H2 Math 2025 Suggested Answers (Paper 2)

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Section A: Pure Mathematics [40 marks]

Question 1. System of Linear Equations

Solution. (a) Using the question,

$\displaystyle \left\{ \begin{matrix} 8c + 11t + 5b = 114, \\ 5c + 14t + 7b = 112, \\ 9c + 9t + 4b = 110. \end{matrix} \right.$

Solving the system of linear equations, $c = 7, t= 3, b = 5$ .

(b) By the question, Sam scored $113$ points. Let $x,y,z$ denote the number of car, train, and boat tokens respectively that Sam has. Hence,

$\displaystyle 7x + 3y + 5z = 113.$

Furthermore, we are given that $x \geq 6, y \geq 6, z \geq 6$ , so that

$\displaystyle 7x + 3y + 5z \geq 90.$

Since the sum is $113$ , at least one of $x,y,z$ must be odd. Furthermore, by fixing $y = z = 6$ for simplicity, we must have $7x \leq 65 \Rightarrow x \leq 9$ . Similarly, $y \leq 13$ and $z \leq 10$ .

By trial-and-error, $x = 6$ , $y = 7$ , $z = 10$ works. Since there is only one solution, Sam had $6$ car tokens, $7$ train tokens, and $10$ boat tokens.

Question 2. Vectors

Solution. (a) Firstly,

$\overrightarrow{PQ} = \begin{bmatrix} 1 \\ 6 \\ 6 \end{bmatrix} - \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 6 \end{bmatrix}.$

Since $\overrightarrow{QR} = \frac 43 \overrightarrow{PQ}$ and $P,Q,R$ are collinear, we have

$\begin{aligned} \overrightarrow{PR} &= \overrightarrow{PQ} + \overrightarrow{QR} \\ &= \overrightarrow{PQ} + {\textstyle \frac 43} \overrightarrow{PQ} \\ &= {\textstyle \frac 73} \overrightarrow{PQ} = {\textstyle \frac 73} \begin{bmatrix} 3 \\ 3 \\ 6 \end{bmatrix} = \begin{bmatrix} 7 \\ 7 \\ 14 \end{bmatrix}.\end{aligned}$

Hence,

$\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{PR} = \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} + \begin{bmatrix} 7 \\ 7 \\ 14 \end{bmatrix} = \begin{bmatrix} 5 \\ 10 \\ 14 \end{bmatrix}.$

(b) Firstly,

$\overrightarrow{ SQ } = \begin{bmatrix} 1\\ 6 \\ 6 \end{bmatrix} - \begin{bmatrix} -1 \\ -2 \\ c \end{bmatrix} .$

Since $\overrightarrow{ SQ } \perp \overrightarrow{ PQ }$ :

$\begin{aligned} \left( \begin{bmatrix} 1\\ 6 \\ 6 \end{bmatrix} - \begin{bmatrix} -1 \\ -2 \\ c \end{bmatrix} \right) \cdot \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} &= 0 \\ (1+6+12) - (-1-2+2c) &= 0 \\ 19 - (2c-3) &= 0 \\ c &= 11 .\end{aligned}$

(c) Hence,

$\overrightarrow{ PS } = \begin{bmatrix} -1 \\ -2 \\ 11 \end{bmatrix} - \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ 11 \end{bmatrix}.$

Therefore the required angle $\theta$ is given by

$\begin{aligned} \begin{bmatrix} 1 \\ -5 \\ 11 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} &= \left\| \begin{bmatrix} - 1 \\ -5 \\ 11 \end{bmatrix} \right\| \left\| \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} \right\| \cos \theta \\ 18 &= \sqrt{147} \sqrt{6} \cos \theta \\ \cos \theta &= \frac{18}{\sqrt{147} \sqrt 6}\\ \theta &\approx 52.7^\circ. \end{aligned}$

Question 3. Differential Equations

Solution. (a) By the method of separable variables,

$\begin{aligned} \int y^{-2}\, \mathrm dy &= \int \sin 3x\, \mathrm dx \\ \frac{y^{-1}}{-1} &= -\frac 13 \cos 3x + C \\ \frac 1y &= \frac 13 \cos 3x - C \\ y &= \frac{1}{-C + \frac 13 \cos 3x}. \end{aligned}$

Since the initial condition is $y(\pi) = 1$ ,

$\displaystyle 1 = \frac{1}{-C + \frac 13 \cos 3\pi} \quad \Rightarrow \quad C = -\frac 43 .$

Therefore, $\displaystyle y = \frac{1}{\frac 43 + \frac 13 \cos 3x} = \frac 3{4 + \cos 3x}$ , where $A = 3, B= 4$ .

(b) (i) By the graph $c = 3/5, 1$ (these values correspond with $\cos(x) = \pm 1$ ).

(b) (ii) We first note that the graph of $y = f(x)$ is defined on all of $\mathbb R$ , not just the domain $[0, 2\pi]$ . By considering reflections about vertical lines, we observe that the lines of symmetry of $y = f(x)$ are all of the form $x = k\pi/3$ , where $k$ is any integer (positive, zero, or negative).

(b) (iii) Since $\cos$ is $2\pi$ -periodic,

$\begin{aligned} \cos(3(x+\pi)) &= \cos(3x + 3\pi) \\ &= \cos(3x + \pi) \\ &= -{ \cos 3x }. \end{aligned}$

Hence,

$\displaystyle f(x+\pi) = \frac{A}{B + \cos(3(x+\pi))} = \frac{A}{B - \cos 3x} = \frac{3}{4 - \cos 3x}.$

(c) Using the product rule and the chain rule,

$\begin{aligned} \frac{\mathrm d^2 y}{\mathrm dx^2} &= \frac{\mathrm d}{\mathrm dx}(y^2 \sin 3x) \\ &= 2y \cdot \frac{\mathrm dy}{\mathrm dx} \cdot \sin 3x + y^2 \cdot 3 \cos 3x \\ &= 2y \cdot y^2 \sin 3x \cdot \sin 3x + y^2 \cdot 3 \cos 3x \\ &= (3 \cos 3x) \cdot y^2 + (2 \sin^2 3x) \cdot y^3,\end{aligned}$

where $P(x) = 3 \cos 3x, Q(x) = 2 \sin^2 3x$ .

Question 4. Rate of Change

Solution. (a) Letting $T$ denote the time taken to fill the bowl,

$10\pi \cdot T = \frac 13 \pi \cdot 15^2 \cdot (45 - 15) \quad \Rightarrow \quad T = 225\, \text{s}.$

(b) Using the chain rule,

$\begin{aligned} \frac{\mathrm dV}{\mathrm dt} &= \frac{\mathrm dV}{\mathrm dh} \cdot \frac{\mathrm dh}{\mathrm dt} \\ &= {\textstyle \frac 13} \pi \cdot (2h \cdot (45-h) + h^2 \cdot (-1)) \cdot \frac{\mathrm dh}{\mathrm dt} \\ &= {\textstyle \frac 13} \pi \cdot h (90-3 h) \cdot \frac{\mathrm dh}{\mathrm dt}. \end{aligned}$

At $h = 12$ ,

$\begin{aligned} 10\pi &= {\textstyle \frac 13} \pi \cdot 12 (90-3 \cdot 12) \cdot \frac{\mathrm dh}{\mathrm dt}\Big|_{h = 12} \quad \Rightarrow \quad \frac{\mathrm dh}{\mathrm dt}\Big|_{h = 12} = \frac{5}{108 }\, \text{cm}\ \text{s}^{-1}.\end{aligned}$

(c) The total volume poured into the bowl after $T$ seconds is

$\displaystyle \int_0^T 10\pi t\, \mathrm dt = 10\pi \cdot \frac{T^2}{2} = 5\pi \cdot T^2.$

Since $T > 0$ , the required time taken is given by solving the equation:

$\displaystyle 5\pi \cdot T^2 = 972 \pi \quad \Rightarrow \quad T = \sqrt{194.4} \approx 13.9\, \text{s}.$

(d) At $V = 972\pi$ ,

$\frac 13 \pi h^2 (45 - h) = 972 \pi \quad \Rightarrow \quad h^3 -45h^2 + 2916= 0.$

Solving the cubic equation, $h = 43.455, 9, -7.4558$ . Since $h \in [0, 15]$ , we reject the first and last option and conclude $h = 9$ .

Therefore,

$\begin{aligned} 10\pi T &= {\textstyle \frac 13} \pi \cdot 9 (90-3 \cdot 9) \cdot \frac{\mathrm dh}{\mathrm dt}\Big|_{h = 9} \quad \Rightarrow \quad \frac{\mathrm dh}{\mathrm dt}\Big|_{h = 9} = \frac{10T}{189} \approx 0.738\, \text{cm}\ \text{s}^{-1}.\end{aligned}$

Section B: Probability and Statistics [60 marks]

Question 5. Probability

Solution. (a) The total number of counters is $3+5+7 = 15$ . The probability that the counters are of the same colours is given by

$\displaystyle \frac{3 \times 2 + 5 \times 4 + 7 \times 6}{15 \times 14} = \frac{34}{105}$

Hence, the required probability is its complement $1 - \frac{34}{105} = \frac{71}{105}$ .

(b) By pattern-recognition, the required equation to solve is:

$\displaystyle \frac{12!}{(13-n)!} \cdot \frac{(16-n)!}{15!} \cdot \frac{3}{16-n} = \frac{36}{455}.$

The left-hand side, using factorials, simplifies to

$\begin{aligned} \frac{12!}{(13-n)!} \cdot \frac{(16-n)!}{15!} \cdot \frac{3}{16-n} &= \frac{(15-n)(14-n)}{ 910 } . \end{aligned}$

Hence, we need to solve the quadratic equation

$\displaystyle \frac{(15-n)(14-n)}{ 910 } = \frac{36}{455}.$

Solving, we get $n = 6$ .

Question 6. Normal Distribution

Solution. (a) Using the symmetry of the normal distribution,

$\mathbb P(X < 7) = \mathbb P(X < \mu) - \mathbb P(7 \leq X < \mu) = 0.5 - 0.45 = 0.05.$

(b) Since $X \sim \mathcal N(\mu, 1.8^2)$ , we have

$\begin{aligned} 0.05 = \mathbb P(X < 7) &= \mathbb P\left(Z < \frac{7 - \mu}{1.8}\right). \end{aligned}$

Using $z$ -scores,

$\displaystyle \frac{7 - \mu}{1.8} = -1.6449 \quad \Rightarrow \quad \mu = 7 + 1.6449 \cdot 1.8 = 9.96082 \approx 9.96.$

(c) Using independence,

$\begin{aligned} 0.38 = \mathbb P((X > 7) \cap (Y > 7)) &= \mathbb P(X > 7) \cdot \mathbb P(Y > 7) \\ &= 0.95 \cdot \mathbb P(Y > 7). \end{aligned}$

By algebruh,

$\displaystyle \mathbb P(Y > 7) = \frac{0.38}{0.95} = 0.4\quad \Rightarrow \quad \mathbb P(Y \leq 7) = 0.6.$

Since $Y \sim \mathcal N(\lambda, 2.8^2)$ ,

$\displaystyle \mathbb P\left(Z \leq \frac{7 - \lambda}{2.8} \right) = \mathbb P(Y \leq 7) = 0.6.$

Using $z$ -scores,

$\displaystyle \frac{7 - \lambda}{2.8} = 0.25334 \quad \Rightarrow \quad \lambda = 7 - 2.8 \cdot 0.25334 = 6.2906 \approx 6.29.$

(d) Using complements,

$\begin{aligned} \mathbb P(Y \leq a) &= 1 - \mathbb P(Y > a) \\ &= 1 - 2 \cdot \mathbb P(Y > 7) \\ &= 1 - 2 \cdot 0.4 = 0.2. \end{aligned}$

Since $Y \sim \mathcal N(6.2906, 2.8^2)$ by (c), $a \approx 3.93$ .

Question 7. Permutations and Combinations

Solution. (a) The total number of arrangements, including repetitions, is

$\displaystyle \frac{8!}{2! \cdot 2!} = 10\, 080.$

(b) We first choose the start and end numbers, then place the third even number in the middle four spaces, and then permute the odd numbers:

$\displaystyle (3 \times 2) \times {4 \choose 1} \times \frac{5!}{2! \cdot 2!} = 720.$

(c) The total number of even-number groups (consisting of $2,4,8$ ) is $3! = 6$ . Hence, the required total is

$\displaystyle 3! \cdot \frac{6!}{2! \cdot 2!} = 1080.$

(d) We first count the number of arrangements where the even digits are all together and the odd digits are all together:

$\displaystyle 3! \cdot \frac{5!}{2! \cdot 2!} \cdot 2! = 360.$

The required probability is given by a conditional probability as follows:

$\displaystyle \frac{ 360 }{ 1080 } = \frac 13.$

Question 8. Correlation and Linear Regression

Solution. (a) Since $r = 0.39 \in (0.3, 0.7)$ , there is a weak positive linear correlation between the best performances of the athletes in the high jump and long jump events.

(b) (i) The diagram suggests that there is a positive, though non-linear, correlation between $x$ and $t$ , where $x$ increases at an increasing rate as $t$ increases.

(ii) For the model $t = ax+ b$ , we obtain a correlation coefficient of $r_1 \approx 0.958$ , and for the model $t = cx^2 + d$ we obtain a correlation coefficient of $r_2 \approx 0.990$ . Since $0 < r_1 < r_2 < 1$ (i.e. $|r_2|$ is closer to $1$ than $|r_1|$ ), the model $t = cx^2 + d$ is a better fit to the data. Furthermore, we obtain the regression equation

$\begin{aligned} t &= 1.24969x^2 + 0.87580 \\ t &= 1.25x^2 + 0.876.\quad (\text{3sf}) \end{aligned}$

(iii) At $x = 11$ ,

$t = 1.24969 \cdot 11^2 + 0.87580 \approx 152\, \text{minutes}.$

Since $r_2 \approx 0.990$ is close to $1$ , the correlation between $t$ and $x^2$ is strongly positive. Hence, an interpolation $x = 11 \in (2.1, 11)$ yields a reliable estimate.

Question 9. Binomial Distribution

Solution. (a) The two assumptions are as follows:

The probability that a randomly chosen refrigerator is faulty is constant.
The faultiness of any refrigerator is independent of other refrigerators.

(b) Since $X \sim \mathrm{B}(90, 0.02)$ , the required probability is

$\mathbb P(X > 1) = 1 - \mathbb P(X \leq 1) = 1 - 0.46043 = 0.53956 \approx 0.540.$

(c) Let $V \sim \mathrm{B}(5, 0.53956)$ denote the number of days in a $5$ -day working week, on which Alan finds more than one faulty refrigerator. The required probability is

$\mathbb P(V < 2) = \mathbb P(V \leq 1) = 0.14195 \approx 0.142.$

(d) Since $90 \times 5 = 450$ , let $W \sim \mathrm B(450, 0.02)$ denote the number of faulty refrigerators that Alan finds in total, in a $5$ -day work week. Then the required probability is

$\mathbb P(W < 10 ) = \mathbb P(W \leq 9) = 0.58741 \approx 0.587.$

(e) Since $X \sim \mathrm{B}(90, 0.02)$ and $X \sim \mathrm{B}(60, 0.03)$ , the required probability is given by three cases:

$\begin{aligned} \mathbb P(X + Y = 2) &= \mathbb P(X = 2) \cdot \mathbb P(Y = 0) + \mathbb P(X = 1) \cdot \mathbb P(Y = 1) + \mathbb P(X = 0) \cdot \mathbb P(Y = 2) \\ &= 0.27074 \cdot 0.16080 + 0.29812 \cdot 0.29840 + 0.16231 \cdot 0.27225 \\ &= 0.17668 \approx 0.177. \end{aligned}$

Question 10. Flour-Production

Solution. (a) In the question, $X$ denotes the mass, in kg, of packets produced by Machine $P$ , and $Y$ denotes the mass, in kg, of packets produced by Machine $Q$ . Since the machines are different and not related in general, the masses of packets of flour produced would also not affect one another. Therefore, $X,Y$ are independent.

(b) Let $X \sim \mathcal N( 2.2, 0.1^2 )$ denote the mass of a randomly chosen packet of flour produced by Machine $P$ . Let $Y \sim \mathcal N( 2.1, 0.05^2 )$ denote the mass of a randomly chosen packet of flour produced by Machine $Q$ . Then

$\begin{aligned} \mathbb E[X - Y] &= 2.2 - 2.1 = 0.1, \\ \mathrm{Var}(X - Y) &= 0.1^2 + 0.05^2 = 0.0125, \\ X- Y &\sim \mathcal N(0.1, 0.0125). \end{aligned}$

Hence, the required probability is

$\mathbb P(X > Y) = \mathbb P(X - Y > 0) = 0.81445 \approx 0.814.$

(c) Defining the random variable $W := X_1+ X_2 + X_3 + Y_1 + \cdots + Y_5$ ,

$\begin{aligned} \mathbb E[W] &= 3 \cdot 2.2 + 5 \cdot 2.1 = 17.1, \\ \mathrm{Var} (W) &= 3 \cdot 0.1^2 + 5 \cdot 0.05^2 = 0.0425. \end{aligned}$

Hence, $W \sim \mathcal N(17.1, 0.0425)$ , so that the required probabiltiy is

$\displaystyle \mathbb P(W > 17) = 0.68616 \approx 0.686.$

(d) Let $\mu$ denote the mean mass of a packet of flour, in kg, produced by the adjusted Machine $P$ . Then the null hypothesis $\mathrm H_0$ and the alternative hypothesis $\mathrm H_1$ are given as follows:

$\mathrm H_0 : \mu = 2.2,\quad \mathrm H_1 : \mu \neq 2.2.$

(e) Let $\bar x$ and $s^2$ denote the unbiased estimates of the population mean and variance of the relevant masses. Then

$\displaystyle \bar x = 2 + \frac{4.5}{30} = 2.15$

and

$\displaystyle s^2 = \frac{1}{30-1} \left( 1.11 - \frac{4.5^2}{30} \right) = 0.015.$

(f) Suppose $\mathrm H_0$ holds. Since the sample size $n = 30 \geq 0$ is sufficiently large, by the central limit theorem,

$\displaystyle \bar X \sim \mathcal N \left( 2.2, \frac{0.015}{30} \right)\quad \text{approximately}.$

Since we are conducting a two-tailed test, the required $p$ -value is given by

$p = 2 \cdot \mathbb P(\bar X \leq 2.15)= 2 \cdot 0.012673 \approx 0.0253 < 0.05.$

Since $p < 0.05$ , there is sufficient evidence to reject $\mathrm H_0$ and conclude that the mean mass of packets produced by the adjusted Machine $P$ differs from $2.2$ kg.

(g) If fewer than $30$ packets has been used, then the central limit theorem would lead to a weaker approximate normal distribution for $\bar X$ , yielding a possibly larger $p$ -value.

(h) If the sample had not been chosen randomly, it is plausible that the packets of flour chosen have masses that are not independent with one another, so that $\bar X$ may not follow a normal distribution, even approximately.

—Joel Kindiak, 7 Nov 25, 1522H

Responses

xx

November 7, 2025 at 12:11 pm

is 9b supposed to be 1-P(X<= 1)?

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1. joelkindiak
  
  November 7, 2025 at 3:00 pm
  
  Yep I believe so; doing bug-fixing now (spotted several arithmetic bugs but now I believe they have been fixed)
  
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