Question1
1: C1.1
Write the number thirty thousand and fifty in figures.
▶️Answer/Explanation
30 050
Detailed Solution:
The number is thirty thousand and fifty.
Thirty thousand $= 30,000.$
Fifty $= 50.$
add these
$30,000 + 50 = 30,050$
Question2
2: C1.9
Write 5926 correct to the nearest 10.
▶️Answer/Explanation
5930
Detailed Solution:
The number is 5926, so the ones digit is 6.
- If the ones digit is 5 or more, round up.
- If the ones digit is 4 or less, round down.
the ones digit is 6 (which is 5 or more), round the number up.
5926 rounded to the nearest 10 becomes 5930.
Question3
3: C4.1
Mark the midpoint of the line ST.
▶️Answer/Explanation
Midpoint of ST marked
Detailed Solution:
- halfway point between points S and T on the line.
- Use a ruler to measure the length of the line, then divide by 2 to locate the midpoint.
- If points S(x₁, y₁) and T(x₂, y₂)
$
M\left( \frac{x₁ + x₂}{2}, \frac{y₁ + y₂}{2} \right)
$
Question4
4(a): C1.4
4(b): C1.13
(a) Shade $\frac{2}{9}$ of this shape.
(b) Write $\frac{2}{9}$ as a percentage.
▶️Answer/Explanation
(a) 8 squares shaded
(b) 22.2 or 22.22…
Detailed Solution:
(a)
The grid is $6 × 6,$ so there are 36 squares in total.
$
\frac{2}{9} \times 36 = 8 \text{ squares}.
$
any 8 squares on the grid to represent \(\frac{2}{9}\).
(b)
$
2 \div 9 \approx 0.2222
$
$
0.2222 \times 100 \approx 22.22\%
$
Question5
5: C1.15
A night bus runs from 21 50 to 05 18 the next day. Work out the number of hours and minutes that the night bus runs.
▶️Answer/Explanation
7h 28min
From $21:50$ to midnight $(00:00)$
$60$ minutes $= 1$ hour,
$
22:00 – 21:50 = 10 \text{ minutes}
$
Then from $22:00$ to $00:00$ is $2$ hours.
Total time in the first part $2$ hours $10$ minutes.
midnight $(00:00)$ to $05:18$
5 hours 18 minutes.
Total time
Hours $2 + 5 = 7$ hours
Minutes $10 + 18 = 28$ minutes
The night bus runs for 7 hours and 28 minutes.
Question6
6(a): C1.1
34 55 76 83 111 121
From this list of numbers, write down all the multiples of 11.
▶️Answer/Explanation
55 121
Multiples of 11 are numbers that can be divided by 11 leaves 0 as remainder.
- $34 ÷ 11 ≈ 3.09 →$ Not a multiple
- $55 ÷ 11 = 5 →$ Multiple of 11
- $76 ÷ 11 ≈ 6.91 →$ Not a multiple
- $83 ÷ 11 ≈ 7.55 →$ Not a multiple
- $111 ÷ 11 = 10.09 →$ Not a multiple
- $121 ÷ 11 = 11 →$ Multiple of 11
Question6 (b)
6(b): C1.1
Zaid has a non-calculator method for working out if a number is a multiple of 11. He shows his method for the number 919281.
Show that the number 918271937 is a multiple of 11 by using Zaid’s method.
▶️Answer/Explanation
$9 − 1+ 8 -2 +7 -1 +9 − 3+ 7$
$33=3 \times 11$
Question7
7: C9.3
The range of eight numbers is 31.
These are seven of the numbers.
28 36 42 24 38 16 21
Find the two possible values of the eighth number.
▶️Answer/Explanation
11, 47
- Minimum value = 16
- Maximum value = 42
$\text{Range} = \text{Maximum number} – \text{Minimum number}$
If the 8th number is the new maximum, the range becomes
$\text{New maximum} – 16 = 31$
$\text{New maximum} = 31 + 16 = 47$
So, the 8th number could be 47.
If the 8th number is the new minimum, the range becomes
$42 – \text{New minimum} = 31$
$\text{New minimum} = 42 – 31 = 11$
So, the 8th number could be 11.
Question8
8: C1.14
Calculate $\sqrt{5.76}+2.8^3$.
▶️Answer/Explanation
24.352
$
\sqrt{5.76} = 2.4
$
$
2.8^3 = 2.8 \times 2.8 \times 2.8
$
$
2.8 \times 2.8 = 7.84
$
$
7.84 \times 2.8 = 21.952
$
$
2.4 + 21.952 = 24.352
$
Question9
9: C2.2
Simplify $4 m+7 k-m+3 k$.
▶️Answer/Explanation
$3m+10k$
Divide equation in two part and then add
$4m-m=3m$——(1)
$7k+3k=10k$——(2)
Add both now,
$3m+10k$
Question10
10(a): C1.6
10(b): C1.6
-9 -7 -3 -1 0 2 5 6 8
From this list of numbers,
find
(a) the highest number possible from the product of two of the numbers
(b) the lowest number possible from the product of three of the numbers.
▶️Answer/Explanation
(a) 63
(b) −432
(a)
Multiply two largest positive numbers
Multiply two smallest negative numbers (since negative × negative = positive)
Largest positive numbers: \( 6 \times 8 = 48 \)
Smallest negative numbers: \( (-9) \times (-7) = 63 \)
$
\text{Highest product} = 63
$
(b)
the smallest possible product, the most negative result, from multiplying
1. The two largest negative numbers and the smallest positive number
\( (-9) \times (-7) \times 2 = 63 \times 2 = 126 \) (positive)
2. The three smallest numbers (most negative product)
\( (-9) \times (-7) \times (-3) = 63 \times (-3) = -189 \)
$\text{Lowest product} = -189$
Question11
11(a): C9.4
11(b): C9.3
Sarah records the number of people who play golf on each of 14 days.
(a) Complete the stem-and-leaf diagram.
(b) Find the median.
▶️Answer/Explanation
(a)
(b) 61
The data:
28, 46, 54, 71, 70, 65, 49, 50, 64, 77, 68, 72, 45, 58
Arranged in order:
$
28, 45, 46, 49, 50, 54, 58, 64, 65, 68, 70, 71, 72, 77
$
(b)
there are 14 numbers, the median is the average of the 7th and 8th values.
The 7th number is 58, and the 8th number is 64.
$
\text{Median} = \frac{58 + 64}{2} = \frac{122}{2} = 61
$
Question12
12(a): C5.4
12(b): C5.4
The diagram shows the net of a cuboid.
(a) Work out the surface area of this cuboid.
(b) Work out the volume of this cuboid.
▶️Answer/Explanation
(a) 220
(b) 200
From the diagram
Length (l) = 10 cm
Height (h) = 4 cm
Width (w) = 5 cm
$\text{Surface Area} = 2(lw + lh + wh)$
$= 2(10(5) + 10(4) + 5(4))$
$= 2(50 + 40 + 20)$
$= 2(110)$
$= 220 \text{ cm}^2$
(b)
$\text{Volume} = l \times w \times h$
$= 10 \times 5 \times 4$
$= 200 \text{ cm}^3$
Question13
13(a): C8.1
13(b): C8.1
There are 20 cars in a car park and 3 of the cars are blue.
(a) James wants to draw a pie chart to show this information.
Find the angle of the sector for the blue cars in this pie chart.
(b) One of the 20 cars is picked at random.
Find the probability that this car is not blue.
▶️Answer/Explanation
(a) 54
(b) $\frac{17}{20}$
The total angle in a circle is 360°, and there are 20 cars in total. Since 3 of the cars are blue
$
\text{Angle for blue cars} = \frac{\text{Number of blue cars}}{\text{Total number of cars}} \times 360°
$
$
= \frac{3}{20} \times 360°
$
$
= 54°
$
Part (b)
If there are 3 blue cars out of 20,
the number of non-blue cars is
$
20 – 3 = 17
$
The probability of picking a non-blue car is the ratio of non-blue cars to the total number of cars
$
P(\text{Not blue}) = \frac{17}{20}
$
Question14
14: C2.2
Factorise.
$3x^3-7xy$
▶️Answer/Explanation
$x(3x^2-7y)$
Both terms have a common factor of x, so we can factor it out:
$
= x(3x^2 – 7y)
$
\( 3x^2 – 7y \), cannot be factored further because it’s a sum of unlike terms
Question15
15: C3.1
Write $\vec{AB}$ as a column vector.
▶️Answer/Explanation
$\binom{-10}{3}$
$
\vec{AB} = \mathbf{B} – \mathbf{A}
$
\( \mathbf{A} = (7, 1) \)
\( \mathbf{B} = (-3, 4) \)
$
\vec{AB} = (-3 – 7, \ 4 – 1)
$
$
\vec{AB} = (-10, 3)
$
$
\vec{AB} = \begin{bmatrix} -10 \\ 3 \end{bmatrix}
$
Question16
16: C1.11
Vani changes x euros into dollars.
She then changes the dollars into 17850 rupees.
Calculate the value of x.
▶️Answer/Explanation
221
Vani changes x euros into dollars
$\text{Dollars} = 1.05x$
$\text{Rupees} = \frac{\text{Dollars}}{0.013}$
she ends up with 17,850 rupees
$\frac{1.05x}{0.013} = 17850$
$1.05x = 17850 \times 0.013$
$1.05x = 232.05$$x = 221$
Question17
17: C3.5
The line $y = 2x- 5$ intersects the line $y = 3$ at the point P.
Find the coordinates of the point P.
▶️Answer/Explanation
(4,3)
the y-coordinates are the same at the point of intersection
substitute \( y = 3 \)
$
3 = 2x – 5
$
$
2x = 3 + 5
$
$
2x = 8
$
$
x = 4
$
$
y = 3
$
the coordinates of point \( P \)
$(4,3)$
Question18
18(a): C7.1
18(b): C7.1
The diagram shows two shapes, A and B, on a grid.
(a) Describe fully the single transformation that maps shape A onto shape B.
(b) On the grid, draw the image of shape A after a reflection in the line $x =-1$.
▶️Answer/Explanation
(a)
Rotation
[centre] (0,0)
90° [anticlockwise]
(b)
Shape drawn correctly
Question19
19: C5.3
The diagram shows a small circle with radius 7cm and a large circle with radius Rcm.
The area of 16 small circles is the same as the area of one large circle.
Calculate the value of R.
▶️Answer/Explanation
28
area of a circle is
$
A = \pi r^2
$
small circle, the radius is 7 cm,
$
\text{Area of small circle} = \pi (7^2) = 49\pi \text{ cm}^2
$
Since the area of 16 small circles equals the area of the large circle,
$
\text{Total area of 16 small circles} = 16 \times 49\pi = 784\pi \text{ cm}^2
$
$
\text{Area of large circle} = \pi R^2
$
$
\pi R^2 = 784\pi
$
$
R = \sqrt{784}
$
$
R = 28
$
Question20
20(a): C2.7
20(b): C2.7
(a) The $n$th term of a sequence is $n^2-3$.
Find the first three terms of this sequence.
(b) These are the first five terms of a different sequence.
$$
\begin{array}{lllll}
2 & 9 & 16 & 23 & 30
\end{array}
$$
Find the $n$th term of this sequence.
▶️Answer/Explanation
(a) −2 1 6
(b) 7n- 5
(a)
$
n^2 – 3
$
When \( n = 1 \)
$
1^2 – 3 = 1 – 3 = -2
$
When \( n = 2 \)
$
2^2 – 3 = 4 – 3 = 1
$
When \( n = 3 \)
$
3^2 – 3 = 9 – 3 = 6
$
First three terms
$
-2, 1, 6
$
Part (b)
$
2, 9, 16, 23, 30
$
difference between consecutive terms
$
9 – 2 = 7, \quad 16 – 9 = 7, \quad 23 – 16 = 7, \quad 30 – 23 = 7
$
common difference is 7,
The first term is 2.
$
\text{nth term} = a + (n-1)d
$
$
\text{nth term} = 2 + (n-1)(7)
$
$
= 2 + 7n – 7
$
$
= 7n – 5
$
Question21
21: C1.10
The length, l m, of a rope is $18.7~m$, correct to the nearest 10 centimetres.
Complete this statement about the value of l.
▶️Answer/Explanation
18.65 to 18.75
The length of the rope is
$
l = 18.7~m
$
Since 10 cm = 0.1 m, the measurement could be anywhere within 0.05 meters above or below 18.7 meters.
Lower bound (LB)
$
\text{Lower bound} = 18.7 – 0.05 = 18.65~m
$
Upper bound (UB)
$
\text{Upper bound} = 18.7 + 0.05 = 18.75~m
$
The value of \( l \) lies within the interval:
$
18.65~m \leq l < 18.75~m
$
Question22
22: C1.8
$$
6.5 \times 10^{19} \times n=5.46 \times 10^{23}
$$
Calculate the value of $n$.
Give your answer in standard form.
▶️Answer/Explanation
$8.4 \times 10^3$
$
6.5 \times 10^{19} \times n = 5.46 \times 10^{23}
$
$
n = \frac{5.46 \times 10^{23}}{6.5 \times 10^{19}}
$
$
\frac{5.46}{6.5} \approx 0.84
$
$
\frac{10^{23}}{10^{19}} = 10^{23-19} = 10^4
$
$
n = 0.84 \times 10^4
$
$
n = 8.4 \times 10^3
$
Question23
23: C4.4
Triangle ABC is mathematically similar to triangle DEF.
Calculate the value of h.
▶️Answer/Explanation
72.5
property of similar triangles
$
\frac{AB}{DE} = \frac{AC}{DF}
$
\( AB = h \)
\( DE = 97.5 \text{ cm} \)
\( AC = 118.9 \text{ cm} \)
\( DF = 159.9 \text{ cm} \)
$
\frac{h}{97.5} = \frac{118.9}{159.9}
$
$
\frac{118.9}{159.9} \approx 0.7437
$
$
\frac{h}{97.5} = 0.7437
$
$
h \approx 72.5 \text{ cm}
$
Question24
24: C1.4
Without using a calculator, work out $1 \frac{1}{4}+\frac{5}{6}$
You must show all your working and give your answer as a fraction in its simplest form.
▶️Answer/Explanation
$\frac{25}{12}$
$
1 \frac{1}{4} + \frac{5}{6}
$
$
1 \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{5}{4}
$
$
\frac{5}{4} + \frac{5}{6}
$
The denominators are 4 and 6. The lowest common denominator (LCD) is 12.
$
\frac{5}{4} = \frac{5 \times 3}{4 \times 3} = \frac{15}{12}
$
$
\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}
$
add the numerators
$
\frac{15}{12} + \frac{10}{12} = \frac{15 + 10}{12} = \frac{25}{12}
$
Question25
25: C1.1
The highest common factor (HCF) of two numbers is 6.
The lowest common multiple (LCM) of the two numbers is 90.
Both numbers are greater than 6.
Work out the two numbers.
▶️Answer/Explanation
18 30
$
\text{HCF} \times \text{LCM} = \text{Number 1} \times \text{Number 2}
$
$
6 \times 90 = \text{Number 1} \times \text{Number 2}
$
$
540 = \text{Number 1} \times \text{Number 2}
$
$(6\times 90)=540$ → Not valid (one number is 6, but both should be greater than 6)
$(10\times 54)=540$
$(18\times 30)=540$
As given HCF $= 6$
HCF$(10, 54) = 2$
HCF$(18, 30) = 6 $