1. [Maximum mark: 3]
Write 1400 as a product of powers of its prime factors. Show your working clearly.
▶️Answer/Explanation
Solution:
To express 1400 as a product of powers of its prime factors, we perform prime factorization:
1400 = 2 × 700
700 = 2 × 350
350 = 2 × 175
175 = 5 × 35
35 = 5 × 7
Therefore, 1400 = 2 × 2 × 2 × 5 × 5 × 7 = \(2^3 \times 5^2 \times 7\).
2. [Maximum mark: 4]
(a) Describe fully the single transformation that maps shape A onto shape B.
(b) On the grid above, rotate shape A 180° about (−1, 0). Label your shape C.
▶️Answer/Explanation
Solution:
(a) The transformation is a translation by the vector \(\begin{pmatrix} 3 \\ -5 \end{pmatrix}\).
(b) To rotate shape A 180° about the point (−1, 0), each point of shape A is rotated 180° around (−1, 0), resulting in shape C.
3. [Maximum mark: 2]
Here is a list of four numbers written in ascending order of size: \( x, x, y, 15 \), where \( x \) and \( y \) are integers. The numbers have a median of 12.5 and a range of 4. Find the value of \( x \) and the value of \( y \).
▶️Answer/Explanation
Solution:
The median of the four numbers is the average of the second and third numbers: \(\frac{x + y}{2} = 12.5\).
This gives \( x + y = 25 \).
The range is \( 15 – x = 4 \), so \( x = 11 \).
Substituting \( x = 11 \) into \( x + y = 25 \), we get \( y = 14 \).
Therefore, \( x = 11 \) and \( y = 14 \).
4. [Maximum mark: 5]
Given the sets:
\(\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\),
\(A = \{factors of 6\}\),
\(B = \{prime numbers\}\).
(a) List the members of the set:
(i) \(A \cup B\)
(ii) \(A’\)
(b) Explain why Harpreet is incorrect in stating that \(A \cap B = \emptyset\).
(c) List the 4 members of set \(C\) such that \(A \cap C\) has 2 members and \(B \cap C\) has 2 members, with no common members.
▶️Answer/Explanation
Solution:
(a)(i) \(A \cup B = \{1, 2, 3, 5, 6, 7\}\)
(a)(ii) \(A’ = \{4, 5, 7, 8, 9, 10\}\)
(b) Harpreet is incorrect because \(A \cap B = \{2, 3\}\), which is not empty.
(c) The 4 members of set \(C\) are \(\{1, 5, 6, 7\}\).
5. [Maximum mark: 4]
The diagram shows the design for a badge, which will be made using wire. The design is a circle inside a square ABCD. The circle touches the square at points E, F, G, and H. The area of the square is 81 cm². Calculate the total length of wire needed to make the square and the circle. Give your answer correct to 3 significant figures.
▶️Answer/Explanation
Solution:
The side length of the square is \(\sqrt{81} = 9\) cm.
The perimeter of the square is \(4 \times 9 = 36\) cm.
The diameter of the circle is equal to the side length of the square, so the radius is \(4.5\) cm.
The circumference of the circle is \(2\pi \times 4.5 = 9\pi\) cm.
The total length of wire needed is \(36 + 9\pi \approx 64.3\) cm.
6. [Maximum mark: 8]
(a) Solve \(\frac{2f}{3} = 4f – 17\).
(b) Simplify \((e + 12)^0\) where \(e > 0\).
(c) Simplify fully \(\frac{12a^4h^6}{4ah^2}\).
(d) Factorise fully \(20x^5y + 12x^3y^4\).
▶️Answer/Explanation
Solution:
(a) \(\frac{2f}{3} = 4f – 17\)
Multiply both sides by 3: \(2f = 12f – 51\)
Rearrange: \(10f = 51\)
\(f = \frac{51}{10}\).
(b) \((e + 12)^0 = 1\).
(c) \(\frac{12a^4h^6}{4ah^2} = 3a^3h^4\).
(d) \(20x^5y + 12x^3y^4 = 4x^3y(5x^2 + 3y^3)\).
7. [Maximum mark: 2]
Find the value of \( n \) in the equation \( 3^{n} \times 3^{5} \div 3^{10} = 3^{-2} \).
▶️Answer/Explanation
Solution:
Using the laws of exponents:
\( 3^{n} \times 3^{5} \div 3^{10} = 3^{n + 5 – 10} = 3^{n – 5} \).
Given \( 3^{n – 5} = 3^{-2} \), we equate the exponents:
\( n – 5 = -2 \), so \( n = 3 \).
8. [Maximum mark: 3]
In a sale, all normal prices are reduced by 17%. The sale price of a fridge is 6225 rupees. Work out the normal price of the fridge.
▶️Answer/Explanation
Solution:
Let the normal price be \( P \).
After a 17% reduction, the sale price is \( 0.83P = 6225 \).
Solving for \( P \):
\( P = \frac{6225}{0.83} = 7500 \) rupees.
9. [Maximum mark: 4]
(a) Write \( 6.04 \times 10^{5} \) as an ordinary number.
(b) Write \( 0.00007 \) in standard form.
(c) Work out \( \frac{7.6 \times 10^{10}}{4 \times 10^{5} – 2 \times 10^{4}} \) and give your answer in standard form.
▶️Answer/Explanation
Solution:
(a) \( 6.04 \times 10^{5} = 604000 \).
(b) \( 0.00007 = 7 \times 10^{-5} \).
(c) Simplify the denominator:
\( 4 \times 10^{5} – 2 \times 10^{4} = 4 \times 10^{5} – 0.2 \times 10^{5} = 3.8 \times 10^{5} \).
Now, \( \frac{7.6 \times 10^{10}}{3.8 \times 10^{5}} = 2 \times 10^{5} \).
10. [Maximum mark: 5]
The diagram shows a hexagon \( ABCDEF \). Angle \( BCF = 30^\circ \), and \( AB, FC, \) and \( ED \) are parallel. Calculate the area of \( ABCDEF \).
▶️Answer/Explanation
Solution:
The hexagon can be divided into a rectangle and two triangles. Using trigonometry, the height of the triangle is \( 5 \sin 30^\circ = 2.5 \) cm.
The area of the rectangle is \( 11 \times 4.7 = 51.7 \) cm².
The area of the two triangles is \( 2 \times \frac{1}{2} \times 5 \times 2.5 = 12.5 \) cm².
The total area is \( 51.7 + 12.5 = 64.2 \) cm².
11. [Maximum mark: 7]
The cumulative frequency table gives information about the time, in hours, that each of 60 workers spent working from home in one week.
(a) Draw a cumulative frequency graph.
(b) Use your graph to estimate the interquartile range.
(c) Estimate the value of \( W \) such that 25 workers spent more than \( W \) hours working from home.
(d) Find the probability that a randomly chosen worker spent between 5 and 10 hours working from home.
▶️Answer/Explanation
Solution:
(a) Plot the cumulative frequency graph using the given data points.
(b) The interquartile range is approximately \( 13.5 – 8.5 = 5 \) hours.
(c) \( W \approx 16.5 \) hours.
(d) The probability is \( \frac{11}{60} \approx 0.183 \).