2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

Multiply numerator and denominator by the conjugate of the denominator, $3 + \sqrt{5}$:

$$\frac{8}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}}$$

$$= \frac{8(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2}$$

$$= \frac{8(3 + \sqrt{5})}{9 - 5}$$

$$= \frac{8(3 + \sqrt{5})}{4}$$

$$= 2(3 + \sqrt{5}) = 6 + 2\sqrt{5}$$

M1 for multiplying by $\frac{3 + \sqrt{5}}{3 + \sqrt{5}}$
M1 for denominator simplified to $9 - 5$ or $4$
A1 for $6 + 2\sqrt{5}$ or $2(3 + \sqrt{5})$

First, analyze the numerators: $2, 5, 8, 11, \dots$
This is an arithmetic sequence with first term $a = 2$ and common difference $d = 3$.
The $n$th term of the numerator is:
$$2 + (n - 1) \times 3 = 3n - 1$$

Next, analyze the denominators: $3, 6, 11, 18, \dots$
The differences between terms are $3, 5, 7, \dots$, which has a constant second difference of $2$.
This suggests a quadratic sequence of the form $n^2 + c$.
For $n = 1$, we have $1^2 + c = 3 \implies c = 2$.
The $n$th term of the denominator is:
$$n^2 + 2$$

Combining these, the $n$th term of the sequence is:
$$\frac{3n - 1}{n^2 + 2}$$

B1 for numerator $3n - 1$ seen
B1 for denominator $n^2 + 2$ seen
A1 for final fraction $\frac{3n - 1}{n^2 + 2}$

To subtract the fractions, find a common denominator:

$$\frac{4}{x+3} - \frac{2}{x-1} = \frac{4(x-1) - 2(x+3)}{(x+3)(x-1)}$$

Expand the numerator:

$$4(x-1) - 2(x+3) = 4x - 4 - 2x - 6 = 2x - 10$$

Therefore, the fraction in its simplest form is:

$$\frac{2x - 10}{(x+3)(x-1)}$$

M1 for writing over a common denominator $4(x-1) - 2(x+3)$
A1 for correct expansion of numerator $2x - 10$
A1 for final answer $\frac{2x-10}{(x+3)(x-1)}$ or $\frac{2(x-5)}{(x+3)(x-1)}$

Express both sides of the equation with a base of $3$:

$$9^{2x-3} = (3^2)^{2x-3} = 3^{4x-6}$$
$$27^{x+1} = (3^3)^{x+1} = 3^{3x+3}$$

Now equate the exponents:

$$4x - 6 = 3x + 3$$
$$x = 9$$

M1 for writing $9$ and $27$ in terms of base $3$
M1 for setting exponents equal: $4x - 6 = 3x + 3$
A1 for $x = 9$

Let $y = f(x)$:

$$y = \frac{2x+3}{x-4}$$

Multiply both sides by $x-4$:

$$y(x-4) = 2x+3$$
$$xy - 4y = 2x+3$$

Rearrange to collect $x$ terms on one side:

$$xy - 2x = 4y + 3$$

Factor out $x$:

$$x(y-2) = 4y + 3$$

Make $x$ the subject:

$$x = \frac{4y+3}{y-2}$$

Therefore, the inverse function is:

$$f^{-1}(x) = \frac{4x+3}{x-2}$$

M1 for making first step to swap variables or write $y(x-4) = 2x+3$
M1 for isolating $x$ terms: $x(y-2) = 4y+3$
A1 for final correct function $\frac{4x+3}{x-2}$

The total number of balls is $5 + 3 = 8$.

To get two balls of different colours, we can either select a Red ball then a Blue ball, or a Blue ball then a Red ball.

Calculate the probability of Red then Blue:

$$P(\text{Red then Blue}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$$

Calculate the probability of Blue then Red:

$$P(\text{Blue then Red}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$$

Add the two probabilities together:

$$P(\text{different colours}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$$

M1 for $\frac{5}{8} \times \frac{3}{7}$ or $\frac{3}{8} \times \frac{5}{7}$
M1 for summing the two possible scenarios: $2 \times \left(\frac{15}{56}\right)$
A1 for $\frac{15}{28}$ oe

First, find the midpoint of $AB$:

$$\text{Midpoint} = \left(\frac{2+6}{2},\ \frac{5-3}{2}\right) = (4, 1)$$

Next, find the gradient of the line $AB$:

$$m_{AB} = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2$$

Since the bisector is perpendicular, its gradient $m_{\perp}$ is the negative reciprocal of $m_{AB}$:

$$m_{\perp} = -\frac{1}{-2} = \frac{1}{2}$$

Now, use the point-gradient formula with midpoint $(4, 1)$ and gradient $\frac{1}{2}$:

$$y - 1 = \frac{1}{2}(x - 4)$$
$$y = \frac{1}{2}x - 2 + 1$$
$$y = \frac{1}{2}x - 1$$

B1 for finding the midpoint $(4, 1)$
M1 for finding the perpendicular gradient $m = \frac{1}{2}$
A1 for correct equation $y = \frac{1}{2}x - 1$ oe

Find the exact values of the individual trigonometric functions using reference angles:

$$\sin(240^\circ) = -\sin(60^\circ) = -\frac{\sqrt{3}}{2}$$
$$\tan(300^\circ) = -\tan(60^\circ) = -\sqrt{3}$$

Calculate their product:

$$\sin(240^\circ) \tan(300^\circ) = \left(-\frac{\sqrt{3}}{2}\right) \times (-\sqrt{3}) = \frac{3}{2}$$

B1 for $\sin(240^\circ) = -\frac{\sqrt{3}}{2}$
B1 for $\tan(300^\circ) = -\sqrt{3}$
B1 for correct product $\frac{3}{2}$ or $1.5$

Multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{7} + \sqrt{2}$:

$$\frac{10(\sqrt{7} + \sqrt{2})}{(\sqrt{7} - \sqrt{2})(\sqrt{7} + \sqrt{2})}$$

Simplify the denominator using the difference of squares:

$$\frac{10(\sqrt{7} + \sqrt{2})}{7 - 2} = \frac{10(\sqrt{7} + \sqrt{2})}{5} = 2(\sqrt{7} + \sqrt{2}) = 2\sqrt{7} + 2\sqrt{2}$$

M1 for multiplying numerator and denominator by $\sqrt{7} + \sqrt{2}$
A1 for denominator of 5
A1 for correct final simplified answer

Let's find the differences between consecutive terms:
First differences: $11-5=6$, $21-11=10$, $35-21=14$, $53-35=18$
Second differences: $10-6=4$, $14-10=4$, $18-14=4$

Since the second difference is constant, the $n$-th term is of the form $an^2 + bn + c$ where $2a = 4 \implies a = 2$.

Subtract $2n^2$ from each term of the sequence:
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 remainder is a constant $3$, the $n$-th term is $2n^2 + 3$.

M1 for identifying second difference is 4
M1 for starting with $2n^2$
A1 for correct final expression

Multiply the first equation by 3 and the second equation by 4 to align the coefficients of $y$:

$$9x - 12y = 57$$
$$8x + 12y = 28$$

Add the two equations together:

$$17x = 85 \implies x = 5$$

Substitute $x = 5$ back into the second equation:

$$2(5) + 3y = 7 \implies 10 + 3y = 7 \implies 3y = -3 \implies y = -1$$

M1 for correct method to eliminate one variable
A1 for $x = 5$
A1 for $y = -1$

The total number of pens is $5 + 3 = 8$.

Probability of picking two red pens:
$$\text{P}(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$$

Probability of picking two blue pens:
$$\text{P}(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$$

Probability that both pens are of the same colour:
$$\text{P}(\text{Same Colour}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}$$

M1 for $\frac{5}{8} \times \frac{4}{7}$ or $\frac{3}{8} \times \frac{2}{7}$
M1 for adding their two probabilities
A1 for $\frac{13}{28}$ or equivalent fraction

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

$$\frac{4(x+2) - 3(x-3)}{(x-3)(x+2)}$$

Expand the numerator:

$$\frac{4x + 8 - 3x + 9}{(x-3)(x+2)} = \frac{x + 17}{(x-3)(x+2)}$$

M1 for common denominator $(x-3)(x+2)$
M1 for correct expansion of the numerator $4(x+2) - 3(x-3)$
A1 for correct final answer

First, evaluate $\mathrm{f}(4)$:
$$\mathrm{f}(4) = 2(4) - 5 = 8 - 5 = 3$$

Next, substitute this result into $\mathrm{g}(x)$:
$$\mathrm{g}(3) = \frac{3}{3+1} = \frac{3}{4}$$

Therefore, $\mathrm{g}(\mathrm{f}(4)) = \frac{3}{4}$.

M1 for finding $\mathrm{f}(4) = 3$
M1 for substituting their value into $\mathrm{g}(x)$
A1 for $\frac{3}{4}$ or $0.75$

The formula for exponential growth is:
$$V = P(1 + r)^t$$

Substitute $P = 800$, $r = 0.05$, and $t = 2$:
$$V = 800(1 + 0.05)^2 = 800(1.05)^2$$

Calculate $1.05^2$:
$$1.05 \times 1.05 = 1.1025$$

Now, multiply by 800:
$$V = 800 \times 1.1025 = 882$$

M1 for $800 \times (1.05)^2$ oe
M1 for evaluation of $(1.05)^2 = 1.1025$ or $800 \times \frac{441}{400}$
A1 for 882

Reflection in the line $y = -x$ maps any point $(x, y)$ to $(-y, -x)$.

Applying this mapping to each vertex:
- $(1, 2) \to (-2, -1)$
- $(3, 2) \to (-2, -3)$
- $(1, 5) \to (-5, -1)$

M1 for applying rule $(x, y) \to (-y, -x)$
A1 for any two coordinates correct
A1 for all three coordinates correct

Find the first and second differences of the sequence:
First differences: \(6, 8, 10, 12\)
Second differences: \(2, 2, 2\)
Since the second differences are constant, the sequence is quadratic with a general term of the form \(an^2 + bn + c\), where \(2a = 2 \Rightarrow a = 1\).
Subtract \(n^2\) from each term of the sequence:
\(n=1: 3 - 1 = 2\)
\(n=2: 9 - 4 = 5\)
\(n=3: 17 - 9 = 8\)
\(n=4: 27 - 16 = 11\)
\(n=5: 39 - 25 = 14\)
The resulting sequence \(2, 5, 8, 11, 14\) is linear with a common difference of 3 and first term of 2, which gives \(3n - 1\).
Thus, the expression for the \(n\)th term is \(n^2 + 3n - 1\).

M1 for finding the second difference is 2 oe
M1 for finding the linear part \(3n - 1\) oe
A1 for the correct final expression \(n^2 + 3n - 1\)

Multiply the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{7} + \sqrt{3}\):
\[\frac{12(\sqrt{7} + \sqrt{3})}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})}\]
\[= \frac{12(\sqrt{7} + \sqrt{3})}{7 - 3}\]
\[= \frac{12(\sqrt{7} + \sqrt{3})}{4}\]
\[= 3(\sqrt{7} + \sqrt{3}) = 3\sqrt{7} + 3\sqrt{3}\]

M1 for multiplying numerator and denominator by \(\sqrt{7} + \sqrt{3}\)
A1 for obtaining the denominator of 4
A1 for the correct simplified expression \(3\sqrt{7} + 3\sqrt{3}\) or \(3(\sqrt{7} + \sqrt{3})\)

From the second equation, express \(y\) in terms of \(x\):
\[y = 4x - 13\]
Substitute this into the first equation:
\[3x + 2(4x - 13) = 7\]
\[3x + 8x - 26 = 7\]
\[11x = 33 \Rightarrow x = 3\]
Now find \(y\) by substituting \(x = 3\) back into the equation for \(y\):
\[y = 4(3) - 13 = -1\]
Thus, the solution is \(x = 3, y = -1\).

M1 for a correct method to eliminate one variable
A1 for \(x = 3\)
A1 for \(y = -1\)

Find the common denominator, which is \((x+1)(x-2)\):
\[\frac{3(x-2) - 2(x+1)}{(x+1)(x-2)}\]
Expand the numerator:
\[= \frac{3x - 6 - 2x - 2}{(x+1)(x-2)}\]
Simplify the numerator:
\[= \frac{x - 8}{(x+1)(x-2)}\]

M1 for finding a common denominator of \((x+1)(x-2)\) with at least one correct numerator term
M1 for correct expansion of numerator terms to \(3x - 6 - 2x - 2\) (allow one sign error)
A1 for the correct simplified fraction \(\frac{x - 8}{(x+1)(x-2)}\) oe

The total number of balls is \(5 + 3 = 8\).
There are two ways to pick two balls of different colours: (Red, Blue) or (Blue, Red).
Calculate each probability:
\(\text{P(Red, Blue)} = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)
\(\text{P(Blue, Red)} = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}\)
Add the two probabilities:
\(\text{P(different colours)} = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}\)

M1 for a correct product representing one of the outcomes, e.g., \(\frac{5}{8} \times \frac{3}{7}\)
M1 for adding the two correct products: \(\frac{15}{56} + \frac{15}{56}\)
A1 for \(\frac{15}{28}\) or equivalent

A quadratic function with vertex at \((h, k)\) and leading coefficient 1 can be written in vertex form as:
\[y = (x - h)^2 + k\]
Substituting the vertex \((3, -4)\):
\[y = (x - 3)^2 - 4\]
Expand the expression:
\[y = x^2 - 6x + 9 - 4\]
\[y = x^2 - 6x + 5\]
Comparing this to \(y = x^2 + px + q\), we find:
\(p = -6\) and \(q = 5\).

M1 for writing the function in vertex form: \((x - 3)^2 - 4\)
A1 for \(p = -6\)
A1 for \(q = 5\)

Substitute \(t = 4\) and \(V = 180\) into the formula:
\[180 = 80 \times 1.5^{4k}\]
Divide both sides by 80:
\[1.5^{4k} = \frac{180}{80} = 2.25\]
Recognize that \(2.25 = 1.5^2\):
\[1.5^{4k} = 1.5^2\]
Equating the exponents gives:
\[4k = 2 \Rightarrow k = \frac{2}{4} = 0.5\]

M1 for substituting \(t = 4\) and \(V = 180\)
M1 for simplifying to \(1.5^{4k} = 2.25\) oe
A1 for \(k = 0.5\) or \(\frac{1}{2}\)

Elena's investment is modeled by the formula: \( 4500 \left(1 + \frac{r}{100}\right)^6 = 5835 \). Rearranging the equation: \( \left(1 + \frac{r}{100}\right)^6 = \frac{5835}{4500} = 1.296667 \). Taking the 6th root on both sides: \( 1 + \frac{r}{100} = (1.296667)^{\frac{1}{6}} \approx 1.04423 \). Solving for \( r \): \( \frac{r}{100} \approx 0.04423 \implies r \approx 4.42\% \).

M1 for setting up the equation \( 4500(1 + r/100)^6 = 5835 \)
M1 for division/rearrangement \( (1+r/100)^6 = 1.296667 \)
M1 for taking the 6th root
A1 for 4.42

The volume of the cylinder is: \( V_{\text{cylinder}} = \pi r^2 h = \pi \times 6^2 \times 15 = 540\pi \approx 1696.46 \text{ cm}^3 \). The volume of one sphere is: \( V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \times 2.5^3 = 20.83\pi \approx 65.45 \text{ cm}^3 \). Dividing the volume of the cylinder by the volume of one sphere: \( \frac{540\pi}{20.83\pi} \approx 25.92 \). Therefore, the maximum number of complete spheres that can be made is 25.

M1 for cylinder volume calculation \( 540\pi \) or \( 1696.46 \)
M1 for sphere volume calculation \( 20.83\pi \) or \( 65.45 \)
M1 for division of cylinder volume by sphere volume
A1 for 25

First, find the midpoint of \(AB\): \( M = \left(\frac{-3+1}{2}, \frac{5-7}{2}\right) = (-1, -1) \). Next, find the gradient of \(AB\): \( m_{AB} = \frac{-7-5}{1 - (-3)} = \frac{-12}{4} = -3 \). The gradient of the perpendicular line is \( m_{\perp} = -\frac{1}{m_{AB}} = \frac{1}{3} \). Using the point-slope form with midpoint \((-1, -1)\): \( y - (-1) = \frac{1}{3}(x - (-1)) \implies y + 1 = \frac{1}{3}x + \frac{1}{3} \implies y = \frac{1}{3}x - \frac{2}{3} \).

B1 for finding midpoint \((-1, -1)\)
M1 for finding gradient of \(AB\) = \(-3\)
M1 for perpendicular gradient = \(\frac{1}{3}\)
A1 for equation \( y = \frac{1}{3}x - \frac{2}{3} \)

Find the common denominator, which is \( (x - 3)(2x + 1) \). Rewrite the expression: \( \frac{4(2x + 1) - 3(x - 3)}{(x - 3)(2x + 1)} \). Expand and simplify the numerator: \( 8x + 4 - 3x + 9 = 5x + 13 \). The simplified single fraction is: \( \frac{5x + 13}{(x - 3)(2x + 1)} \).

M1 for establishing correct common denominator \( (x - 3)(2x + 1) \)
M1 for correct expansion of numerator \( 4(2x + 1) - 3(x - 3) \)
A1 for final simplified answer \( \frac{5x + 13}{(x - 3)(2x + 1)} \)

To find the local minimum, we find the first derivative: \( f'(x) = 3x^2 - 8x + 1 \). Set the derivative equal to 0: \( 3x^2 - 8x + 1 = 0 \). Using the quadratic formula: \( x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 1}}{2 \times 3} = \frac{8 \pm \sqrt{52}}{6} \). The local minimum occurs at the larger root: \( x = \frac{8 + \sqrt{52}}{6} \approx 2.535 \). Substitute \( x \approx 2.535 \) back into \( f(x) \) to find \( y \): \( f(2.535) \approx (2.535)^3 - 4(2.535)^2 + 2.535 + 6 \approx -0.879 \). The coordinates are \( (2.54, -0.879) \).

M1 for finding derivative and setting to 0 or plotting on GDC
A1 for \( x \approx 2.54 \)
A1 for \( y \approx -0.879 \)

First evaluate the inner function: \( g(-2) = \frac{3}{-2 + 1} = \frac{3}{-1} = -3 \). Next, substitute this result into the outer function: \( f(g(-2)) = f(-3) = 2(-3) - 5 = -6 - 5 = -11 \).

M1 for evaluating \( g(-2) = -3 \)
M1 for substituting their result into \( f(x) \)
A1 for \( -11 \)

There are two scenarios based on the color of the marble moved from X to Y:
1. A red marble is moved (probability \( \frac{5}{8} \)). Bag Y now has 5 red and 6 blue marbles (total 11). Probability of choosing red from Y is \( \frac{5}{11} \).
Probability of this path = \( \frac{5}{8} \times \frac{5}{11} = \frac{25}{88} \).

2. A blue marble is moved (probability \( \frac{3}{8} \)). Bag Y now has 4 red and 7 blue marbles (total 11). Probability of choosing red from Y is \( \frac{4}{11} \).
Probability of this path = \( \frac{3}{8} \times \frac{4}{11} = \frac{12}{88} \).

Total probability of choosing red from Y = \( \frac{25}{88} + \frac{12}{88} = \frac{37}{88} \).

M1 for considering two cases (red moved or blue moved)
M1 for finding probability of path 1 (red/red) = \( \frac{25}{88} \)
M1 for finding probability of path 2 (blue/red) = \( \frac{12}{88} \)
A1 for total probability = \( \frac{37}{88} \)

Entering the data points into a graphing display calculator (GDC), we calculate the linear regression coefficients:
\( a \approx 4.51428 \) and \( b \approx 37.40000 \).
Rounding to 3 significant figures, \( a = 4.51 \) and \( b = 37.4 \).
Therefore, the regression line equation is \( y = 4.51x + 37.4 \).

M1 for entering the coordinates correctly into list/GDC
M1 for calculation of regression coefficients using linear regression tool
A1 for the correct equation \( y = 4.51x + 37.4 \)

(a) As the hours of study increase, the test scores also increase, indicating a positive correlation.
(b) Using a graphics display calculator (GDC), the regression line \(y = mx + c\) has coefficients: \(m \approx 4.719\) and \(c \approx 24.837\). Thus, the equation of the regression line is \(y = 4.72x + 24.8\) (to 3 significant figures).
(c) Substituting \(x = 6\) into the regression equation: \(y = 4.719(6) + 24.837 = 53.151\). Rounded to 3 significant figures, the estimated test score is 53.2.

(a) B1 for positive correlation.
(b) M1 for finding either the gradient or y-intercept correct to 3 s.f., A1 for both gradient and intercept correct in the equation \(y = 4.72x + 24.8\).
(c) B1 for 53.2 (or follow-through of their regression line equation).

(a) The value \(V\) after 4 years is given by: \(V = 12000 \times (1 - 0.085)^4 = 12000 \times (0.915)^4 \approx 8412.87\). To the nearest dollar, the value is $8 413.
(b) We want to find the smallest integer \(n\) such that: \(12000 \times (0.915)^n < 5000 \implies 0.915^n < \frac{5}{12}\). Taking logarithms: \(n \log(0.915) < \log\left(\frac{5}{12}\right)\). Since \(\log(0.915)\) is negative, when we divide by it, the inequality sign reverses: \(n > \frac{\log(5/12)}{\log(0.915)} \approx 9.86\). So, it takes 10 complete years for the value to fall below $5 000.

(a) M1 for \(12000 \times 0.915^4\), A1 for 8413.
(b) M1 for setting up the inequality or equation \(12000 \times 0.915^n = 5000\) or trial and error with at least 2 values of \(n\), M1 for finding \(n \approx 9.86\) or showing \(V(9) \approx 5427\) and \(V(10) \approx 4966\), A1 for 10.

Since the curve passes through \((0, 3)\): \(3 = a(0)^2 + b(0) + c \implies c = 3\).
Using the point \((2, -5)\): \(-5 = a(2)^2 + b(2) + 3 \implies 4a + 2b = -8 \implies 2a + b = -4\) (Equation 1).
Using the point \((4, -5)\): \(-5 = a(4)^2 + b(4) + 3 \implies 16a + 4b = -8 \implies 4a + b = -2\) (Equation 2).
Subtracting Equation 1 from Equation 2: \((4a + b) - (2a + b) = -2 - (-4) \implies 2a = 2 \implies a = 1\).
Substituting \(a = 1\) into Equation 1: \(2(1) + b = -4 \implies b = -6\).
Therefore, \(a = 1\), \(b = -6\), and \(c = 3\).

B1 for \(c = 3\).
M1 for substituting \(c\) and points to form two simultaneous equations (e.g., \(4a+2b=-8\) and \(16a+4b=-8\)).
M1 for a correct method to solve the simultaneous equations to find \(a\) or \(b\).
A1 for \(a = 1\).
A1 for \(b = -6\).

(a) Volume of hemisphere = \(\frac{2}{3} \pi r^3 = \frac{2}{3} \pi (3)^3 = 18\pi\) cm\(^3\).
Volume of cylinder = \(\pi r^2 h = \pi (3)^2 (8) = 72\pi\) cm\(^3\).
Total volume = \(18\pi + 72\pi = 90\pi \approx 282.74\) cm\(^3\), which is 283 cm\(^3\) (to 3 significant figures).
(b) Curved surface area of hemisphere = \(2 \pi r^2 = 2 \pi (3)^2 = 18\pi\) cm\(^2\).
Curved surface area of cylinder = \(2 \pi r h = 2 \pi (3)(8) = 48\pi\) cm\(^2\).
Area of circular base = \(\pi r^2 = \pi (3)^2 = 9\pi\) cm\(^2\).
Total surface area = \(18\pi + 48\pi + 9\pi = 75\pi \approx 235.62\) cm\(^2\), which is 236 cm\(^2\) (to 3 significant figures).

(a) M1 for volume of hemisphere formula used correctly, M1 for volume of cylinder formula used correctly, A1 for 283 (or \(90\pi\)).
(b) M1 for curved area of hemisphere plus curved area of cylinder, M1 for adding base area \(\pi r^2\), A1 for 236 (or \(75\pi\)).

(a) Midpoint of \(AB = \left(\frac{-4 + 2}{2}, \frac{1 + 9}{2}\right) = (-1, 5)\).
(b) Gradient of \(AB = \frac{9 - 1}{2 - (-4)} = \frac{8}{6} = \frac{4}{3}\).
(c) The gradient of the perpendicular bisector is the negative reciprocal of the gradient of \(AB\): \(m_{\perp} = -\frac{3}{4} = -0.75\). Since the perpendicular bisector passes through the midpoint \((-1, 5)\): \(y - 5 = -0.75(x - (-1)) \implies y - 5 = -0.75x - 0.75 \implies y = -0.75x + 4.25\).

(a) B1 for \((-1, 5)\).
(b) B1 for \(\frac{4}{3}\) (or equivalent).
(c) M1 for finding gradient of perpendicular line as \(-\frac{1}{\text{gradient of } AB}\), M1 for substituting their midpoint and perpendicular gradient into line equation, A1 for \(y = -0.75x + 4.25\) (or \(y = -\frac{3}{4}x + \frac{17}{4}\)).

(a) Probability the first card is white is \(\frac{6}{10}\). Since it is not replaced, there are now 5 white cards out of 9 remaining cards. Probability both are white is \(\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}\).
(b) The two cards are of different colors if we draw (White, Black) or (Black, White). \(P(\text{White, Black}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90}\) and \(P(\text{Black, White}) = \frac{4}{10} \times \frac{6}{9} = \frac{24}{90}\). Total probability is \(\frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\) (or approximately \(0.533\)).

(a) M1 for \(\frac{6}{10} \times \frac{5}{9}\), A1 for \(\frac{1}{3}\) (or equivalent fraction/decimal like 0.333).
(b) M1 for \(\frac{6}{10} \times \frac{4}{9}\) or \(\frac{4}{10} \times \frac{6}{9}\), M1 for summing both product terms, A1 for \(\frac{8}{15}\) (or equivalent fraction/decimal like 0.533).

(a) The common difference is \(13 - 7 = 6\). The next term is \(25 + 6 = 31\).
(b) An arithmetic sequence with first term \(a = 7\) and common difference \(d = 6\) has the formula \(a + (n-1)d = 7 + (n-1)6 = 6n + 1\).
(c) Substituting \(n = 12\) into the given formula: \(12^2 - 3(12) + 5 = 144 - 36 + 5 = 113\).

(a) B1 for 31.
(b) M1 for finding the common difference of 6 or writing \(6n + k\), A1 for \(6n + 1\).
(c) M1 for substituting 12 into the formula: \(12^2 - 3(12) + 5\), A1 for 113.

(a) Combine the terms over a common denominator: \(\frac{4(2x+1) + 3(x-3)}{(x-3)(2x+1)} = \frac{8x + 4 + 3x - 9}{(x-3)(2x+1)} = \frac{11x - 5}{(x-3)(2x+1)}\).
(b) Setting the fraction equal to 2: \(\frac{11x-5}{(x-3)(2x+1)} = 2 \implies 11x - 5 = 2(2x^2 - 5x - 3) \implies 11x - 5 = 4x^2 - 10x - 6 \implies 4x^2 - 21x - 1 = 0\). Using the quadratic formula: \(x = \frac{21 \pm \sqrt{(-21)^2 - 4(4)(-1)}}{8} = \frac{21 \pm \sqrt{457}}{8}\). This gives \(x_1 \approx 5.297\) and \(x_2 \approx -0.04719\). To 3 significant figures, the solutions are \(x = 5.30\) and \(x = -0.0472\).

(a) M1 for common denominator \((x-3)(2x+1)\), M1 for expanding the numerator correctly to \(8x + 4 + 3x - 9\), A1 for \(\frac{11x - 5}{(x-3)(2x+1)}\).
(b) M1 for forming a quadratic equation from their part (a), e.g., \(4x^2 - 21x - 1 = 0\), M1 for utilizing a GDC or the quadratic formula correctly on their quadratic, A1 for \(x = 5.30\) and \(x = -0.0472\) (or equivalent follow-through of their quadratic equation).

Let \( n \) be the number of years. The value of the land after \( n \) years is given by \( 150000 \times (1.058)^n \). We want to find the smallest integer \( n \) such that:\
\( 150000 \times (1.058)^n \ge 250000 \)\
\( (1.058)^n \ge \frac{250000}{150000} \approx 1.6667 \)\
Taking the natural logarithm on both sides:\
\( n \ln(1.058) \ge \ln(1.6667) \)\
\( n \ge \frac{\ln(1.6667)}{\ln(1.058)} \approx 9.06 \)\
Since \( n \) must be a complete number of years, we round up to the next integer, which is 10.

M1 for setting up the equation or inequality: \( 150000 \times 1.058^n \ge 250000 \) or equivalent\
M1 for taking logarithms correctly: \( n \log(1.058) \ge \log(1.6667) \) or for systematic trial showing values at year 9 and year 10\
A1 for finding the decimal value \( n \approx 9.06 \) or showing value at 9 years is $249,151 and at 10 years is $263,602\
A1 for 10

First, find the midpoint \( M \) of \( CD \):\
\( M = \left(\frac{2+6}{2}, \frac{-3+5}{2}\right) = (4, 1) \)\
Next, find the gradient \( m \) of \( CD \):\
\( m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2 \)\
Since the bisector is perpendicular, its gradient \( m_{\perp} \) is:\
\( m_{\perp} = -\frac{1}{m} = -\frac{1}{2} = -0.5 \)\
Now, use the point-slope form with midpoint \( (4, 1) \) and gradient \( -0.5 \):\
\( y - 1 = -0.5(x - 4) \)\
\( y - 1 = -0.5x + 2 \implies y = -0.5x + 3 \)

B1 for finding the midpoint \( (4, 1) \)\
M1 for finding the gradient of \( CD \) as 2\
M1 for finding the perpendicular gradient as \( -0.5 \) (or negative reciprocal of their gradient)\
A1 for the correct equation \( y = -0.5x + 3 \) (or equivalent form with fraction)

There are two mutually exclusive cases:\
Case 1: A red card is transferred from Box A to Box B.\
Probability of picking a red card from Box A: \( P(R_A) = \frac{5}{8} \).\
Box B now has 5 red cards and 6 blue cards (total of 11 cards).\
Probability of picking a red card from Box B: \( P(R_B | R_A) = \frac{5}{11} \).\
Probability for Case 1: \( \frac{5}{8} \times \frac{5}{11} = \frac{25}{88} \).\
\
Case 2: A blue card is transferred from Box A to Box B.\
Probability of picking a blue card from Box A: \( P(B_A) = \frac{3}{8} \).\
Box B now has 4 red cards and 7 blue cards (total of 11 cards).\
Probability of picking a red card from Box B: \( P(R_B | B_A) = \frac{4}{11} \).\
Probability for Case 2: \( \frac{3}{8} \times \frac{4}{11} = \frac{12}{88} \).\
\
Total probability: \( P(R_B) = \frac{25}{88} + \frac{12}{88} = \frac{37}{88} \).

M1 for calculating the probability of Case 1: \( \frac{5}{8} \times \frac{5}{11} = \frac{25}{88} \)\
M1 for calculating the probability of Case 2: \( \frac{3}{8} \times \frac{4}{11} = \frac{12}{88} \)\
M1 for adding the two case probabilities together\
A1 for the correct simplified fraction \( \frac{37}{88} \)

The total volume is the sum of the volume of the cylinder and the volume of the hemisphere.\
Volume of the cylinder:\
\( V_{\text{cylinder}} = \pi r^2 h = \pi \times 3^2 \times 8 = 72\pi \) cubic metres.\
Volume of the hemisphere:\
\( V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi \times 3^3 = 18\pi \) cubic metres.\
Total volume:\
\( V_{\text{total}} = 72\pi + 18\pi = 90\pi \) cubic metres.

M1 for volume of cylinder formula: \( \pi \times 3^2 \times 8 \)\
M1 for volume of hemisphere formula: \( \frac{2}{3} \times \pi \times 3^3 \)\
A1 for finding either cylinder volume as \( 72\pi \) or hemisphere volume as \( 18\pi \)\
A1 for the total volume \( 90\pi \)

A quadratic equation with a maximum (vertex) at \( (h, k) \) and leading coefficient \( a = -1 \) can be written in vertex form as:\
\( y = -(x - h)^2 + k \)\
Substituting \( h = 2 \) and \( k = 9 \):\
\( y = -(x - 2)^2 + 9 \)\
Expanding the expression:\
\( y = -(x^2 - 4x + 4) + 9 \)\
\( y = -x^2 + 4x - 4 + 9 \)\
\( y = -x^2 + 4x + 5 \)\
Comparing this to the given equation \( y = -x^2 + bx + c \), we find:\
\( b = 4 \) and \( c = 5 \).

M1 for writing the vertex form: \( -(x - 2)^2 + 9 \)\
M1 for expanding correctly: \( -x^2 + 4x + 5 \)\
A1 for \( b = 4 \) and \( c = 5 \)

The values of $H_n$ show a pattern of increasing differences:
$5 - 2 = 3$
$9 - 5 = 4$
Following this pattern, the next differences are 5 and 6:
$a = 9 + 5 = 14$
$b = 14 + 6 = 20$.

B1 for $a = 14$
B1 for $b = 20$

The sequence $2, 5, 9, 14, 20$ has constant second differences equal to 1, indicating a quadratic expression of the form $H_n = An^2 + Bn + C$, where $A = \frac{1}{2}$.
Substituting $n=1, 2$:
$1/2 + B + C = 2 \implies B + C = 1.5$
$2 + 2B + C = 5 \implies 2B + C = 3$
Subtracting the first equation from the second gives $B = 1.5$ and thus $C = 0$.
Therefore, $H_n = \frac{1}{2}n^2 + \frac{3}{2}n = \frac{n(n+3)}{2}$.

M1 for attempting quadratic form $An^2 + Bn + C$ with second difference $1$
A1 for $H_n = \frac{n(n+3)}{2}$ or equivalent

The staircase $S_n$ is symmetric about the diagonal line $y = x$. Reflecting the shape across this diagonal line interchanges every horizontal toothpick with a vertical toothpick and vice versa, while preserving the overall shape of the staircase. Therefore, the total count of horizontal toothpicks must equal the total count of vertical toothpicks.

B1 for identifying symmetry / reflection across the diagonal ($y = x$)
B1 for explaining how this interchanges horizontal and vertical segments

The total number of toothpicks $E_n$ is the sum of the horizontal and vertical toothpicks:
$E_n = H_n + V_n$
Since $V_n = H_n$, we have:
$E_n = 2 H_n = 2 \times \frac{n(n+3)}{2} = n(n+3) = n^2 + 3n$.

M1 for writing $E_n = 2 H_n$
A1 for $E_n = n(n+3)$ or $n^2 + 3n$

Using the formula $E_k = k(k+3)$:
$k(k+3) = 270$
$k^2 + 3k - 270 = 0$
Factorising the quadratic equation:
$(k+18)(k-15) = 0$
Since $k$ must be positive, we get $k = 15$.

M1 for setting up $k(k+3) = 270$
M1 for factorising to $(k+18)(k-15) = 0$ or using quadratic formula
A1 for $k = 15$

The sequence $1, 3, 6, 10, \dots$ represents the sum of the first $n$ positive integers:
$T_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$.

B2 for $\frac{n(n+1)}{2}$ (or equivalent)
(If B2 not awarded, B1 for recognizing relationship to sum of first $n$ integers)

The values $4, 8, 12, \dots$ form a simple arithmetic sequence with a common difference of 4.
Therefore, the $n$-th term is $P_n = 4n$.

B2 for $4n$
(If B2 not awarded, B1 for identifying a common difference of 4)

Substituting our expressions for $E_n$ and $P_n$:
$I_n = (n^2 + 3n) - 4n$
$I_n = n^2 - n$
Factorising this gives:
$I_n = n(n-1)$.

M1 for substituting $E_n$ and $P_n$ into the equation $I_n = E_n - P_n$
A1 for $n^2 - n$ or equivalent simplified expression
A1 for the fully factorised form $n(n-1)$

To find the total number of unit triangles, \(T(n)\), we add \(U(n)\) and \(D(n)\):
$$T(n) = U(n) + D(n) = \frac{n(n+1)}{2} + \frac{n(n-1)}{2}$$

Combine the fractions over the common denominator 2:
$$T(n) = \frac{n(n+1) + n(n-1)}{2}$$

Expand the terms in the numerator:
$$T(n) = \frac{n^2 + n + n^2 - n}{2}$$

Simplify the numerator:
$$T(n) = \frac{2n^2}{2} = n^2$$

- **M1**: For attempting to add the two algebraic fractions with a common denominator: \(\frac{n(n+1) + n(n-1)}{2}\)
- **M1**: For correctly expanding the terms to get \(2n^2\) in the numerator
- **A1**: For the simplified final expression \(n^2\)

First, calculate the number of pointing-down unit triangles for \(n = 10\):
$$D(10) = \frac{10(10-1)}{2} = \frac{10 \times 9}{2} = 45$$

Next, calculate the total number of unit triangles for \(n = 10\):
$$T(10) = 10^2 = 100$$

Now, find the percentage of pointing-down unit triangles:
$$\text{Percentage} = \frac{D(10)}{T(10)} \times 100\% = \frac{45}{100} \times 100\% = 45\%$$

- **M1**: For finding the number of pointing-down triangles (45) and the total number of triangles (100)
- **A1**: For the correct percentage of 45%

Using a graphic display calculator, enter the values of \(t\) in List 1 and \(H\) in List 2. Perform a linear regression (\(ax + b\) or \(mx + c\)) on the data. This gives the gradient \(m = 9\) and vertical intercept \(c = 12\). Thus, the equation of this model is \(H = 9t + 12\).

M1 for attempting to use a graphic display calculator to perform linear regression on the given data. A1 for the gradient \(m = 9\). A0.5 for the intercept \(c = 12\).

The gradient represents the rate of change of height \(H\) with respect to time \(t\). Since \(H\) is measured in cm and \(t\) in weeks, a gradient of 9 means the bamboo plant grows by an average of 9 cm each week.

B1.5 for identifying that the gradient represents the rate of growth or the increase in height. B1 for specifying the correct unit of time (per week).

The intercept is the value of the height \(H\) when the time \(t = 0\). In this context, this represents the initial height of the bamboo plant when it was first planted, which is 12 cm.

B1.5 for stating it represents the initial height or height at the start. B1 for referencing the starting time (0 weeks or at planting).

If we use the linear model to predict the height at \(t = 30\) weeks, the predicted height is \(H = 9(30) + 12 = 282\) cm. However, the actual height at 30 weeks is only 143 cm. Since the predicted height is significantly higher than the actual height, the linear model is not valid for older bamboo plants as growth rate decreases over time.

B1 for 'No' with brief reasoning. B1.5 for a quantitative comparison (e.g., calculating a predicted value like 282 cm vs 143 cm) or explaining that the growth rate slows down over time.

Substitute \(t = 10\) and \(H = 96\): \(96 = a(10)^2 + b(10) + 5 \implies 100a + 10b = 91\). Substitute \(t = 20\) and \(H = 131\): \(131 = a(20)^2 + b(20) + 5 \implies 400a + 20b = 126\).

M1 for substituting \(t = 10\) and \(H = 96\) into the quadratic model. A0.75 for \(100a + 10b = 91\) (or equivalent). A0.75 for \(400a + 20b = 126\) (or equivalent).

Multiply \(100a + 10b = 91\) by 2: \(200a + 20b = 182\). Subtract this from \(400a + 20b = 126\): \(200a = -56 \implies a = -0.28\). Substitute \(a = -0.28\) into \(100a + 10b = 91\): \(100(-0.28) + 10b = 91 \implies -28 + 10b = 91 \implies 10b = 119 \implies b = 11.9\). Thus, Sanjay's quadratic model is \(H = -0.28t^2 + 11.9t + 5\).

M1 for a valid method to solve simultaneous linear equations. A0.5 for \(a = -0.28\). A0.5 for \(b = 11.9\). A0.5 for the completed model expression.

At \(t = 0\), the exponent is zero: \(H = \frac{150}{1 + 11.5 \times 10^{-k(0)}} = \frac{150}{1 + 11.5 \times 10^0} = \frac{150}{1 + 11.5 \times 1} = \frac{150}{12.5} = 12\) cm.

M1.5 for substituting \(t = 0\) and demonstrating that \(10^0 = 1\). A1 for obtaining 12.

Substitute \(H = 96\) and \(t = 10\) into Leila's model: \(96 = \frac{150}{1 + 11.5 \times 10^{-10k}} \implies 1 + 11.5 \times 10^{-10k} = \frac{150}{96} = 1.5625 \implies 11.5 \times 10^{-10k} = 0.5625 \implies 10^{-10k} = \frac{0.5625}{11.5} \approx 0.048913 \implies -10k = \log_{10}(0.048913) \approx -1.31057 \implies k \approx 0.131\).

M1 for substituting \(H = 96\) and \(t = 10\) into the formula correctly. M1 for correctly isolating \(10^{-10k}\) or taking logs. A0.5 for \(k = 0.131\).

Substitute \(t = 0\) and \(T = 225\) into the formula: \(225 = a \times 10^{-k(0)} + 25\). Since \(10^{0} = 1\), this simplifies to \(225 = a + 25\). Subtracting 25 from both sides gives \(a = 200\).

M1 for substituting \(t = 0\) and \(T = 225\) into the given formula, i.e., \(225 = a \times 10^{0} + 25\). A1 for completing the simplification to show \(a = 200\).

Substitute \(t = 5\) and \(T = 45\) into the formula: \(45 = 200 \times 10^{-5k} + 25\). Subtract 25 from both sides to get \(20 = 200 \times 10^{-5k}\). Divide by 200 to get \(10^{-5k} = 0.1\). Since \(0.1 = 10^{-1}\), we have \(-5k = -1\), which gives \(k = 0.2\).

M1 for substituting \(t = 5\) and \(T = 45\) to get \(45 = 200 \times 10^{-5k} + 25\). M1 for isolating the exponential term, e.g., \(10^{-5k} = 0.1\). A1 for \(k = 0.2\) or equivalent.