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

1. Calculate the gradient \(m\):
\(m = \frac{-5 - 5}{2 - (-3)} = \frac{-10}{5} = -2\).

2. Substitute the gradient and one point, for example \((2, -5)\), into the equation \(y = mx + c\):
\(-5 = -2(2) + c\)
\(-5 = -4 + c\)
\(c = -1\).

3. The equation is \(y = -2x - 1\).

M1 for finding the gradient \(m = -2\) or substituting a point to find \(c\)
A1 for the final answer \(y = -2x - 1\)

Write all bases as powers of 3:
\(3^{x+2} \times (3^2)^{x-1} = (3^3)^4\)
\(3^{x+2} \times 3^{2x-2} = 3^{12}\)
Combine the powers on the left hand side:
\(3^{(x+2) + (2x-2)} = 3^{12}\)
\(3^{3x} = 3^{12}\)
Equate the exponents:
\(3x = 12\)
\(x = 4\).

M1 for writing \(9^{x-1} = 3^{2x-2}\) or \(27^4 = 3^{12}\)
M1 for equating exponents to get \(3x = 12\) or equivalent
A1 for \(x = 4\)

Use the laws of logarithms to combine the terms:
\(\log_5(x(x-4)) = 1\)
Convert to exponential form:
\(x(x-4) = 5^1\)
\(x^2 - 4x - 5 = 0\)
Factorise the quadratic equation:
\((x-5)(x+1) = 0\)
This gives \(x = 5\) or \(x = -1\).
Since the logarithm is only defined for positive values, the argument \(x - 4\) must be positive, which means \(x > 4\). Thus, \(x = 5\) is the only valid solution.

M1 for combining the logarithms to obtain \(x^2 - 4x = 5\)
A1 for the final answer \(5\) (with the negative solution rejected)

The magnitude of a vector \(\mathbf{p} = \begin{pmatrix} k \\ 6 \end{pmatrix}\) is given by:
\(|\mathbf{p}| = \sqrt{k^2 + 6^2}\)
Since the magnitude is 10:
\(\sqrt{k^2 + 36} = 10\)
Square both sides:
\(k^2 + 36 = 100\)
\(k^2 = 64\)
\(k = \pm 8\).

M1 for setting up \(k^2 + 6^2 = 10^2\)
M1 for solving to get \(k^2 = 64\)
A1 for \(8\) and \(-8\)

Let the original price be \(P\).
After a 15% discount, the selling price is 85% of the original price:
\(0.85 \times P = 680\)
\(P = \frac{680}{0.85} = 800\).

M1 for setting up the equation \(0.85 \times P = 680\) or \(680 \div 0.85\)
A1 for \(800\)

The total number of balls is 10.
Probability of selecting two red balls:
\(P(RR) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\)

Probability of selecting two blue balls:
\(P(BB) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\)

Probability that both balls are the same colour:
\(P(\text{same colour}) = P(RR) + P(BB) = \frac{12}{90} + \frac{30}{90} = \frac{42}{90} = \frac{7}{15}\).

M1 for finding either \(P(RR) = \frac{12}{90}\) or \(P(BB) = \frac{30}{90}\)
M1 for adding the two correct probabilities
A1 for \(\frac{7}{15}\) (or equivalent fraction/decimal)

Find the first and second differences:
Sequence: 2, 9, 20, 35, 54
First differences: 7, 11, 15, 19
Second differences: 4, 4, 4
Since the second difference is constant at 4, the sequence is quadratic of the form \(an^2 + bn + c\), where \(a = \frac{4}{2} = 2\).
Subtract \(2n^2\) from each term:
For \(n=1\): \(2 - 2(1)^2 = 0\)
For \(n=2\): \(9 - 2(2)^2 = 1\)
For \(n=3\): \(20 - 2(3)^2 = 2\)
For \(n=4\): \(35 - 2(4)^2 = 3\)
The remaining sequence \(0, 1, 2, 3, \dots\) has \(n\)th term \(n - 1\).
Therefore, the \(n\)th term is \(2n^2 + n - 1\).

M1 for finding a second difference of 4 and getting \(2n^2\)
A1 for the complete expression \(2n^2 + n - 1\) (or equivalent)

Multiply the numerator and the denominator by the conjugate \(\sqrt{5} + \sqrt{2}\):
\(\frac{\sqrt{3}(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \frac{\sqrt{15} + \sqrt{6}}{5 - 2} = \frac{\sqrt{15} + \sqrt{6}}{3}\).

M1 for multiplying by \(\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}}\)
M1 for denominator simplified to 3
A1 for the correct answer \(\frac{\sqrt{15} + \sqrt{6}}{3}\) or equivalent

Using the laws of logarithms:
\(\log_3(x(x-2)) = 1\)
\(x(x-2) = 3^1\)
\(x^2 - 2x - 3 = 0\)
\((x-3)(x+1) = 0\)

So \(x = 3\) or \(x = -1\).
Since the argument of a logarithm must be positive, we require \(x > 2\). Therefore, we reject \(x = -1\) and get \(x = 3\).

M1 for applying product rule of logarithms: \(\log_3(x(x-2)) = 1\)
M1 for converting to quadratic form: \(x^2 - 2x - 3 = 0\)
A1 for rejecting the negative solution to find \(x = 3\) only

Find the first and second differences of the terms:
First differences: \(7, 11, 15, 19\)
Second differences: \(4, 4, 4\)

Since the second difference is constant, the sequence is quadratic in the form \(an^2 + bn + c\), where \(2a = 4 \implies a = 2\).

Subtracting \(2n^2\) from the sequence:
\(n=1: 2 - 2(1)^2 = 0\)
\(n=2: 9 - 2(2)^2 = 1\)
\(n=3: 20 - 2(3)^2 = 2\)
\(n=4: 35 - 2(4)^2 = 3\)

The resulting linear sequence is \(0, 1, 2, 3, \dots\), which has the formula \(n-1\).
Combining these parts, the \(n\)th term of the sequence is \(2n^2 + n - 1\).

M1 for finding that the second difference is 4 (so coefficient of \(n^2\) is 2)
M1 for subtracting the quadratic term to find the linear sequence \(0, 1, 2, 3, \dots\) or setting up equations
A1 for the correct expression \(2n^2 + n - 1\) oe

First, find the gradient of \(L_1\):
\(m_1 = \frac{4 - (-2)}{3 - 1} = \frac{6}{2} = 3\)

Since \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) satisfies:
\(m_2 = -\frac{1}{m_1} = -\frac{1}{3}\)

Now, use the point-slope form with \((6, 1)\):
\(y - 1 = -\frac{1}{3}(x - 6)\)
\(y - 1 = -\frac{1}{3}x + 2\)
\(y = -\frac{1}{3}x + 3\)

M1 for finding gradient of \(L_1\) is 3
M1 for finding perpendicular gradient of \(-\frac{1}{3}\)
A1 for the final equation \(y = -\frac{1}{3}x + 3\) oe

First, calculate the vector \(2\mathbf{a} + 3\mathbf{b}\):
\(2\mathbf{a} + 3\mathbf{b} = 2\begin{pmatrix} -3 \\ 4 \end{pmatrix} + 3\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix} + \begin{pmatrix} 6 \\ -3 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\)

Now, calculate the magnitude of this vector:
\(|2\mathbf{a} + 3\mathbf{b}| = \sqrt{0^2 + 5^2} = 5\)

M1 for finding \(2\mathbf{a} + 3\mathbf{b} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\)
A1 for magnitude 5

Factorise the numerator:
\(2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\)

Factorise the denominator using the difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\)

Now simplify the fraction by cancelling the common factor \((2x + 1)\):
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} = \frac{x - 3}{2x - 1}\)

M1 for factorising the numerator as \((2x + 1)(x - 3)\)
M1 for factorising the denominator as \((2x - 1)(2x + 1)\)
A1 for the completely simplified fraction \(\frac{x - 3}{2x - 1}\)

Multiply the numerator and denominator by the conjugate \(\sqrt{5} + \sqrt{2}\):
\(\frac{6}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{6(\sqrt{5} + \sqrt{2})}{5 - 2} = \frac{6(\sqrt{5} + \sqrt{2})}{3} = 2(\sqrt{5} + \sqrt{2}) = 2\sqrt{5} + 2\sqrt{2}\)

M1 for multiplying numerator and denominator by \(\sqrt{5} + \sqrt{2}\)
A1 for the correct simplified expression \(2\sqrt{5} + 2\sqrt{2}\) oe

Since there is an 8% discount, the sale price is 92% of the original price.
Let the original price be \(P\).
\(0.92 P = 621\)
\(P = \frac{621}{0.92} = \frac{62100}{92} = 675\)

M1 for writing \(\frac{621}{0.92}\) oe
A1 for 675

We consider two mutually exclusive cases depending on the ball transferred from Bag A:

**Case 1: Red ball is transferred from Bag A**
Probability of transferring a red ball: \(\frac{3}{8}\)
Now Bag B has 5 red balls and 2 blue balls (7 total).
Probability of choosing a red ball from Bag B: \(\frac{5}{7}\)
Joint probability: \(\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)

**Case 2: Blue ball is transferred from Bag A**
Probability of transferring a blue ball: \(\frac{5}{8}\)
Now Bag B has 4 red balls and 3 blue balls (7 total).
Probability of choosing a red ball from Bag B: \(\frac{4}{7}\)
Joint probability: \(\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}\)

**Total Probability:**
\(P(\text{Red from Bag B}) = \frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}\)

M1 for setting up the sum of probabilities for the two paths: \(\frac{3}{8} \times \frac{a}{7} + \frac{5}{8} \times \frac{b}{7}\)
M1 for substituting correct numbers \(a=5\) and \(b=4\) into the products
A1 for \(\frac{5}{8}\) oe

(a) (i) The sequence differences are:
First differences: \(9 - 2 = 7\), \(20 - 9 = 11\), \(35 - 20 = 15\).
Second differences: \(11 - 7 = 4\), \(15 - 11 = 4\).
Since the second difference is constant, the next first difference is \(15 + 4 = 19\).
Next term = \(35 + 19 = 54\).

(ii) The quadratic sequence has form \(an^2 + bn + c\).
Since the second difference is 4, \(2a = 4 \implies a = 2\).
Subtracting \(2n^2\) from the terms:
- \(n = 1\): \(2 - 2(1) = 0\)
- \(n = 2\): \(9 - 8 = 1\)
- \(n = 3\): \(20 - 18 = 2\)
- \(n = 4\): \(35 - 32 = 3\)
The sequence of remainders \(0, 1, 2, 3, \dots\) is \(n - 1\).
Thus, the \(n\)th term is \(2n^2 + n - 1\).

(b) (i) For \(n = 4\):
\(V_4 = 3^4 - 2(4) = 81 - 8 = 73\).

(ii) For \(n = 5\):
\(V_5 = 3^5 - 2(5) = 243 - 10 = 233\).
Difference = \(V_5 - V_4 = 233 - 73 = 160\).

(c) \(W_n = \frac{2n^2 + n - 1}{(3^n - 2n) + 2n} = \frac{2n^2 + n - 1}{3^n}\) or \(W_n = \frac{(2n - 1)(n + 1)}{3^n}\).

(a)(i) B1 [1] for 54
(a)(ii) M1 for finding second differences is 4
M1 for attempting to find the linear part from \(U_n - 2n^2\)
A1 for \(2n^2 + n - 1\) [3]
(b)(i) M1 for substituting \(n = 4\)
A1 for 73 [2]
(b)(ii) M1 for finding \(V_5 = 233\)
A1 for showing subtraction: \(233 - 73 = 160\) [2]
(c) M1 for setting up the algebraic fraction
M1 for simplifying the denominator
A1 for \(\frac{2n^2 + n - 1}{3^n}\) or \(\frac{(2n - 1)(n + 1)}{3^n}\) [3]

(a) Frequencies are calculated by subtracting consecutive cumulative frequencies:
- \(10 < t \le 20\): \(42 - 15 = 27\)
- \(20 < t \le 30\): \(78 - 42 = 36\)
- \(30 < t \le 40\): \(104 - 78 = 26\)
- \(40 < t \le 50\): \(120 - 104 = 16\)

(b) Use the midpoints of the intervals (5, 15, 25, 35, 45):
Estimated mean = \(\frac{15(5) + 27(15) + 36(25) + 26(35) + 16(45)}{120}\)
= \(\frac{75 + 405 + 900 + 910 + 720}{120} = \frac{3010}{120} \approx 25.1\) minutes.

(c) Total adults with time taken \(> 30\) minutes = \(26 + 16 = 42\).
Probability = \(\frac{42}{120} \times \frac{41}{119} = \frac{1722}{14280} = \frac{41}{340} \approx 0.121\).

(d) A probability of 0.4 corresponds to \(0.4 \times 120 = 48\) adults.
Since cumulative frequency reaches 42 at \(t = 20\) and 78 at \(t = 30\), the 48th value must lie in the interval \(20 < t \le 30\).

(e) Percentage = \(\frac{42}{120} \times 100\% = 35\%\).

(a) B2 for all 4 correct frequencies (B1 for 2 or 3 correct) [2]
(b) M1 for identifying correct midpoints
M1 for \(\sum fx\) calculation
A1 for 25.1 (or 25.08) [3]
(c) M1 for finding 42 adults
M1 for \(\frac{42}{120} \times \frac{41}{119}\)
A1 for \(\frac{41}{340}\) or 0.121 [3]
(d) B1 for \(20 < t \le 30\) [1]
(e) M1 for \(\frac{42}{120} \times 100\)
A1 for 35% [2]

(a) (i) \(3\mathbf{p} - \mathbf{q} = 3\begin{pmatrix} -3 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} -9 - 4 \\ 15 - (-2) \end{pmatrix} = \begin{pmatrix} -13 \\ 17 \end{pmatrix}\).

(ii) \(\mathbf{p} + \mathbf{q} = \begin{pmatrix} -3 + 4 \\ 5 - 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\).
Magnitude = \(\sqrt{1^2 + 3^2} = \sqrt{10} \approx 3.16\).

(b) Coordinates of \(D = (5 + 4, -2 - 2) = (9, -4)\).

(c) (i) Under reflection in \(y = -x\), the coordinates \((x, y)\) are mapped to \((-y, -x)\):
- \((1, 1) \rightarrow (-1, -1)\)
- \((3, 1) \rightarrow (-1, -3)\)
- \((1, 4) \rightarrow (-4, -1)\)
Vertices of \(U\) are \((-1, -1)\), \((-1, -3)\), and \((-4, -1)\).

(ii) The coordinates have all been multiplied by 3:
- \((1, 1) \rightarrow (3, 3)\)
- \((3, 1) \rightarrow (9, 3)\)
- \((1, 4) \rightarrow (3, 12)\)
This is an Enlargement, scale factor 3, centre of enlargement \((0, 0)\).

(a)(i) B2 for \(\begin{pmatrix} -13 \\ 17 \end{pmatrix}\) (B1 for one component correct) [2]
(a)(ii) M1 for finding vector \(\mathbf{p} + \mathbf{q} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}\)
M1 for calculating \(\sqrt{1^2 + 3^2}\)
A1 for 3.16 or \(\sqrt{10}\) [3]
(b) B2 for \((9, -4)\) (B1 for each coordinate) [2]
(c)(i) B2 for all three vertices correct (B1 for 1 or 2 correct) [2]
(c)(ii) B1 for Enlargement
B1 for scale factor 3 and centre \((0,0)\) [2]

(a) Deposit = \(0.15 \times 18500 = 2775\) dollars.

(b) Value after 1st year = \(18500 \times (1 - 0.12) = 18500 \times 0.88 = 16280\) dollars.
Value after 3 years (2 more years of 8% decrease) = \(16280 \times (0.92)^2 = 16280 \times 0.8464 = 13779.39\) dollars.
Correct to 3 significant figures: \(13800\) dollars.

(c) \(6000 \times (1 + \frac{r}{100})^5 = 6750\)
\((1 + \frac{r}{100})^5 = \frac{6750}{6000} = 1.125\)
\(1 + \frac{r}{100} = (1.125)^{0.2} \approx 1.0238\)
\(\frac{r}{100} = 0.0238\)
\(r \approx 2.38\).

(d) Let original price be \(P\).
\(1.05 \times P = 24150\)
\(P = \frac{24150}{1.05} = 23000\) dollars.

(a) M1 for \(0.15 \times 18500\)
A1 for 2775 [2]
(b) M1 for finding value after year 1: 16280
M1 for \(16280 \times 0.92^2\)
A1 for 13800 or 13779.39 [3]
(c) M1 for setting up equation: \(6000(1 + r/100)^5 = 6750\)
M1 for taking the 5th root: \((1.125)^{0.2}\)
A1 for 2.38 (accept 2.37 - 2.39) [3]
(d) M2 for \(\frac{24150}{1.05}\) (M1 for \(1.05x = 24150\))
A1 for 23000 [3]

(a) (i) Maximum occurs when \(\cos(2x) = 1\), which is \(4(1) + 1 = 5\).
This happens when \(2x = 0\) or \(2x = 360\), so \(x = 0\) or \(x = 180\).

(ii) Minimum occurs when \(\cos(2x) = -1\), which is \(4(-1) + 1 = -3\).
This happens when \(2x = 180 \implies x = 90\).
So the coordinates of the minimum point are \((90, -3)\).

(b) \(4 \cos(2x) + 1 = 3\)
\(4 \cos(2x) = 2\)
\(\cos(2x) = 0.5\)
\(2x = 60\) or \(2x = 300\)
\(x = 30\) or \(x = 150\).

(c) The sketch should show a cosine wave with a vertical shift of +1, amplitude 4, starting at \((0, 5)\), reaching a minimum of -3 at \((90, -3)\), and ending at \((180, 5)\).

(d) Since the range of the function is \([-3, 5]\), the line \(y = k\) does not intersect it if \(k < -3\) or \(k > 5\).

(a)(i) B1 for maximum value of 5
B1 for \(x = 0\) or \(180\) [2]
(a)(ii) B1 for \(x = 90\)
B1 for \(y = -3\) [2]
(b) M1 for \(\cos(2x) = 0.5\)
A1 for \(x = 30\)
A1 for \(x = 150\) [3]
(c) B2 for correct shape wave starting at \((0, 5)\), min at \((90, -3)\), ending at \((180, 5)\) (B1 for partially correct shape or wrong amplitude/shift) [2]
(d) B1 for \(k < -3\)
B1 for \(k > 5\) [2]

(a) (i) \(P(\text{No Rain and Walk}) = P(\text{No Rain}) \times P(\text{Walk}|\text{No Rain}) = (1 - 0.4) \times 0.8 = 0.6 \times 0.8 = 0.48\).

(ii) \(P(\text{Liam Drives}) = P(\text{Rain and Drive}) + P(\text{No Rain and Drive})\)
\(= 0.4 \times (1 - 0.3) + 0.6 \times (1 - 0.8) = 0.4 \times 0.7 + 0.6 \times 0.2 = 0.28 + 0.12 = 0.40\).

(b) This is a binomial distribution problem with \(n = 5\), \(p = 0.4\), and \(k = 2\):
\(P(X = 2) = \binom{5}{2} (0.4)^2 (0.6)^3 = 10 \times 0.16 \times 0.216 = 0.3456 \approx 0.346\).

(c) We use Bayes' theorem:
\(P(\text{Rain}|\text{Walk}) = \frac{P(\text{Rain and Walk})}{P(\text{Walk})}\)
\(P(\text{Rain and Walk}) = 0.4 \times 0.3 = 0.12\).
\(P(\text{Walk}) = P(\text{Rain and Walk}) + P(\text{No Rain and Walk}) = 0.12 + 0.48 = 0.60\).
\(P(\text{Rain}|\text{Walk}) = \frac{0.12}{0.60} = 0.2\).

(a)(i) M1 for \(0.6 \times 0.8\)
A1 for 0.48 [2]
(a)(ii) M1 for \(0.4 \times 0.7\) or \(0.6 \times 0.2\)
M1 for summing the two correct probabilities
A1 for 0.40 [3]
(b) M1 for using binomial expansion term with \(\binom{5}{2}\)
M1 for \((0.4)^2 \times (0.6)^3\)
A1 for 0.346 (or 0.3456) [3]
(c) M1 for finding \(P(\text{Rain and Walk}) = 0.12\)
M1 for setting up conditional probability fraction with denominator \(0.60\)
A1 for 0.2 [3]

(a) Area = \(\frac{1}{2} \times AB \times AC \times \sin(BAC) = \frac{1}{2} \times 7 \times 9 \times \sin(58^∘) \approx 26.713\text{ cm}^2\).
Correct to 3 significant figures: \(26.7\text{ cm}^2\).

(b) Using the cosine rule:
\(BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC)\)
\(BC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos(58^∘) = 49 + 81 - 126 \times \cos(58^∘)\)
\(BC^2 \approx 130 - 126 \times 0.52992 = 130 - 66.77 = 63.23\)
\(BC = \sqrt{63.23} \approx 7.95\text{ cm}\).

(c) The shortest distance from \(A\) to \(BC\) is the height, \(h\):
Area = \(\frac{1}{2} \times BC \times h \implies 26.713 = \frac{1}{2} \times 7.9517 \times h\)
\(h = \frac{2 \times 26.713}{7.9517} \approx 6.719\text{ cm}\).
Correct to 3 significant figures: \(6.72\text{ cm}\).

(d) Using the sine rule:
\(\frac{\sin(ABC)}{AC} = \frac{\sin(BAC)}{BC} \implies \frac{\sin(ABC)}{9} = \frac{\sin(58^∘)}{7.9517}\)
\(\sin(ABC) = \frac{9 \times \sin(58^∘)}{7.9517} \approx 0.9598\)
\(ABC \approx 73.7^\circ\).

(a) M1 for \(\frac{1}{2} \times 7 \times 9 \times \sin(58)\)
A1 for 26.7 [2]
(b) M1 for correct cosine rule formula
M1 for substituting correct values: \(49 + 81 - 126\cos(58)\)
A1 for 7.95 [3]
(c) M1 for height formula \(\frac{1}{2} \times \text{base} \times \text{height} = \text{Area}\)
M1 for substituting their values of Area and BC
A1 for 6.72 [3]
(d) M1 for correct sine rule setup
M1 for solving for \(\sin(ABC)\)
A1 for 73.7 [3]

(a) Factorize the numerator: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).
Factorize the denominator: \(x^2 - 9 = (x - 3)(x + 3)\).
Cancel the common factor \((x - 3)\):
\(\frac{(2x + 1)(x - 3)}{(x + 3)(x - 3)} = \frac{2x + 1}{x + 3}\).

(b) Multiply through by \(y(y - 2)\):
\(4(y - 2) + y = 3y(y - 2)\)
\(4y - 8 + y = 3y^2 - 6y\)
\(5y - 8 = 3y^2 - 6y\)
\(3y^2 - 11y + 8 = 0\)
Factorize the quadratic equation:
\((3y - 8)(y - 1) = 0\)
So, \(y = 1\) or \(y = \frac{8}{3}\) (or \(2.67\)).

(c) Find the common denominator, which is \(ab\):
\(\frac{3}{a} - \frac{2}{b} + 1 = \frac{3b}{ab} - \frac{2a}{ab} + \frac{ab}{ab} = \frac{3b - 2a + ab}{ab}\).

(a) M1 for factorizing numerator: \((2x+1)(x-3)\)
M1 for factorizing denominator: \((x+3)(x-3)\)
M1 for canceling common factor \((x-3)\)
A1 for \(\frac{2x+1}{x+3}\) [4]
(b) M1 for expanding/clearing fractions: \(4(y-2) + y = 3y(y-2)\)
M1 for setting up quadratic equation: \(3y^2 - 11y + 8 = 0\)
M1 for solving by factorization or formula
A1 for \(y=1\) and \(y=\frac{8}{3}\) (or 2.67) [4]
(c) M1 for finding common denominator \(ab\)
M1 for correct numerators over common denominator
A1 for \(\frac{3b - 2a + ab}{ab}\) [3]

**(a)** Use the laws of logarithms:
\(\log_{3}((2x + 1)(x - 1)) = 2\)
\((2x + 1)(x - 1) = 3^2\)
\(2x^2 - x - 1 = 9\)
\(2x^2 - x - 10 = 0\)

Factorise the quadratic equation:
\((2x - 5)(x + 2) = 0\)

So, \(x = 2.5\) or \(x = -2\).
Since we must have \(x - 1 > 0\) (i.e. \(x > 1\)) for the logarithms to be defined, the only valid solution is \(x = 2.5\).

**(b)** Let the time be \(n\) years.
\(P(1.024)^n = 2P\)
\(1.024^n = 2\)

Take logarithms of both sides:
\(n \log(1.024) = \log(2)\)
\(n = \frac{\log(2)}{\log(1.024)} \approx 29.22\)

For the investment to double, we need the next complete year, which is \(30\) years.

**(c)**
\(y = a \cdot b^{2x}\)
\(\frac{y}{a} = b^{2x}\)

Take logarithms of both sides:
\(\log\left(\frac{y}{a}\right) = 2x \log(b)\)
\(2x = \frac{\log(y/a)}{\log(b)} \implies x = \frac{\log(y/a)}{2 \log(b)}\) or \(x = \frac{\log(y) - \log(a)}{2\log(b)}\)

**(a)** [4 Marks]
- M1 for \(\log_3((2x+1)(x-1)) = 2\)
- M1 for expansion to quadratic form \(2x^2 - x - 1 = 9\) oe
- M1 for factorization/quadratic formula to get \(x = 2.5\) or \(x = -2\)
- A1 for \(x = 2.5\) only (rejecting \(x = -2\))

**(b)** [3 Marks]
- M1 for setting up the equation \((1.024)^n = 2\)
- M1 for correct use of logs, e.g., \(n = \frac{\log(2)}{\log(1.024)}\)
- A1 for 30 (must be a whole number)

**(c)** [3 Marks]
- M1 for isolating the base: \(\frac{y}{a} = b^{2x}\)
- M1 for taking logs of both sides correctly: \(\log(y/a) = 2x \log(b)\)
- A1 for final correct answer \(x = \frac{\log(y/a)}{2\log(b)}\) or equivalent

**(a)(i)** P(all 3 same color) = P(3 blue) + P(3 red) + P(3 green).
Since there are only 2 green pencils, P(3 green) = 0.
\(P(3 \text{ blue}) = \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8} = \frac{60}{720} = \frac{1}{12}\)
\(P(3 \text{ red}) = \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} = \frac{6}{720} = \frac{1}{120}\)
\(P(\text{same color}) = \frac{60 + 6}{720} = \frac{66}{720} = \frac{11}{120} \approx 0.0917\)

**(a)(ii)** P(at least 1 green) = 1 - P(no green).
\(P(\text{no green}) = \frac{8}{10} \times \frac{7}{9} \times \frac{6}{8} = \frac{336}{720} = \frac{7}{15}\)
\(P(\text{at least 1 green}) = 1 - \frac{7}{15} = \frac{8}{15} \approx 0.533\)

**(b)(i)** Let \(S\) represent playing a sport and \(M\) represent playing a musical instrument.
\(P(M) = P(S)P(M|S) + P(S')P(M|S')\)
\(P(M) = 0.60 \times 0.30 + 0.40 \times 0.15 = 0.18 + 0.06 = 0.24\)

**(b)(ii)** \(P(S'|M) = \frac{P(S' \cap M)}{P(M)}\)
\(P(S' \cap M) = 0.40 \times 0.15 = 0.06\)
\(P(S'|M) = \frac{0.06}{0.24} = 0.25\)

**(a)(i)** [3 Marks]
- M1 for either product: \(\frac{5}{10} \times \frac{4}{9} \times \frac{3}{8}\) or \(\frac{3}{10} \times \frac{2}{9} \times \frac{1}{8}\)
- M1 for adding the two products correctly
- A1 for \(\frac{11}{120}\) oe

**(a)(ii)** [3 Marks]
- M1 for product with no green: \(\frac{8}{10} \times \frac{7}{9} \times \frac{6}{8}\)
- M1 for subtracting their product from 1
- A1 for \(\frac{8}{15}\) oe

**(b)(i)** [2 Marks]
- M1 for \(0.60 \times 0.30 + 0.40 \times 0.15\) oe
- A1 for 0.24 oe

**(b)(ii)** [2 Marks]
- M1 for correct formula or ratio of probabilities: \(\frac{0.06}{\text{their } 0.24}\)
- A1 for 0.25 oe

**(a)** Find the first differences:
\(11 - 5 = 6\)
\(21 - 11 = 10\)
\(35 - 21 = 14\)
\(53 - 35 = 18\)

Find the second differences:
\(10 - 6 = 4\)
\(14 - 10 = 4\)
\(18 - 14 = 4\)

Since the second difference is constant and equals 4, the coefficient of \(n^2\) is \(\frac{4}{2} = 2\).

Subtract \(2n^2\) from the original terms to find any remaining linear/constant parts:
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 difference is a constant 3, the \(n\)th term of the sequence is \(2n^2 + 3\).

**(b)(i)** Form simultaneous equations using \(u_n = an^2 + bn\):
For \(n=2\): \(a(2)^2 + b(2) = 14 \implies 4a + 2b = 14 \implies 2a + b = 7\)
For \(n=4\): \(a(4)^2 + b(4) = 44 \implies 16a + 4b = 44 \implies 4a + b = 11\)

Subtract the first simplified equation from the second:
\((4a + b) - (2a + b) = 11 - 7\)
\(2a = 4 \implies a = 2\)

Substitute \(a = 2\) back into the first equation:
\(2(2) + b = 7 \implies b = 3\).

**(b)(ii)** Substitute \(a = 2\), \(b = 3\), and \(n = 10\) into the formula:
\(u_{10} = 2(10)^2 + 3(10) = 2(100) + 30 = 230\).

**(a)** [4 Marks]
- M1 for finding first differences: 6, 10, 14, 18
- M1 for finding constant second difference of 4
- M1 for identifying the quadratic term as \(2n^2\) and subtracting it from sequence
- A1 for \(2n^2 + 3\)

**(b)(i)** [4 Marks]
- M1 for setting up at least one correct equation: \(4a + 2b = 14\) or \(16a + 4b = 44\) (or simplified versions)
- M1 for a valid method to solve the linear simultaneous equations
- A1 for \(a = 2\)
- A1 for \(b = 3\)

**(b)(ii)** [2 Marks]
- M1 for substituting their values of \(a\), \(b\) and \(n = 10\) into \(an^2 + bn\)
- A1 for 230

(a) Using the rule:
\(G_3 = G_2 + 2G_1 = 3 + 2(1) = 5\)
\(G_4 = G_3 + 2G_2 = 5 + 2(3) = 11\)
\(G_5 = G_4 + 2G_3 = 11 + 2(5) = 21\)

(b) Generalizing with \(G_1 = x\) and \(G_2 = y\):
\(G_3 = G_2 + 2G_1 = y + 2x = 2x + y\)
\(G_4 = G_3 + 2G_2 = (2x + y) + 2y = 2x + 3y\)
\(G_5 = G_4 + 2G_3 = (2x + 3y) + 2(2x + y) = 6x + 5y\)

(c) Substituting the algebraic expressions into the left-hand side:
\(G_5 - 2G_3 = (6x + 5y) - 2(2x + y)\)
\(= 6x + 5y - 4x - 2y\)
\(= 2x + 3y\)
This is equal to the expression for \(G_4\), hence the relationship is always true.

(a) **[2 marks]**
- **B1** for \(G_3 = 5\) or \(G_4 = 11\)
- **B1** for all three correct: 5, 11, 21

(b) **[2 marks]**
- **B1** for \(G_4 = 2x + 3y\)
- **B1** for \(G_5 = 6x + 5y\)

(c) **[2 marks]**
- **M1** for attempting to substitute expressions for \(G_5\) and \(G_3\) into the LHS: \((6x + 5y) - 2(2x + y)\)
- **A1** for simplifying correctly to \(2x + 3y\) and concluding that this equals \(G_4\).

(a) The sequence of Fibonacci numbers is: \(1, 1, 2, 3, 5, 8, 13, 21, 34, \dots\) So \(f_9 = 21 + 13 = 34\).

(b) Completing the calculations:
\(S_5 = 15 + 5^2 = 15 + 25 = 40\)
\(S_6 = 40 + 8^2 = 40 + 64 = 104\)

(c) Observing the values:
\(S_1 = 1 = 1 \times 1 = f_1 \times f_2\)
\(S_2 = 2 = 1 \times 2 = f_2 \times f_3\)
\(S_3 = 6 = 2 \times 3 = f_3 \times f_4\)
\(S_4 = 15 = 3 \times 5 = f_4 \times f_5\)
\(S_5 = 40 = 5 \times 8 = f_5 \times f_6\)
\(S_6 = 104 = 8 \times 13 = f_6 \times f_7\)
So the relationship is \(S_n = f_n \times f_{n+1}\).

(d) For \(n = 8\), the sum is \(S_8 = f_8 \times f_9 = 21 \times 34 = 273\).

(a) **[1 mark]**
- **B1** for 34

(b) **[2 marks]**
- **B1** for \(S_5 = 40\)
- **B1** for \(S_6 = 104\)

(c) **[1 mark]**
- **B1** for \(S_n = f_n \times f_{n+1}\) or equivalent verbal description.

(d) **[2 marks]**
- **M1** for \(f_8 \times f_9 = 21 \times 34\)
- **A1** for 273

(a) \(T_6 = \frac{6(7)}{2} = 21\).

(b)
\(T_5^2 - T_4^2 = 15^2 - 10^2 = 225 - 100 = 125\)
\(T_6^2 - T_5^2 = 21^2 - 15^2 = 441 - 225 = 216\)

(c) Observing the results:
\(8 = 2^3\) (for \(n=2\))
\(27 = 3^3\) (for \(n=3\))
\(64 = 4^3\) (for \(n=4\))
\(125 = 5^3\) (for \(n=5\))
\(216 = 6^3\) (for \(n=6\))
The relationship is \(T_n^2 - T_{n-1}^2 = n^3\).

(d) Substituting \(T_n = \frac{n(n+1)}{2}\) and \(T_{n-1} = \frac{(n-1)n}{2}\):
\(T_n^2 - T_{n-1}^2 = \left(\frac{n(n+1)}{2}\right)^2 - \left(\frac{(n-1)n}{2}\right)^2\)
\(= \frac{n^2(n+1)^2}{4} - \frac{n^2(n-1)^2}{4}\)
\(= \frac{n^2}{4} \left( (n^2 + 2n + 1) - (n^2 - 2n + 1) \right)\)
\(= \frac{n^2}{4} \left( 4n \right) = n^3\).
This shows the relationship is always true.

(a) **[1 mark]**
- **B1** for 21

(b) **[2 marks]**
- **B1** for 125
- **B1** for \(21^2 - 15^2\) and 216

(c) **[1 mark]**
- **B1** for \(T_n^2 - T_{n-1}^2 = n^3\) or equivalent statement.

(d) **[2 marks]**
- **M1** for writing \(\frac{n^2(n+1)^2}{4} - \frac{n^2(n-1)^2}{4}\) or equivalent expansion.
- **A1** for correct algebraic expansion and simplification to \(n^3\).

(a) The sequence increases by 2, then 3, then 4, then 5, then 6...
\(D_7 = 9 + 5 = 14\)
\(D_8 = 14 + 6 = 20\)

(b) Sequence of \(D_n\): 0, 2, 5, 9, 14, 20
First differences: \(2, 3, 4, 5, 6\)
Second differences: \(1, 1, 1, 1\)
Since the second differences are constant and non-zero, the sequence is represented by a quadratic expression.

(c) Using \(n = 20\) in the formula:
\(D_{20} = \frac{20(20-3)}{2} = 10 \times 17 = 170\).

(a) **[2 marks]**
- **B1** for 14
- **B1** for 20

(b) **[2 marks]**
- **M1** for finding first differences: 2, 3, 4, 5, 6 and second differences: 1, 1, 1, 1
- **A1** for stating that "second differences are constant, which means the formula is quadratic".

(c) **[2 marks]**
- **M1** for substituting \(n = 20\) into the formula: \(\frac{20(17)}{2}\)
- **A1** for 170

(a) Following the rule \(f_n = f_{n-1} + f_{n-2}\):
\(f_{11} = 55 + 34 = 89\)
\(f_{12} = 89 + 55 = 144\)

(b)
\(P_4 = 8 + f_7 = 8 + 13 = 21\)
\(P_5 = 21 + f_9 = 21 + 34 = 55\)

(c) Comparing \(P_n\) values with the Fibonacci sequence:
\(P_1 = 1 = f_2\)
\(P_2 = 3 = f_4\)
\(P_3 = 8 = f_6\)
\(P_4 = 21 = f_8\)
\(P_5 = 55 = f_{10}\)
So, \(P_n = f_{2n}\).

(d) The sum \(f_1 + f_3 + f_5 + f_7 + f_9 + f_{11} + f_{13} + f_{15}\) is \(P_8\). Using the relationship, \(P_8 = f_{2(8)} = f_{16} = 987\).

(a) **[1 mark]**
- **B1** for both \(f_{11} = 89\) and \(f_{12} = 144\) correct.

(b) **[2 marks]**
- **B1** for \(P_4 = 21\)
- **B1** for \(P_5 = 55\)

(c) **[1 mark]**
- **B1** for \(P_n = f_{2n}\) or equivalent written description.

(d) **[2 marks]**
- **M1** for identifying that the sum of the first 8 odd-indexed terms is \(P_8 = f_{16}\)
- **A1** for 987

Part (a):
A straight-line model passing through \((0,0)\) and \((12, 84)\) has gradient \(m = \frac{84 - 0}{12 - 0} = 7\).
So, the model is \(H = 7D\).

Part (b):
For \(0 \le D \le 2\), the straight line passes through \((0,0)\) and \((2, 24)\). The gradient is \(m = \frac{24 - 0}{2 - 0} = 12\).
So, the model is \(H = 12D\).

Part (c):
For \(D \ge 2\), the line passes through \((2, 24)\) with a gradient of 5.
Using \(H - H_1 = m(D - D_1)\), we have:
\(H - 24 = 5(D - 2)\)
\(H - 24 = 5D - 10\)
\(H = 5D + 14\)

(a) **M1** for attempting to find the gradient \(\frac{84}{12}\) or equivalent. **A1** for \(H = 7D\).
(b) **A2** for \(H = 12D\) (or **M1** for gradient of 12).
(c) **M1** for using gradient 5 with point \((2, 24)\). **A1** for \(H = 5D + 14\).

Part (a):
Substituting \(D = 1, H = 15\) gives:
\(15 = p \ln(1) + q \Rightarrow q = 15\).
Substituting \(D = 12, H = 75\) gives:
\(75 = p \ln(12) + q\).

Part (b):
From \(q = 15\), we substitute into the second equation:
\(75 = p \ln(12) + 15\)
\(60 = p \ln(12)\)
\(p = \frac{60}{\ln 12} \approx 24.145\).
To 1 decimal place, \(p = 24.1\) and \(q = 15.0\).

Part (c):
For \(D = 5\), using the model parameters:
\(H = 24.145 \ln(5) + 15\)
\(H \approx 24.145(1.6094) + 15\)
\(H \approx 38.86 + 15 = 53.86\).
Rounding to the nearest integer gives 54 human years.

(a) **B1** for \(q = 15\) or equivalent. **B1** for \(p \ln 12 + q = 75\).
(b) **M1** for attempting to solve for \(p\) (e.g. \(\frac{60}{\ln 12}\)). **A1** for both \(p = 24.1\) and \(q = 15.0\) correct to 1 d.p.
(c) **M1** for substituting \(D=5\) into their model. **A1** for 54 (accept 53.8 to 54.2).

Part (a):
The vertex of the parabola is given as \((50, 12.8)\).
In the vertex form \(h(t) = a(t - d)^2 + e\), \(d\) represents the \(t\)-coordinate of the vertex and \(e\) represents the \(h\)-coordinate.
Thus, \(d = 50\) and \(e = 12.8\).

Part (b):
Using \(h(0) = 0.3\), substitute \(t = 0\) and the values of \(d\) and \(e\) into the equation:
\(0.3 = a(0 - 50)^2 + 12.8\)
\(0.3 = 2500a + 12.8\)
\(-12.5 = 2500a\)
\(a = -0.005\) (or \(-\frac{1}{200}\)).

Part (c):
Expand the vertex form equation:
\(h(t) = -0.005(t - 50)^2 + 12.8\)
\(h(t) = -0.005(t^2 - 100t + 2500) + 12.8\)
\(h(t) = -0.005t^2 + 0.5t - 12.5 + 12.8\)
\(h(t) = -0.005t^2 + 0.5t + 0.3\)

(a) **B1** for \(d = 50\). **B1** for \(e = 12.8\).
(b) **M1** for substituting \((0, 0.3)\) into \(a(t-50)^2 + 12.8\). **A1** for \(a = -0.005\) or equivalent.
(c) **M1** for correct expansion of \((t-50)^2\). **A1** for \(h(t) = -0.005t^2 + 0.5t + 0.3\) or equivalent.

Part (a):
Since \(P(0) = A k^0 = 500\), we have \(A = 500\).
At \(t = 3\), the population is 4000, so the equation is:
\(500 k^3 = 4000\).

Part (b):
Solving the equation for \(k\):
\(k^3 = 8 \Rightarrow k = 2\).
The complete model is:
\(P(t) = 500 \cdot 2^t\).

Part (c):
We want \(500 \cdot 2^t > 50000\)
\(2^t > 100\).
Since \(2^6 = 64\) and \(2^7 = 128\), the population first exceeds 50 000 at \(t = 7\) hours.
Thus, the number of complete hours is 7.

(a) **B1** for \(A = 500\). **B1** for \(500 k^3 = 4000\) or equivalent.
(b) **B1** for \(k = 2\). **B1** for \(P(t) = 500 \cdot 2^t\).
(c) **M1** for setting up the inequality \(500 \cdot 2^t > 50000\) or equivalent. **A1** for 7.

Part (a):
Substituting \(M = 8\) and \(R = 200\) into the model:
\(200 = \frac{c}{\sqrt[3]{8}}\)
\(200 = \frac{c}{2} \Rightarrow c = 400\).

Part (b):
For a horse with \(R = 40\):
\(40 = \frac{400}{\sqrt[3]{M}}\)
\(\sqrt[3]{M} = \frac{400}{40} = 10\)
\(M = 10^3 = 1000\) kg.

Part (c):
For a blue whale with \(M = 125000\):
\(R = \frac{400}{\sqrt[3]{125000}} = \frac{400}{50} = 8\) beats per minute.

(a) **M1** for correct substitution of \(M=8\) and \(R=200\) into the proportion formula. **A1** for showing \(c = 400\).
(b) **M1** for rearranging \(40 = \frac{400}{\sqrt[3]{M}}\) to find \(\sqrt[3]{M}\). **A1** for 1000 (unit of kg not strictly required if clear).
(c) **M1** for evaluating \(\sqrt[3]{125000}\) as 50. **A1** for 8.