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

(a) The total number of marbles in the bag is \(7 + n\). The probability of choosing a red marble first is \(\frac{7}{7+n}\). Since there is no replacement, the probability of choosing a second red marble is \(\frac{6}{6+n}\). The probability that both marbles are red is \(\frac{7}{7+n} \times \frac{6}{6+n} = \frac{42}{(7+n)(6+n)}\).

(b)(i) Setting the expression from part (a) equal to \(\frac{7}{22}\):
\(\frac{42}{(7+n)(6+n)} = \frac{7}{22}\)
Divide both sides by 7:
\(\frac{6}{n^2 + 13n + 42} = \frac{1}{22}\)
Cross-multiply:
\(n^2 + 13n + 42 = 132\)
Rearrange to form a quadratic equation equal to 0:
\(n^2 + 13n - 90 = 0\).

(b)(ii) Factorise the quadratic equation:
\((n - 5)(n + 18) = 0\)
This gives \(n = 5\) or \(n = -18\). Since the number of marbles must be positive, we reject \(n = -18\). Therefore, \(n = 5\).

(c) With \(n = 5\), the total number of marbles is 12.
To find the probability of choosing different colours, we can calculate \(\text{P(Red, Blue)} + \text{P(Blue, Red)}\):
\(\text{P(Red, Blue)} = \frac{7}{12} \times \frac{5}{11} = \frac{35}{132}\)
\(\text{P(Blue, Red)} = \frac{5}{12} \times \frac{7}{11} = \frac{35}{132}\)
Adding these together:
\(\text{P(Different Colours)} = \frac{35}{132} + \frac{35}{132} = \frac{70}{132} = \frac{35}{66}\) (or approximately 0.530).

(a) M1 for product of fractions \(\frac{7}{7+n} \times \frac{6}{6+n}\), A1 for \(\frac{42}{(7+n)(6+n)}\) or equivalent.
(b)(i) M1 for setting their expression equal to \(\frac{7}{22}\), M1 for clearing fractions, e.g., \(924 = 7(n^2 + 13n + 42)\), A1 for fully correct working showing all steps leading to the given equation.
(b)(ii) M1 for factorising \((n-5)(n+18) = 0\) or using the quadratic formula, A1 for \(n = 5\) (with explicit or implicit rejection of \(n = -18\)).
(c) M1 for identifying two combined events: (Red, Blue) and (Blue, Red), M1 for substituting \(n=5\) to calculate \(\frac{7}{12} \times \frac{5}{11}\), M1 for adding the two identical outcomes, A1 for \(\frac{35}{66}\) or 0.530.

(a) First, find the derivative \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = 6x^2 - 6x - 12\)
Set the derivative to 0 to find the stationary points:
\(6(x^2 - x - 2) = 0 \Rightarrow 6(x-2)(x+1) = 0\)
This gives \(x = 2\) or \(x = -1\).
Substitute these back into the curve's equation to find the \(y\)-coordinates:
For \(x = -1\): \(y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12\). So local maximum is \((-1, 12)\).
For \(x = 2\): \(y = 2(2)^3 - 3(2)^2 - 12(2) + 5 = 16 - 12 - 24 + 5 = -15\). So local minimum is \((2, -15)\).

(b) Using the graphic display calculator (GDC) to find the x-intercepts of the curve \(y = 2x^3 - 3x^2 - 12x + 5\):
We find the roots are \(x \approx -2.04\), \(x \approx 0.389\), and \(x \approx 3.15\).

(c) The inequality is \(2x^3 - 3x^2 - 15x + 3 < 0\).
Rewrite the equation to match the curve:
\(2x^3 - 3x^2 - 12x + 5 < 3x + 2\)
This shows that the curve lies below the straight line \(y = 3x + 2\).
By drawing the line \(y = 3x + 2\) on the GDC, we find the points of intersection with the curve for \(x > 0\). The intersection points are at \(x \approx 0.195\) and \(x \approx 3.51\).
The inequality holds where the curve is below the line, which occurs in the interval \(0.195 < x < 3.51\).

(a) M1 for derivative \(6x^2 - 6x - 12\), M1 for setting to 0 to find \(x = 2\) and \(x = -1\), A1 for local maximum at \((-1, 12)\), A1 for local minimum at \((2, -15)\).
(b) B1 for each correct root: \(x \approx -2.04\), \(x \approx 0.389\), \(x \approx 3.15\).
(c) B1 for identifying the line \(y = 3x + 2\), M1 for finding the positive intersection values \(x \approx 0.195\) and \(x \approx 3.51\) using a GDC, A1 for \(x > 0.195\), A1 for \(x < 3.51\) (accept interval notation \((0.195, 3.51)\)).

(a)(i) The transformation matrix for reflection in the line \(y = -x\) is \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\).

(a)(ii) We multiply the transformation matrix by the coordinate matrix of triangle \(T\):
\(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 3 & 1 \\ 1 & 1 & 4 \end{pmatrix} = \begin{pmatrix} 0(1) - 1(1) & 0(3) - 1(1) & 0(1) - 1(4) \\ -1(1) + 0(1) & -1(3) + 0(1) & -1(1) + 0(4) \end{pmatrix} = \begin{pmatrix} -1 & -1 & -4 \\ -1 & -3 & -1 \end{pmatrix}\)
So, the vertices of triangle \(U\) are \(A'(-1, -1)\), \(B'(-1, -3)\), and \(C'(-4, -1)\).

(b)(i) The matrix for a shear with the \(x\)-axis invariant and shear factor 2 is \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).
We multiply this matrix by the coordinate matrix of triangle \(T\):
\(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 & 1 \\ 1 & 1 & 4 \end{pmatrix} = \begin{pmatrix} 1(1) + 2(1) & 1(3) + 2(1) & 1(1) + 2(4) \\ 0(1) + 1(1) & 0(3) + 1(1) & 0(1) + 1(4) \end{pmatrix} = \begin{pmatrix} 3 & 5 & 9 \\ 1 & 1 & 4 \end{pmatrix}\)
So, the vertices of triangle \(W\) are \(A''(3, 1)\), \(B''(5, 1)\), and \(C''(9, 4)\).

(b)(ii) To map triangle \(W\) back onto triangle \(T\), the inverse transformation is needed. The inverse of the shear matrix \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\) is \(\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\). This matrix represents a shear with the \(x\)-axis invariant and shear factor \(-2\).

(a)(i) B2 for \(\begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}\) (B1 for one incorrect element or transposed columns).
(a)(ii) M1 for matrix multiplication setup, A2 for all three points correct: \(A'(-1, -1)\), \(B'(-1, -3)\), \(C'(-4, -1)\) (A1 if one or two points are correct).
(b)(i) M1 for using the shear matrix \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\), A2 for all three points correct: \(A''(3, 1)\), \(B''(5, 1)\), \(C''(9, 4)\) (A1 if one or two points are correct).
(b)(ii) B1 for "shear", B1 for "\(x\)-axis invariant" (or "\(y=0\) invariant"), B1 for "shear factor \(-2\)".

(a) In triangle \(ABC\), we can use the Cosine Rule to calculate the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\)
\(AC^2 = 8^2 + 11^2 - 2(8)(11)\cos(115^\circ)\)
\(AC^2 = 64 + 121 - 176(-0.42262)\)
\(AC^2 = 185 + 74.381 = 259.381\)
\(AC = \sqrt{259.381} \approx 16.105\text{ cm}\).
So, \(AC \approx 16.1\text{ cm}\) (to 3 significant figures).

(b) In triangle \(ACD\), we first need to find the length of \(AD\) to find the angle \(CAD\). Using the Cosine Rule in \(\triangle ACD\):
\(AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos(ACD)\)
\(AD^2 = (16.105)^2 + 9^2 - 2(16.105)(9)\cos(42^\circ)\)
\(AD^2 = 259.381 + 81 - 289.89(0.74314)\)
\(AD^2 = 340.381 - 215.431 = 124.950\)
\(AD = \sqrt{124.950} \approx 11.178\text{ cm}\).
Now, use the Sine Rule to calculate the angle \(CAD\):
\(\frac{\sin(CAD)}{CD} =
\frac{\sin(ACD)}{AD} \Rightarrow \sin(CAD) = \frac{9 \cdot \sin(42^\circ)}{11.178}\)
\
sin(CAD) = \frac{9(0.66913)}{11.178} \approx 0.53875\)
\(\text{Angle } CAD = \sin^{-1}(0.53875) \approx 32.597^\circ\).
So, angle \(CAD \approx 32.6^\circ\) (to 3 significant figures).

(c) The area of the quadrilateral \(ABCD\) is the sum of the areas of triangles \(ABC\) and \(ACD\).
\(\text{Area of } \triangle ABC = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(ABC) = \frac{1}{2}(8)(11)\sin(115^\circ) = 44(0.90631) \approx 39.877\text{ cm}^2\).
\(\text{Area of } \triangle ACD = \frac{1}{2} \cdot AC \cdot CD \cdot \sin(ACD) = \frac{1}{2}(16.105)(9)\sin(42^\circ) = 72.4725(0.66913) \approx 48.494\text{ cm}^2\).
\(\text{Total Area} = 39.877 + 48.494 = 88.371\text{ cm}^2\).
So, the area is approximately \(88.4\text{ cm}^2\) (to 3 significant figures).

(a) M1 for correct substitution into Cosine Rule: \(8^2 + 11^2 - 2(8)(11)\cos(115^\circ)\), A1 for \(AC^2 \approx 259.38\), A1 for \(AC \approx 16.1\text{ cm}\).
(b) M1 for using Cosine Rule to find \(AD^2\) or \(AD \approx 11.18\text{ cm}\), M1 for using Sine Rule: \(\frac{\sin(CAD)}{9} = \frac{\sin(42^\circ)}{11.18}\), A1 for \(CAD \approx 32.6^\circ\).
(c) M1 for \(\text{Area of } \triangle ABC = \frac{1}{2}(8)(11)\sin(115^\circ)\), A1 for \(\approx 39.9\), M1 for \(\text{Area of } \triangle ACD = \frac{1}{2}(16.105)(9)\sin(42^\circ)\), A1 for \(\approx 48.5\), A1 for total area \(\approx 88.4\text{ cm}^2\).

(a)(i) The area of a sector of a circle is given by \(\text{Area} = \frac{x}{360} \pi r^2\).

(a)(ii) The perimeter consists of two radii and the arc length:
\(\text{Perimeter} = 2r + \frac{x}{360} \cdot 2\pi r = 2r + \frac{x}{180} \pi r\).

(b)(i) Since the perimeter is 45 cm:
\(2r + \frac{x}{180} \pi r = 45\)
Subtract \(2r\) from both sides:
\(\frac{x\pi r}{180} = 45 - 2r\)
Multiply by 180 and divide by \(\pi r\):
\(x = \frac{180(45 - 2r)}{\pi r}\).

(b)(ii) Substitute this expression for \(x\) into the area formula from part (a)(i):
\(A = \frac{\frac{180(45 - 2r)}{\pi r}}{360} \pi r^2\)
\(A = \frac{180(45 - 2r)}{360 \pi r} \cdot \pi r^2\)
\(A = \frac{45 - 2r}{2} \cdot r\)
\(A = (22.5 - r)r = 22.5r - r^2\).

(c)(i) To find the value of \(r\) for which the area is a maximum, we can find the vertex of the quadratic equation \(A = 22.5r - r^2\). Differentiating with respect to \(r\):
\(\frac{dA}{dr} = 22.5 - 2r = 0 \Rightarrow 2r = 22.5 \Rightarrow r = 11.25\text{ cm}\).

(c)(ii) Substitute \(r = 11.25\) back into the area equation:
\(A = 22.5(11.25) - (11.25)^2 = 253.125 - 126.5625 = 126.5625\text{ cm}^2\) (or approximately \(127\text{ cm}^2\)).

(a)(i) B1 for \(\frac{x}{360} \pi r^2\) or equivalent.
(a)(ii) B1 for arc length \(\frac{x}{360} \cdot 2\pi r\) or \(\frac{x}{180} \pi r\), B1 for adding \(2r\).
(b)(i) M1 for setting their perimeter expression equal to 45: \(2r + \frac{x\pi r}{180} = 45\), A1 for correct rearrangement leading to the given show-that expression.
(b)(ii) M1 for substituting the expression for \(x\) into their area formula, M1 for algebraic simplification showing cancellation of \(\pi\) and \(r\), A1 for fully correct steps leading to \(A = 22.5r - r^2\).
(c)(i) M1 for differentiating and setting to 0 (or using vertex formula \(r = -b/(2a)\)), A1 for \(r = 11.25\text{ cm}\).
(c)(ii) B1 for \(126.5625\) or \(126.6\) or \(127\text{ cm}^2\).

(a) Since \(ABCD\) is a rectangle with \(AB = 12\) cm and \(BC = 8\) cm, the diagonal \(AC\) is:
\(AC^2 = AB^2 + BC^2 = 12^2 + 8^2 = 144 + 64 = 208 \Rightarrow AC = \sqrt{208} \approx 14.422\text{ cm}\).
Since \(O\) is the centre of the base, the length \(AO\) is half of \(AC\):
\(AO = \frac{1}{2} \sqrt{208} = \sqrt{52} \approx 7.211\text{ cm}\).
In the right-angled triangle \(VOA\), using Pythagoras' theorem:
\(VO^2 + AO^2 = VA^2 \Rightarrow VO^2 + 52 = 15^2 = 225\)
\(VO^2 = 225 - 52 = 173 \Rightarrow VO = \sqrt{173} \approx 13.153\text{ cm}\).
So, \(VO \approx 13.2\text{ cm}\) (to 3 significant figures).

(b) The angle between the slant edge \(VA\) and the base is the angle \(VAO\).
In the right-angled triangle \(VOA\):
\(\cos(VAO) = \frac{AO}{VA} = \frac{\sqrt{52}}{15} \approx 0.48074\)
\(\text{Angle } VAO = \cos^{-1}(0.48074) \approx 61.265^\circ\).
So, the angle is approximately \(61.3^\circ\) (to 3 significant figures).

(c) Let \(M\) be the midpoint of the edge \(BC\). The line \(OM\) is parallel to \(AB\) and its length is half of \(AB\), so \(OM = 6\) cm.
The angle between the face \(VBC\) and the base is the angle \(VMO\).
In the right-angled triangle \(VOM\):
\(\tan(VMO) = \frac{VO}{OM} = \frac{\sqrt{173}}{6} \approx \frac{13.153}{6} \approx 2.19217\)
\(\text{Angle } VMO = \tan^{-1}(2.19217) \approx 65.480^\circ\).
So, the angle is approximately \(65.5^\circ\) (to 3 significant figures).

(a) M1 for diagonal \(AC = \sqrt{12^2 + 8^2}\), A1 for \(AO = \sqrt{52}\) or \(7.21\), M1 for using Pythagoras in \(\triangle VOA\): \(VO^2 = 15^2 - 52\), A1 for \(VO \approx 13.2\text{ cm}\).
(b) M1 for identifying the angle \(VAO\), M1 for correct trig ratio, e.g., \(\cos(VAO) = \frac{7.211}{15}\), A1 for \(61.3^\circ\).
(c) B1 for identifying midpoint \(M\) of \(BC\) and stating \(OM = 6\text{ cm}\), M1 for identifying the angle \(VMO\), M1 for using \(\tan(VMO) = \frac{13.153}{6}\) or equivalent, A1 for \(65.5^\circ\).

(a)(i) Since \(P\) is directly proportional to the cube of \(v\):
\(P = k v^3\)
Substitute \(v = 8\) and \(P = 1024\):
\(1024 = k (8)^3 \Rightarrow 1024 = 512k \Rightarrow k = 2\).
So, the formula is \(P = 2v^3\).

(a)(ii) Substitute \(v = 12\) into the formula from part (a)(i):
\(P = 2(12)^3 = 2(1728) = 3456\text{ watts}\).

(b)(i) Since \(v\) is inversely proportional to the square root of \(d\):
\(v = \frac{c}{\sqrt{d}}\)
Substitute \(d = 1.44\) and \(v = 10\):
\(10 = \frac{c}{\sqrt{1.44}} \Rightarrow 10 = \frac{c}{1.2} \Rightarrow c = 12\).
So, the formula is \(v = \frac{12}{\sqrt{d}}\).

(b)(ii) Substitute the formula for \(v\) into the formula for \(P\):
\(P = 2 \left( \frac{12}{\sqrt{d}} \right)^3 = 2 \left( \frac{1728}{d^{1.5}} \right) = \frac{3456}{d^{1.5}}\) or \(P = \frac{3456}{d\sqrt{d}}\).

(a)(i) M1 for \(P = k v^3\), M1 for substituting \(v=8, P=1024\) to find \(k\), A1 for \(P = 2v^3\).
(a)(ii) M1 for substituting \(v=12\) into their formula, A1 for \(3456\).
(b)(i) M1 for \(v = \frac{c}{\sqrt{d}}\), M1 for substituting \(d=1.44, v=10\) to find \(c\), A1 for \(v = \frac{12}{\sqrt{d}}\).
(b)(ii) M1 for substituting their \(v\) expression into their \(P\) formula, M1 for raising \(\frac{12}{\sqrt{d}}\) to power 3, A1 for \(P = \frac{3456}{d^{1.5}}\) or \(P = \frac{3456}{d\sqrt{d}}\).

(a) The initial temperature is when \(t = 0\):
\(85 = 20 + A e^{0} \Rightarrow 85 = 20 + A \Rightarrow A = 65\).

(b)(i) When \(t = 10\), \(T = 45\):
\(45 = 20 + 65 e^{-10k}\)
\(25 = 65 e^{-10k}\)
\(e^{-10k} = \frac{25}{65} = \frac{5}{13}\)
Take natural logarithms on both sides:
\(-10k = \ln\left(\frac{5}{13}\right) = -0.955511\)
\(k = \frac{-0.955511}{-10} \approx 0.095551\).
To 5 decimal places, this is \(k = 0.09555\).

(b)(ii) For \(t = 25\), substitute the values into the formula:
\(T = 20 + 65 e^{-0.09555 \times 25}\)
\(T = 20 + 65 e^{-2.38875} \approx 20 + 65(0.09174) \approx 20 + 5.963 = 25.963^\circ\text{C}\).
So, \(T \approx 26.0^\circ\text{C}\) (to 3 significant figures).

(c) When \(T = 25\):
\(25 = 20 + 65 e^{-0.09555t}\)
\(5 = 65 e^{-0.09555t}\)
\(e^{-0.09555t} = \frac{5}{65} = \frac{1}{13}\)
Take natural logarithms:
\(-0.09555t = \ln\left(\frac{1}{13}\right) = -2.564949\)
\(t = \frac{-2.564949}{-0.09555} \approx 26.844\text{ minutes}\).
So, it takes approximately \(26.8\text{ minutes}\) (to 3 significant figures).

(a) M1 for substituting \(t=0, T=85\) in \(T = 20 + A e^{-kt}\), A1 for \(A = 65\).
(b)(i) M1 for substituting \(t=10, T=45\): \(45 = 20 + 65 e^{-10k}\), M1 for simplifying to \(e^{-10k} = \frac{5}{13}\), M1 for taking natural logs on both sides, A1 for establishing \(k \approx 0.09555\) with clear steps.
(b)(ii) M1 for substituting \(t=25\) into their formula, A1 for \(T \approx 26.0^\circ\text{C}\) (accept \(26\) or \(25.96\)).
(c) M1 for equation \(25 = 20 + 65 e^{-0.09555t}\) leading to \(e^{-0.09555t} = \frac{1}{13}\), M1 for taking natural logs to solve for \(t\), A1 for \(t \approx 26.8\text{ minutes}\).

**(a)**

There are \(n\) marbles, of which 5 are red. The probability of drawing a red marble first is \(\frac{5}{n}\). After removing one red marble, there are \(n-1\) marbles left, with 4 red ones remaining. The probability of drawing a red marble second is \(\frac{4}{n-1}\).

Therefore, the probability that both are red is:

\(P(\text{Red and Red}) = \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}\)

**(b)**

We are given that \(P(\text{Red and Red}) = \frac{2}{21}\).

\(\frac{20}{n(n-1)} = \frac{2}{21}\)

Multiplying both sides by \(21n(n-1)\):

\(420 = 2n(n-1)\)

\(210 = n^2 - n\)

Rearranging to standard quadratic form:

\(n^2 - n - 210 = 0\) (as required)

**(c)**

To solve \(n^2 - n - 210 = 0\), we look for two numbers that multiply to \(-210\) and add to \(-1\). These numbers are \(-15\) and \(14\).

\((n - 15)(n + 14) = 0\)

This gives \(n = 15\) or \(n = -14\). Since the number of marbles must be positive, \(n = 15\).

**(d)**

If \(n = 15\), there are 5 red marbles and \(15 - 5 = 10\) blue marbles.

The probability of drawing two marbles of different colors is:

\(P(\text{different colors}) = P(\text{Red, Blue}) + P(\text{Blue, Red})\)

\(P(\text{Red, Blue}) = \frac{5}{15} \times \frac{10}{14} = \frac{50}{210}\)

\(P(\text{Blue, Red}) = \frac{10}{15} \times \frac{5}{14} = \frac{50}{210}\)

\(P(\text{different colors}) = \frac{50}{210} + \frac{50}{210} = \frac{100}{210} = \frac{10}{21}\)

**(a)**
M1: For product of probabilities with correct denominators, e.g., \(\frac{5}{n} \times \frac{4}{n-1}\)
A1: For \(\frac{20}{n(n-1)}\) or equivalent [2 marks]

**(b)**
M1: For equating their expression in (a) to \(\frac{2}{21}\)
M1: For correct algebraic step to eliminate fractions, e.g., \(420 = 2n(n-1)\)
A1: For complete and correct algebraic steps leading to the given equation [3 marks]

**(c)**
M1: For factorization \((n-15)(n+14) = 0\) or correct use of the quadratic formula
A1: For \(n = 15\) (accept omission or rejection of \(n = -14\)) [2 marks]

**(d)**
M1: For finding the number of blue marbles is 10 (when \(n = 15\))
M1: For calculating the probability of one combination, e.g., \(\frac{5}{15} \times \frac{10}{14}\)
M1: For adding both combinations, e.g., \(2 \times \frac{50}{210}\) (or finding \(1 - P(RR) - P(BB)\))
A1: For \(\frac{10}{21}\) in simplest form [4 marks]

**(a)**

Using \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\):

Outward time = \(\frac{120}{x}\) hours.

**(b)**

The return speed is \(x - 15\) km/h.

Return time = \(\frac{120}{x-15}\) hours.

**(c) (i)**

The return journey is 40 minutes longer than the outward journey. Converting 40 minutes to hours: \(\frac{40}{60} = \frac{2}{3}\) hours.

So, \(\frac{120}{x-15} - \frac{120}{x} = \frac{2}{3}\).

Dividing both sides by 2:

\(\frac{60}{x-15} - \frac{60}{x} = \frac{1}{3}\)

Multiplying the entire equation by the common denominator \(3x(x - 15)\):

\(180x - 180(x - 15) = x(x-15)\)

\(180x - 180x + 2700 = x^2 - 15x\)

\(x^2 - 15x - 2700 = 0\) (as required)

**(c) (ii)**

To solve \(x^2 - 15x - 2700 = 0\), we factorize it as:

\((x - 60)(x + 45) = 0\)

This gives \(x = 60\) or \(x = -45\).

Since speed must be positive, the outward speed is \(x = 60\) km/h.

**(d)**

Outward time = \(\frac{120}{60} = 2\) hours.

Return speed = \(60 - 15 = 45\) km/h.

Return time = \(\frac{120}{45} = 2\frac{2}{3}\) hours = 2 hours and 40 minutes.

Total time = 2 hours + 2 hours 40 minutes = 4 hours 40 minutes.

**(a)**
B1: For \(\frac{120}{x}\) [1 mark]

**(b)**
B1: For \(\frac{120}{x-15}\) [1 mark]

**(c) (i)**
B1: For converting 40 minutes to \(\frac{2}{3}\) hours
M1: For setting up the correct equation: \(\frac{120}{x-15} - \frac{120}{x} = \frac{2}{3}\) (or equivalent)
M1: For algebraic steps to remove fractions, e.g., \(120(3x) - 120(3)(x-15) = 2x(x-15)\)
A1: For fully correct working leading to the given equation [4 marks]

**(c) (ii)**
M1: For factorizing to \((x-60)(x+45) = 0\) or correct use of the quadratic formula
A1: For obtaining \(x = 60\) and \(x = -45\)
A1: For choosing the positive value \(x = 60\) [3 marks]

**(d)**
M1: For calculating return time as \(\frac{120}{45}\) (or 2 hours 40 minutes)
A1: For 4 hours 40 minutes (or 4 h 40 min) [2 marks]

**(a)**

The base \(ABCD\) is a rectangle, so \(\angle ABC = 90^\circ\).

Using Pythagoras' theorem on the right-angled triangle \(ABC\):

\(AC^2 = AB^2 + BC^2 = 16^2 + 12^2 = 256 + 144 = 400\)

\(AC = \sqrt{400} = 20\) cm (as required)

**(b)**

The center \(O\) is the midpoint of the diagonal \(AC\).

\(AO = \frac{1}{2} AC = 10\) cm.

\(VO\) is perpendicular to the base, so triangle \(VOA\) is a right-angled triangle with \(\angle VOA = 90^\circ\).

Using Pythagoras' theorem on \(\triangle VOA\):

\(VA^2 = VO^2 + AO^2 = 24^2 + 10^2 = 576 + 100 = 676\)

\(VA = \sqrt{676} = 26\) cm.

**(c)**

The angle between the edge \(VA\) and the base \(ABCD\) is the angle \(\angle VAO\).

In right-angled triangle \(VOA\):

\(\tan(\angle VAO) = \frac{VO}{AO} = \frac{24}{10} = 2.4\)

\(\angle VAO = \arctan(2.4) \approx 67.3801^\circ\)

To 1 decimal place, the angle is \(67.4^\circ\).

**(d)**

Let \(M\) be the midpoint of the side \(BC\).

The line \(OM\) is parallel to \(AB\) and connects the center \(O\) to the midpoint of \(BC\).

\(OM = \frac{1}{2} AB = 8\) cm.

Since \(OM\) is perpendicular to \(BC\) and \(VM\) is perpendicular to \(BC\), the angle between the face \(VBC\) and the base \(ABCD\) is \(\angle VMO\).

In right-angled triangle \(VOM\):

\(\tan(\angle VMO) = \frac{VO}{OM} = \frac{24}{8} = 3\)

\(\angle VMO = \arctan(3) \approx 71.565^\circ\)

To 1 decimal place, the angle is \(71.6^\circ\).

**(a)**
M1: For using Pythagoras' theorem on the base: \(16^2 + 12^2\)
A1: For showing \(AC = 20\) cm from \(\sqrt{400}\) with complete working [2 marks]

**(b)**
M1: For identifying \(AO = 10\) and using Pythagoras' theorem on \(\triangle VOA\): \(\sqrt{24^2 + 10^2}\)
A1: For \(VA = 26\) cm [2 marks]

**(c)**
M1: For recognizing angle \(\angle VAO\) as the required angle
M1: For a correct trigonometric ratio, e.g., \(\tan(\theta) = \frac{24}{10}\) or \(\sin(\theta) = \frac{24}{26}\)
A1: For \(67.4^\circ\) or \(67.38^\circ\) (accept \(67.38...^\circ\)) [3 marks]

**(d)**
M1: For identifying the midpoint of \(BC\), say \(M\), and finding \(OM = 8\) cm
M1: For recognizing angle \(\angle VMO\) as the required angle
M1: For \(\tan(\phi) = \frac{24}{8}\) (or 3) or alternative correct trig method
A1: For \(71.6^\circ\) or \(71.565...^\circ\) [4 marks]