An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.
Extended Structured 部分
Answer all questions. Calculators are allowed. Show all your working clearly to gain method marks.
11 题目 · 121 分
题目 1 · Structured
11 分
The first four terms of sequence $A$ are $3, 7, 13, 21, \dots$
(a) (i) Find the next two terms of sequence $A$. [2] (ii) Find an expression, in terms of $n$, for the $n$-th term of sequence $A$. [3]
(b) The first four terms of sequence $B$ are $5, 11, 17, 23, \dots$
(i) Find the next term of sequence $B$. [1] (ii) Find an expression, in terms of $n$, for the $n$-th term of sequence $B$. [2]
(c) Find an expression, in terms of $n$ in its simplest form, for the $n$-th term of a sequence $D$, formed by subtracting the $n$-th term of sequence $B$ from the $n$-th term of sequence $A$. [3]
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解题
(a) (i) The differences are $4, 6, 8, \dots$, so the next differences are $10$ and $12$. $21 + 10 = 31$, $31 + 12 = 43$. The next two terms are $31$ and $43$. (ii) The second difference is constant and equals $2$. This means the sequence is quadratic of the form $an^2 + bn + c$, with $a = 2 / 2 = 1$. Subtracting $n^2$ from the sequence: $n=1: 3 - 1 = 2$ $n=2: 7 - 4 = 3$ $n=3: 13 - 9 = 4$ $n=4: 21 - 16 = 5$ This linear remainder is $n+1$. So the $n$-th term is $n^2 + n + 1$.
(b) (i) The difference is constant and equals $6$. $23 + 6 = 29$. (ii) The linear sequence has the form $6n + c$. For $n=1$, $6(1) + c = 5 \implies c = -1$. So the $n$-th term is $6n - 1$.
(c) $n$-th term of $D = (n^2 + n + 1) - (6n - 1) = n^2 + n + 1 - 6n + 1 = n^2 - 5n + 2$.
评分标准
(a) (i) B2 for $31, 43$ (B1 for one correct) (ii) M1 for recognizing quadratic form or second diff is 2 M1 for subtracting $n^2$ from terms A1 for $n^2 + n + 1$ (or equivalent) (b) (i) B1 for $29$ (ii) M1 for $6n + c$ or $6n$ A1 for $6n - 1$ (c) M1 for subtracting their expressions M1 for expanding and simplifying A1 for $n^2 - 5n + 2$ (cao)
题目 2 · Structured
11 分
A surveyor measures a triangular plot of land $PQR$. $PQ = 45\text{ m}$, $QR = 62\text{ m}$, and angle $PQR = 78^\circ$.
(a) Calculate the length of $PR$. [3] (b) Calculate angle $QPR$. [3] (c) Calculate the area of the triangular plot. [2] (d) A vertical flagpole stands at $Q$. The angle of elevation of the top of the flagpole from $P$ is $12^\circ$. Calculate the height of the flagpole. [3]
(b) Using the Sine Rule: $\frac{\sin(QPR)}{QR} = \frac{\sin(PQR)}{PR}$ $\sin(QPR) = \frac{62 \sin(78^\circ)}{68.62} \approx 0.88378$ Angle $QPR = \arcsin(0.88378) \approx 62.1^\circ$ (to 1 d.p.)
(c) Area $= \frac{1}{2} \times PQ \times QR \times \sin(PQR)$ Area $= \frac{1}{2} \times 45 \times 62 \times \sin(78^\circ) \approx 1360\text{ m}^2$ (to 3 s.f.)
(d) Let the height of the flagpole be $h$. In the right-angled triangle formed by $P$, $Q$, and the top of the flagpole $T$: $\tan(12^\circ) = \frac{h}{PQ} = \frac{h}{45}$ $h = 45 \tan(12^\circ) \approx 9.57\text{ m}$.
评分标准
(a) M1 for correct substitution into Cosine Rule formula A1 for $PR^2 \approx 4710$ A1 for $68.6$ or $68.62$ (b) M1 for correct substitution into Sine Rule formula M1 for rearranging to find $\sin(QPR)$ A1 for $62.1^\circ$ (c) M1 for $\frac{1}{2} \times 45 \times 62 \times \sin(78^\circ)$ A1 for $1360$ (accept $1364.5$ - $1365$) (d) M1 for identifying right-angled triangle with base $45$ M1 for $45 \tan(12^\circ)$ A1 for $9.57$ (accept $9.6$)
题目 3 · Structured
11 分
Triangle $T$ has vertices $A(1, 2)$, $B(4, 2)$, and $C(1, 4)$.
(a) Triangle $T$ is reflected in the line $y = -1$ to give triangle $T_1$. Find the coordinates of the vertices of $T_1$. [3] (b) Triangle $T$ is translated by vector $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$ to give triangle $T_2$. Find the coordinates of the vertices of $T_2$. [2] (c) Triangle $T$ is transformed by a shear, with the $x$-axis invariant and shear factor $2$. Find the coordinates of the vertices of the image triangle $T_3$. [3] (d) Describe fully the single transformation that maps triangle $T$ onto triangle $T_4$ with vertices $A'(-2, -4)$, $B'(-8, -4)$, and $C'(-2, -8)$. [3]
(d) Comparing vertices: $A(1, 2) \to A'(-2, -4)$ $B(4, 2) \to B'(-8, -4)$ $C(1, 4) \to C'(-2, -8)$ Each coordinate is multiplied by $-2$. This is an enlargement with scale factor $-2$ and center $(0, 0)$.
评分标准
(a) B3 for $(1, -4), (4, -4), (1, -6)$ (B1 for each correct coordinate pair) (b) B2 for $(-2, 3), (1, 3), (-2, 5)$ (B1 for any two correct) (c) M1 for using the shear transformation formula $(x + ky, y)$ M1 for setting $k=2$ A1 for $(5, 2), (8, 2), (9, 4)$ (all correct) (d) B1 for Enlargement B1 for scale factor $-2$ B1 for center $(0, 0)$
题目 4 · Structured
11 分
Let $f(x) = \frac{2x - 5}{x - 3}$ for $x eq 3$.
(a) Write down the equations of the asymptotes of $y = f(x)$. [2] (b) Find the coordinates of the points where the graph of $y = f(x)$ cuts: (i) the $x$-axis, [2] (ii) the $y$-axis. [1] (c) Solve the equation $f(x) = x$. [3] (d) A line $y = -x + k$ is tangent to the curve $y = f(x)$. Find the possible values of $k$. [3]
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解题
(a) Vertical asymptote when denominator is zero: $x = 3$. Horizontal asymptote as $x \to \pm \infty$: $y = 2$.
(b) (i) Cuts $x$-axis when $y = 0 \implies 2x - 5 = 0 \implies x = 2.5$. Coordinates: $(2.5, 0)$. (ii) Cuts $y$-axis when $x = 0 \implies y = \frac{2(0) - 5}{0 - 3} = \frac{5}{3} \approx 1.67$. Coordinates: $(0, 1.67)$.
(a) B1 for vertical asymptote $x = 3$ B1 for horizontal asymptote $y = 2 (b) (i) M1 for setting $2x - 5 = 0$ A1 for $(2.5, 0)$ (accept $x = 2.5$) (ii) B1 for $(0, 1.67)$ or $(0, \frac{5}{3})$ (c) M1 for setting up the equation $x^2 - 5x + 5 = 0$ M1 for applying the quadratic formula A1 for $3.62$ and $1.38$ (accept exact values $\frac{5 \pm \sqrt{5}}{2}$) (d) M1 for equating and forming $x^2 - (k+1)x + (3k-5) = 0$ M1 for setting discriminant equal to $0$ A1 for $k = 3$ and $k = 7$
题目 5 · Structured
11 分
A solid toy is formed by joining a cone of radius $r$ and height $3r$ to a hemisphere of radius $r$.
(a) Find an expression, in terms of $\pi$ and $r$, for: (i) the volume of the cone, [2] (ii) the volume of the hemisphere. [2] (b) The total volume of the toy is $250\text{ cm}^3$. Calculate the radius $r$, correct to 3 significant figures. [3] (c) Calculate the total surface area of the toy, using the value of $r$ from part (b). [4]
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解题
(a) (i) Volume of cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (3r) = \pi r^3$. (ii) Volume of hemisphere $= \frac{2}{3} \pi r^3$.
(c) Total surface area of the toy consists of the curved surface area of the hemisphere and the curved surface area of the cone. Curved surface area of hemisphere $= 2 \pi r^2$. Curved surface area of cone $= \pi r l$, where $l = \sqrt{r^2 + h^2} = \sqrt{r^2 + (3r)^2} = \sqrt{10r^2} = r\sqrt{10}$. Total surface area $= 2\pi r^2 + \pi r^2 \sqrt{10} = \pi r^2(2 + \sqrt{10})$. Using $r = 3.6278\text{ cm}$: Total surface area $= \pi (3.6278)^2 (2 + \sqrt{10}) \approx 41.347 \times (2 + 3.1623) \approx 213\text{ cm}^2$.
评分标准
(a) (i) M1 for correct formula $\frac{1}{3} \pi r^2 h$ A1 for $\pi r^3$ (ii) B2 for $\frac{2}{3} \pi r^3$ (B1 for sphere volume $\frac{4}{3} \pi r^3$) (b) M1 for setting up $\frac{5}{3} \pi r^3 = 250$ M1 for rearranging to find $r^3$ A1 for $3.63$ (or $3.628$) (c) M1 for curved surface area of hemisphere $2\pi r^2$ M1 for finding slant height $l = r\sqrt{10}$ or $11.47\text{ cm}$ M1 for summing curved surface area of cone and hemisphere A1 for $213$ (accept $213$ - $214$)
题目 6 · Structured
11 分
The table below shows the weekly study hours, $x$, and exam scores, $y$ (in percentages), of 8 students.
(a) Describe the type of correlation shown by the data. [1] (b) Find the equation of the regression line of $y$ on $x$ in the form $y = mx + c$. [3] (c) Use your regression line to: (i) estimate the exam score of a student who studies for 11 hours per week, [2] (ii) comment on the reliability of this estimate. [2] (d) Another student studies for 30 hours per week. Explain why the regression line may not give a reliable estimate for this student's score. [3]
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解题
(a) Strong positive correlation.
(b) Using a calculator to find the linear regression line: $m \approx 3.6428$ $c \approx 34.964$ So $y = 3.64x + 35.0$ (to 3 s.f.)
(c) (i) Substitute $x = 11$: $y = 3.6428(11) + 34.964 = 40.07 + 34.964 = 75.03 \approx 75\%$. (ii) This estimate is highly reliable because $x = 11$ lies well within the range of the experimental data $[2, 16]$ (interpolation), and the data shows strong positive correlation.
(d) Studying for 30 hours is far outside the range of the studied data $[2, 16]$ (extrapolation). Additionally, since the maximum exam score is $100\%$, the model would predict $3.64(30) + 35 = 144.2\%$, which is mathematically impossible.
评分标准
(a) B1 for positive (accept strong positive) (b) M2 for finding gradient $m \approx 3.64$ and intercept $c \approx 35.0$ (M1 for one correct or correct method with calculator) A1 for $y = 3.64x + 35.0$ (c) (i) M1 for substituting $x = 11$ into their regression equation A1 for $75$ (accept $75.03$) (ii) B1 for mentioning interpolation / within the data range B1 for mentioning strong correlation (d) B1 for extrapolation / outside range of data B1 for pointing out maximum score limit of $100\%$ B1 for calculating impossible score $144\%$ under linear model
题目 7 · Structured
11 分
A group of 80 students are surveyed about their favorite sports among Football ($F$), Basketball ($B$), and Tennis ($T$). Every student likes at least one of these sports.
- $35$ students like Football. - $42$ students like Basketball. - $30$ students like Tennis. - $12$ students like Football and Basketball. - $10$ students like Basketball and Tennis. - $8$ students like Football and Tennis. - $x$ students like all three sports.
(a) Write down an expression, in terms of $x$, for the number of students who like: (i) Football and Basketball but not Tennis, [1] (ii) Football only. [2] (b) Show that the total number of students can be expressed as $77 + x$. [3] (c) Find the value of $x$. [2] (d) Find the probability that a student chosen at random from this group likes exactly two of these sports. [3]
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解题
(a) (i) Football and Basketball but not Tennis $= 12 - x$. (ii) Football only $= 35 - (12-x) - (8-x) - x = 15 + x$.
(b) Let us represent the number of students in each of the 7 regions of the Venn Diagram: - $F \cap B \cap T = x$ - $F \cap B$ only $= 12-x$ - $B \cap T$ only $= 10-x$ - $F \cap T$ only $= 8-x$ - $F$ only $= 15+x$ - $B$ only $= 42 - (12-x) - (10-x) - x = 20+x$ - $T$ only $= 30 - (8-x) - (10-x) - x = 12+x$
Total number of students $= x + (12-x) + (10-x) + (8-x) + (15+x) + (20+x) + (12+x) = 77 + x$.
(c) Since total $= 80$: $77 + x = 80 \implies x = 3$.
(d) The number of students who like exactly two sports is: $(12-x) + (10-x) + (8-x) = 30 - 3x$. Since $x=3$, number of students $= 30 - 3(3) = 21$. Probability $= \frac{21}{80} = 0.2625$.
评分标准
(a) (i) B1 for $12 - x$ (ii) M1 for subtracting intersections from $35$ A1 for $15 + x$ (b) M1 for finding expressions for $B$ only ($20+x$) or $T$ only ($12+x$) M1 for writing the sum of all seven regions A1 for showing it simplifies to $77 + x$ (c) M1 for $77 + x = 80$ A1 for $x = 3$ (d) M1 for identifying exactly two sports formula $(12-x) + (10-x) + (8-x)$ M1 for substituting $x = 3$ to get $21$ A1 for $\frac{21}{80}$ or $0.2625$ (accept $26.25\%$)
(a) Show that $5x^2 - 12x - 17 = 0$. [3] (b) Solve the equation $5x^2 - 12x - 17 = 0$ by factorising. [3] (c) Find the corresponding values of $y$. [2] (d) State the geometric interpretation of your answers to part (b) and part (c). [3]
(c) Substitute $x$ back into $y = 2x - 3$: For $x = -1 \implies y = 2(-1) - 3 = -5$. For $x = 3.4 \implies y = 2(3.4) - 3 = 3.8$.
(d) The geometric interpretation is that the straight line $y = 2x - 3$ intersects the circle $x^2 + y^2 = 26$ at exactly two points: $(-1, -5)$ and $(3.4, 3.8)$.
评分标准
(a) M1 for correct substitution of $(2x - 3)^2$ M1 for expanding $(2x - 3)^2 = 4x^2 - 12x + 9$ A1 for showing $5x^2 - 12x - 17 = 0$ without any errors (b) M1 for attempting factorisation (e.g., splitting middle term into $-17x + 5x$) M1 for $(5x - 17)(x + 1) = 0$ A1 for $x = 3.4$ (accept $17/5$) and $x = -1$ (c) M1 for substituting one value of $x$ to find $y$ A1 for $y = -5$ and $y = 3.8$ (d) B1 for line and circle B1 for intersection B1 for listing both points $(-1, -5)$ and $(3.4, 3.8)$
题目 9 · 結構題
11 分
The diagram shows a triangle \(PQR\) with \(PQ = 18.4\text{ cm}\) and angle \(QPR = 62^\circ\). The area of triangle \(PQR\) is \(112\text{ cm}^2\).
(a) Show that \(PR = 13.8\text{ cm}\), correct to 1 decimal place. (b) Find \(QR\). (c) Find angle \(PRQ\). (d) Find the shortest distance from \(P\) to the line \(QR\).
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解题
(a) Area of a triangle is given by \(\frac{1}{2} a b \sin C\). \(\frac{1}{2} \times PQ \times PR \times \sin(QPR) = 112\) \(\frac{1}{2} \times 18.4 \times PR \times \sin(62^\circ) = 112\) \(9.2 \times PR \times \sin(62^\circ) = 112\) \(PR = \frac{112}{9.2 \times \sin(62^\circ)} \approx 13.788\text{ cm}\) PR = \(13.8\text{ cm}\) (correct to 1 d.p.)
(c) Using the sine rule: \(\frac{\sin(PRQ)}{PQ} = \frac{\sin(QPR)}{QR}\) \(\frac{\sin(PRQ)}{18.4} = \frac{\sin(62^\circ)}{17.043}\) \(\sin(PRQ) = \frac{18.4 \times \sin(62^\circ)}{17.043} \approx 0.9533\) \(\text{angle } PRQ = \sin^{-1}(0.9533) \approx 72.4^\circ\)
(d) The shortest distance \(d\) from \(P\) to \(QR\) can be found using the area formula: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\) \(112 = \frac{1}{2} \times QR \times d\) \(112 = \frac{1}{2} \times 17.043 \times d\) \(d = \frac{224}{17.043} \approx 13.1\text{ cm}\)
评分标准
(a) M1 for \(\frac{1}{2} \times 18.4 \times PR \times \sin(62^\circ) = 112\) or equivalent. A1 for obtaining \(13.78...\) and rounding to \(13.8\).
(b) M1 for correct substitution into the cosine rule formula: \(18.4^2 + 13.8^2 - 2 \times 18.4 \times 13.8 \times \cos(62^\circ)\). A1 for \(QR^2 = 290.46\) or equivalent. A1 for \(17.0\text{ cm}\) or \(17.04...\text{ cm}\).
(c) M1 for correct substitution into the sine rule: \(\frac{\sin(PRQ)}{18.4} = \frac{\sin(62^\circ)}{\text{their } QR}\). A1 for \(\sin(PRQ) = 0.953...\) or equivalent. A1 for \(72.4^\circ\) or \(72.42...\) (accept \(72.3^\circ\) - \(72.5^\circ\)).
(d) M1 for using Area = \(\frac{1}{2} \times \text{base} \times \text{height}\) or \(PR \times \sin(PRQ)\). M1 for substituting their values: \(112 = \frac{1}{2} \times 17.04 \times d\) or \(13.79 \times \sin(72.4^\circ)\). A1 for \(13.1\text{ cm}\) or \(13.14...\text{ cm}\).
题目 10 · 結構題
11 分
(a) Consider the sequence: \(5, 11, 21, 35, 53, \dots\) (i) Find the next two terms of this sequence. [2] (ii) Find an expression, in terms of \(n\), for the \(n\)th term of this sequence. [3]
(b) A second sequence has \(n\)th term \(3 \times 2^{n-1}\). (i) Find the first three terms of this sequence. [2] (ii) Find which term in this sequence has a value of 1536. [2]
(c) The \(n\)th term of a third sequence is given by \(T_n = \frac{n}{2n+1}\). Find the difference \(T_{n+1} - T_n\), giving your answer as a single fraction in its simplest form. [2]
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解题
(a) (i) The differences between consecutive terms are: \(11 - 5 = 6\) \(21 - 11 = 10\) \(35 - 21 = 14\) \(53 - 35 = 18\) These differences increase by 4 each time. The next difference is \(18 + 4 = 22\). The next term is \(53 + 22 = 75\). The next difference is \(22 + 4 = 26\). The next term after that is \(75 + 26 = 101\). So the next two terms are 75 and 101.
(ii) Since the second difference is constant at 4, the sequence is quadratic of the form \(a n^2 + b n + c\) where \(a = \frac{4}{2} = 2\). Subtracting \(2n^2\) from each term: For \(n=1\): \(5 - 2(1^2) = 3\) For \(n=2\): \(11 - 2(2^2) = 3\) For \(n=3\): \(21 - 2(3^2) = 3\) Since the subtraction yields a constant 3, the \(n\)th term is \(2n^2 + 3\).
(b) (i) For \(n=1\): \(3 \times 2^0 = 3\) For \(n=2\): \(3 \times 2^1 = 6\) For \(n=3\): \(3 \times 2^2 = 12\) The first three terms are 3, 6, 12.
(ii) Set \(3 \times 2^{n-1} = 1536\): \(2^{n-1} = 512\) Since \(2^9 = 512\), we have \(n - 1 = 9 \implies n = 10\). So it is the 10th term.
(ii) M1 for recognizing a quadratic sequence with \(2n^2\) term. M1 for setting up equations or subtracting \(2n^2\): e.g., \(2n^2 + bn + c\). A1 for \(2n^2 + 3\).
(b) (i) B1 for any two correct terms. B1 for all three correct: 3, 6, 12.
(ii) M1 for \(3 \times 2^{n-1} = 1536\) or \(2^{n-1} = 512\). A1 for \(10\).
(c) M1 for writing \(\frac{n+1}{2n+3} - \frac{n}{2n+1}\). A1 for \(\frac{1}{(2n+3)(2n+1)}\).
题目 11 · 結構題
11 分
A solid cylinder has radius \(r\) and height \(h\). A solid sphere has radius \(r\). The total surface area of the cylinder is equal to the surface area of the sphere.
(a) Show that \(h = r\). [3] (b) Show that the ratio of the volume of the cylinder to the volume of the sphere is \(3 : 4\). [3] (c) The volume of the sphere is \(288\pi\text{ cm}^3\). (i) Find the value of \(r\). [2] (ii) Find the total surface area of the cylinder. Give your answer in terms of \(\pi\). [3]
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解题
(a) The total surface area of a solid cylinder is \(2\pi r^2 + 2\pi r h\). The surface area of a sphere is \(4\pi r^2\). Since they are equal: \(2\pi r^2 + 2\pi r h = 4\pi r^2\) \(2\pi r h = 4\pi r^2 - 2\pi r^2\) \(2\pi r h = 2\pi r^2\) Dividing both sides by \(2\pi r\) (since \(r > 0\)): \(h = r\).
(b) The volume of the cylinder is \(V_{\text{cylinder}} = \pi r^2 h = \pi r^2 (r) = \pi r^3\). The volume of the sphere is \(V_{\text{sphere}} = \frac{4}{3}\pi r^3\). The ratio is: \(V_{\text{cylinder}} : V_{\text{sphere}} = \pi r^3 : \frac{4}{3}\pi r^3\) \(= 1 : \frac{4}{3}\) \(= 3 : 4\).
(c) (i) The volume of the sphere is \(288\pi\text{ cm}^3\). \(\frac{4}{3}\pi r^3 = 288\pi\) \(\frac{4}{3} r^3 = 288\) \(r^3 = 288 \times \frac{3}{4}\) \(r^3 = 216\) \(r = \sqrt[3]{216} = 6\text{ cm}\).
(ii) Since the total surface area of the cylinder is equal to the surface area of the sphere: \(\text{Surface Area} = 4\pi r^2\) Substituting \(r = 6\): \(\text{Surface Area} = 4\pi (6)^2 = 4\pi \times 36 = 144\pi\text{ cm}^2\). (Alternatively, using the cylinder formula: \(2\pi(6)^2 + 2\pi(6)(6) = 72\pi + 72\pi = 144\pi\text{ cm}^2\).)
评分标准
(a) M1 for writing cylinder area: \(2\pi r^2 + 2\pi rh\). M1 for equating: \(2\pi r^2 + 2\pi rh = 4\pi r^2\). A1 for completing the algebraic simplification to get \(h = r\).
(b) M1 for \(V_{\text{cylinder}} = \pi r^3\) (substituting \(h = r\)). M1 for setting up the ratio: \(\pi r^3 : \frac{4}{3}\pi r^3\) or \(\frac{\pi r^3}{\frac{4}{3}\pi r^3}\). A1 for simplifying the ratio to \(3 : 4\).
(c) (i) M1 for \(\frac{4}{3}\pi r^3 = 288\pi\) or \(r^3 = 216\). A1 for \(r = 6\).
(ii) M1 for substituting \(r = 6\) into either sphere or cylinder surface area formula. M1 for calculating \(4\pi \times 36\) or \(2\pi(36) + 2\pi(36)\). A1 for \(144\pi\).