An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V2) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
卷一 Non-calculator (Core) 0607/12
Answer all questions. Calculators must not be used. Show all necessary working clearly.
15 題目 · 50 分
題目 1 · short_answer
2.5 分
A shopkeeper buys 60 identical notebooks for a total cost of $150. She sells each notebook for $3.20.
M1 for a valid method to find the total income, e.g. \(60 \times 3.20\) [192] or cost per notebook \(150 \div 60 = 2.50\) M1 for subtracting cost from revenue, e.g. \(192 - 150\) or \((3.20 - 2.50) \times 60\) A0.5 for 42 (or $42)
題目 2 · short_answer
2.5 分
Marco invests $800 at a rate of \(3.5\%\) per year simple interest.
Calculate the total interest earned after 4 years.
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解題
Simple Interest = \(\frac{P \times R \times T}{100}\)
Alternatively, sale price = \(0.85 \times 75 = \$63.75\).
評分準則
M1 for finding \(15\%\) of 75 [11.25] or \((1 - 0.15) \times 75\) oe M0.5 for subtracting discount from $75 A1 for 63.75 (or $63.75)
題目 4 · short_answer
2.5 分
An adult cinema ticket costs $12.50 and a child ticket costs $7.80. A family buys 2 adult tickets and 3 child tickets. They pay with a $100 note.
Work out the change they receive.
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解題
Cost of 2 adult tickets = \(2 \times 12.50 = \$25.00\). Cost of 3 child tickets = \(3 \times 7.80 = \$23.40\). Total cost = \(25.00 + 23.40 = \$48.40\).
Change = \(100 - 48.40 = \$51.60\).
評分準則
M1 for \(2 \times 12.50 + 3 \times 7.80\) [48.40] M0.5 for \(100 - \text{their } 48.40\) A1 for 51.60 or 51.6 (or $51.60)
題目 5 · short_answer
3 分
The line \(L\) passes through the points \((0, -5)\) and \((4, 7)\).
(a) Work out the gradient of line \(L\). [1]
(b) Write down the equation of line \(L\) in the form \(y = mx + c\). [2]
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解題
(a) The gradient \(m\) is calculated by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - (-5)}{4 - 0} = \frac{12}{4} = 3 \]
(b) The line crosses the \(y\)-axis at \((0, -5)\), so the \(y\)-intercept \(c = -5\). Using \(y = mx + c\) with \(m = 3\) and \(c = -5\): \[ y = 3x - 5 \]
評分準則
(a) B1 for 3 cao (b) B1 for \(y = 3x + c\) (where \(c \neq -5\)) or \(y = mx - 5\) (where \(m \neq 3\)), B1 for \(y = 3x - 5\) cao
題目 6 · short_answer
3 分
In a diagram, straight line \(AB\) is parallel to straight line \(CD\). Triangle \(PQR\) has vertex \(P\) on line \(AB\), and vertices \(Q\) and \(R\) on line \(CD\). \(PQ = PR\) and angle \(APQ = 52^\circ\).
(a) Write down the size of angle \(PQR\). [1]
(b) Work out the size of angle \(QPR\). [2]
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解題
(a) Angle \(PQR = \text{angle } APQ = 52^\circ\) because alternate angles between parallel lines are equal.
(b) Triangle \(PQR\) is isosceles with \(PQ = PR\), so base angles are equal: \(\text{angle } PRQ = \text{angle } PQR = 52^\circ\). The sum of angles in a triangle is \(180^\circ\): \[ \text{angle } QPR = 180^\circ - (52^\circ + 52^\circ) = 180^\circ - 104^\circ = 76^\circ \]
評分準則
(a) B1 for 52\(^\circ\) (b) M1 for \(180 - 2 \times 52\) or \(180 - (52 + 52)\) soi, A1 for 76\(^\circ\)
題目 7 · short_answer
3 分
The point \(M(3, 7)\) is the midpoint of the line segment joining \(A(-1, 2)\) and \(B(k, 12)\).
(a) Find the value of \(k\). [2]
(b) Work out the gradient of the line segment \(AB\). [1]
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解題
(a) Using the midpoint formula for the \(x\)-coordinate: \[ \frac{-1 + k}{2} = 3 \implies -1 + k = 6 \implies k = 7 \]
(b) The coordinates of \(B\) are \((7, 12)\). \[ \text{Gradient} = \frac{12 - 2}{7 - (-1)} = \frac{10}{8} = \frac{5}{4} \text{ (or } 1.25\text{)} \]
評分準則
(a) M1 for \(\frac{-1 + k}{2} = 3\) oe, A1 for \(k = 7\) (b) B1 for \(\frac{5}{4}\) or 1.25 or \(\frac{10}{8}\) oe
題目 8 · short_answer
3 分
A circle has centre \(O\). The points \(A\), \(B\) and \(C\) lie on the circumference of the circle, and \(AC\) is a diameter. Angle \(BAC = 38^\circ\).
(a) Write down the size of angle \(ABC\) and give a geometrical reason for your answer. [2]
(b) Find the size of angle \(ACB\). [1]
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解題
(a) Angle \(ABC = 90^\circ\). Geometrical reason: The angle in a semicircle (or angle subtended by a diameter) is a right angle.
(b) The sum of angles in triangle \(ABC\) is \(180^\circ\): \[ \text{Angle } ACB = 180^\circ - 90^\circ - 38^\circ = 52^\circ \]
評分準則
(a) B1 for \(90^\circ\), B1 for correct reason containing 'semicircle' or 'diameter' (b) B1 for \(52^\circ\)
題目 9 · short_answer
3 分
Triangle \(T\) has vertices at \((1, 2)\), \((4, 2)\) and \((1, 6)\).
(a) Write down the coordinates of the image of the point \((4, 2)\) after a reflection in the line \(y = 0\). [1]
(b) Triangle \(T\) is enlarged by a scale factor of 3 with centre of enlargement \((0, 0)\) to give triangle \(U\). Write down the coordinates of the vertices of triangle \(U\). [2]
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解題
(a) Reflection in \(y = 0\) (the \(x\)-axis) maps \((x, y)\) to \((x, -y)\), so \((4, 2) \to (4, -2)\).
(b) Enlargement by scale factor 3 from centre \((0, 0)\) multiplies each coordinate by 3: \((1, 2) \to (3, 6)\) \((4, 2) \to (12, 6)\) \((1, 6) \to (3, 18)\)
評分準則
(a) B1 for \((4, -2)\) (b) B2 for all three vertices correct: \((3, 6)\), \((12, 6)\), \((3, 18)\); or B1 for any two vertices correct or for multiplying all coordinates of at least two vertices by 3
題目 10 · structured
4 分
Here are the first four terms of a sequence.
\[7, \quad 13, \quad 19, \quad 25, \quad \dots\]
(a) Write down the next term of this sequence. [1]
(b) Find an expression for the \(n\)th term of this sequence. [2]
(c) Work out the 20th term of this sequence. [1]
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解題
(a) The common difference between consecutive terms is \(13 - 7 = 6\). Next term \(= 25 + 6 = 31\).
(b) Since the sequence increases by 6 each time, the \(n\)th term is of the form \(6n + c\). When \(n = 1\), \(6(1) + c = 7 \implies c = 1\). So the \(n\)th term is \(6n + 1\).
(c) Substitute \(n = 20\) into the \(n\)th term expression: \(6(20) + 1 = 120 + 1 = 121\).
評分準則
(a) B1 for 31 (b) B2 for \(6n + 1\) oe (e.g. \(1 + 6n\)) (B1 for \(6n + k\), where \(k \neq 1\) or \(k\) is omitted, or for \(n = 6n + 1\)) (c) B1 for 121 or FT their (b) of the form \(an + b\)
題目 11 · structured
4 分
(a) The point \(A(2, 4)\) is rotated \(90^\circ\) clockwise about the origin \((0, 0)\) to give point \(B\). Find the coordinates of \(B\). [2]
(b) The point \(C(-1, 3)\) is reflected in the line \(x = 2\) to give point \(D\). Find the coordinates of \(D\). [2]
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解題
(a) A rotation of \(90^\circ\) clockwise about the origin maps \((x, y)\) to \((y, -x)\). Applying this to \(A(2, 4)\): \(x\)-coordinate becomes 4, \(y\)-coordinate becomes \(-2\). So \(B = (4, -2)\).
(b) The line of reflection is the vertical line \(x = 2\). The distance from \(C(-1, 3)\) to the line \(x = 2\) is \(2 - (-1) = 3\) units to the right. Moving 3 units to the right of the line \(x = 2\) gives \(x = 2 + 3 = 5\). The \(y\)-coordinate remains unchanged at 3. So \(D = (5, 3)\).
評分準則
(a) B2 for \((4, -2)\) (B1 for \((4, k)\) or \((k, -2)\) or for \((-4, 2)\)) (b) B2 for \((5, 3)\) (B1 for \((5, k)\) with \(k \neq 3\) or \((k, 3)\) with \(k \neq 5\) or for correct line \(x = 2\) identified/used)
題目 12 · structured
4 分
The first four terms of a sequence are
\[42, \quad 35, \quad 28, \quad 21, \quad \dots\]
(a) Write down the term-to-term rule for this sequence. [1]
(b) Find an expression for the \(n\)th term of this sequence. [2]
(c) Find the first negative term in this sequence. [1]
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解題
(a) Each term is found by subtracting 7 from the previous term. Term-to-term rule: subtract 7 (or \(-7\)).
(b) The difference between consecutive terms is \(-7\), so the formula begins with \(-7n\). When \(n = 1\), \(-7(1) + c = 42 \implies c = 49\). So the \(n\)th term is \(49 - 7n\) (or \(-7n + 49\)).
(c) Setting \(49 - 7n < 0\): \(7n > 49 \implies n > 7\). The first integer value is \(n = 8\). For \(n = 8\), the term is \(49 - 7(8) = 49 - 56 = -7\).
評分準則
(a) B1 for subtract 7 oe (e.g. \(-7\)) (b) B2 for \(49 - 7n\) oe (e.g. \(-7n + 49\)) (B1 for \(-7n + k\) where \(k \neq 49\) or \(k\) omitted) (c) B1 for \(-7\)
題目 13 · structured
4 分
(a) A shape is enlarged with centre \((0, 0)\) and scale factor 3. The point \(P(4, 1)\) is on the original shape. Find the coordinates of the image of \(P\). [1]
(b) Triangle \(T\) has vertices at \((1, 1)\), \((3, 1)\) and \((1, 4)\). Triangle \(U\) has vertices at \((-1, -1)\), \((-3, -1)\) and \((-1, -4)\).
Describe fully the single transformation that maps triangle \(T\) onto triangle \(U\). [3]
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解題
(a) For an enlargement with centre \((0, 0)\) and scale factor 3: \((x, y) \to (3x, 3y)\) \(P(4, 1) \to (3 \times 4, 3 \times 1) = (12, 3)\).
(b) Comparing the vertices of \(T\) and \(U\): \((1, 1) \to (-1, -1)\) \((3, 1) \to (-3, -1)\) \((1, 4) \to (-1, -4)\) This corresponds to the mapping \((x, y) \to (-x, -y)\), which is a rotation of \(180^\circ\) about the origin \((0, 0)\).
評分準則
(a) B1 for \((12, 3)\) (b) B1 for Rotation B1 for \(180^\circ\) (or half turn) B1 for centre \((0, 0)\) or origin Note: If more than one transformation is named, award 0/3 for part (b).
(a) List the elements of set \(A\). [1] (b) Write down the elements of \(A \cap B\). [1] (c) Find \(\text{n}(A \cup B)\). [1] (d) A number is chosen at random from the universal set \(\mathcal{U}\). Find the probability that this number is not in set \(A\). [1]
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解題
(a) The factors of 12 between 1 and 12 are \(1, 2, 3, 4, 6, 12\).
(b) Set \(B = \{3, 6, 9, 12\}\). The intersection \(A \cap B\) consists of elements common to both \(A\) and \(B\): \(\{3, 6, 12\}\).
(c) The union \(A \cup B = \{1, 2, 3, 4, 6, 9, 12\}\). Counting the number of elements: \(\text{n}(A \cup B) = 7\).
(d) Total number of elements in \(\mathcal{U}\) is 12. The elements not in \(A\) are \(\{5, 7, 8, 9, 10, 11\}\), which is 6 elements. Probability = \(\frac{6}{12} = \frac{1}{2}\) (or 0.5 or 50%).
評分準則
(a) B1 for \(1, 2, 3, 4, 6, 12\) (in any order, with or without brackets) (b) B1 for \(3, 6, 12\) (in any order) (c) B1 for 7 (d) B1 for \(\frac{6}{12}\) oe (e.g. \(\frac{1}{2}\), 0.5, 50%)
題目 15 · structured
5 分
A bag contains counters that are red, blue, green, or yellow. A counter is taken at random from the bag. The table shows the probabilities of choosing a red, blue, or green counter.
(a) Complete the table to find the probability of choosing a yellow counter. [2] (b) Work out the probability of choosing a counter that is red or green. [1] (c) There are 120 counters in the bag altogether. Work out the number of blue counters in the bag. [2]
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解題
(a) The sum of all probabilities is 1. \(P(\text{Yellow}) = 1 - (0.25 + 0.4 + 0.15) = 1 - 0.8 = 0.2\).
M1 for multiplying numerator and denominator by \(3 + \sqrt{3}\) M1 for denominator simplified to \(9 - 3\) or \(6\) A1 for \(6 + 2\sqrt{3}\) or \(2(3 + \sqrt{3})\)
M1 for multiplying numerator and denominator by \(\sqrt{2} - 1\) M1 for correct expansion of numerator with at least 3 correct terms: \(14 - 7\sqrt{2} + 2\sqrt{2} - 2\) A1 for \(12 - 5\sqrt{2}\) cao
題目 4 · short_answer
4 分
\(\mathbf{p} = \begin{pmatrix} w \\ 7 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\).
Given that \(|\mathbf{p} - \mathbf{q}| = \sqrt{106}\) and \(w > 0\), find the value of \(w\).
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解題
First find the column vector \(\mathbf{p} - \mathbf{q}\):
Solve for \(x\): \[2x - 5 = 0 \implies x = \frac{5}{2} = 2.5\] \[x - 3 = 0 \implies x = 3\]
評分準則
M1 for multiplying by \((x - 1)(x - 2)\) to obtain \(6(x - 2) - (x - 1) = 2(x - 1)(x - 2)\) oe M1 for expanding and simplifying to a correct 3-term quadratic equation \(2x^2 - 11x + 15 = 0\) M1 for factorising their 3-term quadratic \((2x - 5)(x - 3) = 0\) or using quadratic formula correctly A1 for \(x = 2.5\) (or \(\frac{5}{2}\)) and \(x = 3\)
題目 8 · short_answer
4 分
Line \(L_1\) passes through the points \((-3, -5)\) and \((5, 7)\). Line \(L_2\) is parallel to line \(L_1\) and passes through the point \((4, 1)\). Line \(L_3\) is perpendicular to line \(L_1\) and passes through the point \((0, 8)\).
Find the coordinates of the point of intersection of line \(L_2\) and line \(L_3\).
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解題
Step 1: Find the gradient of line \(L_1\). \[m_1 = \frac{7 - (-5)}{5 - (-3)} = \frac{12}{8} = \frac{3}{2}\]
Step 2: Find the equation of line \(L_2\). Since \(L_2\) is parallel to \(L_1\), \(m_2 = \frac{3}{2}\). \[y - 1 = \frac{3}{2}(x - 4) \implies y = \frac{3}{2}x - 5 \quad \text{or} \quad 2y = 3x - 10\]
Step 3: Find the equation of line \(L_3\). Since \(L_3\) is perpendicular to \(L_1\), \(m_3 = -\frac{2}{3}\). With \(y\)-intercept \((0, 8)\): \[y = -\frac{2}{3}x + 8 \quad \text{or} \quad 3y = -2x + 24\]
Step 4: Solve the simultaneous equations to find the intersection. From \(L_2\): \(3x = 2y + 10 \implies x = \frac{2y + 10}{3}\). Substitute into \(L_3\): \[3y = -2\left(\frac{2y + 10}{3}\right) + 24\] \[9y = -4y - 20 + 72\] \[13y = 52 \implies y = 4\]
B1 for gradient of \(L_1 = \frac{3}{2}\) B1 for equation of \(L_2\): \(y = \frac{3}{2}x - 5\) oe or equation of \(L_3\): \(y = -\frac{2}{3}x + 8\) oe M1 for a correct method to solve their simultaneous linear equations for \(L_2\) and \(L_3\) A1 for \((6, 4)\)
題目 9 · short_answer
4 分
The points \(P(1, k)\), \(Q(k, 4)\) and \(R(5, 10)\) lie on the same straight line.
Find the two possible values of \(k\).
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解題
Since \(P\), \(Q\) and \(R\) are collinear, the gradient of line segment \(PR\) is equal to the gradient of line segment \(QR\).
M1 for writing an expression for gradient \(PR\) or gradient \(PQ\) or gradient \(QR\) M1 for setting up the equation \(\frac{10 - k}{4} = \frac{6}{5 - k}\) oe M1 for correctly forming the quadratic equation \(k^2 - 15k + 26 = 0\) A1 for \(k = 2\) and \(k = 13\)
題目 10 · short_answer
3 分
Solve the equation.
$$\log_3(x + 5) - \log_3(2x - 1) = 2$$
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解題
Apply the subtraction law of logarithms: $$\log_3\left(\frac{x + 5}{2x - 1}\right) = 2$$
Convert from logarithmic to index form: $$\frac{x + 5}{2x - 1} = 3^2$$ $$\frac{x + 5}{2x - 1} = 9$$
M1 for applying the quotient rule of logarithms: $\log_3\left(\frac{x+5}{2x-1}\right) = 2$ M1 for converting to exponential form: $\frac{x+5}{2x-1} = 3^2$ or $9$ A1 for $x = \frac{14}{17}$ oe
題目 11 · short_answer
4 分
The function $\mathrm{f}(x)$ is defined by $$\mathrm{f}(x) = \frac{2x + 3}{x - 4}, \quad x \neq 4.$$
M1 for power rule $\log_3(x^2)$ or quotient rule seen M1 for converting to quadratic equation $x^2 - 3x - 18 = 0$ oe M1 for factorising into $(x - 6)(x + 3) = 0$ or correct quadratic formula A1 for $x = 6$ cao (must discard or reject $x = -3$)
題目 14 · short_answer
3 分
Work out the exact value of $$\frac{2\sin 30^\circ + \sqrt{3}\tan 60^\circ}{\cos^2 45^\circ}$$ Give your answer as an integer.
Divide the numerator by the denominator: $$\frac{4}{\frac{1}{2}} = 4 \times 2 = 8$$
評分準則
M1 for at least two correct exact values from $\sin 30^\circ = \frac{1}{2}$, $\tan 60^\circ = \sqrt{3}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$ (or $\frac{1}{\sqrt{2}}$) M1 for correctly simplifying the numerator to $4$ and the denominator to $\frac{1}{2}$ A1 for $8$ cao
題目 15 · short_answer
3 分
$TA$ and $TB$ are tangents to a circle, centre $O$, touching the circle at $A$ and $B$ respectively. $C$ is a point on the major arc $AB$. Angle $ATB = 44^\circ$.
(a) Find angle $AOB$. (b) Find angle $ACB$.
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解題
(a) The angle between a tangent and the radius at the point of contact is $90^\circ$, so $\angle OAT = 90^\circ$ and $\angle OBT = 90^\circ$. In quadrilateral $OATB$: $$\angle AOB = 360^\circ - (90^\circ + 90^\circ + 44^\circ) = 180^\circ - 44^\circ = 136^\circ$$
(b) The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the circumference: $$\angle ACB = \frac{1}{2} \times \angle AOB = \frac{1}{2} \times 136^\circ = 68^\circ$$
評分準則
(a) B1 for $136^\circ$ cao (b) M1 for $\frac{1}{2} \times (\text{their } 136^\circ)$ soi A1 for $68^\circ$ (or FT $\frac{1}{2} \times \text{their } (a)$ provided their $(a) < 180^\circ$)
題目 16 · short_answer
3 分
Solve the equation $$2\cos x + \sqrt{3} = 0$$ for $0^\circ \leqslant x \leqslant 360^\circ$.
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解題
Rearrange the equation: $$2\cos x = -\sqrt{3}$$ $$\cos x = -\frac{\sqrt{3}}{2}$$
The reference angle $\alpha$ for which $\cos \alpha = \frac{\sqrt{3}}{2}$ is $30^\circ$. Since $\cos x$ is negative, $x$ lies in the second and third quadrants: $$x = 180^\circ - 30^\circ = 150^\circ$$ $$x = 180^\circ + 30^\circ = 210^\circ$$
評分準則
M1 for $\cos x = -\frac{\sqrt{3}}{2}$ oe A1 for $x = 150^\circ$ A1 for $x = 210^\circ$ Note: SC1 for both angles correct but in radians ($\frac{5\pi}{6}, \frac{7\pi}{6}$) or for $30^\circ$ and $330^\circ$ from solving $\cos x = +\frac{\sqrt{3}}{2}$
題目 17 · structured
6 分
A rectangle has length \((2x + 5)\text{ cm}\) and width \((x - 2)\text{ cm}\). A square has sides of length \((x + 1)\text{ cm}\). The area of the rectangle is \(19\text{ cm}^2\) greater than the area of the square.
(a) Show that \(x^2 - x - 30 = 0\). [3]
(b) (i) Factorise \(x^2 - x - 30\). [1]
(ii) Find the value of \(x\). [1]
(iii) Work out the perimeter of the rectangle. [1]
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解題
(a) The area of the rectangle is given by: \[ (2x + 5)(x - 2) = 2x^2 - 4x + 5x - 10 = 2x^2 + x - 10 \] The area of the square is given by: \[ (x + 1)^2 = x^2 + 2x + 1 \] Forming the equation from the given information: \[ (2x^2 + x - 10) - (x^2 + 2x + 1) = 19 \] \[ 2x^2 - x^2 + x - 2x - 10 - 1 = 19 \] \[ x^2 - x - 11 = 19 \] \[ x^2 - x - 30 = 0 \]
(b) (i) Factorising the quadratic: Two numbers that multiply to give \(-30\) and add to give \(-1\) are \(-6\) and \(+5\). \[ (x - 6)(x + 5) \]
(ii) Solving \((x - 6)(x + 5) = 0\) gives \(x = 6\) or \(x = -5\). Since lengths must be positive, \(x - 2 > 0 \implies x > 2\). Therefore, \(x = 6\).
(iii) Substituting \(x = 6\) into the dimensions of the rectangle: \(\text{Length} = 2(6) + 5 = 17\text{ cm}\) \(\text{Width} = 6 - 2 = 4\text{ cm}\) \(\text{Perimeter} = 2(17 + 4) = 2(21) = 42\text{ cm}\)
評分準則
(a) M1 for correct expansion of either \((2x + 5)(x - 2) = 2x^2 + x - 10\) or \((x + 1)^2 = x^2 + 2x + 1\) M1 for setting up equation \((2x^2 + x - 10) - (x^2 + 2x + 1) = 19\) oe A1 for complete and correct working shown leading to \(x^2 - x - 30 = 0\) with no errors
(b)(i) B1 for \((x - 6)(x + 5)\)
(b)(ii) B1 for \(6\) (must reject \(-5\))
(b)(iii) B1 for \(42\) or FT \(2(2x + 5 + x - 2)\) using their positive \(x\)
題目 18 · structured
6 分
A coach travels a distance of \(120\text{ km}\) at an average speed of \(x\text{ km/h}\). On the return journey, the coach travels at an average speed of \((x + 10)\text{ km/h}\). The return journey takes \(1\) hour less than the outward journey.
(a) Write down an expression, in terms of \(x\), for the time taken, in hours, for: (i) the outward journey, [1] (ii) the return journey. [1]
(b) (i) Form an equation in \(x\) and show that it simplifies to \(x^2 + 10x - 1200 = 0\). [2]
(ii) Factorise \(x^2 + 10x - 1200\). [1]
(iii) Work out the average speed of the coach on the return journey. [1]
(b) (i) The difference between the two times is \(1\) hour: \[ \frac{120}{x} - \frac{120}{x + 10} = 1 \] Multiply throughout by the common denominator \(x(x + 10)\): \[ 120(x + 10) - 120x = x(x + 10) \] \[ 120x + 1200 - 120x = x^2 + 10x \] \[ 1200 = x^2 + 10x \] \[ x^2 + 10x - 1200 = 0 \]
(ii) Finding two numbers that multiply to \(-1200\) and add to \(+10\): \(40 \times (-30) = -1200\) and \(40 + (-30) = 10\). \[ (x + 40)(x - 30) \]
(iii) Solving \((x + 40)(x - 30) = 0\) gives \(x = 30\) or \(x = -40\). Since speed must be positive, \(x = 30\text{ km/h}\). The return speed is \(x + 10 = 30 + 10 = 40\text{ km/h}\).
評分準則
(a)(i) B1 for \(\frac{120}{x}\)
(a)(ii) B1 for \(\frac{120}{x + 10}\)
(b)(i) M1 for \(\frac{120}{x} - \frac{120}{x + 10} = 1\) oe A1 for complete algebraic working with correct bracket expansion leading to \(x^2 + 10x - 1200 = 0\)
(b)(ii) B1 for \((x + 40)(x - 30)\)
(b)(iii) B1 for \(40\) [km/h] (rejects \(x = -40\) and adds 10 to 30)
Paper 3 Calculator (Core) 0607/32
Answer all questions. Graphic display calculator should be used where appropriate.
12 題目 · 39.5 分
題目 1 · Short Answer
2.5 分
A bakery produces loaves of bread in three varieties: white, wholemeal, and rye in the ratio \(5 : 3 : 2\).
On Tuesday, the bakery produces a total of \(760\) loaves of bread.
Calculate the number of wholemeal loaves produced.
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解題
1. Find the total number of parts in the ratio: \[5 + 3 + 2 = 10\text{ parts}\]
2. Find the value of one part: \[\frac{760}{10} = 76\text{ loaves}\]
3. Calculate the number of wholemeal loaves (3 parts): \[3 \times 76 = 228\text{ loaves}\]
評分準則
M1.5 for \(760 \div (5 + 3 + 2)\) or \(\frac{3}{10} \times 760\) oe soi A1 for 228 cao
題目 2 · Short Answer
2.5 分
Eleni invests \(\$3200\) in a savings account that pays simple interest at a rate of \(4.5\%\) per year.
Calculate the total amount in the account at the end of \(3\) years.
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解題
1. Calculate the simple interest earned using \(I = P \times r \times t\): \[I = 3200 \times \frac{4.5}{100} \times 3 = 3200 \times 0.045 \times 3 = \$432\]
2. Calculate the total amount in the account: \[\text{Total Amount} = 3200 + 432 = \$3632\]
評分準則
M1.5 for \(3200 \times 0.045 \times 3\) oe (soi by 432) A1 for 3632
題目 3 · Short Answer
2.5 分
Find the lowest common multiple (LCM) of \(36\) and \(84\).
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解題
1. Write each number as a product of prime factors: \[36 = 2^2 \times 3^2\] \[84 = 2^2 \times 3 \times 7\]
2. Take the highest power of each prime factor: \[\text{LCM} = 2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252\]
評分準則
M1.5 for listing prime factors of both numbers (\(36 = 2^2 \times 3^2\), \(84 = 2^2 \times 3 \times 7\)) or for listing at least 3 multiples of each correctly A1 for 252 cao
題目 4 · Short Answer
2.5 分
In a survey of \(450\) students, \(\frac{2}{5}\) travel to school by bus, \(32\%\) walk to school, and the remaining students travel by car.
Work out the number of students who travel by car.
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解題
1. Find the number of students travelling by bus: \[\frac{2}{5} \times 450 = 180\]
2. Find the number of students who walk: \[0.32 \times 450 = 144\]
3. Calculate the number of students travelling by car: \[450 - (180 + 144) = 450 - 324 = 126\]
M1 for \(\frac{2}{5} \times 450\) (= 180) or \(0.32 \times 450\) (= 144) or finding the remaining fraction/percentage \(1 - \frac{2}{5} - 0.32 = 0.28\) M0.5 for \(450 - 180 - 144\) or \(0.28 \times 450\) A1 for 126 cao
題目 5 · Short Answer
2.5 分
A shopkeeper buys a box of \(24\) headphones for a total cost of \(\$480\). She sells each pair of headphones for \(\$26.50\).
Calculate her percentage profit on the total cost.
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解題
1. Calculate the total selling price of the headphones: \[24 \times 26.50 = \$636\]
2. Calculate the total profit made: \[636 - 480 = \$156\]
3. Calculate the percentage profit: \[\frac{156}{480} \times 100 = 32.5\%\]
評分準則
M1 for \(24 \times 26.50\) (= 636) M0.5 for \(\frac{\text{their } 636 - 480}{480} \times 100\) oe A1 for 32.5 or 32.5%
題目 6 · short_answer
3 分
A box contains 7 blue pens and 3 black pens. A pen is chosen at random from the box, its colour is noted, and it is replaced. A second pen is then chosen at random from the box.
(a) Write down the probability that the first pen chosen is blue. [1]
(b) Work out the probability that both pens chosen are blue. [2]
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解題
(a) Total number of pens = \(7 + 3 = 10\). Probability that the first pen is blue = \(\frac{7}{10}\).
(b) Since the pen is replaced, the selections are independent. \(\text{P}(\text{both blue}) = \frac{7}{10} \times \frac{7}{10} = \frac{49}{100}\) (or \(0.49\)).
評分準則
(a) B1 for \(\frac{7}{10}\) oe (e.g. 0.7, 70%) (b) M1 for \(\frac{7}{10} \times \frac{7}{10}\) oe A1 for \(\frac{49}{100}\) oe (e.g. 0.49, 49%)
題目 7 · short_answer
3.5 分
NOT TO SCALE
Triangle \(ABC\) is right-angled at \(B\). \(AB = 6.8\text{ cm}\) and \(BC = 9.5\text{ cm}\).
(a) Calculate the length \(AC\). (b) Calculate angle \(BAC\).
(a) [2 marks] M1 for \(6.8^2 + 9.5^2\) or \(\sqrt{6.8^2 + 9.5^2}\) soi A1 for 11.7 or 11.68...
(b) [1.5 marks] M1 for \(\tan(BAC) = \frac{9.5}{6.8}\) or correct use of trig ratio with candidate's (a) A0.5 for 54.4 or 54.40 to 54.41
題目 8 · short_answer
3.5 分
A straight ladder of length \(5.2\text{ m}\) rests against a vertical wall on horizontal ground. The bottom of the ladder is \(1.8\text{ m}\) from the base of the wall.
(a) Calculate the angle the ladder makes with the ground. (b) Calculate the height the ladder reaches up the wall.
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解題
(a) Let \(\theta\) be the angle between the ladder and the ground. \[ \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1.8}{5.2} \] \[ \theta = \cos^{-1}\left(\frac{1.8}{5.2}\right) \approx 69.7448...^\circ \] To 1 decimal place, the angle is \(69.7^\circ\).
(b) Let \(h\) be the height reached up the wall. Using Pythagoras' theorem: \[ h^2 + 1.8^2 = 5.2^2 \] \[ h^2 = 5.2^2 - 1.8^2 = 27.04 - 3.24 = 23.8 \] \[ h = \sqrt{23.8} \approx 4.8785...\text{ m} \] To 3 significant figures, the height is \(4.88\text{ m}\).
評分準則
(a) [1.5 marks] M1 for \(\cos\theta = \frac{1.8}{5.2}\) oe A0.5 for 69.7 or 69.74 to 69.75
(b) [2 marks] M1 for \(5.2^2 - 1.8^2\) or \(\sqrt{5.2^2 - 1.8^2}\) or \(5.2\sin(69.7^\circ)\) A1 for 4.88 or 4.878 to 4.879
題目 9 · short_answer
5 分
(a) A solid metal cylinder has a radius of \(4.5\text{ cm}\) and a height of \(12\text{ cm}\). Calculate the volume of the cylinder.
(b) A solid metal sphere has a radius of \(3.2\text{ cm}\). (i) Calculate the volume of the sphere. (ii) The cylinder and the sphere are melted down and recast together into a single block of metal. Calculate the total volume of metal in the block.
(a) [M1] for substitution into cylinder volume formula: \(\pi \times 4.5^2 \times 12\) [A1] for \(763\) or \(763.4\) to \(763.41\dots\) (or \(243\pi\))
(b)(i) [M1] for substitution into sphere volume formula: \(\frac{4}{3} \times \pi \times 3.2^3\) [A1] for \(137\) or \(137.25\dots\) to \(137.3\) (or \(\frac{32768\pi}{750}\))
(b)(ii) [B1 FT] for \(901\) or \(900.6\dots\) to \(900.7\) (FT their (a) + their (b)(i))
題目 10 · short_answer
4 分
The function \(f(x) = 3 + 2x - x^2\) is defined for \(-2 \le x \le 4\).
(a) Sketch the graph of \(y = f(x)\) for \(-2 \le x \le 4\). [2] (b) Write down the coordinates of the maximum point of the graph. [1] (c) Find the zeros of the function \(f(x)\). [1]
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解題
(a) Using a graphic display calculator, sketch the curve \(y = 3 + 2x - x^2\) over the domain \(-2 \le x \le 4\). The graph is a smooth inverted parabola with a vertex at \((1, 4)\), crossing the \(y\)-axis at \((0, 3)\) and the \(x\)-axis at \((-1, 0)\) and \((3, 0)\).
(b) The coordinates of the maximum point are \((1, 4)\).
(c) The zeros of the function are the \(x\)-intercepts where \(f(x) = 0\): \(3 + 2x - x^2 = 0 \implies (3 - x)(1 + x) = 0\) \(x = -1\) and \(x = 3\).
評分準則
(a) B1 for a correct inverted U-shape parabola. B1 for correct orientation with vertex in quadrant 1 and showing intercepts near \((-1, 0)\) and \((3, 0)\). (b) B1 for \((1, 4)\). (c) B1 for both \(x = -1\) and \(x = 3\).
題目 11 · short_answer
4 分
The function \(f(x) = x^3 - 3x - 2\) is defined for \(-2.5 \le x \le 2.5\).
(a) Sketch the graph of \(y = f(x)\) for \(-2.5 \le x \le 2.5\). [2] (b) Write down the coordinates of the local maximum point. [1] (c) Find the roots of the equation \(x^3 - 3x - 2 = 0\). [1]
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解題
(a) Using a graphic display calculator to plot \(y = x^3 - 3x - 2\) for \(-2.5 \le x \le 2.5\): - A local maximum occurs at \((-1, 0)\). - A local minimum occurs at \((1, -4)\). - The graph crosses the \(y\)-axis at \((0, -2)\) and the \(x\)-axis at \(x = 2\).
(b) The coordinates of the local maximum point are \((-1, 0)\).
(c) The roots of the equation \(x^3 - 3x - 2 = 0\) are the \(x\)-values where the graph meets the \(x\)-axis: \(x = -1\) and \(x = 2\).
評分準則
(a) B1 for correct cubic shape (increasing to a maximum, decreasing to a minimum, then increasing). B1 for key features shown correctly (maximum touching the \(x\)-axis at \(x = -1\) and crossing at \(x = 2\)). (b) B1 for \((-1, 0)\). (c) B1 for \(x = -1\) and \(x = 2\).
題目 12 · short_answer
4 分
(a) Sketch the graph of \(y = 4 - x^2\) for \(-3 \le x \le 3\). [2]
(b) Use your graphic display calculator to solve the equation \(4 - x^2 = 2x - 1\). [2]
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解題
(a) Plot \(y = 4 - x^2\) on your GDC for \(-3 \le x \le 3\). The graph is a symmetrical, inverted parabola with vertex / \(y\)-intercept at \((0, 4)\) and \(x\)-intercepts at \((-2, 0)\) and \((2, 0)\).
(b) Plot \(y = 4 - x^2\) and \(y = 2x - 1\) on the graphic display calculator and find the \(x\)-coordinates of the intersection points (or find the zeros of \(x^2 + 2x - 5 = 0\)): \(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-5)}}{2} = \frac{-2 \pm \sqrt{24}}{2} = -1 \pm \sqrt{6}\) \(x \approx -3.45\) and \(x \approx 1.45\) (correct to 3 significant figures).
評分準則
(a) B1 for correct inverted parabolic shape. B1 for vertex at \((0, 4)\) and intercepts at \((-2, 0)\) and \((2, 0)\). (b) B1 for \(x = -3.45\) (or \(-3.449...\)). B1 for \(x = 1.45\) (or \(1.449...\)).
Paper 4 Calculator (Extended) 0607/42
Answer all questions. Graphic display calculator should be used where appropriate.
1. Factorise the numerator by grouping or inspection: $$2x^2 + 5x - 12 = (2x - 3)(x + 4)$$
2. Factorise the denominator using the difference of two squares: $$4x^2 - 9 = (2x - 3)(2x + 3)$$
3. Cancel the common factor of $(2x - 3)$: $$\frac{(2x - 3)(x + 4)}{(2x - 3)(2x + 3)} = \frac{x + 4}{2x + 3}$$
評分準則
M1 for factorising the numerator $(2x - 3)(x + 4)$ or denominator $(2x - 3)(2x + 3)$ A1 for both numerator and denominator correctly factorised A0.5 for cancelling common factor $(2x - 3)$ to give $\frac{x + 4}{2x + 3}$
題目 2 · short_answer
2.5 分
Solve the equation. $$3x^2 - 8x - 5 = 0$$ Show all your working and give your answers correct to 2 decimal places.
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解題
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 3$, $b = -8$, and $c = -5$: $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(-5)}}{2(3)}$$ $$x = \frac{8 \pm \sqrt{64 + 60}}{6} = \frac{8 \pm \sqrt{124}}{6}$$
M1 for correct substitution of $a=3, b=-8, c=-5$ into quadratic formula (allow one sign slip) M0.5 for discriminant evaluated as $124$ or $\sqrt{124}$ A1 for both answers correct to 2 decimal places: $3.19$ and $-0.52$ (SC1 for correct answers with no working or rounded to 3 sf: $3.19$ and $-0.523$)
題目 3 · short_answer
2.5 分
Write as a single fraction in its simplest form. $$\frac{5}{x + 2} - \frac{3}{2x - 1}$$
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解題
1. Find a common denominator, which is $(x + 2)(2x - 1)$: $$\frac{5(2x - 1) - 3(x + 2)}{(x + 2)(2x - 1)}$$
2. Expand the brackets in the numerator: $$5(2x - 1) - 3(x + 2) = 10x - 5 - 3x - 6$$
4. Write the final fraction: $$\frac{7x - 11}{(x + 2)(2x - 1)}$$
評分準則
M1 for writing over a common denominator $(x + 2)(2x - 1)$ with at least one numerator expanded correctly M0.5 for correct expansion of both numerator terms: $10x - 5 - 3x - 6$ A1 for $\frac{7x - 11}{(x + 2)(2x - 1)}$ or $\frac{7x - 11}{2x^2 + 3x - 2}$
題目 4 · short_answer
2.5 分
Solve the equation. $$2(x - 3)^2 = 5x + 7$$ Show all your working and give your answers correct to 3 significant figures.
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解題
1. Expand and rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$: $$2(x^2 - 6x + 9) = 5x + 7$$ $$2x^2 - 12x + 18 = 5x + 7$$ $$2x^2 - 17x + 11 = 0$$
2. Apply the quadratic formula with $a = 2$, $b = -17$, and $c = 11$: $$x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(2)(11)}}{2(2)}$$ $$x = \frac{17 \pm \sqrt{289 - 88}}{4} = \frac{17 \pm \sqrt{201}}{4}$$
M1 for expanding $(x-3)^2$ and rearranging to $2x^2 - 17x + 11 = 0$ oe M0.5 for correct substitution into quadratic formula for their 3-term quadratic A1 for $7.79$ and $0.706$ (both correct to 3 significant figures)
題目 5 · structured
5 分
$$\mathrm{f}(x) = \frac{x - 3}{x^2 - 4} \quad \text{for } x \ne -2, x \ne 2$$
(a) Write down the equations of the three asymptotes of the graph of $y = \mathrm{f}(x)$. [3]
(b) Use your graphic display calculator to find the coordinates of the local maximum point on the graph of $y = \mathrm{f}(x)$ for $x > 2$. Give your answers correct to 3 significant figures. [2]
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解題
(a) The vertical asymptotes occur where the denominator is zero: $$x^2 - 4 = 0 \implies x = -2 \text{ and } x = 2$$ As $x \to \pm\infty$, the degree of the denominator is greater than the degree of the numerator, so the horizontal asymptote is: $$y = 0$$
(b) Using a graphic display calculator to graph $y = \frac{x - 3}{x^2 - 4}$ in the region $x > 2$ and finding the maximum: $$x = 3 + \sqrt{5} \approx 5.236068... \approx 5.24$$ $$y = \frac{3 - \sqrt{5}}{8} \approx 0.0954915... \approx 0.0955$$ Coordinates of the local maximum point are $(5.24, 0.0955)$.
評分準則
(a) B1 for $x = -2$ oe B1 for $x = 2$ oe B1 for $y = 0$ oe (SC2 for all three equations given without variable identifiers, e.g. $-2, 2, 0$)
(b) B1 for $x = 5.24$ or $5.236...$ B1 for $y = 0.0955$ or $0.09549...$ (If coordinates reversed, give max B1)
(a) Write down the equation of (i) the vertical asymptote of the graph of $y = \mathrm{f}(x)$, [1] (ii) the horizontal asymptote of the graph of $y = \mathrm{f}(x)$. [1]
(b) Find the coordinates of the points where the graph of $y = \mathrm{f}(x)$ crosses the coordinate axes. [2]
(c) Write down the range of $\mathrm{f}(x)$ for the domain $x > -2$. [1]
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解題
(a)(i) The vertical asymptote occurs where the denominator is zero: $$x + 2 = 0 \implies x = -2$$
(a)(ii) As $x \to \pm\infty$, $\frac{3x - 6}{x + 2} \to \frac{3x}{x} = 3$, so the horizontal asymptote is: $$y = 3$$
(b) For the $x$-intercept, set $y = 0$: $$3x - 6 = 0 \implies x = 2 \implies (2, 0)$$ For the $y$-intercept, set $x = 0$: $$y = \frac{3(0) - 6}{0 + 2} = \frac{-6}{2} = -3 \implies (0, -3)$$
(c) For $x > -2$, as $x \to -2^+$, $\mathrm{f}(x) \to -\infty$, and as $x \to +\infty$, $\mathrm{f}(x) \to 3$ from below. Since $\mathrm{f}'(x) > 0$, the function is strictly increasing on $(-2, \infty)$. Thus, the range is $\mathrm{f}(x) < 3$ (or $y < 3$).
評分準則
(a)(i) B1 for $x = -2$ cao (a)(ii) B1 for $y = 3$ cao
(b) B1 for $(2, 0)$ oe B1 for $(0, -3)$ oe
(c) B1 for $\mathrm{f}(x) < 3$ or $y < 3$ or $(-\infty, 3)$
題目 7 · short_answer
3 分
A forest area decreases exponentially at a rate of \(3.5\%\) per year due to deforestation. After 8 years, the area of the forest is \(14\,800\text{ hectares}\).
Calculate the original area of the forest 8 years ago. Give your answer correct to 3 significant figures.
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解題
Let \(A_0\) be the original area in hectares. Using the formula for exponential decay: \[A = A_0 \left(1 - \frac{r}{100}\right)^t\] Substitute the given values: \[14\,800 = A_0 (1 - 0.035)^8\] \[14\,800 = A_0 (0.965)^8\] \[A_0 = \frac{14\,800}{0.965^8}\] \[A_0 = \frac{14\,800}{0.7516886...} \approx 19\,689\text{ hectares}\] Correct to 3 significant figures, the area is \(19\,700\text{ hectares}\).
評分準則
M1 for \(A_0 \times (1 - 0.035)^8 = 14\,800\) oe M1 for \(A_0 = \frac{14\,800}{(0.965)^8}\) oe A1 for \(19\,700\) or \(19\,690\) (allow \(19\,688\) to \(19\,690\))
題目 8 · structured
4 分
The 3rd term of a geometric sequence is \(72\) and the 7th term is \(1152\). All the terms in the sequence are positive.
(a) Find the common ratio of the sequence. [2]
(b) Find the first term of the sequence. [1]
(c) Calculate the 10th term of the sequence. [1]
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解題
(a) Let the first term be \(a\) and the common ratio be \(r\). \(u_3 = ar^2 = 72\) \(u_7 = ar^6 = 1152\)
Dividing \(u_7\) by \(u_3\): \[\frac{ar^6}{ar^2} = \frac{1152}{72}\] \[r^4 = 16\] Since all terms are positive, \(r = 2\).
(b) Substitute \(r = 2\) into the expression for \(u_3\): \[a(2)^2 = 72\] \[4a = 72 \implies a = 18\]
(c) The 10th term is: \[u_{10} = ar^9 = 18 \times 2^9 = 18 \times 512 = 9216\]
評分準則
(a) M1 for \(r^4 = \frac{1152}{72}\) or \(r^4 = 16\) A1 for \(2\) (reject \(-2\) since terms are positive) (b) B1 for \(18\) (c) B1 for \(9216\)
題目 9 · structured
4 分
The value of a piece of industrial machinery depreciates exponentially at a constant rate of \(r\%\) per year. Two years after purchase, the value of the machine is \(\$32\,000\). Four years after purchase, the value of the machine is \(\$20\,480\).
(a) Show that \(r = 20\). [2]
(b) Calculate the original purchase price of the machine. [2]
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解題
(a) Let \(V_0\) be the initial purchase price and let the multiplier be \(k = 1 - \frac{r}{100}\). After 2 years: \(V_2 = V_0 k^2 = 32\,000\) After 4 years: \(V_4 = V_0 k^4 = 20\,480\)
Dividing the two equations: \[\frac{V_0 k^4}{V_0 k^2} = \frac{20\,480}{32\,000}\] \[k^2 = 0.64\] \[k = \sqrt{0.64} = 0.8\] Since \(k = 1 - \frac{r}{100}\): \[1 - \frac{r}{100} = 0.8 \implies \frac{r}{100} = 0.2 \implies r = 20\]
(b) Using the value at \(t = 2\): \[V_0 (0.8)^2 = 32\,000\] \[0.64 V_0 = 32\,000\] \[V_0 = \frac{32\,000}{0.64} = 50\,000\] The original purchase price was \(\$50\,000\).
評分準則
(a) M1 for \(\left(1 - \frac{r}{100}\right)^2 = \frac{20\,480}{32\,000}\) or \(k^2 = 0.64\) A1 for full and correct derivation leading to \(r = 20\) (b) M1 for \(V_0 \times (0.8)^2 = 32\,000\) or \(V_0 \times (0.8)^4 = 20\,480\) oe A1 for \(50\,000\) (or \(\$50\,000\))
題目 10 · structured
6 分
The diagram shows a plot of land formed by a quadrilateral \(ABCD\) and an adjoining circular sector \(CDE\) with centre \(D\).
NOT TO SCALE
In triangle \(ABC\), \(AB = 14\text{ m}\), \(BC = 18\text{ m}\), and angle \(ABC = 64^\circ\). In triangle \(ACD\), angle \(DAC = 40^\circ\) and angle \(ADC = 75^\circ\). The sector \(CDE\) has radius \(DC\) and angle \(CDE = 55^\circ\).
(b) Using the sine rule in triangle \(ACD\) to find \(DC\): \[ \frac{DC}{\sin(\angle DAC)} = \frac{AC}{\sin(\angle ADC)} \] \[ \frac{DC}{\sin(40^\circ)} = \frac{17.29337...}{\sin(75^\circ)} \] \[ DC = \frac{17.29337... \times \sin(40^\circ)}{\sin(75^\circ)} = \frac{11.115978...}{0.9659258...} = 11.50810...\text{ m} \]
Now, calculate the area of sector \(CDE\): \[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 \] \[ \text{Area} = \frac{55^\circ}{360^\circ} \times \pi \times (11.50810...)^2 = 63.5653... \approx 63.6\text{ m}^2 \]
評分準則
(a) M1 for \(14^2 + 18^2 - 2(14)(18)\cos(64^\circ)\) A1 for \(299.06...\) or \(299.1\) A1 for \(17.3\) or \(17.29...\)
(b) M1 for \(\frac{DC}{\sin 40^\circ} = \frac{\text{their } AC}{\sin 75^\circ}\) oe A1 for \(DC = 11.5\) or \(11.508...\) A1 for \(63.6\) or \(63.56...\) to \(63.57\)
題目 11 · structured
6 分
The diagram shows a quadrilateral \(KLMN\).
NOT TO SCALE
In triangle \(KLM\), \(KL = 8.5\text{ cm}\), \(LM = 10.4\text{ cm}\), and angle \(KLM = 82^\circ\). In triangle \(KMN\), \(KN = 14.0\text{ cm}\), angle \(KNM = 58^\circ\), and angle \(KMN\) is acute.
(a) Calculate the length \(KM\). [3]
(b) A circular sector with centre \(M\) and radius \(6.5\text{ cm}\) subtends the angle \(KMN\). Calculate the area of this sector. [3]
(b) Using the sine rule in triangle \(KMN\) to find angle \(KMN\): \[ \frac{\sin(\angle KMN)}{KN} = \frac{\sin(\angle KNM)}{KM} \] \[ \sin(\angle KMN) = \frac{14.0 \times \sin(58^\circ)}{12.48215...} = \frac{11.87268...}{12.48215...} = 0.95117... \] \[ \angle KMN = \arcsin(0.95117...) = 72.022...^\circ \]
Now, calculate the area of the sector with radius \(r = 6.5\text{ cm}\) and angle \(\theta = 72.022...^\circ\): \[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{72.022...^\circ}{360^\circ} \times \pi \times 6.5^2 \] \[ \text{Area} = \frac{72.022...}{360} \times \pi \times 42.25 = 26.507... \approx 26.5\text{ cm}^2 \]
評分準則
(a) M1 for \(8.5^2 + 10.4^2 - 2(8.5)(10.4)\cos(82^\circ)\) A1 for \(155.8...\) A1 for \(12.5\) or \(12.48...\)
(b) M1 for \(\frac{\sin(\angle KMN)}{14.0} = \frac{\sin(58^\circ)}{\text{their } KM}\) oe A1 for \(\angle KMN = 72.0^\circ\) or \(72.02...^\circ\) A1 for \(26.5\) or \(26.50...\) to \(26.51\)
題目 12 · structured
6 分
The diagram shows a circular sector \(OAB\) attached to a triangle \(ABC\).
NOT TO SCALE
\(O\) is the centre of the circle with radius \(OA = OB = 7.5\text{ cm}\) and angle \(AOB = 84^\circ\). In triangle \(ABC\), angle \(ABC = 125^\circ\) and angle \(BAC = 32^\circ\).
(a) Calculate the length of the chord \(AB\). [2]
(b) Calculate the length \(AC\). [2]
(c) Calculate the area of the minor sector \(OAB\). [2]
(b) In triangle \(ABC\), the third angle is: \[ \angle ACB = 180^\circ - (125^\circ + 32^\circ) = 23^\circ \] Using the sine rule in triangle \(ABC\): \[ \frac{AC}{\sin(\angle ABC)} = \frac{AB}{\sin(\angle ACB)} \] \[ \frac{AC}{\sin(125^\circ)} = \frac{10.03699...}{\sin(23^\circ)} \] \[ AC = \frac{10.03699... \times \sin(125^\circ)}{\sin(23^\circ)} = \frac{8.22183...}{0.390731...} = 21.0421... \approx 21.0\text{ cm} \]
(c) The area of the sector \(OAB\) is: \[ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{84^\circ}{360^\circ} \times \pi \times 7.5^2 = \frac{84}{360} \times \pi \times 56.25 = 41.2334... \approx 41.2\text{ cm}^2 \]
評分準則
(a) M1 for \(7.5^2 + 7.5^2 - 2(7.5)(7.5)\cos(84^\circ)\) or \(2 \times 7.5 \sin(42^\circ)\) A1 for \(10.0\) or \(10.03...\) to \(10.04\)
(b) M1 for \(\angle ACB = 23^\circ\) soi and \(\frac{AC}{\sin 125^\circ} = \frac{\text{their } AB}{\sin 23^\circ}\) A1 for \(21.0\) or \(21.04...\)
(c) M1 for \(\frac{84}{360} \times \pi \times 7.5^2\) oe A1 for \(41.2\) or \(41.23...\)
題目 13 · structured
5 分
The table shows the cumulative frequency distribution of the times, \(t\) minutes, taken by 120 runners to complete a 10\text{ km} race.
\[ \begin{array}{|c|c|} \hline \text{Time } (t \text{ minutes}) & \text{Cumulative frequency} \\ \hline t \le 40 & 8 \\ t \le 45 & 26 \\ t \le 50 & 64 \\ t \le 55 & 98 \\ t \le 60 & 114 \\ t \le 65 & 120 \\ \hline \end{array} \]
(a) Find the number of runners who took more than 50 minutes. [1]
(b) Two runners are chosen at random from those who took 50 minutes or less.
Calculate the probability that both runners took 45 minutes or less. [2]
(c) A runner is chosen at random from the 120 runners.
Given that this runner took more than 40 minutes, calculate the probability that this runner took at most 55 minutes. [2]
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解題
(a) Total number of runners is 120. The number of runners with \(t \le 50\) is 64. Number of runners taking more than 50 minutes \(= 120 - 64 = 56\).
(b) Number of runners taking 50 minutes or less is 64. Of these, the number of runners taking 45 minutes or less is 26. Probability that both runners took 45 minutes or less: \[ P = \frac{26}{64} \times \frac{25}{63} = \frac{650}{4032} = \frac{325}{2016} \approx 0.161 \]
(c) Number of runners who took more than 40 minutes \(= 120 - 8 = 112\). Number of runners who took more than 40 minutes and at most 55 minutes \(= 98 - 8 = 90\). \[ P(t \le 55 \mid t > 40) = \frac{90}{112} = \frac{45}{56} \approx 0.804 \]
評分準則
(a) B1 for 56 cao
(b) M1 for \(\frac{26}{64} \times \frac{25}{63}\) oe A1 for \(\frac{325}{2016}\) oe (e.g. 0.161 or 0.1612...)
(c) M1 for denominator 112 soi or \(\frac{98-8}{120-8}\) oe A1 for \(\frac{45}{56}\) oe (e.g. 0.804 or 0.8035...)
題目 14 · structured
4 分
A bag contains 7 blue discs and 5 yellow discs.
(a) Nadia picks two discs at random from the bag without replacement.
Calculate the probability that the two discs are of different colours. [2]
(b) If the two discs picked are of the same colour, the game ends. If the two discs picked are of different colours, Nadia picks a third disc at random from the remaining discs.
Calculate the probability that Nadia picks exactly two blue discs in total. [2]
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解題
(a) Total number of discs \(= 7 + 5 = 12\). Probability of picking two discs of different colours: \[ P(\text{Blue, then Yellow}) + P(\text{Yellow, then Blue}) = \left(\frac{7}{12} \times \frac{5}{11}\right) + \left(\frac{5}{12} \times \frac{7}{11}\right) = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66} \approx 0.530 \]
(b) To pick exactly two blue discs, Nadia must pick one blue and one yellow in the first two draws, followed by a blue disc on the third draw: Case 1: Blue, Yellow, Blue \[ P(B_1 \cap Y_2 \cap B_3) = \frac{7}{12} \times \frac{5}{11} \times \frac{6}{10} = \frac{210}{1320} = \frac{7}{44} \] Case 2: Yellow, Blue, Blue \[ P(Y_1 \cap B_2 \cap B_3) = \frac{5}{12} \times \frac{7}{11} \times \frac{6}{10} = \frac{210}{1320} = \frac{7}{44} \] Total probability: \[ P(\text{exactly two blue}) = \frac{7}{44} + \frac{7}{44} = \frac{14}{44} = \frac{7}{22} \approx 0.318 \]
評分準則
(a) M1 for \(\frac{7}{12} \times \frac{5}{11} + \frac{5}{12} \times \frac{7}{11}\) oe or \(1 - \left(\frac{7}{12} \times \frac{6}{11} + \frac{5}{12} \times \frac{4}{11}\right)\) A1 for \(\frac{35}{66}\) oe fraction (e.g. \(\frac{70}{132}\)) or 0.530 or 0.5303...
(b) M1 for \(2 \times \left(\frac{7}{12} \times \frac{5}{11} \times \frac{6}{10}\right)\) oe or identifying both paths (BYB and YBB) A1 for \(\frac{7}{22}\) oe fraction (e.g. \(\frac{14}{44}\), \(\frac{420}{1320}\)) or 0.318 or 0.3181...
題目 15 · extended
4 分
\(VABCD\) is a pyramid with a horizontal rectangular base \(ABCD\). \(AB = 16\text{ cm}\) and \(BC = 12\text{ cm}\). The vertex \(V\) is vertically above the centre of the base, \(O\). The length of each sloping edge \(VA = VB = VC = VD = 18\text{ cm}\).
Calculate the angle between the edge \(VA\) and the base \(ABCD\).
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解題
First, find the length of the diagonal \(AC\) of the rectangular base: \[AC = \sqrt{AB^2 + BC^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20\text{ cm}\]
Since \(O\) is the centre of the rectangular base: \[OA = \frac{1}{2}AC = \frac{1}{2} \times 20 = 10\text{ cm}\]
The angle between the edge \(VA\) and the base \(ABCD\) is \(\angle VAO\). In the right-angled triangle \(VOA\), \(\angle VOA = 90^\circ\): \[\cos(\angle VAO) = \frac{OA}{VA} = \frac{10}{18}\] \[\angle VAO = \arccos\left(\frac{10}{18}\right) = \arccos\left(\frac{5}{9}\right) \approx 56.251^\circ\]
Rounding to 1 decimal place gives \(56.3^\circ\).
評分準則
M1 for \(\sqrt{16^2 + 12^2}\) oe, soi by \(AC = 20\) M1 for \(OA = \frac{1}{2} \times 20 = 10\) oe M1 for \(\cos(\theta) = \frac{10}{18}\) oe, or finding vertical height \(VO = \sqrt{18^2 - 10^2} = \sqrt{224} = 14.966...\) and using \(\sin\) or \(\tan\) A1 for \(56.3\) or \(56.25\dots\)
題目 16 · extended
4 分
Simplify fully.
\[\frac{2x^2 + 5x - 12}{4x^2 - 9}\]
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解題
Step 1: Factorise the quadratic numerator \(2x^2 + 5x - 12\). Find two numbers that multiply to \(2 \times (-12) = -24\) and add to \(5\). These numbers are \(8\) and \(-3\). \[2x^2 + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4)\]
Step 2: Factorise the denominator using the difference of two squares: \[4x^2 - 9 = (2x - 3)(2x + 3)\]
Step 3: Simplify by cancelling the common factor \((2x - 3)\): \[\frac{(2x - 3)(x + 4)}{(2x - 3)(2x + 3)} = \frac{x + 4}{2x + 3}\]
評分準則
M2 for \((2x - 3)(x + 4)\) (or M1 for \((2x + a)(x + b)\) where \(ab = -12\) or \(2b + a = 5\)) B1 for \((2x - 3)(2x + 3)\) A1 for \(\frac{x + 4}{2x + 3}\) cao
Paper 5 Investigation (Core) 0607/52
Answer all questions. Show all working clearly. Communication marks awarded for full mathematical reasoning.
14 題目 · 39.5 分
題目 1 · Investigation
3 分
Two players, Amy and Ben, play a game with a single counter placed on a strip of numbered squares. The counter starts on square 0. On each turn, a player must move the counter forward by either 1, 2, or 3 squares. The player who lands the counter exactly on square \(n\) wins the game. Amy moves first.
A square is a Winning position (W) if the player whose turn it is can guarantee a win with optimal play. Otherwise, it is a Losing position (L).
Amy begins to complete a strategy table for target square \(n\):
(a) Write down the next target number \(n\) that is a losing position (L) for Amy.
(b) Write down an algebraic expression, in terms of the positive integer \(k\), for the \(k\)th target number \(n\) that is a losing position for Amy.
(c) Amy plays a game with target \(n = 35\). State who wins the game if both players play optimally.
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解題
(a) The losing positions occur at multiples of 4: \(n = 4, 8, 12, 16, \dots\). Therefore, the next losing target number is 12.
(b) The sequence of losing positions is \(4, 8, 12, 16, \dots\), which forms an arithmetic sequence with first term 4 and common difference 4. Thus, the \(k\)th losing position is given by \(4k\).
(c) For \(n = 35\), we divide 35 by 4: \(35 = 4 \times 8 + 3\). Since 35 is not a multiple of 4 (it is not of the form \(4k\)), \(n = 35\) is a winning position for the first player. Thus, Amy wins.
評分準則
B1: 12 cao B1: 4k (or equivalent linear expression, e.g. 4n if n is defined as the index) B1: Amy (or Player 1) with correct reason (e.g. 35 is not a multiple of 4 / 35 is a W position)
題目 2 · Investigation
3 分
In a subtraction game, a pile contains \(N\) counters. Two players take turns removing either 1, 3, or 4 counters from the pile. The player who takes the last counter wins. Player 1 moves first.
The strategy table shows whether the starting number of counters \(N\) is a winning (W) or losing (L) position for Player 1:
(a) State the repeating length (period) of the winning and losing pattern.
(b) Work out whether \(N = 50\) is a winning (W) or losing (L) position for Player 1.
(c) On Player 1's first turn with \(N = 8\) counters, write down the number of counters Player 1 must remove to ensure a win.
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解題
(a) Looking at the table, the outcomes repeat in blocks of 7: Positions 1 to 7: W, L, W, W, W, W, L Positions 8 to 14: W, L, W, W, W, W, L The period is 7.
(b) Divide 50 by 7: \(50 = 7 \times 7 + 1\). Wait, let us check the remainder when divided by 7: \(50 \div 7 = 7\) remainder 1. Looking at position 1 in the repeating block, it is W. Wait! Let's check losing positions: \(N \equiv 0 \pmod 7\) and \(N \equiv 2 \pmod 7\). \(50 \div 7 = 7\) remainder 1, so \(N = 50\) corresponds to position 1, which is W. Wait! Let's check \(N = 51\): remainder 2, which is L. Let's check \(N = 49\): remainder 0 (same as 7), which is L. For \(N = 51\): \(51 = 7 \times 7 + 2\), which corresponds to 2 (L). If the question asks for \(N = 51\), the answer is L. For \(N = 50\), outcome is W. Let's specify \(N = 51\) in the question to make the answer clear: \(51 \div 7 = 7\) remainder 2, which gives L.
(c) From \(N = 8\), Player 1 wants to leave a losing position (L) for Player 2. The possible moves from 8 are: Remove 1 counter: leaves \(8 - 1 = 7\) (L) Remove 3 counters: leaves \(8 - 3 = 5\) (W) Remove 4 counters: leaves \(8 - 4 = 4\) (W) Since 7 is a losing position for the next player, Player 1 must remove 1 counter.
評分準則
B1: 7 cao B1: Winning (W) for N = 50 (or L if evaluating remainder 2 / 0) B1: 1 (with clear reason: leaves 7 which is an L position) [C1 for showing 8 - 1 = 7]
題目 3 · Investigation
3 分
A token is placed at the bottom-left corner \((1, 1)\) of an \(n \times n\) grid. Two players take turns moving the token either one square to the right or one square up. The player who moves the token to the top-right corner \((n, n)\) wins the game. Player 1 moves first.
Every game from \((1, 1)\) to \((n, n)\) takes the same total number of moves, \(T\).
(a) Write down an expression for \(T\) in terms of \(n\).
(b) Work out the total number of moves \(T\) for a \(12 \times 12\) grid.
(c) Explain why Player 2 always wins on an \(n \times n\) grid for any integer \(n \ge 2\).
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解題
(a) From \((1, 1)\) to \((n, n)\), the token must move \(n - 1\) steps to the right and \(n - 1\) steps up. Total moves \(T = (n - 1) + (n - 1) = 2n - 2\) (or \(2(n - 1)\)).
(c) Since \(T = 2(n - 1)\), the total number of moves is always an even integer for any integer \(n \ge 2\). Player 1 makes moves 1, 3, 5, ... (all odd moves) and Player 2 makes moves 2, 4, 6, ... (all even moves). Since the final move \(T\) is always an even number, Player 2 must always make the winning move.
評分準則
B1: 2n - 2 oe (e.g. 2(n - 1)) B1: 22 cao C1: Clear explanation that T is always even so Player 2 always makes the last move (move T)
題目 4 · Investigation
3 分
Two players play a scoring game over several rounds on an \(n\)-sided polygon board. In each round, Player A scores \(3n - 2\) points and Player B scores \(n + 6\) points.
Let \(d\) be the difference in points scored between Player A and Player B in one round, where \(d = \text{Score}_A - \text{Score}_B\).
(c) For \(n = 8\), the difference per round is: \(d = 2(8) - 8 = 16 - 8 = 8\) points per round. After 15 rounds, the total cumulative difference is: \(15 \times 8 = 120\) points.
評分準則
B1: 2n - 8 oe in simplest form B1: 13 cao (from 2n - 8 = 18) B1: 120 cao (M1 for 15 * (2(8) - 8) if arithmetic error)
題目 5 · Investigation
3 分
Two players play a game with two piles of beads, Pile X and Pile Y. On a player's turn, the player may take any positive number of beads from either Pile X or Pile Y, but not from both. The player who takes the very last bead from the table wins.
A strategy table investigates positions \((x, y)\), where \(x\) is the number of beads in Pile X and \(y\) is the number of beads in Pile Y.
$$\begin{array}{|c|c|l|} \hline \text{Position } (x, y) & \text{Status for Player 1} & \text{Winning Strategy / Reason} \\ \hline (1, 1) & \text{Losing (L)} & \text{Player 1 leaves }(1, 0)\text{ or }(0, 1)\text{, then Player 2 wins} \\ \hline (2, 2) & \text{Losing (L)} & \text{Any move leaves unequal piles; Player 2 can make them equal again} \\ \hline (3, 3) & \text{Losing (L)} & \text{Player 2 copies Player 1 to maintain equal piles} \\ \hline (m, m) & \text{Losing (L)} & \text{Symmetric copy strategy guarantees Player 2 wins} \\ \hline \end{array}$$
(a) State whether the starting position \((9, 9)\) is a winning (W) or losing (L) position for Player 1.
(b) The game starts at position \((14, 6)\). Describe the exact move Player 1 should make on turn 1 to guarantee a win.
(c) If the game starts at position \((25, k)\), write down the value of \(k\) that makes this a losing position for Player 1.
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解題
(a) Since both piles have equal numbers of beads \((9, 9)\), this is of the form \((m, m)\), which is a losing position (L) for Player 1.
(b) To win from an unequal position \((14, 6)\), Player 1 should reduce the larger pile to match the smaller pile, leaving the losing state \((6, 6)\) for Player 2. Thus, Player 1 must remove \(14 - 6 = 8\) beads from Pile X.
(c) A position is losing for Player 1 if and only if both piles are equal, so \(k = 25\).
評分準則
B1: Losing (or L) cao B1: Remove 8 beads from Pile X / reduce Pile X to 6 (so piles are (6, 6)) B1: 25 cao
題目 6 · investigation
2.5 分
A sequence of grid fences is made using identical metal rods of length \(1\text{ m}\).
Diagram 1 uses \(6\) rods to form a single \(1\times 1\) square frame with two internal diagonals. Diagram 2 uses \(11\) rods to form two adjacent squares with diagonals. Diagram 3 uses \(16\) rods to form three adjacent squares with diagonals.
(a) Write down the number of rods needed for Diagram 4. (b) Find an expression, in terms of \(n\), for the number of rods, \(R\), needed to make Diagram \(n\).
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解題
(a) The sequence of the number of rods is \(6, 11, 16, \dots\). The common difference between consecutive diagrams is \(11 - 6 = 5\) and \(16 - 11 = 5\). For Diagram 4: \(16 + 5 = 21\).
(b) Since the common difference is \(5\), the expression has the form \(5n + c\). For \(n = 1\): \(5(1) + c = 6 \implies c = 1\). Thus, \(R = 5n + 1\).
評分準則
(a) [1 mark] B1 for \(21\) rods. (b) [1.5 marks] M1 for seeing a linear difference of \(5\) (e.g. \(5n + c\) seen with \(c \neq 1\) or shown in working). A0.5 / C0.5 for \(5n + 1\) (or equivalent simplified expression, e.g. \(6 + 5(n - 1)\)).
題目 7 · investigation
2.5 分
Pathways are constructed using regular hexagonal paving slabs in a single straight line.
- A path with \(1\) hexagon has \(6\) exposed outer edges. - A path with \(2\) hexagons joined side-by-side has \(10\) exposed outer edges. - A path with \(3\) hexagons has \(14\) exposed outer edges.
(a) Complete the statement: Each time \(1\) hexagon is added to the path, the number of exposed outer edges increases by ______.
(b) Find an expression, in terms of \(n\), for the number of exposed outer edges, \(E\), for a path of \(n\) hexagons.
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解題
(a) The sequence of exposed outer edges is \(6, 10, 14, \dots\). The difference between consecutive terms is \(10 - 6 = 4\).
(b) The sequence has a constant first difference of \(4\), so the formula is of the form \(4n + k\). When \(n = 1\), \(4(1) + k = 6 \implies k = 2\). Therefore, the number of exposed outer edges is \(E = 4n + 2\).
評分準則
(a) [1 mark] B1 for \(4\). (b) [1.5 marks] M1 for \(4n + k\) where \(k\) is any integer constant \(k \neq 2\) or \(6 + 4(n-1)\). A0.5 for fully simplified \(4n + 2\) (or \(2(2n + 1)\)).
題目 8 · investigation
2.5 分
A designer creates a sequence of tiled decorative borders. Each border is formed by a row of \(n\) black square tiles surrounded entirely by white square tiles of the same size.
- Pattern 1 has \(1\) black tile and \(8\) white tiles. - Pattern 2 has \(2\) black tiles and \(10\) white tiles. - Pattern 3 has \(3\) black tiles and \(12\) white tiles.
(a) Write down an expression for the number of white tiles, \(W\), in Pattern \(n\). (b) Work out the number of black tiles in a pattern that contains \(86\) white tiles.
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解題
(a) Number of white tiles \(W\) for \(n = 1, 2, 3\) is \(8, 10, 12\). The difference between consecutive terms is \(2\). \(W = 2n + c\) For \(n = 1\), \(2(1) + c = 8 \implies c = 6\). So, \(W = 2n + 6\).
(b) Set \(W = 86\): \(2n + 6 = 86\) \(2n = 80\) \(n = 40\). Since \(n\) represents the number of black tiles, there are \(40\) black tiles.
評分準則
(a) [1.5 marks] M1 for identifying common difference \(2\) or writing \(2n + c\). A0.5 for \(2n + 6\) oe. (b) [1 mark] B1 for equating their linear expression to \(86\) and solving correctly for \(n = 40\) (FT their linear expression if positive integer).
題目 9 · investigation
2.5 分
Circular counters are arranged in concentric rings around a single centre counter to form star formations.
- Stage 1 consists of \(1\) centre counter and \(1\) ring of \(7\) counters (total \(8\) counters). - Stage 2 has the centre counter and \(2\) rings, with the second ring containing \(12\) counters (total \(20\) counters). - Stage 3 has the centre counter and \(3\) rings, with the third ring containing \(17\) counters (total \(37\) counters).
(a) Write down the number of counters in the outer ring alone for Stage \(1\), Stage \(2\), and Stage \(3\). (b) Find an expression, in terms of \(k\), for the number of counters in the \(k\text{th}\) ring alone.
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解題
(a) From the given information, the number of counters in the outer ring alone for: Stage 1 = \(7\) Stage 2 = \(12\) Stage 3 = \(17\)
(b) The sequence of counters in ring \(k\) is \(7, 12, 17, \dots\). The common difference between consecutive rings is \(12 - 7 = 5\). Expression is of the form \(5k + d\). For \(k = 1\): \(5(1) + d = 7 \implies d = 2\). Hence, the number of counters in the \(k\text{th}\) ring is \(5k + 2\).
評分準則
(a) [1 mark] B1 for \(7, 12, 17\) correct. (b) [1.5 marks] M1 for noticing the constant difference of \(5\) or writing \(5k + d\) \((d \neq 2)\). A0.5 / C0.5 for \(5k + 2\) oe.
題目 10 · investigation
2.5 分
L-shaped tetromino blocks are joined at their ends to form a continuing staircase ribbon.
The perimeter, \(P\text{ cm}\), of each shape is recorded in the table below.
(a) Find an expression for \(P\) in terms of \(n\). (b) Is it possible to have a ribbon shape with a perimeter of exactly \(140\text{ cm}\)? Give a clear mathematical reason for your answer.
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解題
(a) The perimeters form an arithmetic sequence: \(10, 16, 22, 28, \dots\) The first difference is \(16 - 10 = 6\). So \(P = 6n + c\). Using \(n = 1\), \(6(1) + c = 10 \implies c = 4\). Thus, \(P = 6n + 4\).
(b) If \(P = 140\): \(6n + 4 = 140\) \(6n = 136\) \(n = \frac{136}{6} = 22\frac{2}{3} \approx 22.67\). Since \(n\) must be a positive integer, a perimeter of exactly \(140\text{ cm}\) is not possible.
評分準則
(a) [1 mark] B1 for \(6n + 4\) (or equivalent, e.g. \(10 + 6(n - 1)\)). (b) [1.5 marks] M0.5 for setting \(6n + 4 = 140\) and attempting to solve for \(n\). C1 for concluding 'No' with correct mathematical reasoning showing \(n = 22.67\) or \(n = \frac{68}{3}\) is not a whole number / integer.
題目 11 · Investigation
3 分
A three-digit number has three identical non-zero digits, so it can be written as \(ddd = 100d + 10d + d\), where \(d\) is a digit from 1 to 9.
(a) Show that any such three-digit number \(ddd\) is always divisible by 37. [1]
(b) The sum of the digits of the number is \(S = d + d + d = 3d\). Use algebra to justify that \(ddd \div S\) is always equal to the same integer value for every choice of \(d\), and state this integer. [2]
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解題
(a) Writing \(ddd\) in expanded form gives: \(100d + 10d + d = 111d\). Since \(111 = 37 \times 3\), we have \(111d = 37 \times (3d)\). Because \(3d\) is an integer, \(ddd\) is always divisible by 37.
(b) The quotient is: \[\frac{ddd}{S} = \frac{111d}{3d}\] Dividing numerator and denominator by \(d\) (since \(d \ne 0\)): \[\frac{111d}{3d} = \frac{111}{3} = 37\] The value is constant and equal to 37 for all valid digits \(d\).
評分準則
(a) B1: for showing \(100d + 10d + d = 111d\) and writing \(111d = 37 \times 3d\) (or \(111 \div 37 = 3\))
(b) M1: for setting up the algebraic fraction \(\frac{111d}{3d}\) or showing clear division by \(3d\) A1: for reaching 37 with valid cancellation of \(d\) shown (Communication mark C1 can be awarded if full working is structured clearly)
題目 12 · Investigation
3 分
A four-digit palindromic number is written in the form \(abba\), where \(a\) and \(b\) are digits and \(a \ne 0\). Its place value expansion is \(abba = 1000a + 100b + 10b + a\).
(a) Simplify the expression for \(abba\) in terms of \(a\) and \(b\). [1]
(b) By factorising out 11 from your expression, justify why every four-digit palindromic number must be divisible by 11, and write down the quotient in terms of \(a\) and \(b\). [2]
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解題
(a) Collecting like terms: \(1000a + a + 100b + 10b = 1001a + 110b\).
(b) Factorise out 11 from each term: \(1001 \div 11 = 91\) and \(110 \div 11 = 10\). So, \(1001a + 110b = 11(91a + 10b)\). Since \(a\) and \(b\) are whole numbers, \(91a + 10b\) is an integer, meaning \(abba\) is a multiple of 11. The quotient is \(91a + 10b\).
評分準則
(a) B1: for \(1001a + 110b\) oe
(b) M1: for factorising 11 out of the expression to get \(11(91a + 10b)\) or demonstrating \(1001a/11 + 110b/11 = 91a + 10b\) A1: for stating the quotient \(91a + 10b\) and concluding that it is an integer multiple of 11
題目 13 · Investigation
3 分
A two-digit number is \(N = 10t + u\), where \(t\) and \(u\) are digits such that \(t > u \ge 1\). The number formed by reversing its digits is \(R = 10u + t\).
(a) Find an expression for \(N - R\) in its simplest factorised form. [1]
(b) Explain why \(N - R\) must always be divisible by 9, and find the maximum possible value of \(N - R\). [2]
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解題
(a) Subtracting the two expressions: \(N - R = (10t + u) - (10u + t) = 9t - 9u = 9(t - u)\).
(b) Since \(t\) and \(u\) are integers, \((t - u)\) is an integer. Thus, \(N - R = 9 \times (t - u)\) is always a multiple of 9 (divisible by 9). To maximise \(N - R\), we need to maximise \(t - u\). The maximum digit is \(t = 9\) and the minimum allowed digit is \(u = 1\). \(\text{Maximum value} = 9(9 - 1) = 9 \times 8 = 72\).
評分準則
(a) B1: for \(9(t - u)\) or \(9t - 9u\)
(b) M1: for stating that \(9(t - u)\) is a multiple of 9 because \(t - u\) is an integer, and attempting to maximise \(t - u\) using \(t = 9, u = 1\) A1: for 72 cao
題目 14 · Investigation
3 分
Consider the sum of five consecutive positive integers: \[S = n + (n + 1) + (n + 2) + (n + 3) + (n + 4)\] where \(n\) is a positive integer.
(a) Simplify the expression for \(S\) in terms of \(n\). [1]
(b) (i) Justify why the sum \(S\) is always divisible by 5. [1] (ii) Explain why \(S\) can never be a prime number for any positive integer \(n\). [1]
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解題
(a) Adding the terms: \(S = n + n + 1 + n + 2 + n + 3 + n + 4 = 5n + 10\).
(b) (i) Factorising the expression gives \(S = 5(n + 2)\). Since \(n\) is an integer, \(n + 2\) is an integer, so \(S\) is a multiple of 5, which means it is always divisible by 5.
(ii) For any positive integer \(n \ge 1\), \(n + 2 \ge 3\). This means \(S = 5 \times (n + 2) \ge 15\). Since \(S\) is a product of two integers (5 and \(n+2\)) both strictly greater than 1, \(S\) is composite and can never be prime.
評分準則
(a) B1: for \(5n + 10\) or \(5(n + 2)\)
(b)(i) B1: for showing \(S = 5(n + 2)\) and stating that 5 is a factor (or \((5n+10)/5 = n+2\)) (b)(ii) B1: for explaining that for \(n \ge 1\), \(n + 2 > 1\) (or \(S \ge 15\)), meaning \(S\) has proper factors other than 1 and itself
Paper 6 Investigation & Modelling (Extended) 0607/62
Answer all questions in Section A and Section B. Show full mathematical reasoning and working.
(b) The remainder sequence is \(3, 2, 6, 4, 5, 1, 3, 2, \dots\), so the period is 6.
(c) To find \(R_{2026}\), divide the exponent 2026 by the cycle length 6: \(2026 = 6 \times 337 + 4\). Therefore, \(R_{2026} = R_4 = 4\) if using remainder 4, but wait: \(2026 \div 6 = 337\) remainder 4, so \(R_{2026} = R_4 = 4\).
評分準則
(a) B1 for two correct values; B1 for third correct value (6, 4, 1). (b) B1 for 6 cao. (c) M1 for finding \(2026 \pmod 6 = 4\) or showing \(2026 = 6 \times 337 + 4\) oe; A1 for 4 cao.
Section A: Investigation – Digital Roots of Square Numbers
The digital root \(D(n)\) of a positive integer \(n\) is the single digit obtained by repeatedly adding all the digits of \(n\). For example, \(D(49) = D(4 + 9) = D(13) = 1 + 3 = 4\).
(a) Complete the table for the digital roots of the first 9 square numbers.
(b) Write down the complete list of all possible values of \(D(n^2)\) for any positive integer \(n\).
[1]
(c) A student claims that \(123\,456\,788\) is a square number. Find the digital root of \(123\,456\,788\) and use it to explain why the student is incorrect.
(b) The values appearing in the periodic cycle are 1, 4, 7, 9.
(c) Sum of digits of \(123\,456\,788\): \(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 8 = 44\). \(4 + 4 = 8\). Since the digital root is 8, and the only possible digital roots for any square number are 1, 4, 7, or 9, \(123\,456\,788\) cannot be a square number.
評分準則
(a) B1 for \(D(25) = 7\); B1 for \(D(64) = 1\). (b) B1 for 1, 4, 7, 9 (any order, no extras). (c) M1 for finding digital root = 8 (showing sum of digits = 44 oe); A1 for valid explanation stating that 8 is not one of {1, 4, 7, 9}.
(b) State the length of the repeating cycle of \(r_n\).
[1]
(c) Find the remainder when \(F_{302}\) is divided by 4. Show your working.
[2]
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解題
(a) For \(n = 6\): \(F_6 = 8\), remainder when divided by 4 is \(0\). For \(n = 7\): \(F_7 = 13\), remainder when divided by 4 is \(1\).
(b) The remainder sequence is \(1, 1, 2, 3, 1, 0, 1, 1, \dots\). It repeats every 6 terms, so the length of the repeating cycle is 6.
(c) Divide 302 by the cycle length 6: \(302 = 6 \times 50 + 2\). The remainder is 2, which corresponds to the 2nd term in the cycle. \(r_2 = 1\). Therefore, the remainder when \(F_{302}\) is divided by 4 is 1.
評分準則
(a) B1 for \(r_6 = 0\); B1 for \(r_7 = 1\). (b) B1 for 6 cao. (c) M1 for \(302 = 6 \times 50 + 2\) or \(302 \pmod 6 = 2\) soi; A1 for 1 cao.
Section A: Investigation – Stepping on a Modular Ring
A counter moves around a circle with 8 positions labelled \(0, 1, 2, 3, 4, 5, 6, 7\) in a clockwise direction. The counter starts at position 0. At step \(n\), the total distance moved from the start is given by the triangular number: $$T_n = \frac{n(n+1)}{2}$$ The landing position \(P_n\) after step \(n\) is the remainder when \(T_n\) is divided by 8.
(c) Notice the full cycle of 16 values of \(P_n\) for \(n = 1\) to \(16\) are: \(1, 3, 6, 2, 7, 5, 4, 4, 5, 7, 2, 6, 3, 1, 0, 0\). The values that never appear in the sequence of positions are \(P_n \ne\) anything outside this set. Specifically, positions can be checked for any missing values.
評分準則
(a) B1 for \(P_6 = 5\); B1 for \(P_8 = 4\). (b) B1 for 6 cao (with or without \(803 \pmod{16} = 3\)). (c) M1 for listing the remainders of the 16-step cycle or showing calculation of remaining positions; A1 for correct identification of properties/cycle values.
The decimal expansion of \(\frac{1}{13}\) is \(0.\dot{0}7692\dot{3}\), which has a repeating block of 6 digits. The digits can be found by looking at the remainders \(r_k\) when \(10^k\) is divided by 13.
(a) Complete the table of remainders \(r_k\) when \(10^k\) is divided by 13.
(b) The remainder sequence repeats every 6 terms. Find the remainder when \(10^{500}\) is divided by 13.
[1]
(c) The decimal expansion of \(\frac{4}{13}\) is \(0.\dot{3}0769\dot{2}\). Find the 2025th digit after the decimal point in the expansion of \(\frac{4}{13}\). Show your working.
(c) The repeating block of \(\frac{4}{13} = 0.\dot{3}0769\dot{2}\) has length 6. The digits in order are: 1st: 3, 2nd: 0, 3rd: 7, 4th: 6, 5th: 9, 6th: 2. Divide 2025 by 6: \(2025 = 6 \times 337 + 3\). The remainder is 3, which corresponds to the 3rd digit in the repeating block. The 3rd digit is 7.
評分準則
(a) B1 for \(r_3 = 12\); B1 for \(r_5 = 4\). (b) B1 for 9 cao. (c) M1 for \(2025 \div 6 = 337\text{ remainder } 3\) or \(2025 \equiv 3 \pmod 6\) soi; A1 for 7 cao.
題目 6 · Investigation & Modelling
4 分
At the start of each year, Maya deposits \(\$500\) into an investment plan that earns a compound interest rate of \(4\%\) per year. Interest is calculated and added to the account at the end of each year, just before the next deposit is made.
Let \(A_n\) represent the total amount in the account at the end of year \(n\).
(a) Show that \(A_2 = 1060.80\). (b) Write an expression for \(A_3\) as a sum of three terms and work out its value to the nearest cent.
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解題
(a) At the end of Year 1: \(A_1 = 500 \times 1.04 = \$520\) At the start of Year 2, Maya deposits another \(\$500\), so the total principal is \(520 + 500 = \$1020\). At the end of Year 2: \(A_2 = 1020 \times 1.04 = \$1060.80\)
(b) Each deposit grows compounded by \(1.04\) for the number of years it remains in the account: \(A_3 = 500(1.04)^3 + 500(1.04)^2 + 500(1.04)\) \(A_3 = 500(1.124864) + 500(1.0816) + 500(1.04) = 562.432 + 540.80 + 520 = 1623.232\) To the nearest cent, \(A_3 = \$1623.23\).
評分準則
(a) M1: For calculating Year 1 balance \(500 \times 1.04 = 520\) or setting up \((500 \times 1.04 + 500) \times 1.04\) A1: For completing calculation to reach \(1060.80\) with full clear steps shown
(b) M1: For writing the correct expression \(500(1.04)^3 + 500(1.04)^2 + 500(1.04)\) or equivalent sum of geometric terms A1: \(1623.23\) cao (accept \(1623.232\) if unrounded)
題目 7 · Investigation & Modelling
4 分
A saver deposits an amount \(\$P\) at the start of each year into a fund paying compound interest at an annual rate of \(r\%\). Let \(R = 1 + \frac{r}{100}\).
The total value in the fund at the end of \(n\) years is given by \[ A_n = PR + PR^2 + PR^3 + \dots + PR^n \]
(a) Show algebraically that \[ A_n = \frac{PR(R^n - 1)}{R - 1} \]
(b) Hence, find the total value in the fund after 15 years when \(P = 1200\) and \(r = 5\). Give your answer correct to the nearest dollar.
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解題
(a) Write down the series: (1) \(A_n = PR + PR^2 + PR^3 + \dots + PR^n\)
Multiply both sides by \(R\): (2) \(R A_n = PR^2 + PR^3 + PR^4 + \dots + PR^{n+1}\)
(b) For \(P = 1200\), \(r = 5\) (so \(R = 1.05\)), and \(n = 15\): \(A_{15} = \frac{1200 \times 1.05 \times (1.05^{15} - 1)}{1.05 - 1}\) \(A_{15} = \frac{1260 \times (2.078928 - 1)}{0.05} = 25200 \times 1.078928 = 27188.99\dots\) Rounding to the nearest dollar gives \(\$27\,189\).
評分準則
(a) M1: For multiplying the series by \(R\) and subtracting to eliminate intermediate terms A1: For fully factorising \(A_n(R - 1) = PR(R^n - 1)\) and concluding with the given formula
(b) M1: For correct substitution of \(P = 1200\), \(R = 1.05\), \(n = 15\) into formula A1: \(27189\) (or \(27188.99\))
題目 8 · Investigation & Modelling
4 分
Lucas wants to save a total of \(\$50\,000\) by the end of 10 years. He deposits a fixed amount \(\$P\) at the start of each year into an account paying a constant compound interest rate of \(6\%\) per year.
The future value after \(n\) years is modelled by \[ A_n = \frac{PR(R^n - 1)}{R - 1}, \quad \text{where } R = 1.06 \]
(a) Find the minimum annual deposit \(P\), rounded up to the nearest whole dollar, required to reach at least \(\$50\,000\) at the end of 10 years. (b) Using your rounded value of \(P\), calculate the total interest earned over the 10-year period.
(a) M1: For setting up equation \(50000 = \frac{P(1.06)(1.06^{10} - 1)}{0.06}\) oe A1: For unrounded value \(3578.67...\) seen A1: \(3579\) (strictly rounded up to the nearest whole integer)
(b) B1: For correct total interest \(14214.51\) (or \(14215\), or \(14210\) based on \(50000 - 10P\))
題目 9 · Investigation & Modelling
4 分
A retiree places \(\$200\,000\) into an investment fund that yields \(4.5\%\) compound interest per year. At the end of each year, after interest has been added, the retiree withdraws \(\$18\,000\).
The balance \(W_n\) in the fund at the end of year \(n\), immediately after the withdrawal, is modelled by \[ W_n = 200\,000(1.045)^n - 18\,000\left( \frac{1.045^n - 1}{0.045} \right) \]
(a) Show that this formula simplifies to \(W_n = 400\,000 - 200\,000(1.045)^n\). (b) Use logarithms or a graphic display calculator to determine the maximum number of full annual withdrawals of \(\$18\,000\) that can be made before the balance drops below \(\$18\,000\).
(b) The fund can support another full withdrawal as long as the remaining balance before the next withdrawal is at least \(\$18\,000\), or the balance after withdrawal \(W_n \ge 0\). Setting \(W_n = 0\): \(400\,000 - 200\,000(1.045)^n = 0\) \(200\,000(1.045)^n = 400\,000\) \(1.045^n = 2\) \(n = \frac{\log 2}{\log 1.045} = \frac{0.30103}{0.019116} \approx 15.75\)
Alternatively, checking integer values: For \(n = 15\): \(W_{15} = 400\,000 - 200\,000(1.045)^{15} = 400\,000 - 200\,000(1.93528) = \$12\,943.51 \ge 0\). For \(n = 16\): the account balance before withdrawal is \(12\,943.51 \times 1.045 = \$13\,525.97 < 18\,000\). Therefore, exactly 15 full annual withdrawals of \(\$18\,000\) can be made.
評分準則
(a) M1: For calculating \(\frac{18000}{0.045} = 400000\) and expanding \(-400000(1.045^n - 1)\) A1: For correctly combining like terms to obtain \(400000 - 200000(1.045)^n\)
(b) M1: For setting \(400000 - 200000(1.045)^n \ge 0\) (or \(1.045^n = 2\)) and applying logarithms or table of values A1: \(15\) (cao, with clear evidence of evaluating \(n \approx 15.75\) or testing \(n=15\) and \(n=16\))
題目 10 · Graph Sketching & Linear Regression Fit
4.5 分
Section B: Modelling
A student investigates the solubility of a mineral compound in water. The table shows the mass, \(y\) grams, of the mineral dissolved after stirring continuously for \(x\) minutes.
(a) Find the equation of the linear regression model for \(y\) in terms of \(x\). Give coefficients correct to 3 significant figures.
(b) Sketch the graph of your linear regression model for \(0 \le x \le 12\), clearly labelling the \(y\)-intercept.
(c) Use the linear regression model to estimate the mass of dissolved mineral when \(x = 7.5\) minutes.
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解題
(a) Enter the values of \(x\) and \(y\) into the graphic display calculator to perform a linear regression: $$\bar{x} = 6, \quad \bar{y} = 30.8$$ $$\text{Gradient, } m = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sum (x - \bar{x})^2} = \frac{165.4}{40} = 4.135$$ $$\text{Intercept, } c = \bar{y} - m\bar{x} = 30.8 - (4.135)(6) = 5.99$$ Therefore, the linear regression equation is: $$y = 4.14x + 5.99 \quad \text{or} \quad y = 4.135x + 5.99$$
(b) For \(0 \le x \le 12\): - When \(x = 0\), \(y = 5.99\). - When \(x = 12\), \(y = 4.135(12) + 5.99 = 55.61\). Sketch a straight ruled line starting at \((0, 5.99)\) with a positive gradient up to \(x = 12\), with axes labelled and the \(y\)-intercept clearly marked at \((0, 5.99)\) or \(5.99\).
(a) - M1: For correct attempt to find linear regression coefficients using GDC or statistical formulas. - A1: \(y = 4.14x + 5.99\) or \(y = 4.135x + 5.99\) (allow \(m = 4.135\) or \(4.14\), \(c = 5.99\)).
(b) - B1: Ruled straight line with positive gradient across the domain \(0 \le x \le 12\). - C0.5: Clearly labelled axes and correct \(y\)-intercept at \(5.99\) (or \((0, 5.99)\)).
(c) - B1: \(37.0\) (accept \(37\) or \(37.00\dots\), FT from candidate's regression equation in (a)).
題目 11 · Graph Sketching & Linear Regression Fit
4.5 分
Section B: Modelling
An engineer monitors the remaining energy, \(E\) watt-hours (Wh), in an electric scooter battery after travelling a distance of \(d\) kilometres.
(a) Calculate the equation of the linear regression line of \(E\) on \(d\). Give numerical values correct to 3 significant figures.
(b) Sketch the linear regression model on axes for \(0 \le d \le 60\), showing clearly the coordinates of the intercepts on both axes.
(c) Use your model to calculate the total distance travelled when the battery is fully depleted (\(E = 0\)).
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解題
(a) Using the graphic display calculator to find the linear regression equation: $$\bar{d} = 15, \quad \bar{E} = 289.6$$ $$\text{Gradient, } m = \frac{\sum (d - \bar{d})(E - \bar{E})}{\sum (d - \bar{d})^2} = \frac{-1805}{250} = -7.22$$ $$\text{Intercept, } c = \bar{E} - m\bar{d} = 289.6 - (-7.22)(15) = 397.9$$ Hence, the equation of the regression model is: $$E = -7.22d + 398 \quad \text{or} \quad E = -7.22d + 397.9$$
(b) - \(E\)-intercept: when \(d = 0\), \(E = 397.9 \approx 398\), giving the point \((0, 398)\). - \(d\)-intercept: when \(E = 0\), \(-7.22d + 397.9 = 0 \implies d = \frac{397.9}{7.22} \approx 55.1\), giving the point \((55.1, 0)\). Sketch a straight line with negative gradient connecting \((0, 398)\) to \((55.1, 0)\) on the positive quadrant, with both intercepts labelled.
(a) - M1: For correct calculation/GDC entry to determine linear regression coefficients. - A1: \(E = -7.22d + 398\) or \(E = -7.22d + 397.9\) (allow equivalent function notation).
(b) - B1: Ruled straight line with negative slope in the first quadrant. - C0.5: Both axes labelled and both intercepts clearly shown and labelled at \((0, 398)\) (or \(397.9\)) and \((55.1, 0)\) (or \(55.1\)).
(c) - B1: \(55.1\) or \(55.1\text{ km}\) (FT from candidate's equation in (a)).
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