Cambridge IGCSE · thinka-original Practice Paper

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

Thinka Nov 2024 (V2) Cambridge IGCSE-Style Mock — International Mathematics (0607)

220 marks280 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V2) Cambridge IGCSE International Mathematics (0607) paper. Not affiliated with or reproduced from Cambridge.

Paper 22 (Extended)

Answer all questions. Calculators must not be used.

18 Question · 42 marks

Question 1 · Short Answer

2 marks

Simplify fully $\sqrt{28} - \sqrt{63} + \sqrt{112}$.

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Worked solution

First, simplify each surd by finding the largest square factor of the number inside the square root:

$\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$
$\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$
$\sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7}$

Substitute these back into the expression:

$2\sqrt{7} - 3\sqrt{7} + 4\sqrt{7} = 3\sqrt{7}$

Marking scheme

M1 for writing at least two of the terms in the form $k\sqrt{7}$ (specifically $2\sqrt{7}$, $3\sqrt{7}$, or $4\sqrt{7}$)
A1 for $3\sqrt{7}$ final answer

Question 2 · Short Answer

3 marks

Solve the equation $\log_4(x+3) + \log_4(2) = 2$.

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Worked solution

Use the laws of logarithms to combine the terms on the left-hand side:

$\log_4(2(x+3)) = 2$

Convert the logarithmic equation into its exponential form:

$2(x+3) = 4^2$

$2x + 6 = 16$

$2x = 10$

$x = 5$

Marking scheme

M1 for using the addition law of logarithms to write $\log_4(2(x+3)) = 2$ oe
M1 for converting to exponential form: $2(x+3) = 16$ oe
A1 for $x = 5$ cao

Question 3 · Short Answer

3 marks

Two mathematically similar solids have volumes of $16\text{ cm}^3$ and $54\text{ cm}^3$. The surface area of the larger solid is $45\text{ cm}^2$. Calculate the surface area of the smaller solid.

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Worked solution

Find the ratio of the volumes of the two similar solids:

$\frac{V_{\text{small}}}{V_{\text{large}}} = \frac{16}{54} = \frac{8}{27}$

Since the ratio of the volumes of similar solids is equal to the cube of the scale factor of their lengths $k$:

$k^3 = \frac{8}{27} \implies k = \frac{2}{3}$

The ratio of their surface areas is equal to the square of the scale factor of their lengths:

$\frac{A_{\text{small}}}{A_{\text{large}}} = k^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

Use this ratio to find the surface area of the smaller solid:

$\frac{A_{\text{small}}}{45} = \frac{4}{9} \implies A_{\text{small}} = 45 \times \frac{4}{9} = 20\text{ cm}^2$

Marking scheme

M1 for simplifying the volume ratio to $\frac{8}{27}$ or finding the linear scale factor $\frac{2}{3}$
M1 for squaring the linear scale factor to obtain $\frac{4}{9}$
A1 for $20$ (accept $20\text{ cm}^2$)

Question 4 · Short Answer

3 marks

Find the equation of the line perpendicular to the line $y = 3x - 5$ that passes through the point $(6, 2)$. Give your answer in the form $ax + by = c$, where $a$, $b$ and $c$ are integers.

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Worked solution

The gradient of the given line is $m_1 = 3$.

The gradient $m_2$ of a line perpendicular to it must satisfy $m_1 \times m_2 = -1$, so:

$m_2 = -\frac{1}{3}$

Use the point-slope formula with point $(6, 2)$:

$y - 2 = -\frac{1}{3}(x - 6)$

Multiply both sides by 3 to eliminate the fraction:

$3(y - 2) = -(x - 6)$

$3y - 6 = -x + 6$

Rearrange the equation into the form $ax + by = c$:

$x + 3y = 12$

Marking scheme

B1 for gradient of perpendicular line being $-\frac{1}{3}$
M1 for substituting point $(6, 2)$ into a linear equation using their perpendicular gradient
A1 for $x + 3y = 12$ oe with integer coefficients

Question 5 · Short Answer

2 marks

Solve the equation $\cos x = -\frac{1}{2}$ for $0^\circ \le x \le 360^\circ$.

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Worked solution

Find the reference angle by solving $\cos \theta = \frac{1}{2}$, which gives $\theta = 60^\circ$.

Since $\cos x$ is negative, $x$ must lie in the second or third quadrant.

In the second quadrant:
$x = 180^\circ - 60^\circ = 120^\circ$

In the third quadrant:
$x = 180^\circ + 60^\circ = 240^\circ$

So the solutions are $x = 120^\circ$ or $x = 240^\circ$.

Marking scheme

B1 for $120$ (or $120^\circ$) or $240$ (or $240^\circ$)
B1 for the other correct solution and no extras in range

Question 6 · Short Answer

3 marks

Write as a single fraction in its simplest form:

\[ \frac{3}{2x-1} - \frac{2}{x+3} \]

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Worked solution

To subtract the fractions, find a common denominator, which is $(2x-1)(x+3)$:

\[ \frac{3(x+3) - 2(2x-1)}{(2x-1)(x+3)} \]

Expand the numerator:

\[ 3x + 9 - (4x - 2) = 3x + 9 - 4x + 2 = -x + 11 \]

Write the final simplified fraction:

\[ \frac{11-x}{(2x-1)(x+3)} \]

Marking scheme

M1 for a common denominator of $(2x-1)(x+3)$ with at least one correct numerator term
M1 for correct expansion of numerator: $3x + 9 - 4x + 2$ or $11 - x$ seen
A1 for $\frac{11-x}{(2x-1)(x+3)}$ oe final answer

Question 7 · Short Answer

3 marks

Solve the inequality $x^2 - 5x - 6 \le 0$.

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Worked solution

First, factorise the quadratic expression on the left-hand side:

$(x - 6)(x + 1) \le 0$

The critical values are the roots of $(x - 6)(x + 1) = 0$, which are $x = 6$ and $x = -1$.

Since the inequality is $\le 0$, the solution is the region between the two critical values:

$-1 \le x \le 6$

Marking scheme

M1 for factorising the quadratic to $(x-6)(x+1)$ or finding the roots $6$ and $-1$
M1 for choosing the region between their critical values
A1 for $-1 \le x \le 6$ oe

Question 8 · Short Answer

3 marks

A sphere of radius $r$ has the same volume as a cone with radius $2r$ and height $h$. Find $h$ in terms of $r$.

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Worked solution

Let the volume of the sphere be $V_{\text{sphere}}$ and the volume of the cone be $V_{\text{cone}}$.

$V_{\text{sphere}} = \frac{4}{3}\pi r^3$

$V_{\text{cone}} = \frac{1}{3}\pi (2r)^2 h = \frac{1}{3}\pi (4r^2) h = \frac{4}{3}\pi r^2 h$

Since the volumes are equal:

$\frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^2 h$

Divide both sides by $\frac{4}{3}\pi r^2$:

$h = r$

Marking scheme

M1 for using correct formula for volume of sphere: $\frac{4}{3}\pi r^3$
M1 for using correct formula for volume of cone: $\frac{1}{3}\pi (2r)^2 h$
A1 for $h = r$

Question 9 · Short Answer

2 marks

Expand and simplify $3x(x - 4) - 2(x^2 - 5x)$.

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Worked solution

Expand the first term:
$3x(x - 4) = 3x^2 - 12x$

Expand the second term:
$-2(x^2 - 5x) = -2x^2 + 10x$

Combine like terms:
$(3x^2 - 2x^2) + (-12x + 10x) = x^2 - 2x$

Marking scheme

M1 for $3x^2 - 12x$ or $-2x^2 + 10x$ seen
A1 for $x^2 - 2x$ final answer

Question 10 · Short Answer

2 marks

Find the value of $x$ when $\log_5(x) - \log_5(2) = 2$.

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Worked solution

Use the subtraction law of logarithms:
$\log_5\left(\frac{x}{2}\right) = 2$

Convert to exponential form:
$\frac{x}{2} = 5^2$
$\frac{x}{2} = 25$

Multiply both sides by 2:
$x = 50$

Marking scheme

M1 for $\log_5\left(\frac{x}{2}\right) = 2$ or $\frac{x}{2} = 5^2$ seen
A1 for 50

Question 11 · Short Answer

2 marks

Simplify fully $\sqrt{12} + \sqrt{75} - \sqrt{27}$.

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Worked solution

Simplify each surd into its simplest form:
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
$\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$
$\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

Combine the terms:
$2\sqrt{3} + 5\sqrt{3} - 3\sqrt{3} = (2 + 5 - 3)\sqrt{3} = 4\sqrt{3}$

Marking scheme

M1 for any of $2\sqrt{3}$, $5\sqrt{3}$, or $3\sqrt{3}$ seen
A1 for $4\sqrt{3}$ as the final simplified answer

Question 12 · Short Answer

2 marks

Simplify $\left(27x^9\right)^{\frac{2}{3}}$.

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Worked solution

Apply the power $\frac{2}{3}$ to each factor inside the parentheses:
$(27)^{\frac{2}{3}} \times \left(x^9\right)^{\frac{2}{3}}$

Calculate the numerical part:
$(27)^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$

Calculate the algebraic part:
$\left(x^9\right)^{\frac{2}{3}} = x^{9 \times \frac{2}{3}} = x^6$

Combine the parts:
$9x^6$

Marking scheme

M1 for $9x^k$ (where $k \neq 0$) or $kx^6$ (where $k \neq 0$) seen
A1 for $9x^6$ final answer

Question 13 · Short Answer

2 marks

Work out $(3 \times 10^5) \times (8 \times 10^{-2})$, giving your answer in standard form.

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Worked solution

Multiply the numbers:
$3 \times 8 = 24$

Multiply the powers of 10:
$10^5 \times 10^{-2} = 10^{5 + (-2)} = 10^3$

Combine:
$24 \times 10^3$

Convert to standard form:
$2.4 \times 10^4$

Marking scheme

M1 for $24 \times 10^3$ or $0.24 \times 10^5$ seen
A1 for $2.4 \times 10^4$ in correct standard form

Question 14 · Short Answer

2 marks

The $n$-th term of a sequence is $n^2 - 3n$. Find the 10th term of this sequence.

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Worked solution

Substitute $n = 10$ into the formula:
$10^2 - 3(10)$
$= 100 - 30$
$= 70$

Marking scheme

M1 for substituting $n = 10$ into the expression: $10^2 - 3(10)$
A1 for 70

Question 15 · Short Answer

2 marks

Solve the inequality $5 - 3x < 17$.

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Worked solution

Subtract 5 from both sides:
$-3x < 12$

Divide both sides by $-3$ and reverse the inequality sign because you are dividing by a negative number:
$x > -4$

Marking scheme

M1 for $-3x < 12$ or for dividing by $-3$ without reversing the inequality sign (i.e. $x < -4$)
A1 for $x > -4$ or $-4 < x$

Question 16 · Short Answer

2 marks

Find the equation of the line that is parallel to $y = 4x - 3$ and passes through the point $(2, 10)$.

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Worked solution

Parallel lines have equal gradients, so the gradient $m = 4$.

The equation of the line is of the form:
$y = 4x + c$

Substitute the point $(2, 10)$ to calculate $c$:
$10 = 4(2) + c$
$10 = 8 + c$
$c = 2$

Thus, the equation is $y = 4x + 2$.

Marking scheme

M1 for stating gradient is 4 or writing $y = 4x + c$
A1 for $y = 4x + 2$ or equivalent equation

Question 17 · Short Answer

2 marks

Simplify $\left(81y^{-8}\right)^{-\frac{3}{4}}$.

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Worked solution

First, apply the power of $-\frac{3}{4}$ to both terms inside the parentheses: $81^{-\frac{3}{4}} \times \left(y^{-8}\right)^{-\frac{3}{4}}$. Simplify the numerical part: $81^{-\frac{3}{4}} = \frac{1}{81^{\frac{3}{4}}} = \frac{1}{(3^4)^{\frac{3}{4}}} = \frac{1}{3^3} = \frac{1}{27}$. Simplify the algebraic part: $\left(y^{-8}\right)^{-\frac{3}{4}} = y^{-8 \times -\frac{3}{4}} = y^6$. Combine the simplified parts to get $\frac{y^6}{27}$.

Marking scheme

M1 for writing $81^{-\frac{3}{4}}$ as $\frac{1}{27}$ or $\left(y^{-8}\right)^{-\frac{3}{4}}$ as $y^6$. A1 for the final answer $\frac{y^6}{27}$ or $\frac{1}{27}y^6$.

Question 18 · Short Answer

2 marks

Simplify fully: $\sqrt{48} - \sqrt{27} + \sqrt{75}$.

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Worked solution

Simplify each surd by finding the largest square factor of each radicand: $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$, $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$, and $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$. Substituting these simplified forms back into the expression gives: $4\sqrt{3} - 3\sqrt{3} + 5\sqrt{3} = 6\sqrt{3}$.

Marking scheme

M1 for simplifying at least two of the surds correctly to the form $k\sqrt{3}$. A1 for the final simplified answer of $6\sqrt{3}$.

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Practice This Topic

Paper 42 (Extended)

Answer all questions. Graphic display calculators should be used where appropriate.

11 Question · 118 marks

Question 1 · Structured

11 marks

The function $ f(x) $ is defined as:

\[ f(x) = \frac{x^2 - 4}{x^2 - 9} \]

(a) Write down the equations of all the asymptotes to the graph of $ y = f(x) $.

(b) Find the coordinates of the local maximum point on the graph of $ y = f(x) $.

(c) Find the range of the function $ f(x) $ for $ x \in \mathbb{R} $.

(d) Solve the equation $ f(x) = 2 $.

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Worked solution

(a) Vertical asymptotes occur where the denominator is zero:
$ x^2 - 9 = 0 \Rightarrow x = \pm 3 $.
Horizontal asymptote is found by considering the limit as $ x \to \pm\infty $:
$ f(x) = \frac{1 - 4/x^2}{1 - 9/x^2} \to 1 $, so $ y = 1 $.

(b) To find the local maximum, differentiate $ f(x) $ using the quotient rule:
$ f'(x) = \frac{2x(x^2 - 9) - 2x(x^2 - 4)}{(x^2 - 9)^2} = \frac{-10x}{(x^2 - 9)^2} $.
Setting $ f'(x) = 0 $ gives $ x = 0 $.
Substituting $ x = 0 $ into $ f(x) $ gives:
$ f(0) = \frac{0 - 4}{0 - 9} = \frac{4}{9} $.
Therefore, the coordinates of the local maximum are $ \left(0, \frac{4}{9}\right) $.

(c) From the asymptotes and local maximum, the graph lies entirely below $ y = \frac{4}{9} $ (for $ -3 < x < 3 $) and entirely above the horizontal asymptote $ y = 1 $ (for $ x < -3 $ and $ x > 3 $).
Thus, the range is $ y \le \frac{4}{9} $ or $ y > 1 $.

(d) Solve $ \frac{x^2 - 4}{x^2 - 9} = 2 $:
$ x^2 - 4 = 2(x^2 - 9) $
$ x^2 - 4 = 2x^2 - 18 $
$ x^2 = 14 $
$ x = \pm \sqrt{14} \approx \pm 3.74 $.

Marking scheme

(a) B1 for both vertical asymptotes $ x = -3 $ and $ x = 3 $.
B1 for horizontal asymptote $ y = 1 $.
(b) M1 for attempting to differentiate (quotient rule or GDC solver).
A1 for $ x = 0 $.
A1 for $ y = \frac{4}{9} $ (or 0.444).
(c) B1 for $ y \le \frac{4}{9} $.
B1 for $ y > 1 $.
(d) M1 for clear algebraic step to clear the fraction (e.g., $ x^2 - 4 = 2x^2 - 18 $).
A1 for $ x = \pm\sqrt{14} $ (or $ \pm 3.74 $).

Question 2 · Structured

11 marks

Three sequences have the following terms for $ n = 1, 2, 3, 4, 5 $:

Sequence P: $ 5, \, 11, \, 17, \, 23, \, 29, \dots $
Sequence Q: $ 3, \, 12, \, 27, \, 48, \, 75, \dots $
Sequence R: $ 5, \, 10, \, 19, \, 36, \, 69, \dots $

(a) Find an expression for the $ n $th term of Sequence P.

(b) Find an expression for the $ n $th term of Sequence Q.

(c) For Sequence R:
(i) Find the next term (6th term) in the sequence.
(ii) Find an expression for the $ n $th term.

(d) Find which term in Sequence Q has a value of 1200.

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Worked solution

(a) Sequence P is arithmetic with first term $ a = 5 $ and common difference $ d = 6 $.
$ U_n = 5 + (n-1)6 = 6n - 1 $.

(b) Sequence Q terms are multiples of perfect squares:
$ 3(1)^2, \, 3(2)^2, \, 3(3)^2, \dots $
$ U_n = 3n^2 $.

(c)(i) Looking at the differences for Sequence R:
$ 10 - 5 = 5 $
$ 19 - 10 = 9 $
$ 36 - 19 = 17 $
$ 69 - 36 = 33 $
The differences are: $ 5, \, 9, \, 17, \, 33, \dots $, which are of the form $ 2^{k} + 1 $ (specifically $ 2^2+1, \, 2^3+1, \, 2^4+1, \, 2^5+1 $).
The next difference will be $ 2^6 + 1 = 65 $.
So the 6th term is $ 69 + 65 = 134 $.

(c)(ii) Testing combinations of powers of 2:
$ 2^{1+1} + 1 = 5 $
$ 2^{2+1} + 2 = 10 $
$ 2^{3+1} + 3 = 19 $
$ 2^{4+1} + 4 = 36 $
$ 2^{5+1} + 5 = 69 $
Thus, the formula is $ 2^{n+1} + n $.

(d) Solve $ 3n^2 = 1200 $:
$ n^2 = 400 $
$ n = 20 $ (since $ n > 0 $).

Marking scheme

(a) B2 for $ 6n - 1 $ (B1 for $ 6n + k $).
(b) B2 for $ 3n^2 $ (B1 for $ kn^2 $ or $ 3n^k $).
(c)(i) B1 for 134.
(c)(ii) B3 for $ 2^{n+1} + n $ (B2 for recognizing the power of 2 part as $ 2^{n+1} $ or the linear part as $ n $).
(d) M1 for $ 3n^2 = 1200 $.
A1 for $ n = 20 $.

Question 3 · Structured

11 marks

A solid toy consists of a cylinder of radius $ r $ and height $ 3r $. On one end of the cylinder is a cone of radius $ r $ and slant height $ 2r $. On the other end is a hemisphere of radius $ r $.

(a) Show that the total external surface area of the toy is $ 10\pi r^2 $.

(b) The total external surface area of the toy is $ 250\text{ cm}^2 $. Find the value of $ r $.

(c) Calculate the total volume of the toy when $ r = 5\text{ cm} $.

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Worked solution

(a) The total external surface area of the toy is composed of:
1. Curved surface area of the cone: $ \pi r l = \pi r (2r) = 2\pi r^2 $.
2. Curved surface area of the cylinder: $ 2\pi r h = 2\pi r (3r) = 6\pi r^2 $.
3. Curved surface area of the hemisphere: $ 2\pi r^2 $.

Total Surface Area = $ 2\pi r^2 + 6\pi r^2 + 2\pi r^2 = 10\pi r^2 $.

(b) Set $ 10\pi r^2 = 250 $:
$ r^2 = \frac{25}{\pi} \approx 7.9577 $
$ r = \sqrt{\frac{25}{\pi}} \approx 2.8209 \approx 2.82\text{ cm} $.

(c) To find the volume:
- Volume of cylinder = $ \pi r^2 h = \pi r^2 (3r) = 3\pi r^3 $.
- Volume of hemisphere = $ \frac{2}{3}\pi r^3 $.
- Volume of cone = $ \frac{1}{3}\pi r^2 h_{\text{cone}} $.
First, find the vertical height of the cone, $ h_{\text{cone}} $, using Pythagoras:
$ h_{\text{cone}} = \sqrt{l^2 - r^2} = \sqrt{(2r)^2 - r^2} = r\sqrt{3} $.
Thus, Volume of cone = $ \frac{1}{3}\pi r^3 \sqrt{3} $.

Total Volume = $ 3\pi r^3 + \frac{2}{3}\pi r^3 + \frac{\sqrt{3}}{3}\pi r^3 = \pi r^3 \left( \frac{11 + \sqrt{3}}{3} \right) $.

For $ r = 5 $:
Total Volume = $ \pi (5)^3 \left( \frac{11 + \sqrt{3}}{3} \right) = 125\pi \left( \frac{11 + 1.73205}{3} \right) \approx 125 \times 3.14159 \times 4.24402 \approx 1666.6 \approx 1670\text{ cm}^3 $.

Marking scheme

(a) M1 for curved surface area of cone = $ 2\pi r^2 $.
M1 for curved surface area of cylinder = $ 6\pi r^2 $.
M1 for curved surface area of hemisphere = $ 2\pi r^2 $.
A1 for obtaining $ 10\pi r^2 $ with clear steps.

(b) M1 for $ 10\pi r^2 = 250 $.
A1 for $ r^2 \approx 7.96 $.
A1 for $ 2.82 $ (or 2.821).

(c) M1 for volume of cylinder + volume of hemisphere = $ \frac{11}{3}\pi r^3 $ (or substituting $ r=5 $ to get 1178.1 or 1256.6).
M1 for finding the height of the cone: $ h = r\sqrt{3} $ (or 8.66).
M1 for calculating the volume of the cone when $ r=5 $ (or $ 226.7 $).
A1 for $ 1670 $ (or 1666 to 1667).

Question 4 · Structured

11 marks

The weights, $ w $ grams, of 120 apples are recorded in the table below:

| Weight ($ w $ grams) | Frequency |
| :--- | :--- |
| $ 80 < w \le 100 $ | 15 |
| $ 100 < w \le 120 $ | 28 |
| $ 120 < w \le 140 $ | 42 |
| $ 140 < w \le 160 $ | 25 |
| $ 160 < w \le 180 $ | 10 |

(a) Calculate an estimate of the mean weight of these apples.

(b) (i) Complete the cumulative frequency table below:

| Weight ($ w $ grams) | Cumulative Frequency |
| :--- | :--- |
| $ w \le 100 $ | 15 |
| $ w \le 120 $ | |
| $ w \le 140 $ | |
| $ w \le 160 $ | |
| $ w \le 180 $ | 120 |

(ii) Use your cumulative frequency table to estimate the median weight.

(c) Two apples are chosen at random from the 120 apples, without replacement. Find the probability that both apples weigh more than 140g. Give your answer as a simplified fraction.

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Worked solution

(a) Find the midpoints of the groups: 90, 110, 130, 150, 170.
Estimate of the mean:
$ \text{Mean} = \frac{15(90) + 28(110) + 42(130) + 25(150) + 10(170)}{120} $
$ = \frac{1350 + 3080 + 5460 + 3750 + 1700}{120} = \frac{15340}{120} \approx 127.83\text{ g} \approx 128\text{ g} $.

(b)(i) Cumulative frequencies are:
- $ w \le 100 $: 15
- $ w \le 120 $: $ 15 + 28 = 43 $
- $ w \le 140 $: $ 43 + 42 = 85 $
- $ w \le 160 $: $ 85 + 25 = 110 $
- $ w \le 180 $: $ 110 + 10 = 120 $.

(b)(ii) The median corresponds to the 60th value.
Using linear interpolation in the class $ 120 < w \le 140 $:
$ \text{Median} = 120 + \left(\frac{60 - 43}{85 - 43}\right) \times (140 - 120) $
$ = 120 + \frac{17}{42} \times 20 \approx 120 + 8.10 = 128.1\text{ g} \approx 128\text{ g} $.

(c) Number of apples weighing more than 140g is $ 25 + 10 = 35 $.
Probability of first apple weighing more than 140g is $ \frac{35}{120} $.
Probability of second apple weighing more than 140g without replacement is $ \frac{34}{119} $.
$ \text{Total Probability} = \frac{35}{120} \times \frac{34}{119} = \frac{7}{24} \times \frac{2}{7} = \frac{14}{168} = \frac{1}{12} $.

Marking scheme

(a) M1 for finding at least 3 midpoints correctly.
M1 for calculating $ \sum f \cdot x $ (obtained 15340).
A1 for $ 127.83 $ or 128.

(b)(i) B2 for all three values correct (43, 85, 110). B1 for one or two correct.
(b)(ii) M1 for identifying the correct interval $ 120 < w \le 140 $ or using 60th value on a curve.
A1 for 128 (or 128.1).

(c) M1 for the first fraction $ \frac{35}{120} $ (or simplified $ \frac{7}{24} $).
M1 for multiplying by $ \frac{34}{119} $ (or simplified $ \frac{2}{7} $).
A1 for $ \frac{1}{12} $ (accept decimal equivalent 0.0833).

Question 5 · Structured

11 marks

A triangular plot of land $ ABC $ has boundaries $ AB = 120\text{ m} $ and $ BC = 150\text{ m} $. The angle between these boundaries, $ \angle ABC $, is $ 65^\circ $.

(a) Calculate the length of the boundary $ AC $.

(b) Calculate the angle $ \angle ACB $.

(c) Find the area of the plot of land.

(d) A straight path is built from point $ B $ to the nearest point on the boundary $ AC $. Calculate the length of this path.

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Worked solution

(a) Using the Cosine Rule on triangle $ ABC $:
$ AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle ABC) $
$ AC^2 = 120^2 + 150^2 - 2(120)(150)\cos(65^\circ) $
$ AC^2 = 14400 + 22500 - 36000\cos(65^\circ) $
$ AC^2 = 36900 - 36000(0.422618) \approx 21685.74 $
$ AC = \sqrt{21685.74} \approx 147.26\text{ m} \approx 147\text{ m} $.

(b) Using the Sine Rule to find $ \angle ACB $:
$ \frac{\sin(\angle ACB)}{120} = \frac{\sin(65^\circ)}{147.26} $
$ \sin(\angle ACB) = \frac{120 \sin(65^\circ)}{147.26} \approx 0.7385 $
$ \angle ACB = \arcsin(0.7385) \approx 47.60^\circ \approx 47.6^\circ $.

(c) Area of the triangle:
$ \text{Area} = \frac{1}{2} a c \sin(B) = \frac{1}{2} (120)(150)\sin(65^\circ) = 9000\sin(65^\circ) \approx 8156.7\text{ m}^2 \approx 8160\text{ m}^2 $.

(d) The shortest path from $ B $ to $ AC $ is the perpendicular height, $ d $, of the triangle with base $ AC $.
$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $
$ 8156.7 = \frac{1}{2} \times 147.26 \times d $
$ d = \frac{2 \times 8156.7}{147.26} \approx 110.78\text{ m} \approx 111\text{ m} $.

Marking scheme

(a) M1 for substituting correctly into the cosine rule: $ 120^2 + 150^2 - 2(120)(150)\cos 65^\circ $.
A1 for $ AC^2 \approx 21686 $.
A1 for $ 147 $ or $ 147.26 $.

(b) M1 for setting up the sine rule correctly: $ \frac{\sin(\angle ACB)}{120} = \frac{\sin(65^\circ)}{AC} $.
M1 for isolating $ \sin(\angle ACB) $.
A1 for $ 47.6^\circ $ (or 47.60).

(c) M1 for $ \frac{1}{2} (120)(150)\sin(65^\circ) $.
A1 for $ 8160 $ (or 8157).

(d) M1 for recognizing that the path length is the perpendicular distance.
M1 for setting up $ \frac{1}{2} \times \text{their } AC \times d = \text{their Area} $ (or equivalent trig: $ 150 \sin(47.6^\circ) $).
A1 for $ 111 $ (or 110.8).

Question 6 · Structured

11 marks

The coordinates of two points are $ P(-3, 5) $ and $ Q(5, -1) $.

(a) Find the coordinates of the midpoint of the line segment $ PQ $.

(b) Find the equation of the perpendicular bisector of $ PQ $, giving your answer in the form $ y = mx + c $.

(c) Express the vector $ \vec{PQ} $ as a column vector.

(d) Find the magnitude of the vector $ \vec{PQ} $.

Show answer & marking scheme

Worked solution

(a) Midpoint $ M $:
$ M = \left( \frac{-3 + 5}{2}, \, \frac{5 + (-1)}{2} \right) = (1, 2) $.

(b) Gradient of line $ PQ $ ($ m_1 $):
$ m_1 = \frac{-1 - 5}{5 - (-3)} = \frac{-6}{8} = -\frac{3}{4} $.
Gradient of the perpendicular bisector ($ m_2 $):
$ m_2 = -\frac{1}{m_1} = \frac{4}{3} $.
Using point-slope form with midpoint $ (1, 2) $:
$ y - 2 = \frac{4}{3}(x - 1) $
$ y - 2 = \frac{4}{3}x - \frac{4}{3} $
$ y = \frac{4}{3}x + \frac{2}{3} $.

(c) Vector $ \vec{PQ} $:
$ \vec{PQ} = \begin{pmatrix} x_Q - x_P \\ y_Q - y_P \end{pmatrix} = \begin{pmatrix} 5 - (-3) \\ -1 - 5 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \end{pmatrix} $.

(d) Magnitude of $ \vec{PQ} $:
$ |\vec{PQ}| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 $.

Marking scheme

(a) B2 for $ (1, 2) $ (B1 for one coordinate correct).

(b) M1 for finding the gradient of $ PQ $ as $ -\frac{3}{4} $.
M1 for finding the gradient of the perpendicular line as $ \frac{4}{3} $ (FT their gradient).
M1 for substituting midpoint coordinates and perpendicular gradient into a linear equation.
A1 for $ y = \frac{4}{3}x + \frac{2}{3} $.

(c) B2 for $ \begin{pmatrix} 8 \\ -6 \end{pmatrix} $ (B1 for one component correct).

(d) M1 for $ \sqrt{8^2 + (-6)^2} $ (FT their vector components).
A1 for 10.

Question 7 · Structured

11 marks

The population, $ P $, of a bacterial culture at time $ t $ hours is modeled by the equation:

\[ P(t) = A \cdot b^t \]

At $ t = 0 $, the population is 500.
At $ t = 3 $, the population is 4000.

(a) Show that $ A = 500 $ and find the value of $ b $.

(b) Find the population of the culture when $ t = 5 $.

(c) Use logarithms to solve the equation $ P(t) = 1,000,000 $. Give your answer to 3 significant figures.

(d) Write an expression for $ \log_2(P(t)) $ in terms of $ t $.

Show answer & marking scheme

Worked solution

(a) Substitute $ t = 0 $ and $ P = 500 $:
$ 500 = A \cdot b^0 \Rightarrow A = 500 $.
Substitute $ t = 3 $ and $ P = 4000 $:
$ 4000 = 500 \cdot b^3 \Rightarrow b^3 = 8 \Rightarrow b = 2 $.

(b) Using the formula $ P(t) = 500 \cdot 2^t $, substitute $ t = 5 $:
$ P(5) = 500 \cdot 2^5 = 500 \cdot 32 = 16000 $.

(c) Solve $ 500 \cdot 2^t = 1,000,000 $:
$ 2^t = 2000 $
$ t \log(2) = \log(2000) $
$ t = \frac{\log(2000)}{\log(2)} \approx 10.9658 \approx 11.0 $ (to 3 s.f.).

(d) Simplify $ \log_2(P(t)) $:
$ \log_2(500 \cdot 2^t) = \log_2(500) + \log_2(2^t) = \log_2(500) + t $.

Marking scheme

(a) B1 for showing $ A = 500 $.
M1 for substituting $ A=500 $, $ t=3 $, and $ P=4000 $ to get $ b^3 = 8 $.
A1 for $ b = 2 $.

(b) M1 for substituting $ t=5 $ into $ 500 \cdot 2^t $.
A1 for 16000.

(c) M1 for $ 2^t = 2000 $.
M1 for using log rules (e.g. $ t \log 2 = \log 2000 $).
A1 for 11.0 (or 11).

(d) M1 for split log into sum: $ \log_2 500 + \log_2 2^t $.
A1 for $ \log_2 500 + t $.

Question 8 · Structured

11 marks

A survey of 80 high school students was conducted to see which clubs they belonged to: Drama ($ D $), Music ($ M $), or Sports ($ S $).

- 5 students belong to all three clubs.
- 8 students belong to Drama and Music only.
- 12 students belong to Music and Sports only.
- 7 students belong to Drama and Sports only.
- 22 students belong to Drama in total.
- 30 students belong to Music in total.
- 35 students belong to Sports in total.

(a) Find:
(i) the number of students who belong to the Drama club only,
(ii) the number of students who do not belong to any of these three clubs.

(b) A student is chosen at random from the 80 students. Find the probability that this student belongs to at least two clubs.

(c) Two students are chosen at random from those who belong to the Sports club. Find the probability that they both belong to the Music club as well. Give your answer as a simplified fraction.

Show answer & marking scheme

Worked solution

(a)(i) Drama total is 22.
Drama only = $ 22 - (8 + 7 + 5) = 22 - 20 = 2 $.

(a)(ii) First calculate the other single-club regions:
Music total is 30.
Music only = $ 30 - (8 + 12 + 5) = 30 - 25 = 5 $.

Sports total is 35.
Sports only = $ 35 - (7 + 12 + 5) = 35 - 24 = 11 $.

Total students in at least one club = $ 2\text{ (D only)} + 5\text{ (M only)} + 11\text{ (S only)} + 8\text{ (D\cap M only)} + 12\text{ (M\cap S only)} + 7\text{ (D\cap S only)} + 5\text{ (All three)} = 50 $.

Students in no clubs = $ 80 - 50 = 30 $.

(b) Students belonging to at least two clubs:
$ 8 + 12 + 7 + 5 = 32 $.
Probability = $ \frac{32}{80} = \frac{2}{5} = 0.4 $.

(c) Number of students in the Sports club is 35.
Number of Sports students who also belong to Music (which means they are in $ M \cap S $) is $ 12\text{ (M\cap S only)} + 5\text{ (All three)} = 17 $.

We select two students from the 35 Sports students without replacement.
Probability that both are also in Music:
$ \text{Probability} = \frac{17}{35} \times \frac{16}{34} = \frac{17}{35} \times \frac{8}{17} = \frac{8}{35} $.

Marking scheme

(a)(i) B2 for 2 (B1 for showing Drama total minus intersection sum: $ 22 - 20 $).
(a)(ii) M1 for finding Music only = 5 and Sports only = 11.
M1 for calculating the sum of all union regions (obtained 50).
A1 for 30.

(b) M1 for summing the correct overlapping regions: $ 8 + 12 + 7 + 5 = 32 $.
A1 for $ \frac{2}{5} $ (or 0.4).

(c) M1 for identifying 17 Sports students are in Music.
M1 for first fraction $ \frac{17}{35} $.
M1 for multiplying by $ \frac{16}{34} $.
A1 for $ \frac{8}{35} $.

Question 9 · Structured

10 marks

Part (a)
Solve the equation:
$$2 \log_2(x) - \log_2(x+4) = 1$$

Part (b)
Find the value of $y$ if:
$$\log_a(y) = 2 \log_a(5) + \frac{1}{2} \log_a(16) - \log_a(2)$$

Part (c)
Solve the equation:
$$3 \times 5^{2x-1} = 42$$
Give your answer correct to 3 significant figures.

Show answer & marking scheme

Worked solution

Part (a)
Using laws of logarithms:
$$2 \log_2(x) - \log_2(x+4) = 1$$
$$\log_2(x^2) - \log_2(x+4) = 1$$
$$\log_2\left(\frac{x^2}{x+4}\right) = 1$$
Converting to exponential form:
$$\frac{x^2}{x+4} = 2^1 = 2$$
$$x^2 = 2(x+4)$$
$$x^2 - 2x - 8 = 0$$
$$(x-4)(x+2) = 0$$
This gives $x = 4$ or $x = -2$.
Since $\log_2(x)$ requires $x > 0$, we reject $x = -2$.
Thus, $x = 4$.

Part (b)
Using laws of logarithms:
$$\log_a(y) = \log_a(5^2) + \log_a(16^{1/2}) - \log_a(2)$$
$$\log_a(y) = \log_a(25) + \log_a(4) - \log_a(2)$$
$$\log_a(y) = \log_a\left(\frac{25 \times 4}{2}\right)$$
$$\log_a(y) = \log_a(50)$$
Thus, $y = 50$.

Part (c)
$$3 \times 5^{2x-1} = 42$$
$$5^{2x-1} = 14$$
Taking the natural logarithm of both sides:
$$(2x-1) \ln(5) = \ln(14)$$
$$2x-1 = \frac{\ln(14)}{\ln(5)}$$
$$2x-1 \approx 1.6397$$
$$2x \approx 2.6397$$
$$x \approx 1.32$$

Marking scheme

Part (a) [3 marks]
M1 for $\log_2\left(\frac{x^2}{x+4}\right) = 1$ oe
M1 for $x^2 - 2x - 8 = 0$
A1 for $x = 4$ (rejecting $x = -2$)

Part (b) [3 marks]
M1 for $2 \log_a(5) = \log_a(25)$ or $\frac{1}{2} \log_a(16) = \log_a(4)$
M1 for $\log_a(25 \times 4 / 2)$ oe
A1 for $y = 50$

Part (c) [4 marks]
M1 for dividing by 3 to get $5^{2x-1} = 14$
M1 for taking logs on both sides, e.g., $(2x-1) \log(5) = \log(14)$
M1 for isolating $x$, e.g., $x = \frac{1}{2}\left(\frac{\log(14)}{\log(5)} + 1\right)$
A1 for $1.32$ (accept $1.3198...$)

Question 10 · Structured

10 marks

A sector of a circle has a radius of $12$ cm and an area of $48\pi$ $\text{cm}^2$.

Part (a)
Find the sector angle, $\theta$.

Part (b)
Find the exact perimeter of the sector, giving your answer in terms of $\pi$.

Part (c)
The sector is folded so that the two straight edges meet, forming the curved surface of a cone.
(i) Show that the radius of the base of the cone is $4$ cm.
(ii) Calculate the height of the cone. Give your answer correct to 3 significant figures.

Show answer & marking scheme

Worked solution

Part (a)
$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$$
$$48\pi = \frac{\theta}{360} \times \pi (12^2)$$
$$48 = \frac{144\theta}{360}$$
$$\theta = \frac{48 \times 360}{144} = 120^\circ$$

Part (b)
$$\text{Arc length} = \frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2\pi(12) = 8\pi \text{ cm}$$
$$\text{Perimeter} = \text{Arc length} + 2 \times \text{radius}$$
$$\text{Perimeter} = 8\pi + 2(12) = 8\pi + 24 \text{ cm}$$

Part (c)
(i) The arc length of the sector ($8\pi$ cm) becomes the circumference of the base of the cone:
$$2\pi r_{\text{cone}} = 8\pi$$
$$r_{\text{cone}} = 4 \text{ cm}$$

(ii) The slant height $l$ of the cone is the radius of the sector, so $l = 12$ cm.
Using Pythagoras' theorem:
$$h^2 + r_{\text{cone}}^2 = l^2$$
$$h^2 + 4^2 = 12^2$$
$$h^2 + 16 = 144$$
$$h^2 = 128$$
$$h = \sqrt{128} \approx 11.3 \text{ cm}$$

Marking scheme

Part (a) [3 marks]
M1 for expression $\frac{\theta}{360} \times \pi (12)^2$
M1 for setting up equation $\frac{\theta}{360} \times 144\pi = 48\pi$ oe
A1 for $120$

Part (b) [3 marks]
M1 for finding arc length $\frac{120}{360} \times 2\pi \times 12$ oe
M1 for adding $2 \times 12$ to their arc length
A1 for $8\pi + 24$

Part (c)(i) [2 marks]
M1 for equating base circumference of cone to arc length of sector: $2\pi r = 8\pi$ oe
A1 for obtaining $r = 4$

Part (c)(ii) [2 marks]
M1 for $h^2 + 4^2 = 12^2$ oe
A1 for $11.3$ or $11.31...$ (or $8\sqrt{2}$)

Question 11 · Structured

10 marks

The table shows the first four terms of three different sequences, $A$, $B$, and $C$.

| Sequence | 1st term | 2nd term | 3rd term | 4th term |
| :--- | :--- | :--- | :--- | :--- |
| $A$ | 7 | 12 | 17 | 22 |
| $B$ | 3 | 9 | 19 | 33 |
| $C$ | 16 | 8 | 4 | 2 |

Part (a)
Find the $n$-th term of sequence $A$.

Part (b)
Find the $n$-th term of sequence $B$.

Part (c)
Find the $n$-th term of sequence $C$.

Part (d)
Find the value of $n$ for which the term in sequence $B$ is $149$ more than the term in sequence $A$.

Show answer & marking scheme

Worked solution

Part (a)
Sequence $A$ is linear: $7, 12, 17, 22$.
Common difference is $5$.
$n$-th term is $5n + c$.
For $n=1$, $5(1) + c = 7 \implies c = 2$.
$n$-th term of $A$ is $5n + 2$.

Part (b)
Sequence $B$: $3, 9, 19, 33$.
First differences: $6, 10, 14$.
Second differences: $4, 4$.
Since the second difference is constant, it is quadratic with $a = \frac{4}{2} = 2$, so $2n^2$.
Subtracting $2n^2$ from Sequence $B$:
For $n=1$: $3 - 2(1)^2 = 1$
For $n=2$: $9 - 2(2)^2 = 1$
For $n=3$: $19 - 2(3)^2 = 1$
For $n=4$: $33 - 2(4)^2 = 1$
Thus, the $n$-th term of $B$ is $2n^2 + 1$.

Part (c)
Sequence $C$: $16, 8, 4, 2$.
This is a geometric sequence with first term $a = 16$ and common ratio $r = 0.5$.
$n$-th term is $16 \times (0.5)^{n-1}$ (or $2^{5-n}$).

Part (d)
We require:
$$\text{Term of } B = \text{Term of } A + 149$$
$$(2n^2 + 1) = (5n + 2) + 149$$
$$2n^2 + 1 = 5n + 151$$
$$2n^2 - 5n - 150 = 0$$
Factoring the quadratic:
$$(2n + 15)(n - 10) = 0$$
Since $n$ must be a positive integer, $n = 10$.

Marking scheme

Part (a) [2 marks]
M1 for finding common difference of $5$ (e.g., $5n + k$)
A1 for $5n + 2$

Part (b) [3 marks]
M1 for identifying a quadratic sequence (first differences $6, 10, 14$, second difference $4$)
M1 for finding the coefficient of $n^2$ is $2$ (e.g., $2n^2 + bn + c$)
A1 for $2n^2 + 1$

Part (c) [3 marks]
M1 for identifying geometric sequence with ratio $0.5$ (or $\frac{1}{2}$)
M1 for formula $a \times r^{n-1}$
A1 for $16 \times (0.5)^{n-1}$ or $2^{5-n}$ oe

Part (d) [2 marks]
M1 for setting up the equation $(2n^2 + 1) - (5n + 2) = 149$ oe
A1 for $n = 10$ (must reject $n = -7.5$)

Paper 62 (Extended)

Answer both Part A (Investigation) and Part B (Modelling).

2 Question · 60 marks

Question 1 · Structured Tasks

30 marks

### PART A: INVESTIGATION (30 marks)

**DIFFERENCE TRIANGLES**

In this investigation, we look at **difference triangles**. To construct a difference triangle of size $ k $:
- Row 1 contains $ k $ positive numbers.
- Each entry in the next row is the absolute difference between the two adjacent numbers directly above it.
- This process is repeated until the bottom row contains a single number.

For example, here is a difference triangle of size 4:

$$\begin{array}{ccccccc}
\text{Row 1:} & 15 & & 8 & & 3 & & 9 \\
\text{Row 2:} & & 7 & & 5 & & 6 & \\
\text{Row 3:} & & & 2 & & 1 & & \\
\text{Row 4:} & & & & 1 & & &
\end{array}$$

Where:
- $|15 - 8| = 7$, $|8 - 3| = 5$, $|3 - 9| = 6$
- $|7 - 5| = 2$, $|5 - 6| = 1$
- $|2 - 1| = 1$

---

**(a)** (i) Complete this difference triangle of size 4:
$$\begin{array}{ccccccc}
12 & & 5 & & 9 & & 2 \\
& ? & & ? & & ? & \\
& & ? & & ? & & \\
& & & ? & & &
\end{array}$$

(ii) Complete this difference triangle of size 5:
$$\begin{array}{ccccccccc}
20 & & 14 & & 9 & & 5 & & 2 \\
& ? & & ? & & ? & & ? & \\
& & ? & & ? & & ? & & \\
& & & ? & & ? & & &
\end{array}$$

**(b)** (i) Complete the difference triangle where Row 1 is an arithmetic progression with first term $ a $ and common difference $ d > 0 $:
$$\begin{array}{ccccccccc}
a & & a+d & & a+2d & & a+3d & & a+4d \\
& ? & & ? & & ? & & ? & \\
& & ? & & ? & & ? & & \\
& & & ? & & ? & & &
\end{array}$$

(ii) Write down the value at the bottom of a difference triangle of size $ n $ (where $ n \ge 3 $) when Row 1 is any arithmetic progression with common difference $ d > 0 $.

**(c)** (i) Complete the difference triangle of size 4 where Row 1 is:
$$\begin{array}{ccccccc}
1 & & 3 & & 9 & & 27 \\
& ? & & ? & & ? & \\
& & ? & & ? & & \\
& & & ? & & &
\end{array}$$

(ii) Find the bottom value of a difference triangle of size 5 when Row 1 is $ 1, 3, 9, 27, 81 $.

(iii) In general, if Row 1 of a difference triangle of size $ k $ is $ 1, 3, 9, \dots, 3^{k-1} $, find an expression in terms of $ k $ for the bottom value.

**(d)** Now we restrict the numbers in our triangles to only $ 0 $ and $ 1 $.

(i) Complete the difference triangle of size 4 for Row 1: $ 1, 1, 0, 1 $.

(ii) Let $ T(n) $ be the total number of $ 1 $s in a difference triangle of size $ n $. Find $ T(3) $ for the triangle with Row 1: $ 1, 0, 1 $.

(iii) For a difference triangle of size 4, find a Row 1 of $ 0 $s and $ 1 $s (not all $ 0 $s) that results in a bottom value of $ 1 $ and has the minimum possible number of $ 1 $s in the entire triangle.

(iv) Explain why, if Row 1 of a size $ n $ triangle has only one $ 1 $ at the far left (i.e., $ 1, 0, 0, \dots, 0 $), the entire left boundary of the triangle consists of $ 1 $s.

Show answer & marking scheme

Worked solution

**(a)**
(i)
- Row 1: $12, 5, 9, 2$
- Row 2: $|12-5|=7$, $|5-9|=4$, $|9-2|=7$ $\Rightarrow 7, 4, 7$
- Row 3: $|7-4|=3$, $|4-7|=3$ $\Rightarrow 3, 3$
- Row 4: $|3-3|=0$ $\Rightarrow 0$

(ii)
- Row 1: $20, 14, 9, 5, 2$
- Row 2: $6, 5, 4, 3$
- Row 3: $1, 1, 1$
- Row 4: $0, 0$
- Row 5: $0$

**(b)**
(i)
- Row 1: $ a \quad a+d \quad a+2d \quad a+3d \quad a+4d $
- Row 2: $ d \quad d \quad d \quad d $
- Row 3: $ 0 \quad 0 \quad 0 $
- Row 4: $ 0 \quad 0 $
- Row 5: $ 0 $

(ii) Since Row 3 onwards consists of all zeros, the bottom value is $ 0 $.

**(c)**
(i)
- Row 1: $1, 3, 9, 27$
- Row 2: $|1-3|=2$, $|3-9|=6$, $|9-27|=18$ $\Rightarrow 2, 6, 18$
- Row 3: $|2-6|=4$, $|6-18|=12$ $\Rightarrow 4, 12$
- Row 4: $|4-12|=8$ $\Rightarrow 8$

(ii)
- Row 1: $1, 3, 9, 27, 81$
- Row 2: $2, 6, 18, 54$
- Row 3: $4, 12, 36$
- Row 4: $8, 24$
- Row 5: $16$

(iii) The bottom values for sizes 3, 4, 5 are $ 4 $, $ 8 $, $ 16 $, which are $ 2^2 $, $ 2^3 $, $ 2^4 $. Thus, for size $ k $, the bottom value is $ 2^{k-1} $.

**(d)**
(i)
- Row 1: $ 1, 1, 0, 1 $
- Row 2: $ 0, 1, 1 $
- Row 3: $ 1, 0 $
- Row 4: $ 1 $

(ii)
Triangle:
$$\begin{array}{ccccc}
1 & & 0 & & 1 \\
& 1 & & 1 & \\
& & 0 & &
\end{array}$$
Total number of $ 1 $s is $ 4 $.

(iii)
For a size 4 triangle to have bottom value $ 1 $, Row 3 must contain a $ 1 $, Row 2 must contain a $ 1 $, and Row 1 must contain a $ 1 $. Together with the bottom $ 1 $, there must be at least $ 4 $ ones in the entire triangle. A row of $ 1, 0, 0, 0 $ achieves this:
$$\begin{array}{ccccccc}
1 & & 0 & & 0 & & 0 \\
& 1 & & 0 & & 0 & \\
& & 1 & & 0 & & \\
& & & 1 & & &
\end{array}$$
Total number of $ 1 $s is $ 4 $.

(iv) If Row 1 is $ 1, 0, 0, \dots, 0 $, the first element of Row 2 is $ |1-0|=1 $, and the rest of Row 2 consists of $ |0-0|=0 $. Thus Row 2 is $ 1, 0, 0, \dots, 0 $. By mathematical induction, every subsequent row $ i $ starts with a $ 1 $ and is followed by zeros, ensuring the left boundary is entirely composed of $ 1 $s.

Marking scheme

**(a)**
(i) **[4 marks]**
- **B1** for Row 2: $ 7, 4, 7 $
- **B2** for Row 3: $ 3, 3 $ (or **B1** for one correct)
- **B1** for Row 4: $ 0 $

(ii) **[4 marks]**
- **B1** for Row 2: $ 6, 5, 4, 3 $
- **B1** for Row 3: $ 1, 1, 1 $
- **B1** for Row 4: $ 0, 0 $
- **B1** for Row 5: $ 0 $

**(b)**
(i) **[3 marks]**
- **B1** for Row 2: $ d, d, d, d $
- **B1** for Row 3: $ 0, 0, 0 $
- **B1** for remaining rows of $ 0 $s

(ii) **[2 marks]**
- **B2** for $ 0 $ (or **B1** if correctly identified for a specific size)

**(c)**
(i) **[3 marks]**
- **B1** for Row 2: $ 2, 6, 18 $
- **B1** for Row 3: $ 4, 12 $
- **B1** for Row 4: $ 8 $

(ii) **[2 marks]**
- **B2** for $ 16 $ (or **M1** for correct partial working showing Row 4: $ 8, 24 $)

(iii) **[3 marks]**
- **M1** for identifying the geometric progression of bottom values (e.g. 4, 8, 16)
- **A2** for $ 2^{k-1} $ (or **A1** for $ 2^k / 2 $)

**(d)**
(i) **[2 marks]**
- **B1** for Row 2 and Row 3 correct
- **B1** for Row 4: $ 1 $

(ii) **[2 marks]**
- **B2** for $ 4 $ (or **M1** for drawing the correct full triangle)

(iii) **[3 marks]**
- **M1** for attempting a valid binary row starting/ending with a single 1
- **A2** for $ 1, 0, 0, 0 $ or $ 0, 0, 0, 1 $

(iv) **[2 marks]**
- **C1** for showing that $|1-0|=1$ and $|0-0|=0$ preserves the pattern
- **C1** for concluding that this propagates down the left boundary

Question 2 · Structured Tasks

30 marks

### PART B: MODELLING (30 marks)

**DESIGNING A ROAD TUNNEL**

A civil engineering company is planning a one-way road tunnel through a mountain.

$$\text{Parabolic model: } y = H - a x^2$$

Where:
- $ y $ is the height of the tunnel (in metres) at a horizontal distance of $ x $ metres from the centre of the road.
- $ H $ is the maximum height of the tunnel (in metres) at the centre ($ x = 0 $).
- $ w $ is the total width of the tunnel at ground level (so $ y = 0 $ at $ x = \pm \frac{w}{2} $).

---

**(a)** In the first design, the maximum height of the tunnel is $ 6 \text{ m} $ and the width at ground level is $ 8 \text{ m} $.

(i) Show that the equation of this parabolic tunnel is $ y = 6 - 0.375 x^2 $.

(ii) Sketch the cross-section of this tunnel on a coordinate grid, labeling the coordinates of the intercepts with both axes.

(iii) A rectangular cargo truck has a width of $ 2.6 \text{ m} $ and a height of $ 4.2 \text{ m} $.
If the truck drives exactly down the centre of the tunnel, calculate the vertical clearance between the top corners of the truck and the tunnel roof.

**(b)** To accommodate two lanes of traffic, the engineers propose a semi-elliptical model with a width of $ 10 \text{ m} $ and a maximum height of $ 6 \text{ m} $:

$$\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \quad \text{for } y \ge 0$$

(i) Write down the values of $ A $ and $ B $.

(ii) Write down the equation of this semi-elliptical tunnel.

(iii) Sketch this new tunnel shape on a new coordinate grid, showing the coordinates of the intercepts with both axes.

(iv) The area of a parabolic segment is given by $ \text{Area} = \frac{2}{3} \times \text{width} \times \text{height} $.
The area of a semi-ellipse is given by $ \text{Area} = \frac{1}{2} \pi A B $.
Calculate the cross-sectional area of both a parabolic tunnel and a semi-elliptical tunnel, both having a width of $ 10 \text{ m} $ and a maximum height of $ 6 \text{ m} $.
Determine which tunnel has the larger cross-sectional area and by how many square metres.

**(c)** Safety regulations state that for a two-lane tunnel, a rectangular emergency vehicle of width $ 3.2 \text{ m} $ and height $ 3.5 \text{ m} $ must be able to pass through the tunnel even if it is off-centre by up to $ 1.5 \text{ m} $.

(i) If the vehicle's centre is positioned at $ x = 1.5 $, find the range of $ x $-coordinates occupied by the vehicle.

(ii) Determine whether the semi-elliptical tunnel designed in part **(b)** complies with this safety regulation. Show all your working.

(iii) State one limitation of using simple mathematical curves (like parabolas or ellipses) to model the cross-section of a physical road tunnel.

Show answer & marking scheme

Worked solution

**(a)**
(i) The maximum height is $ H = 6 $, so $ y = 6 - a x^2 $.
Since the width is $ 8 \text{ m} $, the ground level is reached at $ x = \pm 4 $. Thus, $ y = 0 $ when $ x = 4 $.
$$0 = 6 - a (4^2) \Rightarrow 16a = 6 \Rightarrow a = \frac{6}{16} = 0.375$$
Thus, the equation is $ y = 6 - 0.375 x^2 $.

(ii) The sketch is a downward-opening parabola with y-intercept at $ (0, 6) $ and x-intercepts at $ (-4, 0) $ and $ (4, 0) $.

(iii) A truck of width $ 2.6 \text{ m} $ centred at $ x = 0 $ occupies $ x $ from $ -1.3 $ to $ 1.3 $. The critical height of the tunnel roof is at the edges of the truck ($ x = \pm 1.3 $):
$$y = 6 - 0.375 (1.3)^2 = 6 - 0.375 (1.69) = 6 - 0.63375 = 5.36625 \text{ m}$$
Since the height of the truck is $ 4.2 \text{ m} $, the vertical clearance is:
$$\text{Clearance} = 5.36625 - 4.2 = 1.16625 \text{ m} \approx 1.17 \text{ m}$$

**(b)**
(i) The total width is $ 10 \text{ m} $, so $ A = 5 $. The maximum height is $ 6 \text{ m} $, so $ B = 6 $.

(ii) Substituting the values into the equation:
$$\frac{x^2}{25} + \frac{y^2}{36} = 1$$

(iii) The sketch is a semi-ellipse with y-intercept at $ (0, 6) $ and x-intercepts at $ (-5, 0) $ and $ (5, 0) $.

(iv)
- Parabolic tunnel area: $ \frac{2}{3} \times 10 \times 6 = 40 \text{ m}^2 $
- Semi-elliptical tunnel area: $ \frac{1}{2} \pi (5)(6) = 15\pi \approx 47.12 \text{ m}^2 $
The semi-elliptical tunnel has the larger area by:
$$15\pi - 40 \approx 47.12 - 40 = 7.12 \text{ m}^2 \text{ (or } 7.1 \text{ m}^2\text{)}$$

**(c)**
(i) Since the vehicle is $ 3.2 \text{ m} $ wide and centred at $ x = 1.5 $, it occupies $ x $ from $ 1.5 - 1.6 = -0.1 $ to $ 1.5 + 1.6 = 3.1 $.

(ii) The critical point of the vehicle closest to the tunnel wall is at $ x = 3.1 $.
At $ x = 3.1 $, the height of the semi-elliptical tunnel is:
$$\frac{3.1^2}{25} + \frac{y^2}{36} = 1 \Rightarrow \frac{9.61}{25} + \frac{y^2}{36} = 1$$
$$0.3844 + \frac{y^2}{36} = 1 \Rightarrow \frac{y^2}{36} = 0.6156$$
$$y^2 = 22.1616 \Rightarrow y = \sqrt{22.1616} \approx 4.71 \text{ m}$$
Since the height of the tunnel roof at the outer edge ($ 4.71 \text{ m} $) is greater than the vehicle height ($ 3.5 \text{ m} $), the vehicle can safely pass. Yes, the tunnel complies with the regulation.

(iii) Sensible limitations include:
- Physical tunnels do not have perfectly smooth walls or mathematically exact shapes.
- The road surface is rarely perfectly flat at the edges due to drainage gutters or curbs.
- It does not account for the structural thickness of the tunnel lining.

Marking scheme

**(a)**
(i) **[3 marks]**
- **M1** for substituting $ H = 6 $ into $ y = H - a x^2 $
- **M1** for substituting $ (4, 0) $ to find $ a $
- **A1** for obtaining $ y = 6 - 0.375 x^2 $ with full steps shown

(ii) **[3 marks]**
- **B1** for drawing a correct downward-opening curve
- **B1** for labeling the y-intercept at $ (0, 6) $
- **B1** for labeling x-intercepts at $ (-4, 0) $ and $ (4, 0) $

(iii) **[4 marks]**
- **M1** for identifying the critical horizontal position at $ x = 1.3 $
- **M1** for substituting $ x = 1.3 $ into the parabolic equation
- **A1** for calculating roof height $ y = 5.37 \text{ m} $ (or $ 5.366... $)
- **A1** for clearance of $ 1.17 \text{ m} $ (or $ 1.166... $)

**(b)**
(i) **[2 marks]**
- **B1** for $ A = 5 $
- **B1** for $ B = 6 $

(ii) **[2 marks]**
- **B2** for $ \frac{x^2}{25} + \frac{y^2}{36} = 1 $ (or equivalent)

(iii) **[2 marks]**
- **B1** for drawing a correct semi-elliptical curve
- **B1** for labeling intercepts at $ (0, 6) $, $ (-5, 0) $, and $ (5, 0) $

(iv) **[5 marks]**
- **B1** for parabolic area $ = 40 $
- **M1** for semi-elliptical area calculation $ \frac{1}{2} \pi (5)(6) $
- **A1** for semi-elliptical area $ \approx 47.1 $ (or $ 47.12 $)
- **B1** for stating the semi-elliptical tunnel is larger
- **A1** for difference $ \approx 7.12 \text{ m}^2 $ (or $ 7.1 \text{ m}^2 $)

**(c)**
(i) **[2 marks]**
- **B1** for range limits $ [-0.1, 3.1] $ (or equivalent explanation of vehicle limits)

(ii) **[5 marks]**
- **M1** for identifying $ x = 3.1 $ as the critical outer edge
- **M1** for substituting $ x = 3.1 $ into the elliptical model equation
- **A1** for calculating roof height $ y \approx 4.71 \text{ m} $ (or $ 4.707... $)
- **M1** for comparing $ 4.71 \text{ m} > 3.5 \text{ m} $
- **A1** for final correct conclusion (Yes, it complies)

(iii) **[2 marks]**
- **R2** for any sensible practical limitation (e.g., thickness of lining, construction tolerances, presence of walkways/gutters, etc.)

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