2023 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

(a) The sketch should show a cubic curve with a local maximum in the second quadrant and a local minimum in the fourth quadrant, passing through the y-intercept at \((0,5)\).

(b) Using a graphic display calculator:
(i) The local maximum is at \(x = -1\).
\(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = 10\).
Coordinates: \((-1, 10)\).
(ii) The local minimum is at \(x = 3\).
\(f(3) = 3^3 - 3(3)^2 - 9(3) + 5 = -22\).
Coordinates: \((3, -22)\).

(c) The line \(y = 2x - 3\) is a straight line with gradient 2 and y-intercept at \((0, -3)\).

(d) Set \(f(x) = g(x)\):
\(x^3 - 3x^2 - 9x + 5 = 2x - 3 \implies x^3 - 3x^2 - 11x + 8 = 0\).
Using the intersection or solver function on the GDC, the \(x\)-coordinates are:
\(x \approx -2.12\), \(x \approx 0.73\), \(x \approx 4.39\) (correct to 3 significant figures).

(a)
**B1** for correct cubic shape.
**B1** for correct intercepts or key features shown.

(b)(i)
**B2** for \((-1, 10)\).
(If not fully correct, **B1** for \(x = -1\) or \(y = 10\)).

(b)(ii)
**B2** for \((3, -22)\).
(If not fully correct, **B1** for \(x = 3\) or \(y = -22\)).

(c)
**B2** for a straight line with positive gradient and negative y-intercept.
(If not fully correct, **B1** for a line through \((0, -3)\) or with positive gradient).

(d)
**M1** for setting \(x^3 - 3x^2 - 9x + 5 = 2x - 3\).
**A2** for all three values: \(-2.12\), \(0.73\), \(4.39\).
(If only one or two correct, **A1\).

(a) (i) The probability of selecting the first blue sock is \(\frac{b}{12}\).
The probability of selecting the second blue sock without replacement is \(\frac{b-1}{11}\).
The probability that both are blue is \(\frac{b}{12} \times \frac{b-1}{11} = \frac{b(b-1)}{132}\).

(ii) Set the probability equal to \(\frac{5}{33}\):
\(\frac{b(b-1)}{132} = \frac{5}{33} \implies b(b-1) = \frac{5 \times 132}{33} = 20\).
Thus, \(b^2 - b = 20 \implies b^2 - b - 20 = 0\).

(iii) Solve the quadratic equation:
\((b - 5)(b + 4) = 0 \implies b = 5\) or \(b = -4\).
Since the number of socks must be positive, \(b = 5\).

(b) (i) Total socks = 15.
P(both Blue) = \(\frac{6}{15} \times \frac{5}{14} = \frac{30}{210}\).
P(both Red) = \(\frac{4}{15} \times \frac{3}{14} = \frac{12}{210}\).
P(both Green) = \(\frac{5}{15} \times \frac{4}{14} = \frac{20}{210}\).
P(same colour) = \(\frac{30 + 12 + 20}{210} = \frac{62}{210} = \frac{31}{105}\).

(ii) P(no green sock) = \(\frac{10}{15} \times \frac{9}{14} = \frac{90}{210} = \frac{3}{7}\).
P(at least one green sock) = \(1 - \frac{3}{7} = \frac{4}{7}\).

(a)(i)
**M1** for \(\frac{b}{12} \times \frac{b-1}{11}\).
**A1** for fully convincing show.

(a)(ii)
**M1** for \(\frac{b(b-1)}{132} = \frac{5}{33}\).
**M1** for \(b(b-1) = 20\).
**A1** for \(b^2 - b - 20 = 0\).

(a)(iii)
**M1** for factorising \((b-5)(b+4) = 0\) or using formula.
**A1** for \(b = 5\).

(b)(i)
**M1** for \(\frac{6}{15}\times\frac{5}{14} + \frac{4}{15}\times\frac{3}{14} + \frac{5}{15}\times\frac{4}{14}\).
**A1** for \(\frac{31}{105}\) or equivalent decimal \(0.295\).

(b)(ii)
**M1** for \(1 - \frac{10}{15}\times\frac{9}{14}\).
**A1** for \(\frac{4}{7}\) or equivalent decimal \(0.571\).

(a) Find the diagonal \(AC\) of the base:
\(AC^2 = 10^2 + 10^2 = 200 \implies AC = 10\sqrt{2}\) cm.
Since \(M\) is the midpoint of \(AC\), \(AM = 5\sqrt{2}\) cm.
In the right-angled triangle \(VMA\):
\(VM^2 = VA^2 - AM^2 = 13^2 - (5\sqrt{2})^2 = 169 - 50 = 119\).
Therefore, \(VM = \sqrt{119}\) cm.

(b) Let \(N\) be the midpoint of \(BC\). Then \(MN = 5\) cm (half of the side length of the base).
The angle between the sloping face \(VBC\) and the base \(ABCD\) is the angle \(VNM\) in the right-angled triangle \(VMN\).
\(\tan(\angle VNM) = \frac{VM}{MN} = \frac{\sqrt{119}}{5}\).
\(\angle VNM = \tan^{-1}\left(\frac{\sqrt{119}}{5}\right) \approx 65.4^\circ\).

(c) The angle between the sloping edge \(VA\) and the base \(ABCD\) is the angle \(VAM\).
In the right-angled triangle \(VMA\):
\(\sin(\angle VAM) = \frac{VM}{VA} = \frac{\sqrt{119}}{13}\).
\(\angle VAM = \sin^{-1}\left(\frac{\sqrt{119}}{13}\right) \approx 57.0^\circ\).

(a)
**M1** for \(AC^2 = 10^2 + 10^2\) or \(AM = \sqrt{5^2 + 5^2}\).
**M1** for \(VM^2 = 13^2 - AM^2\).
**A1** for reaching \(\sqrt{119}\) with no errors.

(b)
**M1** for identifying the angle \(VNM\).
**B1** for \(MN = 5\) cm.
**M1** for \(\tan \theta = \frac{\sqrt{119}}{5}\).
**A1** for \(65.4^\circ\) or \(65.37^\circ\) to \(65.38^\circ\).

(c)
**M1** for identifying the angle \(VAM\).
**M1** for \(\sin \phi = \frac{\sqrt{119}}{13}\) or \(\cos \phi = \frac{5\sqrt{2}}{13}\).
**A1** for \(5\sqrt{2}\) or \(7.07\).
**A1** for \(57.0^\circ\) or \(57.04^\circ\) to \(57.05^\circ\).

(a) (i) A reflection in the line \(y = -x\) maps \((x, y) \to (-y, -x)\).
Vertices: \((-1, -1)\), \((-1, -3)\), \((-4, -1)\).
(ii) Translation by \(\begin{pmatrix} -3 \\ 2 \end{pmatrix}\) maps \((x, y) \to (x-3, y+2)\).
Vertices: \((-2, 3)\), \((0, 3)\), \((-2, 6)\).
(iii) Rotation of \(90^\circ\) anticlockwise about the origin maps \((x, y) \to (-y, x)\).
Vertices: \((-1, 1)\), \((-1, 3)\), \((-4, 1)\).

(b) (i) Enlargement with scale factor 2 and centre at the origin \((0, 0)\), since all coordinates are doubled.
(ii) Reflection in the x-axis (or the line \(y = 0\)), since the y-coordinates are negated while the x-coordinates remain unchanged.

(a)(i)
**B2** for all three vertices correct. (If 1 or 2 correct, **B1**).

(a)(ii)
**B2** for all three vertices correct. (If 1 or 2 correct, **B1**).

(a)(iii)
**B2** for all three vertices correct. (If 1 or 2 correct, **B1**).

(b)(i)
**B1** for "Enlargement, scale factor 2".
**B1** for "centre \((0, 0)\)" or "origin".

(b)(ii)
**B2** for "Reflection in the x-axis" or "Reflection in the line \(y = 0\)".
(If not fully correct, **B1** for "Reflection" but wrong or missing line).

(a) Use the Cosine Rule to find \(AC\):
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(B)\)
\(AC^2 = 75^2 + 110^2 - 2(75)(110)\cos(64^circ) \approx 10491.9\)
\(AC \approx 102.43 \approx 102\) m.

(b) Use the Sine Rule to find angle \(ACB\) (let it be \(C\)):
\(\frac{\sin(C)}{AB} = \frac{\sin(B)}{AC}\)
\(\sin(C) = \frac{75 \times \sin(64^\circ)}{102.43} \approx 0.6581\)
\(C = \sin^{-1}(0.6581) \approx 41.2^\circ\).
(If using rounded value of \(102\) m for \(AC\), the answer is \(41.4^\circ\). Answers between \(41.1^\circ\) and \(41.4^\circ\) are accepted).

(c) \(\text{Area} = \frac{1}{2} a c \sin(B) = \frac{1}{2} \times 110 \times 75 \times \sin(64^\circ) \approx 3710\) m\(^2\).

(d) The triangle \(AB\text{Top}\) is right-angled at \(A\):
\(\tan(18^\circ) = \frac{h}{AB} = \frac{h}{75}\)
\(h = 75 \tan(18^\circ) \approx 24.4\) m.

(a)
**M1** for \(75^2 + 110^2 - 2(75)(110)\cos 64^\circ\).
**A1** for \(10490\) to \(10500\).
**A1** for \(102\) or \(102.4\).

(b)
**M1** for \(\frac{\sin C}{75} = \frac{\sin 64^\circ}{\text{their } AC}\).
**M1** for \(\sin C = \frac{75\sin 64^\circ}{\text{their } AC}\).
**A1** for \(41.1^\circ\) to \(41.4^\circ\).

(c)
**M1** for \(\frac{1}{2} \times 110 \times 75 \sin 64^\circ\).
**A1** for \(3710\) or \(3708\).

(d)
**M1** for identifying the right-angled triangle.
**M1** for \(\tan 18^\circ = \frac{h}{75}\).
**A1** for \(24.4\) or \(24.37\).

(a) (i) Value = \(5000 \times (1 + 0.032)^5 = 5000 \times (1.032)^5 \approx 5852.86 \approx \$5853\).
(ii) \(5000 \times (1.032)^n \ge 10000 \implies (1.032)^n \ge 2\).
Taking logarithms:
\(n \ge \frac{\log(2)}{\log(1.032)} \approx 21.999\) years.
Thus, the number of complete years is 22.

(b) (i) \(15746.40 = 24000 \times (1 - \frac{r}{100})^4\)
\((1 - \frac{r}{100})^4 = 0.6561\)
\(1 - \frac{r}{100} = \sqrt[4]{0.6561} = 0.9 \implies r = 10\).
(ii) Value at 8 years = \(24000 \times (0.9)^8 = \$10331.21\).

(a)(i)
**M2** for \(5000 \times 1.032^5\).
(If not M2, **M1** for \(5000 \times 1.032^k\) with \(k \ne 5\)).
**A1** for \(5853\).

(a)(ii)
**M1** for \(1.032^n = 2\).
**M1** for \(n = \frac{\log 2}{\log 1.032}\).
**A1** for \(22\).

(b)(i)
**M1** for \(24000 \times (1 - \frac{r}{100})^4 = 15746.40\).
**M1** for \((1 - \frac{r}{100}) = \sqrt[4]{0.6561}\).
**A1** for convincing conclusion \(r = 10\).

(b)(ii)
**M1** for \(24000 \times 0.9^8\) or \(15746.40 \times 0.9^4\).
**A1** for \(10331.21\) (or \(10330\)).

(a) (i) \(f(2) = 3(2) - 5 = 1\).
(ii) \(g(f(x)) = g(3x-5) = \frac{(3x-5) + 2}{2(3x-5) - 1} = \frac{3x-3}{6x-11}\).

(b) Let \(y = 3x - 5 \implies y+5 = 3x \implies x = \frac{y+5}{3}\).
Therefore, \(f^{-1}(x) = \frac{x+5}{3}\).

(c) Let \(y = \frac{x+2}{2x-1} \implies 2xy - y = x + 2 \implies 2xy - x = y + 2 \implies x(2y - 1) = y + 2 \implies x = \frac{y+2}{2y-1}\).
Therefore, \(g^{-1}(x) = \frac{x+2}{2x-1}\).

(d) \(g(x) = 2 \implies \frac{x+2}{2x-1} = 2 \implies x+2 = 4x-2 \implies 3x = 4 \implies x = \frac{4}{3}\).

(a)(i)
**B1** for 1.

(a)(ii)
**M1** for substituting \(3x-5\) into \(g(x)\).
**A1** for numerator \(3x-3\) correct.
**A1** for denominator \(6x-11\) correct.

(b)
**M1** for \(x = 3y-5\) or \(y+5 = 3x\).
**A1** for \(\frac{x+5}{3\}. (c) **M1** for \)y(2x-1) = x+2\).
**M1** for collecting \(x\) terms: \(x(2y-1) = y+2\).
**A1** for \(\frac{x+2}{2x-1}\).

(d)
**M1** for \(x+2 = 2(2x-1)\).
**A1** for \(\frac{4}{3}\) or \(1.33\).

(a) (i) Area of sector = \(\frac{120}{360} \times \pi r^2 = \frac{\pi r^2}{3}\).
(ii) Arc length = \(\frac{120}{360} \times 2 \pi r = \frac{2\pi r}{3}\).

(b) Area of equilateral triangle \(OBC\) is \(\frac{1}{2} \times r \times r \times \sin(60^\circ) = \frac{\sqrt{3}}{4}r^2\).
Total Area = Sector Area + Triangle Area
\(\text{Total Area} = \frac{\pi r^2}{3} + \frac{\sqrt{3}}{4}r^2 = r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\).

(c) (i) \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) = 150 \implies 1.48021 r^2 = 150 \implies r^2 = 101.337 \implies r \approx 10.1\) cm.
(ii) The perimeter consists of the arc \(AB\), and the three straight edges \(OA\), \(BC\), and \(CO\) of length \(r\).
\(\text{Perimeter} = \frac{2\pi r}{3} + 3r \approx 10.0666 \times (2.0944 + 3) \approx 51.3\) cm.

(a)(i)
**M1** for \(\frac{120}{360} \times \pi r^2\).
**A1** for \(\frac{\pi r^2}{3}\).

(a)(ii)
**M1** for \(\frac{120}{360} \times 2 \pi r\).
**A1** for \(\frac{2\pi r}{3}\).

(b)
**M1** for Area of triangle = \(\frac{\sqrt{3}}{4}r^2\).
**M1** for summing the two areas.
**A1** for factoring out \(r^2\) to get \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\).

(c)(i)
**M1** for \(r^2\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) = 150\).
**M1** for \(1.48... r^2 = 150\).
**A1** for \(10.1\) or \(10.07\).

(c)(ii)
**M1** for Perimeter = \(\text{their } AB + 3r\).
**A1** for \(51.3\) or \(51.28\).

(a) Since \(ABCD\) is a rectangle, angle \(ABC = 90^\circ\). Applying Pythagoras' theorem in triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100\)
\(AC = \sqrt{100} = 10\text{ cm}\).

(b) Since \(O\) is the center of the rectangular base, \(AO = \frac{1}{2} AC = 5\text{ cm}\).
In the right-angled triangle \(VOA\):
\(VA^2 = VO^2 + AO^2 = 12^2 + 5^2 = 144 + 25 = 169\)
\(VA = \sqrt{169} = 13\text{ cm}\).

(c) The angle between the sloping edge \(VA\) and the base \(ABCD\) is the angle \(\angle VAO\). Let this angle be \(\theta\).
In triangle \(VOA\):
\(\tan(\theta) = \frac{VO}{AO} = \frac{12}{5} = 2.4\)
\[\theta = \arctan(2.4) \approx 67.38^\circ \approx 67.4^\circ\ (to 1 d.p.)\]

(d) Let \(M\) be the midpoint of \(BC\). The line \(VM\) is the perpendicular height of the triangular face \(VBC\). Since \(ABCD\) is a rectangle with center \(O\), \(OM\) is parallel to \(AB\) and \(OM = \frac{1}{2} AB = 4\text{ cm}\).
The angle between the face \(VBC\) and the base \(ABCD\) is \(\angle VMO\). Let this angle be \(\phi\).
In the right-angled triangle \(VOM\):
\(\tan(\phi) = \frac{VO}{OM} = \frac{12}{4} = 3\)
\[\phi = \arctan(3) \approx 71.57^\circ \approx 71.6^\circ\ (to 1 d.p.)\]

(a) [2 marks]
M1 for \(8^2 + 6^2\) seen
A1 for \(\sqrt{100} = 10\) clearly shown

(b) [2 marks]
M1 for \(12^2 + 5^2\) (where \(AO = 5\) is either stated or used)
A1 for \(13\text{ cm}\)

(c) [3 marks]
M1 for identifying angle \(VAO\) as the required angle
M1 for correct trig ratio, e.g. \(\tan(\theta) = \frac{12}{5}\) or \(\sin(\theta) = \frac{12}{13}\)
A1 for \(67.4^\circ\) (accept \(67.38^\circ\) to \(67.4^\circ\))

(d) [4 marks]
M1 for identifying angle \(VMO\) as the required angle (where \(M\) is the midpoint of \(BC\))
B1 for stating or using \(OM = 4\)
M1 for correct trig ratio, e.g. \(\tan(\phi) = \frac{12}{4}\)
A1 for \(71.6^\circ\) (accept \(71.56^\circ\) to \(71.6^\circ\))

(a) (i) \(f(-2) = 3(-2) - 5 = -6 - 5 = -11\).
(ii) \(g(4) = \frac{12}{4+2} = \frac{12}{6} = 2\).
(iii) First find \(f(3) = 3(3) - 5 = 4\).
Then \(h(f(3)) = h(4) = 4^2 - 2(4) = 16 - 8 = 8\).

(b) Let \(y = \frac{12}{x+2}\).
Multiply by \(x+2\):
\(y(x+2) = 12\)
\(xy + 2y = 12\)
\(xy = 12 - 2y\)
\(x = \frac{12 - 2y}{y}\)
Replacing \(y\) with \(x\):
\(g^{-1}(x) = \frac{12 - 2x}{x}\) (or \(\frac{12}{x} - 2\)).

(c) Set \(f(x) = h(x)\):
\(3x - 5 = x^2 - 2x\)
Rearrange to form a quadratic equation:
\(x^2 - 5x + 5 = 0\)
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)} = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}\)
\(x_1 = \frac{5 + \sqrt{5}}{2} \approx 3.618 \approx 3.62\ (to 3 s.f.)\)
\(x_2 = \frac{5 - \sqrt{5}}{2} \approx 1.382 \approx 1.38\ (to 3 s.f.)\)

(a) [4 marks]
(i) B1 for \(-11\)
(ii) B1 for \(2\)
(iii) M1 for finding \(f(3) = 4\) seen or implied, A1 for \(8\)

(b) [3 marks]
M1 for multiplying by \(x+2\) to get \(y(x+2) = 12\) or \(x(y+2) = 12\)
M1 for isolating \(x\) (or \(y\)) correctly
A1 for \(g^{-1}(x) = \frac{12-2x}{x}\) or equivalent, e.g. \(\frac{12}{x}-2\)

(c) [4 marks]
M1 for setting up equation \(3x - 5 = x^2 - 2x\)
M1 for rearranging into quadratic standard form \(x^2 - 5x + 5 = 0\)
A1 for \(1.38\) (accept \(1.381\) to \(1.382\))
A1 for \(3.62\) (accept \(3.617\) to \(3.618\))

(a) (i) The probability of drawing the first green marble is \(\frac{6}{n}\).
Since the first green marble is not replaced, the probability of drawing a second green marble is \(\frac{5}{n-1}\).
Therefore, the probability of both being green is:
\(P(G, G) = \frac{6}{n} \times \frac{5}{n-1} = \frac{30}{n(n-1)}\)
Set this equal to \(\frac{1}{3}\):
\(\frac{30}{n(n-1)} = \frac{1}{3}\)
\(n(n-1) = 90\)
\(n^2 - n = 90\)
\(n^2 - n - 90 = 0\).

(ii) Factorise the quadratic equation:
\((n - 10)(n + 9) = 0\)
So, \(n = 10\) or \(n = -9\).
Since the total number of marbles must be positive, \(n = 10\).

(b) With \(n = 10\), the bag contains 10 marbles in total: 6 green and \(10 - 6 = 4\) yellow.
(i) The two drawn marbles have different colors if the outcome is Green then Yellow (GY) or Yellow then Green (YG).
\(P(G, Y) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\)
\(P(Y, G) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\)
\(P(\text{different}) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\) (or \(\approx 0.533\)).

(ii) The probability of drawing at least one green marble is:
\(P(\text{at least one green}) = 1 - P(Y, Y)\)
\(P(Y, Y) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\)
\(P(\text{at least one green}) = 1 - \frac{2}{15} = \frac{13}{15}\) (or \(\approx 0.867\)).

(a) (i) [4 marks]
M1 for probability of first green being \(\frac{6}{n}\)
M1 for probability of second green being \(\frac{5}{n-1}\)
M1 for setting product equal to \(\frac{1}{3}\)
A1 for correct expansion and algebraic steps leading to \(n^2 - n - 90 = 0\)

(ii) [2 marks]
M1 for factorisation \((n-10)(n+9) = 0\) or correct use of quadratic formula
A1 for \(n = 10\) (rejecting \(-9\))

(b) [5 marks]
(i) [3 marks]
M1 for correct calculation of one branch, e.g. \(\frac{6}{10} \times \frac{4}{9}\)
M1 for adding the two possible different outcomes: \(\frac{24}{90} + \frac{24}{90}\)
A1 for \(\frac{8}{15}\) or \(\frac{48}{90}\) or \(0.533\) (accept \(0.5333...\))

(ii) [2 marks]
M1 for using \(1 - P(Y, Y)\) or \(P(G, G) + P(G, Y) + P(Y, G)\)
A1 for \(\frac{13}{15}\) or \(\frac{78}{90}\) or \(0.867\) (accept \(0.8666...\))