2024 HKDSE 數學 模擬試題連答案詳解

\(\begin{aligned} \dfrac{5}{3u - 2} - \dfrac{2}{u + 4} &= \dfrac{5(u + 4) - 2(3u - 2)}{(3u - 2)(u + 4)} \\ &= \dfrac{5u + 20 - 6u + 4}{(3u - 2)(u + 4)} \\ &= \dfrac{24 - u}{(3u - 2)(u + 4)} \end{aligned}\)

1M for putting expressions under a common denominator: \(\dfrac{5(u + 4) - 2(3u - 2)}{(3u - 2)(u + 4)}\); 1M for expanding the numerator correctly: \(5u + 20 - 6u + 4\); 1A for \(\dfrac{24 - u}{(3u - 2)(u + 4)}\) or \(\dfrac{24 - u}{3u^2 + 10u - 8}\) or equivalent.

\(\begin{aligned} \dfrac{2y + 5h}{k} &= 4y - 3 \\ 2y + 5h &= k(4y - 3) \\ 2y + 5h &= 4ky - 3k \\ 5h + 3k &= 4ky - 2y \\ 5h + 3k &= y(4k - 2) \\ y &= \dfrac{5h + 3k}{4k - 2} \end{aligned}\)

1M for removing fraction: \(2y + 5h = 4ky - 3k\); 1M for grouping terms involving \(y\) on one side: \(4ky - 2y = 5h + 3k\) or \(2y - 4ky = -5h - 3k\); 1A for \(y = \dfrac{5h + 3k}{4k - 2}\) or \(y = \dfrac{5h + 3k}{2(2k - 1)}\) or equivalent.

(a) \(4p^2 - 9q^2 = (2p - 3q)(2p + 3q)\)

(b) \(\begin{aligned} 4p^2 - 9q^2 - 6p + 9q &= (2p - 3q)(2p + 3q) - 3(2p - 3q) \\ &= (2p - 3q)(2p + 3q - 3) \end{aligned}\)

(a) 1A for \((2p - 3q)(2p + 3q)\)
(b) 1M for using the result of (a): \((2p - 3q)(2p + 3q) - 3(2p - 3q)\); 1A for \((2p - 3q)(2p + 3q - 3)\)

(a) For \(\dfrac{3x - 1}{5} + 2 \ge x\):
\(3x - 1 + 10 \ge 5x \implies 9 \ge 2x \implies x \le 4.5\).

For \(4(x - 2) < 16\):
\(x - 2 < 4 \implies x < 6\).

Combining both inequalities: \(x \le 4.5\).

(b) The positive integers satisfying \(x \le 4.5\) are \(1, 2, 3, 4\). Therefore, there are 4 positive integers.

(a) 1M for solving \(\dfrac{3x - 1}{5} + 2 \ge x\) to obtain \(x \le 4.5\); 1M for solving \(4(x - 2) < 16\) to obtain \(x < 6\); 1A for \(x \le 4.5\) (or \(x \le \dfrac{9}{2}\))
(b) 1A for 4

From \(3u = 4v\), we have \(u : v = 4 : 3\).
Let \(u = 4k\) and \(v = 3k\), where \(k \ne 0\).

Substituting into \(\dfrac{u + w}{2v - w} = 2\):
\(\begin{aligned} \dfrac{4k + w}{2(3k) - w} &= 2 \\ 4k + w &= 2(6k - w) \\ 4k + w &= 12k - 2w \\ 3w &= 8k \\ w &= \dfrac{8}{3}k \end{aligned}\)

Thus, \(u : v : w = 4 : 3 : \dfrac{8}{3} = 12 : 9 : 8\).

Let \(u = 12m\), \(v = 9m\), and \(w = 8m\), where \(m \ne 0\):
\(\dfrac{u + 2v}{v + 2w} = \dfrac{12m + 2(9m)}{9m + 2(8m)} = \dfrac{12m + 18m}{9m + 16m} = \dfrac{30m}{25m} = \dfrac{6}{5}\).

1M for expressing \(u\) and \(v\) in terms of a single parameter (e.g., \(u = 4k, v = 3k\)); 1M for expressing \(w\) in terms of \(k\) (e.g., \(w = \dfrac{8}{3}k\)); 1M for substituting expressions of \(u, v, w\) into \(\dfrac{u + 2v}{v + 2w}\); 1A for \(\dfrac{6}{5}\) (or \(1.2\)).

Let \(\$C\) be the cost of the jacket.
The marked price of the jacket is \(\$(1 + 60\%)C = \$1.6C\).

The selling price of the jacket is:
\(1.6C \times (1 - 35\%) = 1.6C \times 0.65 = \$1.04C\).

Since the profit is \(\$48\):
\(\begin{aligned} 1.04C - C &= 48 \\ 0.04C &= 48 \\ C &= 1200 \end{aligned}\)

Therefore, the marked price is \(1.6 \times 1200 = \$1920\).

1M for expressing marked price or selling price in terms of cost (e.g. \(\text{Selling price} = (1 + 60\%)(1 - 35\%)C = 1.04C\)); 1M for setting up profit equation \(1.04C - C = 48\) or equivalent; 1M for finding cost \(C = 1200\); 1A for marked price \(\$1920\).

(a) \(\angle AOB = 164^\circ - 74^\circ = 90^\circ\)

(b) Since \(\angle AOB = 90^\circ\), \(\triangle AOB\) is a right-angled triangle.
By Pythagoras' theorem:
\(AB = \sqrt{OA^2 + OB^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\).

The perimeter of \(\triangle AOB\) is:
\(OA + OB + AB = 15 + 20 + 25 = 60\).

(a) 1A for \(90^\circ\)
(b) 1M for applying Pythagoras' theorem \(AB = \sqrt{15^2 + 20^2}\); 1A for \(AB = 25\); 1A for perimeter \(= 60\).

(a) The total number of students is 30:
\(5 + 9 + 8 + m + n = 30 \implies m + n = 8\) --- (1)

The mean number of books is \(2.8\):
\(\begin{aligned} \dfrac{1(5) + 2(9) + 3(8) + 4m + 5n}{30} &= 2.8 \\ 5 + 18 + 24 + 4m + 5n &= 84 \\ 47 + 4m + 5n &= 84 \\ 4m + 5n &= 37 \quad \text{--- (2)} \end{aligned}\)

From (1), \(4m + 4n = 32\). Subtracting this from (2):
\(n = 37 - 32 = 5\).
Substituting \(n = 5\) into (1):
\(m = 8 - 5 = 3\).

(b) The highest frequency is 9, which corresponds to 2 books. Thus, \(\text{Mode} = 2\).
The median is the average of the 15th and 16th values.
Cumulative frequencies: 5 (for 1), 14 (for 2), 22 (for 3).
Both the 15th and 16th values are 3. Thus, \(\text{Median} = 3\).

(a) 1M for setting up \(m + n = 8\); 1M for setting up mean equation \(\dfrac{47 + 4m + 5n}{30} = 2.8\); 1A for both \(m = 3\) and \(n = 5\)
(b) 1A for mode \(= 2\); 1A for median \(= 3\).

(a)
Solving the first inequality:
\[ \begin{aligned} \dfrac{3x-5}{2} &\le x + 1 \\ 3x - 5 &\le 2(x + 1) \\ 3x - 5 &\le 2x + 2 \\ x &\le 7 \end{aligned} \]

Solving the second inequality:
\[ \begin{aligned} 4 - 2x &< 10 \\ -2x &< 6 \\ x &> -3 \end{aligned} \]

Combining both inequalities, the required range of values of \(x\) is \(-3 < x \le 7\).

(b)
The integers that satisfy the compound inequality are \(-2, -1, 0, 1, 2, 3, 4, 5, 6, 7\).
Among these, the negative integers are \(-2\) and \(-1\).
Thus, there are 2 negative integers satisfying the compound inequality.

(a)
\(3x - 5 \le 2x + 2 \implies x \le 7\) [1M for solving one inequality]
\(-2x < 6 \implies x > -3\) [1M for solving the other inequality]
\(-3 < x \le 7\) [1A]

(b)
2 [1A]

(a) Let \( f(x) = ax + bx^2 \), where \( a \) and \( b \) are non-zero constants.
Since \( f(2) = 10 \) and \( f(3) = 24 \), we have:
\(\begin{cases} 2a + 4b = 10 \\ 3a + 9b = 24 \end{cases}\)
\(\begin{cases} a + 2b = 5 \\ a + 3b = 8 \end{cases}\)
Solving the system of linear equations:
\( b = 3 \)
\( a = -1 \)
Thus, \( f(x) = 3x^2 - x \).

(b) \( h(x) = (3x^2 - x) - (cx + d) = 3x^2 - (1 + c)x - d \).
The equation \( h(x) = 0 \) is \( 3x^2 - (1 + c)x - d = 0 \).
Using the relations between roots and coefficients:
Sum of roots \( = \alpha + \beta = \dfrac{1 + c}{3} \)
\( \dfrac{1 + c}{3} = 5 \implies 1 + c = 15 \implies c = 14 \)

Product of roots \( = \alpha\beta = \dfrac{-d}{3} \)
\( \dfrac{-d}{3} = -2 \implies d = 6 \).

(a)
- Let \( f(x) = ax + bx^2 \) and set up the system of linear equations: [1M]
- Correct values of \( a = -1 \) and \( b = 3 \): [1M]
- \( f(x) = 3x^2 - x \): [1A]

(b)
- Form the equation \( 3x^2 - (1 + c)x - d = 0 \): [1M]
- Use sum of roots \( \dfrac{1 + c}{3} = 5 \) or product of roots \( \dfrac{-d}{3} = -2 \): [1M]
- \( c = 14 \): [1A]
- \( d = 6 \): [1A]

(a) Since \( x - 2 \) is a factor of \( p(x) \), \( p(2) = 0 \):
\( 2(2)^3 + a(2)^2 + b(2) - 12 = 0 \)
\( 16 + 4a + 2b - 12 = 0 \)
\( 4a + 2b = -4 \implies 2a + b = -2 \quad \cdots (1) \)

Since the remainder when \( p(x) \) is divided by \( x + 1 \) is \( -18 \), \( p(-1) = -18 \):
\( 2(-1)^3 + a(-1)^2 + b(-1) - 12 = -18 \)
\( -2 + a - b - 12 = -18 \)
\( a - b = -4 \quad \cdots (2) \)

Adding (1) and (2):
\( 3a = -6 \implies a = -2 \)
Substituting \( a = -2 \) into (2):
\( -2 - b = -4 \implies b = 2 \).

(b) \( p(x) = 2x^3 - 2x^2 + 2x - 12 \).
Since \( x - 2 \) is a factor of \( p(x) \):
\( p(x) = (x - 2)(2x^2 + 2x + 6) = 2(x - 2)(x^2 + x + 3) \).

Consider the quadratic equation \( x^2 + x + 3 = 0 \).
Discriminant \( \Delta = 1^2 - 4(1)(3) = 1 - 12 = -11 < 0 \).
Thus, \( x^2 + x + 3 = 0 \) has no real roots.
Therefore, the equation \( p(x) = 0 \) has only one real root, so the claim is disagreed.

(a)
- Using \( p(2) = 0 \) or \( p(-1) = -18 \): [1M]
- Setting up both equations correctly: [1M]
- \( a = -2 \) and \( b = 2 \): [1A]

(b)
- Factorizing \( p(x) \) as \( (x - 2)(2x^2 + 2x + 6) \) or equivalent: [1M]
- Considering the discriminant \( \Delta = 1^2 - 4(1)(3) \): [1M]
- Obtaining \( \Delta = -11 < 0 \): [1A]
- Conclusion with correct reasoning: [1A]

(a) Sum of the 20 scores \( = 32.5 \times 20 = 650 \).
Sum of the given scores without \( (30 + a) \) and \( (40 + b) \):
\( (16 + 18) + (20 + 22 + 23 + 25 + 27 + 29) + (31 + 32 + 34 + 37 + 38 + 39) + (40 + 43 + 45 + 46) = 34 + 146 + 211 + 174 = 565 \).
Thus, \( 565 + (30 + a) + (40 + b) = 650 \implies a + b = 15 \).

Since the unique mode is \( 37 \) and all other existing values appear at most once, \( 30 + a \) must be \( 37 \), so \( a = 7 \).
Then \( b = 15 - 7 = 8 \).
(Note: \( 4 \le a \le 7 \) and \( 6 \le b \le 9 \) are both satisfied).

(b) The 20 scores in ascending order are:
16, 18, 20, 22, 23, 25, 27, 29, 31, 32, 34, 37, 37, 38, 39, 40, 43, 45, 46, 48.

Lower quartile \( Q_1 = \dfrac{23 + 25}{2} = 24 \).
Upper quartile \( Q_3 = \dfrac{39 + 40}{2} = 39.5 \).
Interquartile range \( = Q_3 - Q_1 = 39.5 - 24 = 15.5 \).

Standard deviation \( = \sqrt{\dfrac{\sum (x_i - \bar{x})^2}{20}} = \sqrt{90.05} \approx 9.49 \) (or \( 9.4895 \)).

(a)
- Setting up the equation for the mean: \( 565 + (30 + a) + (40 + b) = 650 \): [1M]
- \( a + b = 15 \): [1A]
- Explaining \( a = 7 \) using the mode condition: [1M]
- \( a = 7 \) and \( b = 8 \): [1A]

(b)
- Finding \( Q_1 = 24 \) and \( Q_3 = 39.5 \): [1M]
- Interquartile range \( = 15.5 \): [1A]
- Standard deviation \( = 9.49 \) (accept \( 9.4895 \)): [1A]

(a) Midpoint of \( AB = \left( \dfrac{2 + 8}{2}, \dfrac{9 + 1}{2} \right) = (5, 5) \).
Slope of \( AB = \dfrac{1 - 9}{8 - 2} = \dfrac{-8}{6} = -\dfrac{4}{3} \).
Slope of the perpendicular bisector \( = -\dfrac{1}{-\frac{4}{3}} = \dfrac{3}{4} \).
The equation of the perpendicular bisector is:
\( y - 5 = \dfrac{3}{4}(x - 5) \)
\( 4y - 20 = 3x - 15 \)
\( 3x - 4y + 5 = 0 \).

(b)(i) The centre of \( C \) lies on the perpendicular bisector of \( AB \) and on the line \( L: x + 2y - 25 = 0 \).
\(\begin{cases} 3x - 4y = -5 \\ x + 2y = 25 \end{cases}\)
From the second equation, \( x = 25 - 2y \).
\( 3(25 - 2y) - 4y = -5 \implies 75 - 10y = -5 \implies 10y = 80 \implies y = 8 \).
\( x = 25 - 2(8) = 9 \).
Thus, the coordinates of the centre of \( C \) are \( (9, 8) \).

(ii) Radius \( r = \sqrt{(9 - 2)^2 + (8 - 9)^2} = \sqrt{7^2 + (-1)^2} = \sqrt{50} \).
The equation of \( C \) is:
\( (x - 9)^2 + (y - 8)^2 = 50 \)
(or \( x^2 + y^2 - 18x - 16y + 95 = 0 \)).

(a)
- Midpoint \( (5, 5) \) and slope of \( AB = -\dfrac{4}{3} \): [1M]
- Using perpendicular slope condition: [1M]
- Equation \( 3x - 4y + 5 = 0 \) (or equivalent): [1A]

(b)(i)
- Solving \( \begin{cases} 3x - 4y + 5 = 0 \\ x + 2y - 25 = 0 \end{cases} \): [1M]
- Centre \( = (9, 8) \): [1A]

(b)(ii)
- Finding radius \( r^2 = (9-2)^2 + (8-9)^2 = 50 \): [1M]
- Equation \( (x - 9)^2 + (y - 8)^2 = 50 \) or \( x^2 + y^2 - 18x - 16y + 95 = 0 \): [1A]

(a) By the cosine formula in \( \triangle ABC \):
\( \cos \angle BAC = \dfrac{AB^2 + AC^2 - BC^2}{2(AB)(AC)} \)
\( \cos \angle BAC = \dfrac{15^2 + 14^2 - 13^2}{2(15)(14)} = \dfrac{225 + 196 - 169}{420} = \dfrac{252}{420} = 0.6 \)
\( \angle BAC = \arccos(0.6) \approx 53.1301^\circ \approx 53.1^\circ \).

(b)(i) In the right-angled triangle \( \triangle ACD \):
\( \sin \angle BAC = \sqrt{1 - 0.6^2} = 0.8 \)
\( CD = AC \sin \angle BAC = 14 \times 0.8 = 11.2\text{ cm} \).

(ii) In \( \triangle ACD \), \( AD = AC \cos \angle BAC = 14 \times 0.6 = 8.4\text{ cm} \).
\( BD = AB - AD = 15 - 8.4 = 6.6\text{ cm} \).

Since \( E \) lies on \( AC \) produced with \( CE = 6\text{ cm} \), \( AE = AC + CE = 14 + 6 = 20\text{ cm} \).
The perpendicular distance from \( E \) to the line containing \( AB \) is:
\( h_E = AE \sin \angle BAC = 20 \times 0.8 = 16\text{ cm} \).

Taking \( BD \) as the base of \( \triangle BDE \):
\( \text{Area of } \triangle BDE = \dfrac{1}{2} \times BD \times h_E = \dfrac{1}{2} \times 6.6 \times 16 = 52.8\text{ cm}^2 \).

(a)
- Applying the cosine formula: \( \cos \angle BAC = \dfrac{15^2 + 14^2 - 13^2}{2(15)(14)} \): [1M]
- \( \angle BAC \approx 53.1^\circ \): [1A]

(b)(i)
- \( CD = 14 \sin 53.1301^\circ \) (or \( 14 \times 0.8 \)): [1M]
- \( CD = 11.2\text{ cm} \): [1A]

(b)(ii)
- Finding \( BD = 15 - 14(0.6) = 6.6\text{ cm} \): [1M]
- Finding the height from \( E \) to \( AB \) (or using \( \text{Area}(\triangle ABE) - \text{Area}(\triangle ADE) \)): [1M]
- Area of \( \triangle BDE = 52.8\text{ cm}^2 \): [1A]

(a) Since \( x - 2 \) is a factor of \( p(x) \), \( p(2) = 0 \).
\( 2(2)^3 + a(2)^2 + b(2) - 30 = 0 \)
\( 16 + 4a + 2b - 30 = 0 \)
\( 2a + b = 7 \quad \cdots (1) \)
Since the remainder of \( p(x) \div (x + 3) \) is \( -150 \), \( p(-3) = -150 \).
\( 2(-3)^3 + a(-3)^2 + b(-3) - 30 = -150 \)
\( -54 + 9a - 3b - 30 = -150 \)
\( 9a - 3b = -66 \implies 3a - b = -22 \quad \cdots (2) \)
\((1) + (2)\): \( 5a = -15 \implies a = -3 \).
Substitute \( a = -3 \) into (1): \( 2(-3) + b = 7 \implies b = 13 \).

(b) With \( a = -3 \) and \( b = 13 \), \( p(x) = 2x^3 - 3x^2 + 13x - 30 \).
Since \( x - 2 \) is a factor, performing polynomial division gives:
\( p(x) = (x - 2)(2x^2 + x + 15) \).
Setting \( p(x) = 0 \) gives \( x = 2 \) or \( 2x^2 + x + 15 = 0 \).
For \( 2x^2 + x + 15 = 0 \), the discriminant is:
\( \Delta = 1^2 - 4(2)(15) = 1 - 120 = -119 < 0 \).
Thus, \( 2x^2 + x + 15 = 0 \) has no real roots.
Therefore, \( p(x) = 0 \) has only 1 real root and 2 non-real roots, so the claim is disagreed with.

(a) For \( p(2) = 0 \) or \( p(-3) = -150 \) [1M]
For setting up simultaneous equations \( 2a + b = 7 \) and \( 3a - b = -22 \) [1M]
For \( a = -3 \) and \( b = 13 \) [1A]

(b) For factorizing \( p(x) = (x - 2)(2x^2 + x + 15) \) [1M]
For considering the discriminant of \( 2x^2 + x + 15 = 0 \) [1M]
For \( \Delta = -119 < 0 \) [1A]
For correct conclusion with explanation [1A]

(a) Since \( \angle AOB = 90^\circ \), \( AB \) is a diameter of \( C \).
Center of \( C = \left(\frac{0+6}{2}, \frac{8+0}{2}\right) = (3, 4) \).
Radius of \( C = \frac{1}{2}\sqrt{(6-0)^2 + (0-8)^2} = \frac{1}{2}(10) = 5 \).
Equation of \( C \): \( (x - 3)^2 + (y - 4)^2 = 25 \) (or \( x^2 + y^2 - 6x - 8y = 0 \)).

(b)(i) The perpendicular distance from the center \( (3, 4) \) to \( 4x + 3y + k = 0 \) is equal to radius 5:
\( \frac{|4(3) + 3(4) + k|}{\sqrt{4^2 + 3^2}} = 5 \)
\( \frac{|24 + k|}{5} = 5 \implies |24 + k| = 25 \)
\( 24 + k = 25 \implies k = 1 \) or \( 24 + k = -25 \implies k = -49 \).

(b)(ii) The smaller value of \( k \) is \( -49 \), so \( L_1: 4x + 3y - 49 = 0 \).
The normal line passing through the center \( (3, 4) \) is perpendicular to \( L_1 \).
Slope of \( L_1 = -\frac{4}{3} \implies \text{slope of normal} = \frac{3}{4} \).
Equation of the normal: \( y - 4 = \frac{3}{4}(x - 3) \implies 3x - 4y + 7 = 0 \).
Solving the system \( 4x + 3y = 49 \) and \( 3x - 4y = -7 \):
From \( 4(4x + 3y) + 3(3x - 4y) = 4(49) + 3(-7) \implies 25x = 175 \implies x = 7 \).
Substitute \( x = 7 \) into \( 3(7) - 4y = -7 \implies 4y = 28 \implies y = 7 \).
Thus, \( T = (7, 7) \).

(a) For center \( (3, 4) \) or radius 5 [1M]
For correct equation of circle [1A]

(b)(i) For using distance formula from center to line \( = 5 \) [1M]
For \( k = 1 \) or \( k = -49 \) [1A]

(b)(ii) For finding the equation of the normal line through \( (3,4) \) [1M]
For solving the simultaneous equations of the tangent and normal [1M]
For \( T = (7, 7) \) [1A]

(a) In \( \triangle BCD \), by the cosine formula:
\( BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos 60^\circ \)
\( BD^2 = 10^2 + 14^2 - 2(10)(14)(0.5) = 100 + 196 - 140 = 156 \)
\( BD = \sqrt{156} = 2\sqrt{39} \approx 12.5\text{ cm} \).

(b) Let \( N \) be a point on \( CD \) such that \( BN \perp CD \).
Since \( AB \perp \text{plane } BCD \), by the three perpendiculars theorem, \( AN \perp CD \).
Thus, the angle between plane \( ACD \) and plane \( BCD \) is \( \angle ANB \).
In right-angled \( \triangle BCN \):
\( BN = BC \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\text{ cm} \).
In right-angled \( \triangle ABN \):
\( \tan \angle ANB = \frac{AB}{BN} = \frac{12}{5\sqrt{3}} = \frac{4\sqrt{3}}{5} \)
\( \angle ANB = \tan^{-1}\left(\frac{4\sqrt{3}}{5}\right) \approx 54.1835^\circ \approx 54.2^\circ \).

(c) In right-angled \( \triangle ABN \):
\( AN = \sqrt{AB^2 + BN^2} = \sqrt{12^2 + (5\sqrt{3})^2} = \sqrt{144 + 75} = \sqrt{219}\text{ cm} \).
Area of \( \triangle ACD = \frac{1}{2} \times CD \times AN = \frac{1}{2} \times 14 \times \sqrt{219} = 7\sqrt{219} \approx 103.5905\text{ cm}^2 \approx 104\text{ cm}^2 \).

(a) For applying cosine formula in \( \triangle BCD \) [1M]
For \( BD = \sqrt{156}\text{ cm} \approx 12.5\text{ cm} \) [1A]

(b) For identifying \( \angle ANB \) as the angle between the planes [1M]
For finding \( BN = 5\sqrt{3}\text{ cm} \) [1M]
For \( \angle ANB \approx 54.2^\circ \) (r.t. \( 54.2^\circ \)) [1A]

(c) For finding \( AN = \sqrt{219}\text{ cm} \) [1M]
For area \( \approx 104\text{ cm}^2 \) (r.t. \( 104\text{ cm}^2 \)) [1A]

(a) Standard score in Test \( X = \frac{79 - 64}{12} = 1.25 \).

(b)(i) Let \( Y = 1.2X - 8 \).
Mean score of Test \( Y = 1.2(64) - 8 = 76.8 - 8 = 68.8 \) marks.
Standard deviation of Test \( Y = 1.2 \times 12 = 14.4 \) marks.

(b)(ii) Kelvin's standard score in Test \( Y = \frac{85 - 68.8}{14.4} = \frac{16.2}{14.4} = 1.125 \).
Since his standard score in Test \( X \) (1.25) is greater than his standard score in Test \( Y \) (1.125), he performed relatively better in Test \( X \).

(c) Since both new scores equal the original mean (64 marks), the new mean remains 64 marks.
The sum of squared deviations from the mean remains unchanged: \( \sum (x_i - \bar{x})^2 = 25 \times 12^2 = 3600 \).
The new variance is \( \frac{3600 + (64-64)^2 + (64-64)^2}{27} = \frac{3600}{27} \approx 133.33 < 144 \).
Therefore, the variance of the scores in Test \( X \) will decrease.

(a) For standard score \( = 1.25 \) [1A]

(b)(i) For mean \( = 68.8 \) marks and SD \( = 14.4 \) marks [1A+1A]

(b)(ii) For calculating Kelvin's standard score in Test \( Y \) \( = 1.125 \) [1M]
For comparing standard scores and concluding Test \( X \) [1A]

(c) For showing new variance \( = \frac{3600}{27} < 144 \) or reasoning that sum of squares is divided by a larger number [1M]
For concluding that variance decreases [1A]

(a) Let \( r \) be the common ratio.
\( T_1 = 54 \) and \( T_4 = T_1 r^3 = 54r^3 \).
\( 54r^3 = 16 \implies r^3 = \frac{16}{54} = \frac{8}{27} \implies r = \frac{2}{3} \).

(b) The sum of the first \( n \) terms is:
\( S(n) = \frac{54\left(1 - \left(\frac{2}{3}\right)^n\right)}{1 - \frac{2}{3}} = \frac{54\left(1 - \left(\frac{2}{3}\right)^n\right)}{\frac{1}{3}} = 162\left(1 - \left(\frac{2}{3}\right)^n\right) = 162 - 162\left(\frac{2}{3}\right)^n \).
We are given \( 162 - S(n) < 10^{-3} \):
\( 162\left(\frac{2}{3}\right)^n < 10^{-3} \)
\( \left(\frac{2}{3}\right)^n < \frac{10^{-3}}{162} = \frac{1}{162000} \)
Taking common logarithm on both sides:
\( \log_{10}\left(\left(\frac{2}{3}\right)^n\right) < \log_{10}\left(\frac{1}{162000}\right) \)
\( n \log_{10}\left(\frac{2}{3}\right) < -\log_{10}(162000) \)
Since \( \log_{10}\left(\frac{2}{3}\right) < 0 \), dividing by \( \log_{10}\left(\frac{2}{3}\right) \) reverses the inequality:
\( n > \frac{-\log_{10}(162000)}{\log_{10}\left(\frac{2}{3}\right)} \approx \frac{-5.209515}{-0.176091} \approx 29.584 \).
Therefore, the least integer \( n \) satisfying the inequality is 30.

(a) For setting up \( 54r^3 = 16 \) [1M]
For \( r = \frac{2}{3} \) [1A]

(b) For using formula \( S(n) = \frac{a(1-r^n)}{1-r} \) to get \( S(n) = 162 - 162\left(\frac{2}{3}\right)^n \) [1M]
For setting up \( 162\left(\frac{2}{3}\right)^n < 10^{-3} \) [1M]
For taking logarithms and correctly solving inequality for \( n \) [1M]
For \( n > 29.584 \) [1A]
For least integer \( n = 30 \) [1A]