2025 HKDSE Mathematics Practice Paper with Answers

\begin{aligned} \frac{(u^4 v^{-3})^3}{u^{-6} v^5} &= \frac{u^{12} v^{-9}}{u^{-6} v^5} \[6pt] &= u^{12 - (-6)} v^{-9 - 5} \[6pt] &= u^{18} v^{-14} \[6pt] &= \frac{u^{18}}{v^{14}} \end{aligned}

For \((u^4)^3 = u^{12}\) or \((v^{-3})^3 = v^{-9}\): 1M
For \(u^{12 - (-6)} = u^{18}\) or \(v^{-9 - 5} = v^{-14}\): 1M
For expressing with positive index for \(v\): 1M
For correct final answer \(\frac{u^{18}}{v^{14}}\): 1A

\begin{aligned} 7(5h - 2k) &= 2(3h + 4k) \[4pt] 35h - 14k &= 6h + 8k \[4pt] 35h - 6h &= 8k + 14k \[4pt] 29h &= 22k \[4pt] k &= \frac{29h}{22} \end{aligned}

For clearing fractions: \(7(5h - 2k) = 2(3h + 4k)\): 1M
For expanding and collecting like terms in \(k\) on one side: 1M
For \(22k = 29h\): 1M
For correct answer \(k = \frac{29h}{22}\) (or equivalent): 1A

(a) \(6xy - 15xz = 3x(2y - 5z)\)

(b) \(4y^2 - 25z^2 = (2y)^2 - (5z)^2 = (2y - 5z)(2y + 5z)\)

(c) \begin{aligned} 4y^2 - 25z^2 - 6xy + 15xz &= (4y^2 - 25z^2) - (6xy - 15xz) \\ &= (2y - 5z)(2y + 5z) - 3x(2y - 5z) \\ &= (2y - 5z)(2y + 5z - 3x) \end{aligned}

(a) 1A for \(3x(2y - 5z)\)
(b) 1A for \((2y - 5z)(2y + 5z)\)
(c) 1M for utilizing results of (a) and (b)
1A for \((2y - 5z)(2y + 5z - 3x)\)

(a) \begin{aligned} \frac{3x - 5}{2} &\le x + 1 \\ 3x - 5 &\le 2x + 2 \\ x &\le 7 \end{aligned}

\begin{aligned} 4 - 3x &< 19 \\ -3x &< 15 \\ x &> -5 \end{aligned}

Thus, the solution of \((*)\) is \(-5 < x \le 7\).

(b) The integers satisfying \((*)\) are \(-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7\).
Therefore, the number of integers is \(7 - (-4) + 1 = 12\).

(a) 1M for solving either inequality correctly (\(x \le 7\) or \(x > -5\))
1M for combining with "and"
1A for \(-5 < x \le 7\)
(b) 1A for 12

(a) Marked price \(= 600 \times (1 + 40\%) = 600 \times 1.4 = \$840\).
Selling price \(= 840 \times (1 - 25\%) = 840 \times 0.75 = \$630\).

(b) Profit \(= 630 - 600 = \$30\).
Profit percentage \(= \frac{30}{600} \times 100\% = 5\%\).

(a) 1M for finding marked price: \(600 \times 1.4 = 840\)
1A for selling price: \(\$630\)
(b) 1M for \(\frac{630 - 600}{600} \times 100\%\)
1A for profit percentage of \(5\%\) (or profit \(5\%\))

(a) Let the original number of red pens and blue pens be \(5k\) and \(3k\) respectively, where \(k > 0\).
According to the question:
\[ 5k - 8 = 3k + 12 \]
\[ 2k = 20 \]
\[ k = 10 \]
Original number of red pens \(= 5(10) = 50\).

(b) Original number of blue pens \(= 3(10) = 30\).
Total number of pens after change \(= (50 - 8) + (30 + 12) = 42 + 42 = 84\).

(a) 1M for setting up the equation \(5k - 8 = 3k + 12\) (or a pair of simultaneous equations in two variables)
1M for solving \(k = 10\)
1A for 50
(b) 1A for 84

(a) Coordinates of the mid-point \(= \left(\frac{-4 + 2}{2}, \frac{6 + (-2)}{2}\right) = (-1, 2)\).

(b) After rotating \((-4, 6)\) anticlockwise about \(O\) through \(90^\circ\), the coordinates of \(A'\) are \((-6, -4)\).
Slope of \(A'B = \frac{-2 - (-4)}{2 - (-6)} = \frac{2}{8} = \frac{1}{4}\).
Equation of the line passing through \(A'\) and \(B\):
\begin{aligned} y - (-2) &= \frac{1}{4}(x - 2) \\ 4(y + 2) &= x - 2 \\ 4y + 8 &= x - 2 \\ x - 4y - 10 &= 0 \end{aligned}

(a) 1A for \((-1, 2)\)
(b) 1A for coordinates of \(A'(-6, -4)\)
1M for method of finding line equation using slope and a point
1A for \(x - 4y - 10 = 0\) (or equivalent form)

(a) Median \(= \frac{x + y}{2} = 19.5 \implies x + y = 39\).
Mean \(= \frac{8 + 11 + 14 + x + y + 25 + 28 + 31}{8} = 20\).
Sum \(= 117 + x + y = 160 \implies x + y = 43\)... wait, let's calculate:
\(8 + 11 + 14 + 25 + 28 + 31 = 117\).
With \(x + y = 39\), sum \(= 117 + 39 = 156\), so mean \(= \frac{156}{8} = 19.5\).
Let the mean be \(19.5\) and median be \(19.5\), then if \(y - x = 5\):
\(x + y = 39\) and \(y - x = 5 \implies x = 17, y = 22\).

Let's re-verify with: \(x + y = 39\) and sum \(= 156\) (mean \(= 19.5\)).

(b) Lower quartile \(Q_1 = \frac{11 + 14}{2} = 12.5\).
Upper quartile \(Q_3 = \frac{25 + 28}{2} = 26.5\).
Interquartile range \(= Q_3 - Q_1 = 26.5 - 12.5 = 14\).

(a) 1M for \(\frac{x+y}{2} = 19.5\) and equating sum or mean
1A for \(x = 17\) and \(y = 22\)
(b) 1M for finding \(Q_1 = 12.5\) and \(Q_3 = 26.5\)
1A for \(\text{interquartile range} = 14\)

(a) \(\frac{2x+5}{3} \le x-1\)
\(2x + 5 \le 3(x - 1)\)
\(2x + 5 \le 3x - 3\)
\(-x \le -8\)
\(x \ge 8\)

\(4x + 15 < 3\)
\(4x < -12\)
\(x < -3\)

Thus, the solution of \((*)\) is \(x < -3\) or \(x \ge 8\).

(b) The greatest negative integer satisfying \(x < -3\) is \(-4\).

(a) \(\frac{2x+5}{3} \le x-1 \implies x \ge 8\) (1M)
\(4x + 15 < 3 \implies x < -3\) (1M)
Solution: \(x < -3\) or \(x \ge 8\) (1A)

(b) \(-4\) (1A)

(a) By the Remainder Theorem, \(f(2) = -12\).
\(3(2)^3 + p(2)^2 - 19(2) + q = -12\)
\(24 + 4p - 38 + q = -12\)
\(4p + q = 2\) ... (1)

By the Factor Theorem, \(f(-3) = 0\).
\(3(-3)^3 + p(-3)^2 - 19(-3) + q = 0\)
\(-81 + 9p + 57 + q = 0\)
\(9p + q = 24\) ... (2)

Subtracting (1) from (2):
\(5p = 22 \implies p = \frac{22}{5}\) -- wait, let's adjust integers:
Let's check: (2)-(1): \(5p = 22\).
Let's choose nice integer values: if \(p = 4\), then (1) gives \(q = 2 - 16 = -14\). Then \(9(4) + (-14) = 36 - 14 = 22 \neq 24\).
To make integer: \(4p+q = -2\) and \(9p+q = 18 \implies 5p = 20 \implies p=4, q=-18\).
Let's adjust stem: when \(f(x)\) is divided by \(x - 2\), the remainder is \(-16\).
Then \(24 + 4p - 38 + q = -16 \implies 4p + q = -2\).
When \(x + 3\) is a factor, \(-81 + 9p + 57 + q = 0 \implies 9p + q = 24\).
Then \(5p = 26\), still not integer.
Let's rework:
\(f(x) = (x+3)(3x-1)(x+2) = (3x^2+8x-3)(x+2) = 3x^3 + 14x^2 + 13x - 6\).
Then \(f(2) = 3(8)+14(4)+13(2)-6 = 24+56+26-6 = 100\).
Let \(f(x) = 3x^3 + px^2 + 13x + q\).
\(f(-3) = 3(-27) + 9p - 39 + q = -120 + 9p + q = 0 \implies 9p + q = 120\).
\(f(2) = 24 + 4p + 26 + q = 50 + 4p + q = 100 \implies 4p + q = 50\).
Then \(5p = 70 \implies p = 14, q = -6\).
Roots of \(f(x) = 0\) are \(-3, -2, \frac{1}{3}\), all rational.

Let's write this cleanly:
\(f(x) = 3x^3 + px^2 + 13x + q\).
(a) By Remainder Theorem: \(f(2) = 100\)
\(3(2)^3 + p(2)^2 + 13(2) + q = 100 \implies 4p + q = 50\)
By Factor Theorem: \(f(-3) = 0\)
\(3(-3)^3 + p(-3)^2 + 13(-3) + q = 0 \implies 9p + q = 120\)
Solving gives \(p = 14\) and \(q = -6\).

(b) \(f(x) = 3x^3 + 14x^2 + 13x - 6 = 0\)
Since \(x+3\) is a factor, by long division:
\(f(x) = (x+3)(3x^2 + 5x - 2) = 0\)
\((x+3)(3x - 1)(x + 2) = 0\)
\(x = -3\), \(x = \frac{1}{3}\), or \(x = -2\).
Since \(-3\), \(\frac{1}{3}\), and \(-2\) are all rational numbers, the claim is agreed.

(a) \(f(2) = 100 \implies 4p + q = 50\) [1M]
\(f(-3) = 0 \implies 9p + q = 120\) [1M]
\(p = 14, q = -6\) [1A]

(b) Factorizing \(f(x) = (x+3)(3x^2+5x-2)\) [1M]
\((x+3)(3x-1)(x+2) = 0\) [1M]
Roots are \(-3, \frac{1}{3}, -2\) [1A]
Conclusion: The roots are all rational numbers, so the claim is agreed. [1A (f.t.)]

(a) Let \(W = k_1 + \frac{k_2}{x^2}\), where \(k_1\) and \(k_2\) are non-zero constants.
When \(x = 2\), \(W = 19\):
\(k_1 + \frac{k_2}{4} = 19 \implies 4k_1 + k_2 = 76\) ... (1)
When \(x = 3\), \(W = 14\):
\(k_1 + \frac{k_2}{9} = 14 \implies 9k_1 + k_2 = 126\) ... (2)
Subtracting (1) from (2):
\(5k_1 = 50 \implies k_1 = 10\)
Substituting \(k_1 = 10\) into (1):
\(4(10) + k_2 = 76 \implies k_2 = 36\)
Thus, \(W = 10 + \frac{36}{x^2}\).

(b) (i) When \(x = 6\):
\(W = 10 + \frac{36}{6^2} = 10 + 1 = 11\).

(ii) For any non-zero real number \(x\), \(x^2 > 0\).
Since \(36 > 0\) and \(x^2 > 0\), \(\frac{36}{x^2} > 0\).
Therefore, \(W = 10 + \frac{36}{x^2} > 10\).
Thus, \(W\) can never be less than or equal to \(10\).
The claim is correct.

(a) Let \(W = k_1 + \frac{k_2}{x^2}\) [1A]
Setting up simultaneous equations \(k_1 + \frac{k_2}{4} = 19\) and \(k_1 + \frac{k_2}{9} = 14\) [1M]
Solving to get \(k_1 = 10, k_2 = 36\) and \(W = 10 + \frac{36}{x^2}\) [1A]

(b)(i) \(W = 10 + \frac{36}{36} = 11\) [1A]
(b)(ii) Noting \(x^2 > 0\) for any real \(x \neq 0\) [1M]
Arguing \(\frac{36}{x^2} > 0 \implies W > 10\) [1M]
Conclusion: The claim is correct [1A (f.t.)]

(a) There are \(20\) participants. The median is the average of the 10th and 11th values.
The 10th value is \(30 + a\) and the 11th value is \(36\).
\(\frac{(30 + a) + 36}{2} = 37.5\)
\(66 + a = 75 \implies a = 5\).

The range is the difference between the maximum value and the minimum value.
Minimum value \(= 21\).
Maximum value \(= 59\).
Range \(= 59 - 21 = 38\), which matches.
Since \(b\) must satisfy the ordering of the stem \(4\): the leaves are \(1, 3, 5, b, 8\), so \(5 \le b \le 8\).
Wait, let's provide a condition to uniquely determine \(b\):
Let the mean of the distribution be \(39.0\).
Sum of all 20 ages:
\(21+23+24+27+28 + 30+32+35+36+38+39 + 41+43+45+(40+b)+48 + 52+54+57+59 = 733 + b\).
\(\frac{733 + b}{20} = 39.0 \implies 733 + b = 780 \implies b = 47 - 40 = 7\).
So \(a = 5\) and \(b = 7\).

(b) Lower quartile \(Q_1\) is the average of the 5th and 6th values:
5th value \(= 28\), 6th value \(= 30\).
\(Q_1 = \frac{28 + 30}{2} = 29\).
Upper quartile \(Q_3\) is the average of the 15th and 16th values:
15th value \(= 47\), 16th value \(= 48\).
\(Q_3 = \frac{47 + 48}{2} = 47.5\).
Inter-quartile range \(= Q_3 - Q_1 = 47.5 - 29 = 18.5\) years.

(c) The mean of the original 20 participants is \(39\).
The two new ages are \(35\) and \(42\).
The mean of the two new ages is \(\frac{35+42}{2} = 38.5\), which is very close to \(39\).
Both \(35\) and \(42\) are closer to the mean \(39\) than most existing data points (the original variance is large, with data spread from 21 to 59).
Adding data values close to the mean reduces the dispersion of the distribution from the mean.
Therefore, the standard deviation decreases.

(a) Using median: \(\frac{(30+a)+36}{2} = 37.5 \implies a = 5\) [1M+1A]
Using mean/sum to find \(b = 7\) [1A]

(b) Finding \(Q_1 = 29\) and \(Q_3 = 47.5\) [1M]
\(\text{IQR} = 47.5 - 29 = 18.5\) years [1A]

(c) Explaining that the new data points (35 and 42) lie very close to the mean (39) compared to the overall spread [1M]
Conclusion: The standard deviation decreases [1A]

(a) The center of circle \(C\) is the mid-point of \(AB\):
Center \(= \left(\frac{-2 + 6}{2}, \frac{1 + 7}{2}\right) = (2, 4)\).
Radius \(r = \frac{1}{2}AB = \frac{1}{2}\sqrt{(6 - (-2))^2 + (7 - 1)^2} = \frac{1}{2}\sqrt{8^2 + 6^2} = \frac{1}{2}(10) = 5\).
Equation of \(C\):
\((x - 2)^2 + (y - 4)^2 = 5^2\)
\(x^2 - 4x + 4 + y^2 - 8y + 16 = 25\)
\(x^2 + y^2 - 4x - 8y - 5 = 0\).

(b) Since \(L\) passes through the center \((2, 4)\):
\(4(2) - 3(4) + k = 0\)
\(8 - 12 + k = 0\)
\(k = 4\).

(c) Since \(L\) passes through the center of \(C\), \(PQ\) is a diameter of \(C\).
Therefore, the length of base \(PQ = 2r = 10\).
Since \(R\) lies on the circle and \(\triangle PQR\) is an isosceles right-angled triangle with hypotenuse \(PQ\) (as \(\angle PRQ = 90^\circ\) by angle in a semicircle), the perpendicular height from \(R\) to \(PQ\) is equal to the radius \(r = 5\).
Therefore, the area of \(\triangle PQR\) is:
\(\text{Area} = \frac{1}{2} \times PQ \times r = \frac{1}{2} \times 10 \times 5 = 25\).

(a) Finding center \((2, 4)\) and radius \(r = 5\) [1M]
Equation of circle: \((x - 2)^2 + (y - 4)^2 = 25\) (or \(x^2 + y^2 - 4x - 8y - 5 = 0\)) [1A]

(b) Substituting \((2, 4)\) into \(4x - 3y + k = 0\) [1M]
\(k = 4\) [1A]

(c) Identifying \(PQ\) as the diameter so \(PQ = 10\) [1M]
Recognizing the height of \(\triangle PQR\) is \(r = 5\) [1M]
Area \(= \frac{1}{2}(10)(5) = 25\) [1A]

(a) \(f(x) = -2x^2 + 8kx - 6k^2 - 8\)
\(= -2(x^2 - 4kx) - 6k^2 - 8\)
\(= -2[(x - 2k)^2 - 4k^2] - 6k^2 - 8\)
\(= -2(x - 2k)^2 + 8k^2 - 6k^2 - 8\)
\(= -2(x - 2k)^2 + 2k^2 - 8\)
Thus, the coordinates of \(V\) are \((2k, 2k^2 - 8)\).

(b) (i) Translating leftwards by 3 units: \(x\)-coordinate becomes \(2k - 3\).
Translating upwards by 10 units: \(y\)-coordinate becomes \((2k^2 - 8) + 10 = 2k^2 + 2\).
Thus, the coordinates of \(U\) are \((2k - 3, 2k^2 + 2)\).

(ii) The slope of \(OU\) is:
\(\frac{2k^2 + 2 - 0}{2k - 3 - 0} = 2\)
\(2k^2 + 2 = 2(2k - 3)\)
\(2k^2 + 2 = 4k - 6\)
\(2k^2 - 4k + 8 = 0 \implies k^2 - 2k + 4 = 0\) (no real roots, let's adjust).
Let slope of \(OU\) be \(4\):
\(2k^2 + 2 = 4(2k - 3) = 8k - 12\)
\(2k^2 - 8k + 14 = 0 \implies k^2 - 4k + 7 = 0\) (still no real roots).
Let slope of \(OU\) be \(3\) and translation upwards by \(-2\) units? Let's check:
\(\frac{2k^2 + 2}{2k - 3} = m\).
If \(k = 2\), \(2k - 3 = 1\), \(2k^2 + 2 = 10\), slope is \(10\).
Let the slope of \(OU\) be \(10\).
Then \(\frac{2k^2 + 2}{2k - 3} = 10\)
\(2k^2 + 2 = 10(2k - 3)\)
\(2k^2 + 2 = 20k - 30\)
\(2k^2 - 20k + 32 = 0\)
\(k^2 - 10k + 16 = 0\)
\((k - 2)(k - 8) = 0\)
\(k = 2\) or \(k = 8\).
If the question specifies \(0 < k < 5\), then \(k = 2\).

(a) Completing the square: \(f(x) = -2(x - 2k)^2 + 2k^2 - 8\) [1M]
Coordinates of \(V = (2k, 2k^2 - 8)\) [1A]

(b)(i) \(U = (2k - 3, 2k^2 + 2)\) [2A]

(b)(ii) Setting up slope equation: \(\frac{2k^2 + 2}{2k - 3} = 10\) [1M]
\(k^2 - 10k + 16 = 0 \implies (k - 2)(k - 8) = 0\) [1M]
\(k = 2\) (or \(k = 8\)) [1A]

(a) Let \(a\) be the 1st term and \(r\) be the common ratio (where \(r > 0\)).
We have:
\(a + ar^2 = 50 \implies a(1 + r^2) = 50\) ... (1)
\(ar + ar^3 = 150 \implies ar(1 + r^2) = 150\) ... (2)

Dividing (2) by (1):
\(\frac{ar(1 + r^2)}{a(1 + r^2)} = \frac{150}{50} \implies r = 3\)

Substituting \(r = 3\) into (1):
\(a(1 + 3^2) = 50 \implies 10a = 50 \implies a = 5\)

Thus, the 1st term is 5 and the common ratio is 3.

(b) \(S_k = \frac{a(r^k - 1)}{r - 1} = \frac{5(3^k - 1)}{3 - 1} = \frac{5}{2}(3^k - 1)\).

Since \(S_k > 5 \times 10^7\):
\(\frac{5}{2}(3^k - 1) > 5 \times 10^7 \implies 3^k - 1 > 2 \times 10^7 \implies 3^k > 20000001 \implies \log(3^k) > \log(20000001) \implies k\log 3 > \log(20000001) \implies k > \frac{\log 20000001}{\log 3} \approx 15.302\)

Since \(k\) is an integer, the least integer \(k\) is 16.

(a)
For setting up \(a(1+r^2)=50\) and \(ar(1+r^2)=150\): 1M
For \(r = 3\): 1A
For \(a = 5\): 1A

(b)
For using formula for sum of geometric sequence \(\frac{5(3^k - 1)}{3 - 1}\): 1M
For \(3^k > 20000001\): 1M
For taking logarithm and solving \(k > 15.302\): 1M
For \(k = 16\): 1A

(a) Total number of possible committees of 5 selected from 14 people is \(C^5_{14} = 2002\).
The number of ways to select 3 boys from 6 and 2 girls from 8 is \(C^3_6 \times C^2_8 = 20 \times 28 = 560\).
Therefore, the required probability is:
\(P = \frac{560}{2002} = \frac{40}{143}\).

(b) The complement event is that all 5 members are boys or all 5 members are girls.
Number of committees with all boys: \(C^5_6 = 6\).
Number of committees with all girls: \(C^5_8 = 56\).
Therefore, the required probability is:
\(1 - \frac{6 + 56}{2002} = 1 - \frac{62}{2002} = \frac{1940}{2002} = \frac{970}{1001}\).

(c) The number of possible committees given that Alan is selected is \(C^4_{13} = 715\).
Among these committees, Alan is already 1 boy. To have more girls than boys in a committee of 5:
- Case 1: 1 boy (Alan only) and 4 girls (excluding Betty).
Number of ways = \(C^0_5 \times C^4_7 = 1 \times 35 = 35\).
- Case 2: 2 boys (Alan and 1 other boy) and 3 girls (excluding Betty).
Number of ways = \(C^1_5 \times C^3_7 = 5 \times 35 = 175\).

Total number of favorable committees = \(35 + 175 = 210\).
Therefore, the conditional probability is:
\(P = \frac{210}{715} = \frac{42}{143}\).

(a)
For \(\frac{C^3_6 \times C^2_8}{C^5_{14}}\): 1M
For \(\frac{40}{143}\) (or r.t. 0.280): 1A

(b)
For \(1 - \frac{C^5_6 + C^5_8}{C^5_{14}}\): 1M
For \(\frac{970}{1001}\) (or r.t. 0.969): 1A

(c)
For total number of outcomes with Alan selected = \(C^4_{13} = 715\): 1M
For numerator \(C^0_5 C^4_7 + C^1_5 C^3_7 = 210\): 1M
For \(\frac{42}{143}\) (or r.t. 0.294): 1A

(a) The straight line passes through \((3, 0)\) and \((0, -6)\).
Slope of the line \(m = \frac{0 - (-6)}{3 - 0} = 2\).

The equation of the line is:
\(\log_4 y = 2\log_4 x - 6 \implies \log_4 y = \log_4 (x^2) - 6 \implies \log_4 y - \log_4 (x^2) = -6 \implies \log_4 \left(\frac{y}{x^2}\right) = -6 \implies \frac{y}{x^2} = 4^{-6} = \frac{1}{4096} \implies y = \frac{x^2}{4096}\).

(b) From \(\log_2 x + \log_2 y = 6\), we have:
\(\log_2(xy) = 6 \implies xy = 2^6 = 64\).

Substituting \(y = \frac{x^2}{4096}\) into \(xy = 64\):
\(x \left(\frac{x^2}{4096}\right) = 64 \implies \frac{x^3}{4096} = 64 \implies x^3 = 64 \times 4096 = 262144 \implies x = \sqrt[3]{262144} = 64\).

Substituting \(x = 64\) into \(y = \frac{x^2}{4096}\):
\(y = \frac{64^2}{4096} = \frac{4096}{4096} = 1\).

Thus, the solution is \(x = 64\) and \(y = 1\).

(a)
For finding the slope \(m = 2\) or setting up \(\frac{\log_4 y - 0}{\log_4 x - 3} = \frac{0 - (-6)}{3 - 0}\): 1M
For \(\log_4 y = 2\log_4 x - 6\): 1M
For \(y = \frac{x^2}{4096}\) (or \(y = 4^{-6}x^2\)): 1A

(b)
For \(xy = 64\) or \(\log_2 x + \log_2\left(\frac{x^2}{4096}\right) = 6\): 1M
For substituting to form an equation in \(x\): \(x^3 = 262144\) or \(3\log_2 x = 18\): 1M
For \(x = 64\): 1A
For \(y = 1\): 1A

(a) Centre of \(C = \left(-\frac{-12}{2}, -\frac{6}{2}\right) = (6, -3)\).
Radius of \(C = \sqrt{6^2 + (-3)^2 - 20} = \sqrt{36 + 9 - 20} = \sqrt{25} = 5\).

(b)(i) Since \(L\) is tangent to \(C\), the perpendicular distance from the centre \((6, -3)\) to \(L\) is equal to the radius \(5\).
\(\frac{|3(6) + 4(-3) - k|}{\sqrt{3^2 + 4^2}} = 5 \implies \frac{|18 - 12 - k|}{5} = 5 \implies |6 - k| = 25\)

So \(6 - k = 25\) or \(6 - k = -25\).
\(k = -19\) or \(k = 31\).
Since \(k > 0\), we have \(k = 31\).

(ii) Let \(P\) be the point of contact.
The slope of \(L\) is \(-\frac{3}{4}\).
Since the normal through the centre \(G(6, -3)\) is perpendicular to \(L\), the slope of \(GP\) is \(\frac{4}{3}\).

The equation of the line passing through \(G\) and \(P\) is:
\(y - (-3) = \frac{4}{3}(x - 6) \implies 3(y + 3) = 4(x - 6) \implies 4x - 3y - 33 = 0\)

Solving the system of linear equations:
\(\begin{cases} 3x + 4y - 31 = 0 & \text{--- (1)} \\ 4x - 3y - 33 = 0 & \text{--- (2)} \end{cases}\)

Multiply (1) by 3: \(9x + 12y - 93 = 0\).
Multiply (2) by 4: \(16x - 12y - 132 = 0\).
Adding the two equations:
\(25x - 225 = 0 \implies x = 9\).

Substitute \(x = 9\) into (2):
\(4(9) - 3y - 33 = 0 \implies 36 - 3y - 33 = 0 \implies 3y = 3 \implies y = 1\).

Thus, the coordinates of the point of contact are \((9, 1)\).

(a)
For centre \((6, -3)\): 1A
For radius \(= 5\): 1A

(b)(i)
For using perpendicular distance formula \(\frac{|3(6) + 4(-3) - k|}{\sqrt{3^2 + 4^2}} = 5\): 1M
For \(|6 - k| = 25\): 1M
For \(k = 31\): 1A

(b)(ii)
For finding equation of normal line \(4x - 3y - 33 = 0\) (or substituting \(y = \frac{31-3x}{4}\) into the circle): 1M
For coordinates \((9, 1)\): 1A

(a) Kelvin's standard score \(z = \frac{77 - 62}{10} = \frac{15}{10} = 1.5\).

(b)(i) Original sum of test scores \(\sum x = 30 \times 62 = 1860\).
Correct sum of test scores \(\sum x_{\text{new}} = 1860 - 45 + 75 = 1890\).

Correct mean \(\bar{x}_{\text{new}} = \frac{1890}{30} = 63\) marks.

(ii) Let \(\sigma = 10\) and \(\bar{x} = 62\).
\(\sigma^2 = \frac{\sum x^2}{30} - \bar{x}^2 \implies 10^2 = \frac{\sum x^2}{30} - 62^2 \implies 100 = \frac{\sum x^2}{30} - 3844 \implies \frac{\sum x^2}{30} = 3944 \implies \sum x^2 = 30 \times 3944 = 118320\).

Correct sum of squares of test scores:
\(\sum x_{\text{new}}^2 = 118320 - 45^2 + 75^2 = 118320 - 2025 + 5625 = 121920\).

Correct variance:
\(\sigma_{\text{new}}^2 = \frac{\sum x_{\text{new}}^2}{30} - \bar{x}_{\text{new}}^2 = \frac{121920}{30} - 63^2 = 4064 - 3969 = 95\).

Correct standard deviation:
\(\sigma_{\text{new}} = \sqrt{95} \approx 9.75\) marks.

(a)
For \(\frac{77 - 62}{10}\): 1M
For \(1.5\): 1A

(b)(i)
For \(\sum x_{\text{new}} = 30 \times 62 - 45 + 75 = 1890\): 1M
For correct mean \(= 63\) marks: 1A

(b)(ii)
For original \(\sum x^2 = 30(10^2 + 62^2) = 118320\): 1M
For new \(\sum x^2 = 118320 - 45^2 + 75^2 = 121920\): 1M
For correct standard deviation \(= \sqrt{95}\) marks (or r.t. 9.75 marks): 1A