$km in the form $1:n.$
▶️Answer/Explanation
<p7(a) 48
7(b) The position of I correctly marked on diagram
7(c) The position of C correctly marked on diagram with correct working seen
7(d) 141
7(e) 800000
Detailed Solution
(a)
The scale given is 1 cm represents 8 km.
So, if the measured distance on the map is \( d \) cm,
$
\text{Actual distance} = d \times 8 \text{ km}
$
(b)
Using a protractor,
From \( R \), measure 163° clockwise from North and extend the line.
From \( T \), measure 215° clockwise from North and extend the line.
The intersection is the position of \( V \).
(c)
The cyclist travels from \( S \) to \( T \) at 24 km/h for 1 hour and 50 minutes.
$
1 \text{ hr } 50 \text{ min } = 1 + \frac{50}{60} = 1.8333 \text{ hr}
$
$
\text{Distance} = \text{speed} \times \text{time} = 24 \times 1.8333 = 44 \text{ km}
$
$
\text{Map distance} = \frac{44}{8} = 5.5 \text{ cm}
$
Measure 5.5 cm from \( S \) along \( ST \) and mark this as \( C \).
(d)
The bearing of \( H \) from \( R \) is 321°.
The bearing of \( R \) from \( H \)
$
\text{Bearing} = 321° – 180° = 141°
$
(e)
8 km to cm
$
8 \times 1000 \times 100 = 800,000 \text{ cm}
$
Scale in the form \( 1:n \)
$
1 : 800,000
Question8
8(a)(i): C4.5
8(a)(ii): C4.5
8(b)(i): C2.7
8(b)(ii)(a): C2.7
8(b)(ii)(b): C2.7
8(b)(ii)(c): C2.7
8(b)(iii)(a): C2.7
8(b)(iii)(b): C2.7
8(b)(iii)(c): C2.7
8(b)(iii)(d): C2.7
(a)
(i) Write down the order of rotational symmetry of the diagram.
(ii) On the diagram, draw all the lines of symmetry.
(b) The grid shows the first three diagrams in a sequence.
Each diagram is made using small grey and small white squares to make grey and white columns.
(i) On the grid, draw Diagram 4. [1]
(ii)
(a) Complete this statement.
Diagram $n$ has ……………………….. grey columns. [1]
(b) Find an expression, in terms of n, for the total number of columns in Diagram $n.$
(c) Find an expression, in terms of n, for the fraction of columns that are grey in Diagram $n.$
(iii)
(a) Complete the table. [3]
(b) Write an expression, in terms of $n$, for the number of grey squares in Diagram n.
(c) The number of white squares in Diagram $n$ is $n ( n+2)$ .
Work out the number of white squares in Diagram 30.
(d) Diagram $k$ has a total of $1296$ squares.
Work out the value of $k$.
▶️Answer/Explanation
8(a)(i) 2
8(a)(ii) 2 lines of symmetry drawn
8(b)(i) Correct diagram drawn
8(b)(ii)(a) 2
8(b)(ii)(b) n + 2 oe final answer
8(b)(ii)(c) 2 / (n + 2) oe final answer
8(b)(iii)(a) 12, 14, 24, 35, 36, 49
8(b)(iii)(b) 2n + 4 oe final answer
8(b)(iii)(c) 960
8(b)(iii)(d) 34
Detailed Solution
(a)(i)
(a)(ii)
(b)(i)
(b)(ii)(a)
For diagram 1 total column = 3
For diagram 2 total column = 4
For diagram 3 total column = 5
For diagram n total column = n+2
(b)(ii)(c)
in each diagram only 2 grey column , for n , total column = n+2
so grey will be $\frac{2}{ (n + 2)}$
(b)(iii)(a)
(b)(iii)(b)
Grey will be $2\times (n + 2)$
(b)(iii)(c)
The number of white squares in Diagram $n$ is $n ( n+2)$
For n=30
$30\times (30+2)\Rightarrow 960$
(b)(iii)(d)
Diagram $k$ has a total of $1296$ squares.
$
\text{Total squares} = (n+2)^2
$
$
(n+2)^2 = 1296
$
$
n+2 = \sqrt{1296}
$
$
\sqrt{1296} = 36
$
$
n+2 = 36
$
$
n = 34
$
Question9
9(a): C2.2
9(b): C2.5
(a) 
Write down an expression for the area of this rectangle.
Give your answer in its simplest form.
(b) In this part, all measurements are in centimetres.
The perimeter of the triangle is $526$cm.
Find the value of x.
▶️Answer/Explanation
<p9(a) 6y² cao
9(b) 64.5
Detailed Solution
(a):
$
A = \text{length} \times \text{width}
$
the length is \( 3y \) and the width is \( 2y \).
$
A = (3y) \times (2y)
$
$
A = 6y^2
$
(b):
From the diagram, the side lengths
\( 3x – 10 \)
\( x + 70 \)
\( 4x – 50 \)
Perimeter is 526 cm,
$
(3x – 10) + (x + 70) + (4x – 50) = 526
$
$
3x + x + 4x – 10 + 70 – 50 = 526
$
$
8x + 10 = 526
$
$
8x = 516
$
$
x = 64.5
$