An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V3) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
卷一 (Core Non-Calculator)
Answer all questions. Calculators must not be used. Show all necessary working clearly.
17 題目 · 45 分
題目 1 · short_answer
2 分
Work out \(\frac{5}{6} - \frac{3}{8}\). Give your answer as a fraction in its simplest form.
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解題
Find a common denominator for 6 and 8, which is 24. Write each fraction with denominator 24: \(\frac{5}{6} = \frac{20}{24}\) and \(\frac{3}{8} = \frac{9}{24}\). Subtract the numerators: \(\frac{20}{24} - \frac{9}{24} = \frac{11}{24}\). The fraction is already in simplest form.
評分準則
M1 for finding a common denominator and writing at least one correct equivalent fraction, e.g. \(\frac{20}{24} - \frac{9}{24}\) or \(\frac{40}{48} - \frac{18}{48}\). A1 for \(\frac{11}{24}\) cao.
題目 2 · short_answer
2 分
Work out \(18 - 3 \times (4 - 7)^2\).
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解題
First evaluate inside the brackets: \(4 - 7 = -3\). Next evaluate the power: \((-3)^2 = 9\). Perform multiplication before subtraction: \(3 \times 9 = 27\). Finally subtract: \(18 - 27 = -9\).
評分準則
M1 for correct evaluation of bracket and square, e.g. seeing \((-3)^2 = 9\) or \(18 - 3 \times 9\) or \(18 - 27\). A1 for \(-9\) cao.
題目 3 · short_answer
2 分
Work out \(0.3 \times (1.4 - 0.65)\).
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解題
First subtract inside the brackets: \(1.40 - 0.65 = 0.75\). Then multiply: \(0.3 \times 0.75 = 0.225\).
評分準則
M1 for \(0.75\) seen or for a complete method to multiply \(3 \times 75 = 225\) with correct decimal placement attempted. A1 for \(0.225\) or equivalent fraction \(\frac{9}{40}\).
題目 4 · short_answer
2 分
Work out \(35\%\) of \(\$480\).
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解題
Calculate \(10\%\) of \(480 = 48\). Then \(30\% = 3 \times 48 = 144\), and \(5\% = \frac{48}{2} = 24\). Adding these together gives \(35\% = 144 + 24 = 168\).
評分準則
M1 for a correct method to find \(35\%\), e.g. \(\frac{35}{100} \times 480\) or \(3 \times 48 + 24\) or \(0.35 \times 480\). A1 for \(168\) (accept \(\$168\)).
題目 5 · short_answer
2 分
Find the lowest common multiple (LCM) of 24 and 36.
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解題
Write the prime factorisations: \(24 = 2^3 \times 3\) and \(36 = 2^2 \times 3^2\). The LCM is the product of the highest power of each prime factor: \(2^3 \times 3^2 = 8 \times 9 = 72\).
評分準則
M1 for prime factor decomposition of both numbers (e.g. \(24 = 2^3 \times 3\) and \(36 = 2^2 \times 3^2\)) or for listing at least 3 multiples of each number correctly. A1 for \(72\) cao.
題目 6 · short_answer
3 分
NOT TO SCALE
In the diagram, the line \(AB\) is parallel to the line \(CD\). The line \(EF\) intersects \(AB\) at \(E\) and \(CD\) at \(F\). \(G\) is a point on the line \(CD\) such that \(EF = EG\). Angle \(AEF = 112^\circ\).
(a) Find the size of angle \(EFD\). Give a geometric reason for your answer. [2] (b) Work out the size of angle \(FEG\). [1]
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解題
(a) Since \(AB\) is parallel to \(CD\), angle \(EFD\) and angle \(AEF\) are alternate angles. Therefore, angle \(EFD = 112^\circ\). Reason: Alternate angles are equal.
(b) Points \(C\), \(F\), and \(G\) lie on a straight line, so angle \(EFG = 180^\circ - 112^\circ = 68^\circ\). Since triangle \(EFG\) is isosceles with \(EF = EG\), the base angles are equal: angle \(EGF = \text{angle } EFG = 68^\circ\).
The sum of angles in triangle \(EFG\) is \(180^\circ\): angle \(FEG = 180^\circ - (68^\circ + 68^\circ) = 180^\circ - 136^\circ = 44^\circ\).
評分準則
(a) B1 for 112 [cao] B1 for reason: alternate (angles) [dep on 112 or correct identification of angle pair] (b) B1 for 44 [cao]
題目 7 · short_answer
3 分
NOT TO SCALE
\(ABCDEF\) is a regular hexagon. The line segment \(AB\) is extended to the point \(X\).
(a) Work out the size of one interior angle of the regular hexagon. [2] (b) Find the size of angle \(CBX\). [1]
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解題
(a) The sum of the interior angles of an \(n\)-sided polygon is \((n - 2) \times 180^\circ\). For a hexagon (\(n = 6\)): \(\text{Sum} = (6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ\). Since the hexagon is regular, each interior angle is: \(\frac{720^\circ}{6} = 120^\circ\). (Alternatively: \(\text{Exterior angle} = \frac{360^\circ}{6} = 60^\circ\), so \(\text{Interior angle} = 180^\circ - 60^\circ = 120^\circ\).)
(b) Angles on a straight line add up to \(180^\circ\): \(\text{Angle } CBX = 180^\circ - 120^\circ = 60^\circ\).
評分準則
(a) M1 for (6 - 2) * 180 / 6 or 180 - (360 / 6) soi A1 for 120 [cao] (b) B1 for 60 or FT (180 - their interior angle)
題目 8 · short_answer
3 分
NOT TO SCALE
In the diagram, \(PT\) is a tangent to the circle, centre \(O\), at the point \(T\). \(PO\) is a straight line that passes through the centre \(O\). Angle \(TPO = 34^\circ\).
(a) Write down the size of angle \(PTO\). Give a geometric reason for your answer. [2] (b) Work out the size of angle \(POT\). [1]
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解題
(a) The angle between a tangent and a radius at the point of contact is \(90^\circ\). Therefore, angle \(PTO = 90^\circ\). Reason: Tangent is perpendicular to radius (at point of contact).
(b) In right-angled triangle \(PTO\), the sum of angles is \(180^\circ\): angle \(POT = 180^\circ - 90^\circ - 34^\circ = 56^\circ\).
評分準則
(a) B1 for 90 [cao] B1 for reason: tangent (is) perpendicular to radius / radius meets tangent at 90° (b) B1 for 56 [cao]
題目 9 · short_answer
3 分
The coordinates of point \(A\) are \((1, -2)\) and the coordinates of point \(B\) are \((7, 10)\).
(a) Find the coordinates of the midpoint of the line segment \(AB\).
(b) Find the gradient of the line \(AB\).
(c) Write down the equation of the line parallel to \(AB\) that passes through the point \((0, 5)\).
(c) Parallel lines have the same gradient, so \(m = 2\). The line passes through \((0, 5)\), which gives the \(y\)-intercept \(c = 5\). Therefore, the equation of the line is \(y = 2x + 5\).
評分準則
(a) B1 for \((4, 4)\) or correct working shown. (b) B1 for 2. (c) B1 for \(y = 2x + 5\) or \(f(x) = 2x + 5\).
題目 10 · short_answer
4 分
A straight line \(L\) has the equation \(2y - 6x = -8\).
(a) Find the gradient of line \(L\).
(b) Write down the coordinates of the point where line \(L\) crosses the \(y\)-axis.
(c) Find the equation of the line parallel to \(L\) that passes through the point \((3, 1)\). Give your answer in the form \(y = mx + c\).
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解題
(a) Rearrange the equation into the form \(y = mx + c\): \(2y = 6x - 8 \implies y = 3x - 4\). The gradient is \(m = 3\).
(b) When \(x = 0\), \(y = -4\). So the coordinates are \((0, -4)\).
(c) The gradient of the parallel line is \(m = 3\). Substitute \((3, 1)\) into \(y = mx + c\): \(1 = 3(3) + c \implies 1 = 9 + c \implies c = -8\). So the equation is \(y = 3x - 8\).
評分準則
(a) B1 for 3. (b) B1 for \((0, -4)\). (c) M1 for substituting \((3, 1)\) into \(y = 3x + c\) or using \(y - 1 = 3(x - 3)\). A1 for \(y = 3x - 8\).
題目 11 · short_answer
2.5 分
Solve the equation.
\[5(2x - 3) = 3x + 6\]
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解題
First expand the bracket on the left-hand side: \[10x - 15 = 3x + 6\]
Subtract \(3x\) from both sides: \[7x - 15 = 6\]
Add 15 to both sides: \[7x = 21\]
Divide both sides by 7: \[x = 3\]
評分準則
M1 for correctly expanding the bracket to get \(10x - 15\) M0.5 for correctly rearranging terms into the form \(7x = 21\) (or equivalent) A1 for \(x = 3\) cao
題目 12 · short_answer
2.5 分
Expand and simplify.
\[(2x - 5)(x + 4)\]
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解題
Expand the product of the two brackets: \[(2x - 5)(x + 4) = 2x(x) + 2x(4) - 5(x) - 5(4)\] \[= 2x^2 + 8x - 5x - 20\]
Combine like terms: \[= 2x^2 + 3x - 20\]
評分準則
M1.5 for at least 3 correct terms out of 4 in the expansion \(2x^2 + 8x - 5x - 20\) A1 for fully simplified expression \(2x^2 + 3x - 20\) cao
題目 13 · short_answer
2.5 分
Make \(w\) the subject of the formula.
\[P = 4w - 7\]
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解題
Add 7 to both sides of the formula: \[P + 7 = 4w\]
Divide both sides by 4: \[w = \frac{P + 7}{4}\]
評分準則
M1.5 for correctly isolating the term containing \(w\) (e.g. \(P + 7 = 4w\) or \(\frac{P}{4} = w - \frac{7}{4}\)) A1 for \(w = \frac{P + 7}{4}\) or \(w = \frac{P}{4} + \frac{7}{4}\) cao
題目 14 · short_answer
2.5 分
Factorise fully.
\[12a^2b - 18ab^2\]
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解題
Find the highest common factor of 12 and 18, which is 6. Find the highest common power of \(a\), which is \(a\). Find the highest common power of \(b\), which is \(b\).
M1.5 for partially factorising, extracting any two of 6, \(a\), or \(b\) (e.g. \(6(2a^2b - 3ab^2)\), \(ab(12a - 18b)\), or \(3ab(4a - 6b)\)) A1 for the completely factorised expression \(6ab(2a - 3b)\) cao
題目 15 · Probability, Sets & Statistics
3 分
The universal set \(\mathscr{U} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\). \(A = \{\text{multiples of } 3\}\) \(B = \{\text{factors of } 12\}\)
(a) List the elements of the set \(A \cap B\). [1] (b) Find \(n((A \cup B)')\). [2]
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解題
(a) The elements of \(A\) are \(\{3, 6, 9, 12\}\). The elements of \(B\) are \(\{1, 2, 3, 4, 6, 12\}\). The intersection \(A \cap B\) contains the common elements: \(\{3, 6, 12\}\).
(b) The union is \(A \cup B = \{1, 2, 3, 4, 6, 9, 12\}\), which contains 7 elements. The complement \((A \cup B)'\) contains the elements in \(\mathscr{U}\) not in \(A \cup B\): \((A \cup B)' = \{5, 7, 8, 10, 11\}\). So, \(n((A \cup B)') = 12 - 7 = 5\).
評分準則
(a) B1 for 3, 6, 12 (brackets/order not required, no extra elements). (b) M1 for identifying the 5 elements \(\{5, 7, 8, 10, 11\}\) or for calculation \(12 - 7\) soi. A1 for 5.
題目 16 · Probability, Sets & Statistics
3 分
A bag contains 7 red beads, 5 blue beads, and 8 green beads only. A bead is chosen at random from the bag.
(a) Find the probability that the bead is blue. Give your answer as a fraction in its simplest form. [1]
(b) The bead is replaced in the bag. A bead is drawn at random and replaced 160 times. Work out the expected number of times a red bead is drawn. [2]
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解題
(a) Total number of beads = \(7 + 5 + 8 = 20\). \(\text{P(blue)} = \frac{5}{20} = \frac{1}{4}\).
(b) The probability of drawing a red bead is \(\text{P(red)} = \frac{7}{20}\). Expected number of red beads = \(160 \times \frac{7}{20} = 8 \times 7 = 56\).
評分準則
(a) B1 for \(\frac{1}{4}\) cao (must be in lowest terms). (b) M1 for \(160 \times \frac{7}{20}\) or \(160 \times 0.35\) oe. A1 for 56 cao.
題目 17 · Probability, Sets & Statistics
3 分
Here is a list of the numbers of goals scored by a team in 8 matches: \[2, \quad 3, \quad 5, \quad 6, \quad 6, \quad 7, \quad 10, \quad 17\]
(a) Write down the mode. [1] (b) Find the range. [1] (c) Calculate the mean. [1]
Combine into a single fraction: \[\frac{16\sqrt{2} - 13}{7}\]
評分準則
M1 for multiplying numerator and denominator by \(3 - \sqrt{2}\) M1 for correctly expanding numerator to \(16\sqrt{2} - 13\) (allow 1 sign error) A1 for \(\frac{16\sqrt{2} - 13}{7}\) or \(\frac{-13 + 16\sqrt{2}}{7}\) cao
Substitute these values into the first term: \[\frac{\sin 60^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{3} \times \sqrt{2}}{2} = \frac{\sqrt{6}}{2}\]
Add the two terms together: \[\frac{\sqrt{6}}{2} + 2\sqrt{6} = \left(\frac{1}{2} + 2\right)\sqrt{6} = \frac{5}{2}\sqrt{6}\]
評分準則
M1 for substituting correct exact values \(\sin 60^\circ = \frac{\sqrt{3}}{2}\) and \(\cos 45^\circ = \frac{\sqrt{2}}{2}\) (or \(\frac{1}{\sqrt{2}}\)) M1 for simplifying the first term to \(\frac{\sqrt{6}}{2}\) OR simplifying \(\sqrt{24}\) to \(2\sqrt{6}\) A1 for \(\frac{5}{2}\sqrt{6}\) or \(\frac{5\sqrt{6}}{2}\) or \(2.5\sqrt{6}\)
題目 3 · short_answer
3 分
Expand and simplify fully.
\[(3\sqrt{5} - 2\sqrt{3})^2 + \sqrt{240}\]
Give your answer in the form \(a + b\sqrt{15}\), where \(a\) and \(b\) are integers.
M1 for expanding \((3\sqrt{5} - 2\sqrt{3})^2\) correctly to \(45 - 12\sqrt{15} + 12\) (allow one arithmetic slip) M1 for simplifying \(\sqrt{240}\) to \(4\sqrt{15}\) A1 for \(57 - 8\sqrt{15}\) cao
Calculate the full fraction: \[\frac{2}{\frac{7}{2}} = 2 \times \frac{2}{7} = \frac{4}{7}\]
評分準則
M1 for correctly stating at least three of: \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\tan 45^\circ = 1\), \(\sin 30^\circ = \frac{1}{2}\), \(\cos 60^\circ = \frac{1}{2}\), \(\tan 60^\circ = \sqrt{3}\) M1 for evaluating the numerator to \(2\) OR the denominator to \(\frac{7}{2}\) A1 for \(\frac{4}{7}\) cao
題目 5 · short-answer
3.25 分
Rearrange the formula to make \(w\) the subject.
\[T = \sqrt{\frac{3w + 5k}{w - 2k}}\]
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解題
Square both sides of the formula: \[T^2 = \frac{3w + 5k}{w - 2k}\]
Multiply both sides by \((w - 2k)\): \[T^2(w - 2k) = 3w + 5k\]
Expand the bracket: \[T^2w - 2kT^2 = 3w + 5k\]
Collect all terms containing \(w\) on one side and terms without \(w\) on the other side: \[T^2w - 3w = 5k + 2kT^2\]
Factorise \(w\) from the left-hand side: \[w(T^2 - 3) = k(5 + 2T^2)\]
M1 for squaring both sides: \(T^2 = \frac{3w + 5k}{w - 2k}\) seen or implied M1 for correctly removing the fraction and expanding: \(T^2w - 2kT^2 = 3w + 5k\) oe M0.25 for isolating terms in \(w\) on one side and factorising \(w\): \(w(T^2 - 3) = ...\) A1 for \(w = \frac{2kT^2 + 5k}{T^2 - 3}\) or \(w = \frac{-2kT^2 - 5k}{3 - T^2}\) oe
題目 6 · open-ended
3.5 分
The line \(L\) is the perpendicular bisector of the line segment joining the points \(A(-3, 1)\) and \(B(5, 7)\).
Find the equation of line \(L\). Give your answer in the form \(y = mx + c\).
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解題
1. Find the midpoint, \(M\), of the line segment \(AB\): \[ M = \left( \frac{-3 + 5}{2}, \frac{1 + 7}{2} \right) = \left( \frac{2}{2}, \frac{8}{2} \right) = (1, 4) \]
2. Find the gradient of line segment \(AB\): \[ m_{AB} = \frac{7 - 1}{5 - (-3)} = \frac{6}{8} = \frac{3}{4} \]
3. Find the gradient of the perpendicular bisector \(L\): \[ m_L = -\frac{1}{m_{AB}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3} \]
4. Use the point-slope formula with point \((1, 4)\) and gradient \(m = -\frac{4}{3}\): \[ y - 4 = -\frac{4}{3}(x - 1) \] \[ y - 4 = -\frac{4}{3}x + \frac{4}{3} \] \[ y = -\frac{4}{3}x + \frac{4}{3} + \frac{12}{3} \] \[ y = -\frac{4}{3}x + \frac{16}{3} \]
評分準則
B1: for midpoint \((1, 4)\) seen or implied. M1: for finding gradient of \(AB = \frac{3}{4}\) and perpendicular gradient \(-\frac{4}{3}\). M1: for substituting their midpoint and their perpendicular gradient into \(y - y_1 = m(x - x_1)\) or \(y = mx + c\). A0.5: for correct final equation \(y = -\frac{4}{3}x + \frac{16}{3}\) or exact equivalent in the form \(y = mx + c\).
題目 7 · open-ended
3.5 分
The line \(L_1\) has the equation \(2x + 3y = 12\). The line \(L_2\) is perpendicular to \(L_1\) and passes through the point \(P(1, -2)\). Line \(L_2\) intersects the \(x\)-axis at the point \(Q\).
Find the exact length of the line segment \(PQ\). Give your answer as a single fraction in simplest surd form.
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解題
1. Find the gradient of \(L_1\): \[ 2x + 3y = 12 \implies 3y = -2x + 12 \implies y = -\frac{2}{3}x + 4 \] So, \(m_1 = -\frac{2}{3}\).
2. Determine the gradient of \(L_2\): \[ m_2 = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \]
3. Write the equation of \(L_2\) passing through \(P(1, -2)\): \[ y - (-2) = \frac{3}{2}(x - 1) \implies y + 2 = \frac{3}{2}x - \frac{3}{2} \]
4. Find the coordinates of \(Q\) where \(L_2\) intersects the \(x\)-axis (set \(y = 0\)): \[ 0 + 2 = \frac{3}{2}(x - 1) \implies 2 = \frac{3}{2}(x - 1) \implies \frac{4}{3} = x - 1 \implies x = \frac{7}{3} \] Thus, \(Q = \left(\frac{7}{3}, 0\right)\).
5. Calculate the length of the line segment \(PQ\): \[ PQ = \sqrt{\left(\frac{7}{3} - 1\right)^2 + (0 - (-2))^2} = \sqrt{\left(\frac{4}{3}\right)^2 + (2)^2} = \sqrt{\frac{16}{9} + 4} = \sqrt{\frac{16 + 36}{9}} = \sqrt{\frac{52}{9}} = \frac{\sqrt{52}}{3} = \frac{2\sqrt{13}}{3} \]
評分準則
B1: for gradient of \(L_1 = -\frac{2}{3}\) and perpendicular gradient \(m_2 = \frac{3}{2}\). M1: for setting \(y = 0\) in their equation of \(L_2\) to find the \(x\)-coordinate of \(Q\) as \(\frac{7}{3}\). M1: for correct method to find the length \(PQ = \sqrt{(\Delta x)^2 + (\Delta y)^2}\). A0.5: for \(\frac{2\sqrt{13}}{3}\) (or \(\frac{\sqrt{52}}{3}\)).
題目 8 · short-answer
3 分
NOT TO SCALE
\(A\), \(B\), \(C\) and \(D\) lie on the circumference of a circle, centre \(O\). \(BD\) is a diameter of the circle. Angle \(ADB = 38^\circ\) and angle \(CBD = 24^\circ\).
Find angle \(ADC\).
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解題
\(\angle BAD = 90^\circ\) (angle in a semicircle). In \(\triangle ABD\): \(\angle ABD = 180^\circ - 90^\circ - 38^\circ = 52^\circ\).
\(ABCD\) is a cyclic quadrilateral, so opposite angles sum to \(180^\circ\): \(\angle ADC = 180^\circ - \angle ABC = 180^\circ - 76^\circ = 104^\circ\).
評分準則
M1 for \(\angle BAD = 90^\circ\) soi, leading to \(\angle ABD = 52^\circ\) M1 for \(\angle ABC = 52^\circ + 24^\circ = 76^\circ\) or finding \(\angle CAD = 24^\circ\) and \(\angle ACD = 52^\circ\) A1 for \(104^\circ\) cao
題目 9 · structured
3 分
The interior angle of a regular polygon with \(n\) sides is \(7\) times its exterior angle.
(a) Find the value of \(n\). [2] (b) Work out the sum of the interior angles of this polygon. [1]
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解題
(a) Let \(e\) be the exterior angle. The interior angle is \(7e\). Since interior and exterior angles sum to \(180^\circ\): \(7e + e = 180^\circ\) \(8e = 180^\circ\) \(e = 22.5^\circ\)
The sum of exterior angles of any convex polygon is \(360^\circ\): \(n = \frac{360^\circ}{22.5^\circ} = 16\).
(a) M1 for \(7e + e = 180^\circ\) or \(e = 22.5^\circ\) or \(\frac{(n-2) \times 180}{n} = 7 \times \frac{360}{n}\) A1 for \(16\) (b) B1 for \(2520^\circ\) or FT \((n - 2) \times 180^\circ\)
題目 10 · structured
3 分
NOT TO SCALE
\(A\), \(B\) and \(C\) lie on the circumference of a circle, centre \(O\). The straight lines \(TA\) and \(TB\) are tangents to the circle. Angle \(ATB = 44^\circ\).
(a) Work out angle \(AOB\). [1] (b) Work out angle \(ACB\), where \(C\) lies on the major arc \(AB\). [1] (c) Work out angle \(OAB\). [1]
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解題
(a) Tangents are perpendicular to radii at the points of contact, so \(\angle OAT = \angle OBT = 90^\circ\). In quadrilateral \(OATB\): \(\angle AOB = 360^\circ - 90^\circ - 90^\circ - 44^\circ = 180^\circ - 44^\circ = 136^\circ\).
(b) The angle at the centre is twice the angle at the circumference subtended by the same arc: \(\angle ACB = \frac{136^\circ}{2} = 68^\circ\).
(c) In \(\triangle OAB\), \(OA = OB\) (radii), so the triangle is isosceles: \(\angle OAB = \frac{180^\circ - 136^\circ}{2} = \frac{44^\circ}{2} = 22^\circ\).
評分準則
(a) B1 for \(136^\circ\) (b) B1 for \(68^\circ\) or FT \(\frac{\text{their (a)}}{2}\) (c) B1 for \(22^\circ\) or FT \(\frac{180^\circ - \text{their (a)}}{2}\)
題目 11 · free_text
3 分
The functions \(\mathrm{f}\) and \(\mathrm{g}\) are defined by \[\mathrm{f}(x) = 3x - 4\] \[\mathrm{g}(x) = \frac{6}{x - 2}, \quad x \ne 2.\] (a) Find \(\mathrm{g}(\mathrm{f}(3))\). (b) Find \(\mathrm{f}^{-1}(x)\).
(b) Let \(y = 3x - 4\). Rearrange to make \(x\) the subject: \[y + 4 = 3x\] \[x = \frac{y + 4}{3}\] Therefore, \(\mathrm{f}^{-1}(x) = \frac{x + 4}{3}\).
評分準則
(a) [1 mark] B1 for 2 cao
(b) [2 marks] M1 for correct algebraic step to isolate \(x\), e.g. \(y + 4 = 3x\) or \(x + 4 = 3y\) A1 for \(\frac{x+4}{3}\) or \(\frac{1}{3}x + \frac{4}{3}\) cao
Apply the logarithm subtraction rule \(\log_{10} a - \log_{10} b = \log_{10}\left(\frac{a}{b}\right)\): \[\log_{10}\left(\frac{5x + 15}{x - 1}\right) = 1\] Rewrite in exponential form: \[\frac{5x + 15}{x - 1} = 10^1\] Multiply both sides by \(x - 1\): \[5x + 15 = 10(x - 1)\] \[5x + 15 = 10x - 10\] Rearrange to solve for \(x\): \[15 + 10 = 10x - 5x\] \[25 = 5x\] \[x = 5\]
評分準則
M1 for applying logarithm division rule correctly: \(\log_{10}\left(\frac{5x + 15}{x - 1}\right) = 1\) soi M1 for converting logarithmic equation to linear form: \(\frac{5x + 15}{x - 1} = 10\) A1 for \(x = 5\) cao
題目 13 · free_text
4 分
The function \(\mathrm{h}\) is defined by \(\mathrm{h}(x) = 4 \times 8^{x - 1}\). (a) Find the exact value of \(\mathrm{h}\left(\frac{5}{3}\right)\). (b) Find the value of \(x\) when \(\mathrm{h}(x) = \frac{1}{16}\).
(a) [2 marks] M1 for evaluating index to \(8^{2/3}\) or converting to base 2 as \(2^2 \times 2^{3(2/3)}\) A1 for 16 cao
(b) [2 marks] M1 for converting all terms to base 2: \(2^{3x-1} = 2^{-4}\) or dividing first to give \(8^{x-1} = 2^{-6}\) oe A1 for \(x = -1\) cao
題目 14 · short_answer
2 分
A bag contains 5 green discs and 4 yellow discs. Two discs are chosen at random without replacement. Find the probability that both discs are green. Give your answer as a fraction in its simplest form.
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解題
The total number of discs is \(5 + 4 = 9\). The probability that the first disc chosen is green is \(\frac{5}{9}\). After one green disc is removed, there are 4 green discs remaining out of a total of 8 discs. The probability that the second disc chosen is green is \(\frac{4}{8}\). Therefore, \(P(\text{both green}) = \frac{5}{9} \times \frac{4}{8} = \frac{5}{9} \times \frac{1}{2} = \frac{5}{18}\).
評分準則
M1 for \(\frac{5}{9} \times \frac{4}{8}\) oe A1 for \(\frac{5}{18}\) cao
Paper 3 (Core Calculator)
Answer all questions. Graphic display calculator should be used where appropriate.
20 題目 · 50 分
題目 1 · short-answer
2 分
Elena changes €450 into Japanese yen (¥). The exchange rate is €1 = ¥162.40. During her trip, she spends ¥52 000.
Work out the amount of Japanese yen Elena has left.
(a) B2 for fully correct ordered diagram (all leaves correct and ordered in increasing order) (B1 for correct leaves but unordered or 1 error/omission in ordered leaves) (b) B1 for 37 cao (or FT their correct middle values)
題目 10 · free-response
3 分
A wire of length \(14.5\text{ m}\) is attached to the top of a vertical mast and fixed to a point on horizontal ground. The wire makes an angle of \(38^\circ\) with the ground. Calculate the height of the mast.
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解題
Let \(h\) be the height of the mast in metres. In the right-angled triangle formed by the mast, ground, and wire: \(\sin(38^\circ) = \frac{h}{14.5}\). Solving for \(h\): \(h = 14.5 \times \sin(38^\circ) \approx 14.5 \times 0.615661 = 8.92709\text{ m}\). Correct to 3 significant figures, the height is \(8.93\text{ m}\).
評分準則
M1 for \(\sin(38^\circ) = \frac{h}{14.5}\) or equivalent correct trigonometrical statement. M1 for \(h = 14.5 \times \sin(38^\circ)\). A1 for 8.93 or 8.927... (awarded for answer rounding to 8.93).
題目 11 · free-response
3 分
The size of each interior angle of a regular polygon is \(156^\circ\). Calculate the number of sides of this polygon.
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解題
The size of each exterior angle of the regular polygon is \(180^\circ - 156^\circ = 24^\circ\). The sum of the exterior angles of any polygon is \(360^\circ\). Therefore, the number of sides is \(n = \frac{360^\circ}{24^\circ} = 15\).
評分準則
M1 for finding the exterior angle \(180 - 156 = 24\) (soi). M1 for \(\frac{360}{24}\) or for setting up \((n - 2) \times 180 = 156n\). A1 for 15.
題目 12 · free-response
3 分
Triangle \(P\) has vertices at \((1, 2)\), \((3, 2)\), and \((1, 5)\). Triangle \(Q\) has vertices at \((2, 4)\), \((6, 4)\), and \((2, 10)\). Describe fully the single transformation that maps triangle \(P\) onto triangle \(Q\).
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解題
Comparing corresponding vertices: \((1, 2) \to (2, 4)\), \((3, 2) \to (6, 4)\), and \((1, 5) \to (2, 10)\). Each coordinate is doubled. Therefore, the transformation is an enlargement with a scale factor of 2 and centre of enlargement at the origin \((0, 0)\).
評分準則
B1 for Enlargement (do not accept stretch/resize). B1 for scale factor 2. B1 for centre \((0, 0)\) or origin. Note: If more than one transformation is stated, award 0 marks.
題目 13 · free-response
3 分
A rectangular garden has a length of \(84\text{ m}\) and a width of \(35\text{ m}\). A straight path runs diagonally across the garden from one corner to the opposite corner. Calculate the length of the path.
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解題
Let \(d\) be the length of the diagonal path in metres. By Pythagoras' theorem in the right-angled triangle formed by the length and width of the garden: \(d^2 = 84^2 + 35^2 = 7056 + 1225 = 8281\). Taking the square root: \(d = \sqrt{8281} = 91\text{ m}\).
評分準則
M1 for \(84^2 + 35^2\) or \(7056 + 1225\). M1 for \(\sqrt{84^2 + 35^2}\) or \(\sqrt{8281}\). A1 for 91.
題目 14 · free-response
3 分
\(AB\) is a diameter of a circle with centre \(O\). \(C\) is a point on the circumference such that angle \(BAC = 28^\circ\). The straight line \(TB\) is a tangent to the circle at point \(B\). (a) Write down the size of angle \(ACB\). (b) Find the size of angle \(ABC\). (c) Find the size of angle \(CBT\).
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解題
(a) Angle \(ACB = 90^\circ\) because the angle in a semicircle is a right angle. (b) In triangle \(ABC\), angles sum to \(180^\circ\), so angle \(ABC = 180^\circ - 90^\circ - 28^\circ = 62^\circ\). (c) The tangent \(TB\) is perpendicular to the diameter \(AB\), so angle \(ABT = 90^\circ\). Hence, angle \(CBT = 90^\circ - 62^\circ = 28^\circ\).
評分準則
B1 for \(90^\circ\) (in (a)). B1 for \(62^\circ\) (in (b)). B1 for \(28^\circ\) (in (c)) or FT \(90^\circ - \text{their (b)}\).
題目 15 · short_answer
3 分
\(f(x) = x^2 - 2x - 8\)
(a) Write down the coordinates of the \(y\)-intercept. [1]
(b) Write down the zeros of \(f(x)\). [1]
(c) Write down the coordinates of the local minimum. [1]
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解題
(a) The \(y\)-intercept occurs when \(x = 0\): \(f(0) = 0^2 - 2(0) - 8 = -8\) Coordinates are \((0, -8)\).
(b) The zeros occur when \(f(x) = 0\): \(x^2 - 2x - 8 = 0\) \((x - 4)(x + 2) = 0\) \(x = -2\) and \(x = 4\).
(c) Using the graphic display calculator or vertex formula \(x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1\): \(f(1) = 1^2 - 2(1) - 8 = -9\) Coordinates of the local minimum are \((1, -9)\).
評分準則
(a) B1 for \((0, -8)\) (b) B1 for \(-2\) and \(4\) (or \(x = -2, x = 4\)) (c) B1 for \((1, -9)\)
題目 16 · structured
3 分
The function \(\mathrm{f}(x) = x^2 - 4x - 5\) is defined for all real numbers.
(a) Write down the coordinates of the point where the graph of \(y = \mathrm{f}(x)\) crosses the \(y\)-axis. [1]
(b) Find the coordinates of the local minimum point of the graph. [1]
(c) Write down the zeros of \(\mathrm{f}(x)\). [1]
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解題
(a) When \(x = 0\), \(y = 0^2 - 4(0) - 5 = -5\). The coordinates of the \(y\)-intercept are \((0, -5)\).
(b) Using a graphic display calculator or completing the square, \(\mathrm{f}(x) = (x - 2)^2 - 9\). The minimum turning point is at \((2, -9)\).
(c) The zeros occur where \(\mathrm{f}(x) = 0\): \(x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0\) So the zeros are \(x = -1\) and \(x = 5\).
評分準則
(a) B1 for \((0, -5)\) or \(x = 0, y = -5\) (b) B1 for \((2, -9)\) (c) B1 for \(-1\) and \(5\) (accept as separate values or coordinate points \((-1, 0)\) and \((5, 0)\))
題目 17 · short_answer
2 分
Find the \(n\text{th}\) term of the sequence \(7,\; 12,\; 17,\; 22,\; 27,\; \dots\)
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解題
The sequence increases by \(5\) each time, so the \(n\text{th}\) term contains \(5n\). When \(n = 1\), \(5(1) + c = 7 \implies c = 2\). Therefore, the \(n\text{th}\) term is \(5n + 2\).
評分準則
B1 for \(5n + k\) (where \(k\) is any constant, \(k \neq 2\)) or for \(5n\) seen B1 for \(5n + 2\) or \(2 + 5n\) oe
題目 18 · short_answer
2 分
The \(n\text{th}\) term of a sequence is given by \(80 - 6n\).
Find the value of \(n\) when the term is \(-22\).
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解題
Set the expression for the \(n\text{th}\) term equal to \(-22\): \[80 - 6n = -22\] Subtract 80 from both sides: \[-6n = -102\] Divide by \(-6\): \[n = \frac{-102}{-6} = 17\]
評分準則
M1 for \(80 - 6n = -22\) or \(6n = 102\) or equivalent linear equation A1 for 17 cao
題目 19 · short_answer
2 分
Here are the first four terms of a sequence: \[43,\; 37,\; 31,\; 25,\; \dots\]
Find the \(n\text{th}\) term of this sequence.
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解題
The terms decrease by \(6\) each time, so the common difference is \(-6\). The term rule contains \(-6n\). For \(n = 1\), \(-6(1) + c = 43 \implies c = 49\). Hence, the \(n\text{th}\) term is \(49 - 6n\) (or \(-6n + 49\)).
評分準則
B1 for \(-6n + k\) (where \(k \neq 49\)) or \(-6n\) seen B1 for \(49 - 6n\) or \(-6n + 49\) oe
題目 20 · short_answer
2 分
The number of counters used to make each pattern forms an arithmetic sequence:
Pattern 1 has 5 counters. Pattern 2 has 9 counters. Pattern 3 has 13 counters.
Find the number of counters in Pattern 24.
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解題
The sequence of counters is \(5,\; 9,\; 13,\; \dots\) The difference between consecutive patterns is \(4\). The \(n\text{th}\) term rule is \(4n + 1\). For Pattern 24, substitute \(n = 24\): \[4(24) + 1 = 96 + 1 = 97\]
評分準則
M1 for \(4n + 1\) oe seen or for \(5 + 23 \times 4\) or correct step method A1 for 97 cao
Paper 4 (Extended Calculator)
Answer all questions. Graphic display calculator should be used where appropriate. Give non-exact answers to 3 significant figures.
17 題目 · 59 分
題目 1 · short_answer
4 分
\(f(x) = \frac{4x - 6}{x + 3}\), for \(x \neq -3\). (a) Write down the equations of the asymptotes of the graph of \(y = f(x)\). (b) Find the coordinates of the point where the graph of \(y = f(x)\) crosses the \(x\)-axis. (c) Solve the equation \(f(x) = -2\).
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解題
(a) Vertical asymptote occurs where the denominator is zero: \(x + 3 = 0 \implies x = -3\). Horizontal asymptote: as \(x \to \pm\infty\), \(y \to \frac{4x}{x} = 4\), so \(y = 4\). (b) Setting \(y = 0\): \(4x - 6 = 0 \implies 4x = 6 \implies x = 1.5\). The coordinates are \((1.5, 0)\). (c) \(\frac{4x - 6}{x + 3} = -2 \implies 4x - 6 = -2(x + 3) \implies 4x - 6 = -2x - 6 \implies 6x = 0 \implies x = 0\).
評分準則
(a) B1 for \(x = -3\), B1 for \(y = 4\) [2 marks]. (b) B1 for \((1.5, 0)\) or \(x = 1.5, y = 0\) [1 mark]. (c) B1 for \(x = 0\) [1 mark].
題目 2 · short_answer
4 分
\(g(x) = \frac{2x^2 + 5}{x^2 - 4}\), for \(x \neq -2\) and \(x \neq 2\). (a) Write down the equations of the vertical asymptotes of the graph of \(y = g(x)\). (b) Write down the equation of the horizontal asymptote of the graph of \(y = g(x)\). (c) Write down the coordinates of the local maximum point. (d) Solve the equation \(g(x) = 3\).
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解題
(a) Denominator is zero when \(x^2 - 4 = 0 \implies x = 2\) and \(x = -2\). (b) As \(x \to \pm\infty\), \(g(x) \to \frac{2x^2}{x^2} = 2\), so the horizontal asymptote is \(y = 2\). (c) Using GDC maximum function or evaluating at \(x = 0\): \(g(0) = \frac{5}{-4} = -1.25\), so the local maximum point is \((0, -1.25)\). (d) \(\frac{2x^2 + 5}{x^2 - 4} = 3 \implies 2x^2 + 5 = 3(x^2 - 4) \implies 2x^2 + 5 = 3x^2 - 12 \implies x^2 = 17 \implies x = \pm\sqrt{17} \approx \pm 4.12\).
評分準則
(a) B1 for \(x = -2\) and \(x = 2\) (both required) [1 mark]. (b) B1 for \(y = 2\) [1 mark]. (c) B1 for \((0, -1.25)\) or \((0, -\frac{5}{4})\) [1 mark]. (d) B1 for \(x = -4.12\) and \(x = 4.12\) (or \(\pm 4.12\), accept exact \(\pm\sqrt{17}\)) [1 mark].
題目 3 · short_answer
4 分
\(f(x) = \frac{12}{x^2 - 2x - 8}\), for \(x \neq -2\) and \(x \neq 4\). (a) Write down the equations of the vertical asymptotes of the graph of \(y = f(x)\). (b) Write down the equation of the horizontal asymptote of the graph of \(y = f(x)\). (c) Write down the coordinates of the local maximum point. (d) Solve the equation \(f(x) = -2\).
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解題
(a) The denominator factorises as \((x - 4)(x + 2) = 0\), giving vertical asymptotes \(x = -2\) and \(x = 4\). (b) As \(x \to \pm\infty\), \(f(x) \to 0\), so the horizontal asymptote is \(y = 0\). (c) By symmetry or using GDC, the turning point occurs at \(x = 1\). \(f(1) = \frac{12}{1 - 2 - 8} = -\frac{12}{9} = -\frac{4}{3} \approx -1.33\). Coordinates are \((1, -1.33)\). (d) \(\frac{12}{x^2 - 2x - 8} = -2 \implies x^2 - 2x - 8 = -6 \implies x^2 - 2x - 2 = 0\). Using the quadratic formula or GDC solver: \(x = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3} \approx -0.732\) and \(2.73\).
評分準則
(a) B1 for \(x = -2\) and \(x = 4\) (both required) [1 mark]. (b) B1 for \(y = 0\) [1 mark]. (c) B1 for \((1, -1.33)\) or \((1, -\frac{4}{3})\) [1 mark]. (d) B1 for \(x = -0.732\) and \(x = 2.73\) (accept \(1 \pm \sqrt{3}\)) [1 mark].
題目 4 · structured
4 分
The table shows the body mass (fit{x} ext{ grams}) and the resting metabolic rate (fit{y} ext{ units}) of a sample of six small rodents.
(a) Find the coordinates of the mean point $(\bar{x}, \bar{y})$. [1] (b) Find the equation of the linear regression line for $y$ in terms of $x$. [2] (c) Use your equation from part (b) to estimate the resting metabolic rate of a rodent with a body mass of $52\text{ g}$. [1]
(a) B1: cao for $(42.5, 3.55)$ or $\bar{x} = 42.5, \bar{y} = 3.55$.
(b) B1: for gradient $0.0615$ (accept $0.06145\dots$) B1: for $y$-intercept $0.938$ (accept $0.9381\dots$) (Award B1B1 if written as $y = 0.0615x + 0.938$ or full equation given; allow SC1 for correct values of $m$ and $c$ reversed).
(c) B1: for $4.13$ or $4.14$ or FT their linear equation from (b).
題目 5 · structured
4 分
A college survey recorded the preferences of 120 applicants regarding their field of study and whether they requested on-campus accommodation.
(a) One applicant is chosen at random from those who prefer STEM. Find the probability that this applicant requested on-campus accommodation. [1] (b) Two applicants are chosen at random without replacement from the 60 applicants who prefer Humanities. Find the probability that exactly one of the two applicants requested on-campus accommodation. [3]
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解題
(a) The total number of applicants who prefer STEM is $60$. Among them, $42$ requested on-campus accommodation. $$\text{P}(\text{On-campus} \mid \text{STEM}) = \frac{42}{60} = \frac{7}{10} = 0.7$$
(b) Out of the $60$ applicants who prefer Humanities, $28$ requested on-campus and $32$ requested off-campus accommodation.
The probability that the first requested on-campus and the second off-campus is: $$\text{P}(\text{On}, \text{Off}) = \frac{28}{60} \times \frac{32}{59} = \frac{896}{3540}$$
The probability that the first requested off-campus and the second on-campus is: $$\text{P}(\text{Off}, \text{On}) = \frac{32}{60} \times \frac{28}{59} = \frac{896}{3540}$$
Total probability: $$\text{P}(\text{exactly one on-campus}) = 2 \times \left(\frac{28}{60} \times \frac{32}{59}\right) = \frac{1792}{3540} = \frac{448}{885} \approx 0.506$$
評分準則
(a) B1: for $\frac{42}{60}$ oe (e.g. $\frac{7}{10}$ or $0.7$).
(b) M1: for $\frac{28}{60} \times \frac{32}{59}$ oe (seen or implied). M1: for identifying two mutually exclusive outcomes: $2 \times \left(\frac{28}{60} \times \frac{32}{59}\right)$ oe. A1: for $\frac{448}{885}$ or exact fraction in lowest terms, or decimal $0.506$ (or $0.5062\dots$ to 3 or more s.f.).
題目 6 · free_response
4 分
A solid frustum of a cone is formed by removing a small cone from the top of a larger cone. The bottom circular base of the frustum has radius \(8\text{ cm}\) and the top circular face has radius \(5\text{ cm}\). The vertical height of the frustum is \(9\text{ cm}\).
Calculate the volume of the frustum.
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解題
Let \(h\) be the vertical height of the smaller cone that was removed, and \(H = h + 9\) be the height of the original larger cone.
Using similar triangles: \[\frac{h}{h + 9} = \frac{5}{8}\] \[8h = 5(h + 9)\] \[8h = 5h + 45\] \[3h = 45 \implies h = 15\text{ cm}\]
The height of the larger cone is: \[H = 15 + 9 = 24\text{ cm}\]
The volume of a cone is given by \(V = \frac{1}{3}\pi r^2 h\).
Volume of the large cone: \[V_{\text{large}} = \frac{1}{3}\pi (8)^2(24) = 512\pi\text{ cm}^3\]
Volume of the small cone: \[V_{\text{small}} = \frac{1}{3}\pi (5)^2(15) = 125\pi\text{ cm}^3\]
Volume of the frustum: \[V = 512\pi - 125\pi = 387\pi \approx 1215.84\dots\text{ cm}^3\]
To 3 significant figures, the volume is \(1220\text{ cm}^3\).
評分準則
M1 for setting up a correct equation using similar triangles to find the height, e.g. \(\frac{h}{h+9} = \frac{5}{8}\) or \(\frac{H-9}{H} = \frac{5}{8}\) oe A1 for \(h = 15\) or \(H = 24\) M1 for correct method to find the volume of the frustum by subtraction: \(\frac{1}{3}\pi(8)^2(24) - \frac{1}{3}\pi(5)^2(15)\) or by direct formula \(\frac{1}{3}\pi(9)(8^2 + 8 \times 5 + 5^2)\) A1 for 1220 or 1216 or \(387\pi\)
題目 7 · free_response
4 分
A solid right circular cone has a base radius of \(12\text{ cm}\) and a vertical height of \(16\text{ cm}\). A smaller cone with a base radius of \(6\text{ cm}\) is removed from the top of this cone to form a solid frustum.
Calculate the total surface area of the solid frustum.
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解題
1. Find the slant height \(L\) of the original cone: \[L = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ cm}\]
2. The scale factor between the smaller cone and the original cone is: \[k = \frac{6}{12} = \frac{1}{2}\] So the slant height of the smaller cone is: \[l = \frac{1}{2} \times 20 = 10\text{ cm}\]
3. Calculate the curved surface area of the frustum using \(\text{CSA} = \pi R L - \pi r l\): \[\text{CSA} = \pi (12)(20) - \pi (6)(10) = 240\pi - 60\pi = 180\pi\text{ cm}^2\]
4. Calculate the area of the two circular faces: \[\text{Area of base} = \pi (12)^2 = 144\pi\text{ cm}^2\] \[\text{Area of top face} = \pi (6)^2 = 36\pi\text{ cm}^2\]
5. Total surface area of the solid frustum: \[\text{Total Surface Area} = 180\pi + 144\pi + 36\pi = 360\pi \approx 1130.97\dots\text{ cm}^2\]
To 3 significant figures, the total surface area is \(1130\text{ cm}^2\).
評分準則
M1 for finding slant height of large cone \(L = \sqrt{12^2 + 16^2} = 20\) or small cone \(l = 10\) M1 for curved surface area of frustum \(\pi(12)(20) - \pi(6)(10) [= 180\pi]\) or \(\pi(12+6)(10) [= 180\pi]\) M1 for adding both circular end areas: \(\text{curved surface area} + \pi(12)^2 + \pi(6)^2\) A1 for 1130 or 1131 or \(360\pi\)
題目 8 · short_answer
3 分
A sailing yacht travels from a harbour \(A\) to a buoy \(B\) on a bearing of \(048^\circ\) for a distance of \(18.5\text{ km}\). It then sails from buoy \(B\) to an island \(C\) on a bearing of \(132^\circ\) for a distance of \(24.0\text{ km}\).
NOT TO SCALE
Calculate the direct distance from \(A\) to \(C\).
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解題
1. Find the interior angle \(\angle ABC\): - The back-bearing of \(A\) from \(B\) is \(048^\circ + 180^\circ = 228^\circ\). - The bearing of \(C\) from \(B\) is \(132^\circ\). - Therefore, \(\angle ABC = 228^\circ - 132^\circ = 96^\circ\).
[B1] for finding angle \(\angle ABC = 96^\circ\) (seen or implied) [M1] for correct substitution into the cosine rule: \(18.5^2 + 24.0^2 - 2(18.5)(24.0)\cos(96^\circ)\) [A1] for \(31.8\) or \(31.80\text{ km}\) (accept values in range \([31.79, 31.80]\))
題目 9 · short_answer
3 分
Two observation stations \(P\) and \(Q\) are \(14.2\text{ km}\) apart on flat ground, with \(Q\) on a bearing of \(075^\circ\) from \(P\). A fire tower \(R\) is on a bearing of \(020^\circ\) from \(P\) and on a bearing of \(310^\circ\) from \(Q\).
NOT TO SCALE
Calculate the distance \(PR\).
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解題
1. Calculate the angles within \(\triangle PQR\): - At point \(P\): \(\angle QPR = 075^\circ - 020^\circ = 55^\circ\). - At point \(Q\): the back-bearing of \(P\) from \(Q\) is \(075^\circ + 180^\circ = 255^\circ\). The angle between line \(QP\) and line \(QR\) is \(\angle PQR = 310^\circ - 255^\circ = 55^\circ\). - At point \(R\): \(\angle PRQ = 180^\circ - (55^\circ + 55^\circ) = 70^\circ\).
2. Use the sine rule to find \(PR\): \[ \frac{PR}{\sin(\angle PQR)} = \frac{PQ}{\sin(\angle PRQ)} \] \[ \frac{PR}{\sin(55^\circ)} = \frac{14.2}{\sin(70^\circ)} \] \[ PR = \frac{14.2 \times \sin(55^\circ)}{\sin(70^\circ)} \approx 12.378\text{ km} \] Rounding to 3 significant figures gives \(12.4\text{ km}\).
評分準則
[B1] for finding angle \(\angle PRQ = 70^\circ\) or angle \(\angle PQR = 55^\circ\) [M1] for correct substitution into the sine rule: \(\frac{PR}{\sin(55^\circ)} = \frac{14.2}{\sin(70^\circ)}\) [A1] for \(12.4\text{ km}\) (accept values in range \([12.37, 12.38]\))
題目 10 · short_answer
3 分
A helicopter flies from a base \(H\) to a checkpoint \(K\) on a bearing of \(110^\circ\) for \(35\text{ km}\). It then flies from \(K\) to a landing pad \(L\) on a bearing of \(215^\circ\) for \(42\text{ km}\).
NOT TO SCALE
Calculate the bearing of \(H\) from \(L\).
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解題
1. Find angle \(\angle HKL\): - Back-bearing of \(H\) from \(K = 110^\circ + 180^\circ = 290^\circ\). - Bearing of \(L\) from \(K = 215^\circ\). - \(\angle HKL = 290^\circ - 215^\circ = 75^\circ\).
3. Find angle \(\angle HLK\) using the sine rule: \[ \frac{\sin(\angle HLK)}{35} = \frac{\sin(75^\circ)}{47.202} \] \[ \sin(\angle HLK) = \frac{35\sin(75^\circ)}{47.202} \approx 0.71622 \implies \angle HLK \approx 45.74^\circ \]
4. Calculate the bearing of \(H\) from \(L\): - Bearing of \(K\) from \(L = 215^\circ - 180^\circ = 035^\circ\). - Line \(LH\) lies counter-clockwise from \(LK\), so the bearing is: \[ 035^\circ - 45.74^\circ = -10.74^\circ \equiv 360^\circ - 10.74^\circ = 349.26^\circ \approx 349^\circ \text{ (or } 349.3^\circ\text{)} \]
評分準則
[M1] for calculating \(HL = 47.2\text{ km}\) using the cosine rule with \(\angle HKL = 75^\circ\) [M1] for calculating \(\angle HLK \approx 45.7^\circ\) using the sine rule or cosine rule [A1] for bearing \(349^\circ\) or \(349.3^\circ\) (accept \(349.2^\circ\) to \(349.3^\circ\))
題目 11 · structured
4 分
A painting was bought for $15\,000 in the year 2018. Its value increases exponentially at a rate of 4.5% per year. (a) Calculate the value of the painting in the year 2026. Give your answer correct to the nearest dollar. [2] (b) Find the year in which the value of the painting first exceeds $28\,000. [2]
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解題
(a) Number of years from 2018 to 2026 is \(2026 - 2018 = 8\) years. \(\text{Value} = 15000 \times (1 + 0.045)^8 = 15000 \times 1.045^8 = 21331.516...\) To the nearest dollar, the value is $21\,332. (b) We solve the inequality \(15000 \times 1.045^n > 28000\). \(1.045^n > \frac{28000}{15000} = \frac{28}{15}\). Taking logarithms: \(n > \frac{\log(28/15)}{\log(1.045)} = \frac{0.27107...}{0.019116...} \approx 14.18\) years. Since \(n = 14\) gives $27\,779.67 and \(n = 15\) gives $29\,029.75, the value first exceeds $28\,000 after 15 complete years. \(2018 + 15 = 2033\).
評分準則
(a) M1 for \(15000 \times 1.045^8\) oe; A1 for 21 332 (or 21 330 from 3 s.f.). (b) M1 for \(15000 \times 1.045^n = 28000\) or \(n = \frac{\log(28/15)}{\log(1.045)}\) or value 14.18 seen; A1 for 2033 cao.
題目 12 · structured
3 分
The mass, \(M\) grams, of a sample of a radioactive isotope decays according to the formula \(M = M_0 (0.88)^t\), where \(M_0\) is the initial mass in grams and \(t\) is the time in days. (a) Write down the percentage decrease in the mass of the sample each day. [1] (b) The initial mass of the sample is 450 g. Calculate the least number of whole days it takes for the mass to fall below 50 g. [2]
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解題
(a) Each day the mass is multiplied by 0.88, which represents \(100\% - 88\% = 12\%\) decrease per day. (b) We set up the inequality \(450 \times (0.88)^t < 50\). Dividing by 450 gives \(0.88^t < \frac{50}{450} = \frac{1}{9}\). Solving for \(t\): \(t > \frac{\log(1/9)}{\log(0.88)} \approx \frac{-0.95424}{-0.055517} \approx 17.19\). Testing integer values: at \(t = 17\), \(M = 450 \times 0.88^{17} \approx 50.84\text{ g} > 50\text{ g}\); at \(t = 18\), \(M = 450 \times 0.88^{18} \approx 44.74\text{ g} < 50\text{ g}\). Therefore, the least number of whole days is 18.
評分準則
(a) B1 for 12[%]. (b) M1 for \(450 \times 0.88^t = 50\) or \(t = \frac{\log(1/9)}{\log(0.88)}\) or 17.18... seen or evaluating \(M\) at \(t=17\) and \(t=18\); A1 for 18 cao.
M1 for correct expansion of any two brackets, seen or implied (e.g. \(2x^2 + 5x - 12\) or \(3x^2 + 11x - 4\) or \(6x^2 - 11x + 3\)) M1 for multiplying their quadratic by the third bracket, with at least 4 terms correct A1 for fully correct and simplified expression \(6x^3 + 13x^2 - 41x + 12\) cao
題目 14 · short_answer
3 分
Rearrange the formula to make \(w\) the subject. \[ P = \frac{5w + 2}{3 - 4w} \]
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解題
Multiply both sides by the denominator \((3 - 4w)\): \[ P(3 - 4w) = 5w + 2 \]
B1 for \(9x^2 - 12x + 4\) M1 for correct expansion of \((x + 4)(3x - 5) = 3x^2 + 7x - 20\) soi or correctly multiplying by \(-2\) A1 for \(3x^2 - 26x + 44\) cao
題目 16 · short_answer
3 分
NOT TO SCALE
The points \(A\), \(B\), \(C\), and \(D\) lie on the circumference of a circle.
M1 for angle \(AOB = 180 - 2(2x + 11)\) or \(158 - 4x\) oe seen M1 for setting up \(158 - 4x = 2(3x - 1)\) oe A1 for \(16\)
Paper 5 (Core Investigation)
Answer all questions. Full reasons and steps must be communicated clearly.
12 題目 · 40 分
題目 1 · investigation
3 分
Square frames are constructed using small square tiles. A frame of side length $n$ has shaded tiles along its outer border and an unshaded $(n-2) \times (n-2)$ square in the centre. The table shows the number of shaded border tiles, $B$, for frames of different side lengths $n$.
B1 for 20 (for $n=6$) B1 for 24 (for $n=7$) B1 for 28 (for $n=8$) [M1 for noticing constant difference of $+4$ if at least two values are correct from wrong arithmetic]
題目 2 · investigation
3 分
A sequence of honeycomb patterns is formed using equal-length rods. - Pattern 1 consists of $1$ regular hexagon and uses $6$ rods. - Pattern 2 adds a second hexagon sharing $1$ edge and uses $11$ rods. - Pattern 3 adds a third hexagon in the chain sharing $1$ edge and uses $16$ rods.
B1 for 21 (for $n=4$) B1 for 26 (for $n=5$) B1 for 31 (for $n=6$) [Allow C1 for identifying a common difference of $+5$ soi]
題目 3 · investigation
3 分
Straight cuts are made across a flat circular paper disc. No two cuts are parallel and no three cuts intersect at a single point. The table shows the maximum number of non-overlapping regions, $R$, created by $c$ cuts.
Complete the table for $c = 5$, $c = 6$, and $c = 7$.
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解題
Find the difference between consecutive numbers of regions: - From $c = 1$ to $c = 2$: $4 - 2 = +2$ - From $c = 2$ to $c = 3$: $7 - 4 = +3$ - From $c = 3$ to $c = 4$: $11 - 7 = +4$
The difference increases by $1$ with each additional cut: - For $c = 5$: $R = 11 + 5 = 16$ - For $c = 6$: $R = 16 + 6 = 22$ - For $c = 7$: $R = 22 + 7 = 29$
評分準則
B1 for 16 (for $c=5$) B1 for 22 (for $c=6$) B1 for 29 (for $c=7$) [M1 for sequence of differences $+5, +6, +7$ seen or implied]
題目 4 · investigation
3 分
A sequence of triangular dot patterns is formed. - Pattern 1 has $1$ dot. - Pattern 2 has $3$ dots. - Pattern 3 has $6$ dots. - Pattern 4 has $10$ dots.
Complete the table by finding the value of $D$ when $n = 5$, $n = 6$, and $n = 7$.
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解題
The successive differences between total dots are: - $3 - 1 = 2$ - $6 - 3 = 3$ - $10 - 6 = 4$
Continuing the pattern by adding consecutive integers: - For $n = 5$: $D = 10 + 5 = 15$ - For $n = 6$: $D = 15 + 6 = 21$ - For $n = 7$: $D = 21 + 7 = 28$
評分準則
B1 for 15 (for $n=5$) B1 for 21 (for $n=6$) B1 for 28 (for $n=7$)
題目 5 · investigation
3 分
L-shaped figures are constructed on a square grid using unit squares. Pattern $k$ has two perpendicular arms of length $k$ joined at a single corner square. The table shows the pattern number $k$, the total number of unit squares $S$, and the outer perimeter $P$.
Looking at the rows: 1. The number of squares $S$ increases by $2$ each time ($1, 3, 5, \dots$): - When $k = 4$: $S = 5 + 2 = 7$ - When $k = 5$: $S = 7 + 2 = 9$
2. The perimeter $P$ increases by $4$ each time ($4, 8, 12, \dots$): - When $k = 4$: $P = 12 + 4 = 16$ (Checking $k = 5$: $P = 16 + 4 = 20$, which matches the given table value).
B1 for $S = 7$ (when $k = 4$) B1 for $P = 16$ (when $k = 4$) B1 for $S = 9$ (when $k = 5$)
題目 6 · Pattern Rule Formulation & Communication
2.5 分
A sequence of fence designs is constructed using wooden slats. Design 1 uses 6 slats. Design 2 uses 11 slats. Design 3 uses 16 slats. Design 4 uses 21 slats. (a) State the difference between consecutive terms. (b) Find a formula for the number of slats, \(S\), needed for Design \(n\). Clearly communicate your reasoning.
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解題
First, calculate the first differences between consecutive terms: 11 - 6 = 5, 16 - 11 = 5, 21 - 16 = 5. Since the first difference is constant (d = 5), the formula is linear and has the form S = 5n + c. To find c, substitute n = 1: S = 5(1) + c = 6, which gives c = 1. Check with n = 2: S = 5(2) + 1 = 11, which matches. Therefore, the formula is S = 5n + 1.
評分準則
B1: for stating constant difference is 5 (or showing 5n in expression). M0.5: for setting up linear form S = 5n + c and substituting a known term to find c = 1 (or communication of adjusting zero-term). A1: for correct final formula S = 5n + 1 (or equivalent algebraic expression in terms of n).
題目 7 · Pattern Rule Formulation & Communication
2.5 分
A series of rectangular dot grids is investigated. Grid \(n\) has 3 rows and \(n + 2\) columns of dots. The total number of dots is \(T = 3(n + 2)\). The dots on the outer boundary of the grid are called border dots, \(B\). For Grid 1 (3 by 3), \(B = 8\). For Grid 2 (3 by 4), \(B = 10\). For Grid 3 (3 by 5), \(B = 12\). (a) Write down an expression for the number of border dots, \(B\), in terms of \(n\). (b) Using \(I = T - B\), communicate the algebraic steps to show that the number of interior dots, \(I\), is always equal to \(n\).
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解題
The border dots B form a linear sequence: 8, 10, 12, ... with a common difference of 2. For n = 1, B = 2(1) + 6 = 8, so B = 2n + 6. The total number of dots is T = 3(n + 2) = 3n + 6. The number of interior dots is given by I = T - B = (3n + 6) - (2n + 6). Expanding and simplifying: I = 3n + 6 - 2n - 6 = n. Thus, the number of interior dots is shown to equal n.
評分準則
B1: for finding B = 2n + 6 (or equivalent unsimplified expression for perimeter dots). M0.5: for setting up I = (3n + 6) - (2n + 6) with correct bracket usage. A1: for fully showing expansion 3n + 6 - 2n - 6 and arriving at I = n.
題目 8 · Pattern Rule Formulation & Communication
2.5 分
A sequence of total sums of consecutive even positive integers is given as follows: Row 1: 2 = 2. Row 2: 2 + 4 = 6. Row 3: 2 + 4 + 6 = 12. Row 4: 2 + 4 + 6 + 8 = 20. Let \(S_n\) be the sum for Row \(n\). (a) Write down the first two differences of the sequence 2, 6, 12, 20. (b) Formulate an expression for \(S_n\) in terms of \(n\), showing clearly how you established the formula.
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解題
First differences: 6 - 2 = 4, 12 - 6 = 6, 20 - 12 = 8. Second differences: 6 - 4 = 2, 8 - 6 = 2. Since the second difference is constant (2), the sequence is quadratic of the form S_n = an^2 + bn + c with 2a = 2, so a = 1. The values of S_n - n^2 are: for n=1: 2 - 1 = 1; for n=2: 6 - 4 = 2; for n=3: 12 - 9 = 3; for n=4: 20 - 16 = 4. This linear difference is n, so S_n - n^2 = n, which gives S_n = n^2 + n (or n(n + 1)).
評分準則
B1: for identifying first differences (4, 6, 8) and constant second difference (2). M0.5: for deducing the coefficient of n^2 is 2/2 = 1. A1: for complete communication leading to S_n = n^2 + n or n(n + 1).
題目 9 · Pattern Rule Formulation & Communication
2.5 分
The number of diagonals, \(D\), in a regular polygon with \(n\) sides is investigated for \(n \ge 3\): For a triangle (\(n = 3\)), \(D = 0\). For a quadrilateral (\(n = 4\)), \(D = 2\). For a pentagon (\(n = 5\)), \(D = 5\). For a hexagon (\(n = 6\)), \(D = 9\). By examining differences or the geometry of vertices, find the formula for \(D\) in terms of \(n\). Communicate your reasoning clearly.
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解題
Method 1 (Differences): The sequence of values for n = 3, 4, 5, 6 is 0, 2, 5, 9. First differences: 2, 3, 4. Second difference: 1. Constant second difference = 1 implies a quadratic expression with coefficient of n^2 equal to 1/2 = 0.5. Let D = 0.5n^2 + bn + c. Using n = 3: 0.5(9) + 3b + c = 0 => 3b + c = -4.5. Using n = 4: 0.5(16) + 4b + c = 2 => 4b + c = -6. Subtracting gives b = -1.5, and c = 0. Therefore, D = 0.5n^2 - 1.5n = 0.5n(n - 3). Method 2 (Geometric Communication): Each of the n vertices connects to (n - 3) non-adjacent vertices. Since each diagonal connects 2 vertices, total diagonals = n(n - 3)/2.
評分準則
M1: for showing second difference is 1 (so a = 0.5) OR communicating that each vertex connects to (n - 3) other vertices to form diagonals. M0.5: for substituting values to determine linear coefficients OR showing division by 2 to avoid double counting. A1: for the correct formula D = (n(n - 3))/2 or 0.5n^2 - 1.5n.
題目 10 · Investigation
5 分
This question is about investigating the number of factors of positive integers using their prime factorisations.
(a) Write 72 as a product of its prime factors in index form. [1]
(b) The number of factors of an integer written in the form \(N = 2^a \times 3^b\) is given by the rule: $$\text{Number of factors} = (a + 1)(b + 1)$$ Show that this rule gives 12 factors for 72. [1]
(c) An integer \(M\) has only two distinct prime factors, 2 and 3, so that \(M = 2^a \times 3^b\) where \(a \ge 1\) and \(b \ge 1\). \(M\) has exactly 10 factors. Find the two possible values of \(M\). [3]
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解題
(a) Prime factorisation of 72: \(72 = 8 \times 9 = 2^3 \times 3^2\).
(b) For \(72 = 2^3 \times 3^2\), we have \(a = 3\) and \(b = 2\). Using the formula: \((3 + 1)(2 + 1) = 4 \times 3 = 12\).
(c) We are given \((a + 1)(b + 1) = 10\), with integers \(a \ge 1\) and \(b \ge 1\). The factor pairs of 10 are \(1 \times 10\) and \(2 \times 5\). Since \(a \ge 1\) and \(b \ge 1\), \(a + 1 \ge 2\) and \(b + 1 \ge 2\). Case 1: \(a + 1 = 2 \implies a = 1\) \(b + 1 = 5 \implies b = 4\) \(M = 2^1 \times 3^4 = 2 \times 81 = 162\)
Therefore, the two possible values of \(M\) are 48 and 162.
評分準則
(a) B1 for \(2^3 \times 3^2\) cao
(b) C1 for substituting \(a=3\) and \(b=2\) to show \((3+1)(2+1) = 4 \times 3 = 12\)
(c) M1 for setting \((a+1)(b+1) = 10\) and identifying factor pairs \(2 \times 5\) or \(5 \times 2\) A1 for finding one correct value (48 or 162) A1 for finding both correct values (48 and 162)
題目 11 · Investigation
5 分
A student investigates the smallest positive integer multipliers needed to convert a given number into a square number or a cube number.
(a) Write 540 as a product of its prime factors in index form. [1]
(b) Find the smallest positive integer \(k\) such that \(540k\) is a square number. [2]
(c) Find the smallest positive integer \(m\) such that \(540m\) is a cube number. [2]
(b) For \(540k\) to be a square number, the power of every prime factor must be an even integer. \(540 = 2^2 \times 3^3 \times 5^1\) - Power of 2 is 2 (already even) - Power of 3 is 3 (needs at least one more factor of 3 to make \(3^4\)) - Power of 5 is 1 (needs at least one more factor of 5 to make \(5^2\)) Thus, the smallest multiplier is: \(k = 3^1 \times 5^1 = 15\).
(c) For \(540m\) to be a cube number, the power of every prime factor must be a multiple of 3. \(540 = 2^2 \times 3^3 \times 5^1\) - Power of 2 is 2 (needs one more factor of 2 to make \(2^3\)) - Power of 3 is 3 (already a multiple of 3) - Power of 5 is 1 (needs two more factors of 5 to make \(5^3\)) Thus, the smallest multiplier is: \(m = 2^1 \times 5^2 = 2 \times 25 = 50\).
評分準則
(a) B1 for \(2^2 \times 3^3 \times 5\) cao
(b) M1 for identifying missing odd powers \(3^1 \times 5^1\) soi A1 for 15 cao
(c) M1 for identifying powers required to reach multiples of 3, i.e. \(2^1 \times 5^2\) soi A1 for 50 cao
題目 12 · Investigation
5 分
The prime factorisation of 168 is \(2^3 \times 3 \times 7\).
(a) Write down the total number of prime factors of 168, counting repeats. [1]
(b) Two positive integers \(A\) and \(B\) satisfy the relationship: $$\text{HCF}(A, B) \times \text{LCM}(A, B) = A \times B$$ The highest common factor of 168 and an unknown integer \(W\) is 24. The lowest common multiple of 168 and \(W\) is 504. Show that \(W = 72\). [2]
(c) Find the smallest integer \(V\) greater than 24 such that \(\text{HCF}(168, V) = 24\). [2]
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解題
(a) The prime factorisation is \(2^3 \times 3^1 \times 7^1\). Total number of prime factors = \(3 + 1 + 1 = 5\).
(c) Since \(\text{HCF}(168, V) = 24\), \(V\) must be a multiple of 24. The multiples of 24 are 24, 48, 72, 96, ... We check the next multiple greater than 24, which is \(V = 48\): - Prime factorisation of 168: \(2^3 \times 3 \times 7\) - Prime factorisation of 48: \(2^4 \times 3\) - \(\text{HCF}(168, 48) = 2^{\min(3,4)} \times 3^{\min(1,1)} = 2^3 \times 3 = 24\). Since this satisfies the condition, the smallest integer is 48.
評分準則
(a) B1 for 5 cao
(b) M1 for correct substitution into formula: \(24 \times 504 = 168 \times W\) or \(W = \frac{24 \times 504}{168}\) A1 for showing calculation yields 72 with full steps shown
(c) M1 for considering multiples of 24 (e.g. \(24 \times 2 = 48\)) and testing HCF with 168 A1 for 48 cao
This investigation looks at the number of unit squares crossed by the diagonal of an integer grid.
A straight diagonal line is drawn from \((0,0)\) to \((w, h)\) across a grid of \(w \times h\) unit squares, where \(w\) and \(h\) are positive integers. Let \(N(w, h)\) be the number of \(1 \times 1\) unit squares whose interiors are crossed by the diagonal line.
(a) When \(w\) and \(h\) have no common factor greater than 1 (i.e. \(\gcd(w, h) = 1\)), the diagonal does not pass through any interior grid intersection.
Complete the table for cases where \(\gcd(w, h) = 1\):
(b) When \(\gcd(w, h) = g > 1\), the diagonal crosses through \(g - 1\) interior grid intersection points.
Write down an expression for \(N(w, h)\) in terms of \(w\), \(h\), and \(\gcd(w, h)\).
(c) Use your expression from part (b) to find the number of unit squares crossed by the diagonal of a \(84 \times 126\) grid.
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解題
(a) When \(\gcd(w, h) = 1\), each time the diagonal crosses a vertical grid line or horizontal grid line, it enters a new square. There are \(w - 1\) vertical interior lines and \(h - 1\) horizontal interior lines. Total squares crossed \(= 1 + (w - 1) + (h - 1) = w + h - 1\). - For \(w = 3, h = 5\): \(N(3, 5) = 3 + 5 - 1 = 7\). - For \(w = 4, h = 7\): \(N(4, 7) = 4 + 7 - 1 = 10\). - In terms of \(w\) and \(h\): \(N(w, h) = w + h - 1\).
(b) When \(\gcd(w, h) = g\), the diagonal is composed of \(g\) identical segments, each crossing a \(\frac{w}{g} \times \frac{h}{g}\) subgrid. Inside each subgrid, \(\gcd\left(\frac{w}{g}, \frac{h}{g}\right) = 1\), so the number of squares crossed per subgrid is \(\frac{w}{g} + \frac{h}{g} - 1\). Total squares crossed: $$N(w, h) = g \left(\frac{w}{g} + \frac{h}{g} - 1\right) = w + h - g = w + h - \gcd(w, h)$$
This investigation looks at the number of bounces a laser beam makes inside a rectangular grid table.
A laser beam is fired at a \(45^\circ\) angle from the bottom-left corner \((0,0)\) of a rectangle of width \(a\) and height \(b\), where \(a\) and \(b\) are integers with no common factors (i.e. \(\gcd(a, b) = 1\)). The beam reflects off the boundaries at \(45^\circ\) until it lands in one of the other three corners.
Let \(B(a, b)\) be the total number of reflections (bounces off the sides) before the beam hits a corner.
(a) Complete the table for coprime dimensions (\(\gcd(a, b) = 1\)):
$$\begin{array}{|c|c|c|} \hline a & b & B(a, b) \\ \hline 1 & 3 & 2 \\ 2 & 3 & 3 \\ 3 & 4 & 5 \\ 3 & 5 & \text{[i]} \\ 5 & 8 & \text{[ii]} \\ \hline a & b & \text{[iii]} \\ \hline \end{array}$$
(b) When \(a\) and \(b\) have a greatest common divisor \(\gcd(a, b) = g > 1\), the path scales.
Write down an expression for \(B(a, b)\) in terms of \(a\), \(b\), and \(\gcd(a, b)\).
(c) Calculate the total number of reflections \(B(60, 84)\).
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解題
(a) When \(\gcd(a, b) = 1\), the beam travels until the total horizontal distance is \(\text{lcm}(a, b) = ab\) and vertical distance is \(ab\). Number of horizontal reflections \(= b - 1\). Number of vertical reflections \(= a - 1\). Total reflections \(B(a, b) = (a - 1) + (b - 1) = a + b - 2\). - For \(a = 3, b = 5\): \(B(3, 5) = 3 + 5 - 2 = 6\). - For \(a = 5, b = 8\): \(B(5, 8) = 5 + 8 - 2 = 11\). - General expression for \(\gcd(a, b) = 1\): \(a + b - 2\).
(b) When \(\gcd(a, b) = g\), the grid reduces to an equivalent simplified grid of dimensions \(a' = \frac{a}{g}\) and \(b' = \frac{b}{g}\). The number of reflections is determined by the reduced dimensions: $$B(a, b) = a' + b' - 2 = \frac{a}{\gcd(a, b)} + \frac{b}{\gcd(a, b)} - 2 = \frac{a + b}{\gcd(a, b)} - 2$$
This investigation looks at the distinct points visited by a cyclic hopper.
A circle has \(n\) equally spaced markers numbered \(0, 1, 2, \dots, n - 1\) in clockwise order. A hopper starts at marker \(0\) and jumps clockwise in steps of size \(s\), visiting markers \((k \times s) \pmod n\) for \(k = 0, 1, 2, \dots\) until it returns to \(0\).
Let \(V(n, s)\) be the number of distinct markers visited by the hopper.
(b) Write down an expression for \(V(n, s)\) in terms of \(n\) and \(\gcd(n, s)\).
(c) A circle has \(n = 360\) markers. The hopper jumps in steps of \(s = 48\). Find the number of markers that are never visited by the hopper.
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解題
(a) The sequence of markers visited repeats when \(k \cdot s\) is a multiple of \(n\). The smallest positive integer \(k\) is \(\frac{\text{lcm}(n, s)}{s} = \frac{n}{\gcd(n, s)}\). - For \(n = 24, s = 9\): \(\gcd(24, 9) = 3\), so \(V(24, 9) = \frac{24}{3} = 8\). - For \(n = 24, s = 14\): \(\gcd(24, 14) = 2\), so \(V(24, 14) = \frac{24}{2} = 12\).
(b) $$V(n, s) = \frac{n}{\gcd(n, s)}$$
(c) For \(n = 360\) and \(s = 48\): Find \(\gcd(360, 48)\): $$360 = 24 \times 15$$ $$48 = 24 \times 2$$ $$\gcd(360, 48) = 24$$ Number of visited markers: $$V(360, 48) = \frac{360}{24} = 15$$ Number of unvisited markers: $$\text{Unvisited} = n - V(n, s) = 360 - 15 = 345$$
評分準則
(a) - B1: [i] \(8\) cao. - B1: [ii] \(2\) and [iii] \(12\) both correct.
This investigation looks at integer coordinate points lying on straight line segments.
A line segment is drawn from the origin \((0,0)\) to the point \((x, y)\), where \(x\) and \(y\) are positive integers. Let \(L(x, y)\) be the total number of integer coordinate points \((p, q)\) lying on this line segment, including both endpoints \((0,0)\) and \((x, y)\).
(b) Write down a formula for \(L(x, y)\) in terms of \(x\) and \(y\).
(c) A line segment connects \((0,0)\) to \((144, 216)\). Find the number of interior integer coordinate points on this segment (excluding both endpoints).
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解題
(a) The vector from \((0,0)\) to \((x, y)\) can be divided into \(g = \gcd(x, y)\) equal steps of \(\left(\frac{x}{g}, \frac{y}{g}\right)\). The points on the line segment are \(k \left(\frac{x}{g}, \frac{y}{g}\right)\) for \(k = 0, 1, 2, \dots, g\). The total number of points is \(g + 1 = \gcd(x, y) + 1\). - For \((8, 12)\): \(\gcd(8, 12) = 4\), so \(L(8, 12) = 4 + 1 = 5\). - For \((15, 20)\): \(\gcd(15, 20) = 5\), so \(L(15, 20) = 5 + 1 = 6\).
(b) $$L(x, y) = \gcd(x, y) + 1$$
(c) For \(x = 144, y = 216\): $$\gcd(144, 216) = 72$$ Total points including endpoints \(= 72 + 1 = 73\). Interior points exclude the 2 endpoints \((0,0)\) and \((144, 216)\): $$\text{Interior points} = \gcd(x, y) - 1 = 72 - 1 = 71$$
評分準則
(a) - B1: Rows for \((8, 12)\) and \((15, 20)\) all correct ([i]=4, [ii]=5, [iii]=5, [iv]=6).
This investigation explores the synchronization of two flashing beacons.
Beacon A flashes every \(p\) seconds and Beacon B flashes every \(q\) seconds, where \(p\) and \(q\) are positive integers. Both beacons flash together at time \(t = 0\). Let \(T(p, q)\) be the time in seconds between consecutive simultaneous flashes, and let \(S(p, q)\) be the total number of individual flashes (from either Beacon A or Beacon B or both) that occur in the half-open time interval \(0 < t \le T(p, q)\).
*(Note: When both beacons flash at the exact same second \(t\), it counts as 1 simultaneous event.)*
A single strand of decorative copper wire is wound in a uniform cylindrical helix around a solid pillar of radius \(r = 3\text{ cm}\) and vertical height \(h = 8\pi\text{ cm}\). The wire begins at the bottom base, completes exactly 1 full turn around the pillar, and ends at the top rim.
(a) By considering the unrolled curved surface of the cylinder as a rectangle, find the exact length of the wire, \(L_1\), in terms of \(\pi\). (b) Calculate the angle of inclination, \(\alpha\), that the wire makes with the horizontal base. Give your answer correct to 1 decimal place.
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解題
1. Unroll the curved surface of the cylinder into a rectangle: - The horizontal base width equals the cylinder's circumference: \(C = 2\pi r = 2\pi(3) = 6\pi\text{ cm}\). - The vertical height equals \(h = 8\pi\text{ cm}\).
2. The wire forms the diagonal hypotenuse of this right-angled rectangle: \[L_1 = \sqrt{C^2 + h^2} = \sqrt{(6\pi)^2 + (8\pi)^2} = \sqrt{36\pi^2 + 64\pi^2} = \sqrt{100\pi^2} = 10\pi\text{ cm}\]
M1 for calculating circumference \(C = 2\pi(3) = 6\pi\) soi M1 for applying Pythagoras' theorem \(L_1 = \sqrt{(6\pi)^2 + (8\pi)^2}\) A1 for exact length \(10\pi\) (accept \(31.4\text{ cm}\) if \(10\pi\) seen) B1.17 (or M1 A0.17) for \(\tan \alpha = \frac{8\pi}{6\pi}\) leading to \(53.1^\circ\) (accept 53.13...)
題目 7 · Investigation & Modelling
4.17 分
A flexible cable is wound in a uniform helix of \(n\) identical complete turns around a cylindrical column of height \(h\) and radius \(r\).
(a) Show that the total length of the cable, \(L(n)\), is given by: \[L(n) = \sqrt{h^2 + 4\pi^2 n^2 r^2}\] (b) A support column has height \(h = 24\text{ cm}\) and radius \(r = \frac{7}{\pi}\text{ cm}\). Find the total length of cable required to make \(n = 5\) complete turns.
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解題
1. When the curved surface of the cylinder is unrolled, each complete turn covers a horizontal distance equal to one circumference \(C = 2\pi r\). 2. For \(n\) turns, the total unwrapped horizontal distance is \(n \times C = 2\pi n r\). 3. The vertical height spanned is \(h\). 4. By Pythagoras' theorem on the unrolled right triangle: \[L(n) = \sqrt{h^2 + (2\pi n r)^2} = \sqrt{h^2 + 4\pi^2 n^2 r^2}\] 5. For \(h = 24\), \(r = \frac{7}{\pi}\), and \(n = 5\): \[2\pi n r = 2\pi(5)\left(\frac{7}{\pi}\right) = 70\text{ cm}\] \[L(5) = \sqrt{24^2 + 70^2} = \sqrt{576 + 4900} = \sqrt{5476} = 74\text{ cm}\]
評分準則
M1 for expressing total unwrapped horizontal length as \(2\pi n r\) or \(n C\) A1 for clear algebraic derivation: \(L(n) = \sqrt{h^2 + (2\pi n r)^2} = \sqrt{h^2 + 4\pi^2 n^2 r^2}\) M1 for substituting \(h=24\), \(r=7/\pi\), \(n=5\) into formula giving horizontal base \(= 70\) A1.17 for final answer \(74\text{ cm}\) nfww
題目 8 · Investigation & Modelling
4.17 分
A flat protective ribbon of constant width \(w\text{ cm}\) is wound spirally around a pipe of radius \(R = 5\text{ cm}\) and length \(H = 60\text{ cm}\) in \(n = 4\) complete turns so that adjacent edges touch exactly with no overlap and no gaps.
(a) Find the pitch \(p\) (vertical rise per turn) of the helix. (b) Using the relation \(w = p \cos \theta\), where \(\theta\) is the angle between the helical path and the horizontal circumference, calculate the required ribbon width \(w\). Give your answer correct to 3 significant figures.
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解題
1. The vertical pitch per turn is: \[p = \frac{H}{n} = \frac{60}{4} = 15\text{ cm}\]
2. The circumference of the pipe is: \[C = 2\pi R = 2\pi(5) = 10\pi \approx 31.4159\text{ cm}\]
3. The length of one turn of the helical midline is: \[L_1 = \sqrt{C^2 + p^2} = \sqrt{(10\pi)^2 + 15^2} = \sqrt{100\pi^2 + 225} \approx \sqrt{986.9604 + 225} = \sqrt{1211.9604} \approx 34.8132\text{ cm}\]
4. The cosine of the helix angle \(\theta\) is: \[\cos \theta = \frac{C}{L_1} = \frac{10\pi}{34.8132} \approx 0.902414\]
5. The perpendicular width \(w\) of the ribbon is: \[w = p \cos \theta = 15 \times 0.902414 \approx 13.536\text{ cm} \approx 13.5\text{ cm}\]
評分準則
B1 for pitch \(p = \frac{60}{4} = 15\text{ cm}\) M1 for calculating \(C = 10\pi\) and single-turn hypotenuse \(L_1 = \sqrt{(10\pi)^2 + 15^2} \approx 34.81\) M1 for \(\cos \theta = \frac{10\pi}{L_1}\) or \(\theta = \tan^{-1}\left(\frac{15}{10\pi}\right) \approx 25.52^\circ\) A1.17 for \(w = 13.5\text{ cm}\) (or \(13.54\text{ cm}\))
題目 9 · Investigation & Modelling
4.17 分
An insulated sensor wire has a fixed maximum length of \(L = 150\text{ cm}\). It must be coiled in a uniform cylindrical helix around a pipe of outer radius \(r = 4\text{ cm}\).
The total vertical height \(h\) spanned along the pipe must be at least \(90\text{ cm}\).
Find the maximum integer number of complete turns, \(n\), that can be made.
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解題
1. The helical length model is \(L^2 = h^2 + (2\pi n r)^2\). 2. Rearranging for \(n\): \[(2\pi n r)^2 = L^2 - h^2\] 3. Since \(h \ge 90\) and \(L = 150\): \[(2\pi n r)^2 \le 150^2 - 90^2 = 22500 - 8100 = 14400\] 4. Taking the square root on both sides: \[2\pi n r \le \sqrt{14400} = 120\] 5. Substitute \(r = 4\): \[2\pi (4) n \le 120 \implies 8\pi n \le 120 \implies n \le \frac{120}{8\pi} = \frac{15}{\pi} \approx 4.7746\] 6. Since \(n\) must be a whole number of complete turns, the maximum integer value is \(n = 4\).
評分準則
M1 for setting up inequality \(h^2 + (2\pi n r)^2 \le 150^2\) with \(h = 90\) M1 for \((2\pi n (4))^2 \le 150^2 - 90^2 = 14400\) soi A1 for obtaining \(n \le \frac{15}{\pi}\) or \(n \le 4.77\dots\) B1.17 for correctly stating maximum integer \(n = 4\)
題目 10 · Investigation & Modelling
4.17 分
A square-threaded screw operates mechanically as an inclined plane wrapped around a cylinder. The screw has mean radius \(r = 6\text{ mm}\) and pitch \(p = 4\text{ mm}\).
(a) The helix lead angle \(\psi\) satisfies \(\tan \psi = \frac{p}{2\pi r}\). Calculate \(\psi\) in degrees, correct to 2 decimal places. (b) The mechanical efficiency \(\eta\) of the screw is given by the formula: \[\eta = \frac{\tan \psi}{\tan(\psi + \phi)}\] where \(\phi\) is the friction angle such that \(\tan \phi = 0.15\). Calculate the efficiency \(\eta\) as a percentage, correct to 1 decimal place.
M1 for \(\tan \psi = \frac{4}{12\pi}\) leading to \(\psi = 6.06^\circ\) (accept 6.056...) M1 for finding \(\phi = \tan^{-1}(0.15) \approx 8.53^\circ\) M1 for evaluating \(\tan(\psi + \phi) \approx 0.260\) A1.17 for \(\eta = 40.8\%\) or \(40.9\%\) (accept decimal \(0.408\) to \(0.409\))
題目 11 · Investigation & Modelling
4.17 分
A conical helical path runs along the curved surface of a right circular cone with base radius \(R = 12\text{ cm}\) and slant height \(s = 20\text{ cm}\). When the curved surface of the cone is cut along a generator and unrolled flat, it forms a sector of a circle of radius \(s = 20\text{ cm}\) and central angle \(\theta\).
(a) Show that the sector angle \(\theta\) is \(216^\circ\). (b) A path goes from a point \(A\) at the base of the seam to a point \(B\) halfway up the opposite generator (so \(OB = 10\text{ cm}\) where \(O\) is the vertex, and the angle between \(OA\) and \(OB\) on the unrolled sector is \(216^\circ\)). Find the shortest length of this path along the surface of the cone. Give your answer correct to 3 significant figures.
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解題
1. The circumference of the cone base is \(C = 2\pi R = 2\pi(12) = 24\pi\text{ cm}\). 2. The circumference of the complete circle of radius \(s = 20\text{ cm}\) is \(2\pi(20) = 40\pi\text{ cm}\). 3. The sector angle is therefore: \[\theta = \frac{24\pi}{40\pi} \times 360^\circ = \frac{12}{20} \times 360^\circ = 0.6 \times 360^\circ = 216^\circ\]
4. On the flattened sector, the shortest path between \(A\) and \(B\) is the straight line segment \(AB\). In triangle \(OAB\): - \(OA = s = 20\text{ cm}\) - \(OB = 10\text{ cm}\) - \(\angle AOB = 216^\circ\)