2024 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

(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 $D = (n^2 + n + 1) - (6n - 1) = n^2 + n + 1 - 6n + 1 = n^2 - 5n + 2$.

(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)

(a) Using the Cosine Rule:
$PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(PQR)$
$PR^2 = 45^2 + 62^2 - 2(45)(62)\cos(78^\circ)$
$PR^2 = 2025 + 3844 - 5580(0.20791) \approx 4708.85$
$PR = \sqrt{4708.85} \approx 68.6\text{ m}$ (to 3 s.f.)

(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 $T$:
$\tan(12^\circ) = \frac{h}{PQ} = \frac{h}{45}$
$h = 45 \tan(12^\circ) \approx 9.57\text{ m}$.

(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 $9.6$)

(a) Reflection in $y = -1$:
$y$-coordinate is transformed by $y' = -1 - (y - (-1)) = -2 - y$.
$A(1, 2) \to (1, -4)$
$B(4, 2) \to (4, -4)$
$C(1, 4) \to (1, -6)$

(b) Translation by $\begin{pmatrix} -3 \\ 1 \end{pmatrix}$:
$A(1, 2) \to (1-3, 2+1) = (-2, 3)$
$B(4, 2) \to (4-3, 2+1) = (1, 3)$
$C(1, 4) \to (1-3, 4+1) = (-2, 5)$

(c) Shear parallel to $x$-axis, invariant line $y=0$ (the $x$-axis), shear factor $k = 2$:
$(x, y) \to (x + 2y, y)$
$A(1, 2) \to (1 + 2(2), 2) = (5, 2)$
$B(4, 2) \to (4 + 2(2), 2) = (8, 2)$
$C(1, 4) \to (1 + 2(4), 4) = (9, 4)$

(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 $(0, 0)$.

(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 $(0, 0)$

(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: $(0, 1.67)$.

(c) $f(x) = x \implies \frac{2x - 5}{x - 3} = x \implies 2x - 5 = x(x - 3) \implies x^2 - 5x + 5 = 0$.
Using quadratic formula:
$x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(5)}}{2} = \frac{5 \pm \sqrt{5}}{2}$.
$x \approx 3.62$ or $x \approx 1.38$.

(d) $\frac{2x-5}{x-3} = -x + k \implies 2x - 5 = -x^2 + (k+3)x - 3k \implies x^2 + (2 - k - 3)x + (3k - 5) = 0 \implies x^2 - (k+1)x + (3k-5) = 0$.
For tangent, discriminant $\Delta = 0$:
$(k+1)^2 - 4(3k-5) = 0 \implies k^2 + 2k + 1 - 12k + 20 = 0 \implies k^2 - 10k + 21 = 0 \implies (k-3)(k-7) = 0$.
$k = 3$ or $k = 7$.

(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 $k = 7$

(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 $= \frac{2}{3} \pi r^3$.

(b) Total volume $= \pi r^3 + \frac{2}{3} \pi r^3 = \frac{5}{3} \pi r^3$.
Given total volume $= 250\text{ cm}^3$:
$\frac{5}{3} \pi r^3 = 250 \implies r^3 = \frac{750}{5\pi} = \frac{150}{\text{\pi}} \approx 47.7465$.
$r = \sqrt[3]{47.7465} \approx 3.63\text{ cm}$.

(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 $= \pi (3.6278)^2 (2 + \sqrt{10}) \approx 41.347 \times (2 + 3.1623) \approx 213\text{ cm}^2$.

(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 $213$ - $214$)

(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.

(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

(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 $= \frac{21}{80} = 0.2625$.

(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) Substitute $y = 2x - 3$ into $x^2 + y^2 = 26$:
$x^2 + (2x - 3)^2 = 26$
$x^2 + 4x^2 - 12x + 9 = 26$
$5x^2 - 12x + 9 - 26 = 0$
$5x^2 - 12x - 17 = 0$ (Proven).

(b) Factorising $5x^2 - 12x - 17 = 0$:
$5x^2 - 17x + 5x - 17 = 0 \implies x(5x - 17) + 1(5x - 17) = 0 \implies (5x - 17)(x + 1) = 0$.
So, $x = 3.4$ or $x = -1$.

(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 $(3.4, 3.8)$.

(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 $(3.4, 3.8)$

(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.)

(b) Using the cosine rule:
\(QR^2 = PQ^2 + PR^2 - 2 \times PQ \times PR \times \cos(QPR)\)
\(QR^2 = 18.4^2 + 13.788^2 - 2 \times 18.4 \times 13.788 \times \cos(62^\circ)\)
\(QR^2 = 338.56 + 190.11 - 507.40 \times \cos(62^\circ)\)
\(QR^2 = 528.67 - 238.21 = 290.46\)
\(QR = \sqrt{290.46} \approx 17.0\text{ cm}\)

(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:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(112 = \frac{1}{2} \times QR \times d\)
\(112 = \frac{1}{2} \times 17.043 \times d\)
\(d = \frac{224}{17.043} \approx 13.1\text{ cm}\)

(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 \(13.14...\text{ cm}\).

(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.

(c) \(T_{n+1} - T_n = \frac{n+1}{2(n+1)+1} - \frac{n}{2n+1} = \frac{n+1}{2n+3} - \frac{n}{2n+1}\)
\(= \frac{(n+1)(2n+1) - n(2n+3)}{(2n+3)(2n+1)} = \frac{(2n^2 + 3n + 1) - (2n^2 + 3n)}{(2n+3)(2n+1)} = \frac{1}{(2n+3)(2n+1)}\)

(a) (i) B1 for 75.
B1 for 101.

(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 \(\frac{1}{(2n+3)(2n+1)}\).

(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: \(2\pi(6)^2 + 2\pi(6)(6) = 72\pi + 72\pi = 144\pi\text{ cm}^2\).)

(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 \(144\pi\).