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(a) \[\tfrac12\log_{10}25-2\log_{10}3+\log_{10}18=\log_{10}5-\log_{10}9+\log_{10}18=\log_{10}\!\left(\frac{5\times18}{9}\right)=\log_{10}10=1.\]

(b) \(123_{y}=1\cdot y^{2}+2\cdot y+3\). So \[y^{2}+2y+3=83\Rightarrow y^{2}+2y-80=0\Rightarrow(y+10)(y-8)=0.\] Since a base must be positive, \(y=8\).

(c) \[\frac{9^{2x-3}}{3^{x+3}}=1.\] Write \(9=3^{2}\): numerator \(=3^{2(2x-3)}=3^{4x-6}\). Then \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]

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Let \(P(A)=\tfrac23,\;P(K)=\tfrac58,\;P(F)=\tfrac34\), so the failure probabilities are \(\tfrac13,\tfrac38,\tfrac14\). The three events are independent, so multiply.

(a) All three pass: \[\frac23\times\frac58\times\frac34=\frac{30}{96}=\frac{5}{16}.\]

(b) None passes: \[\frac13\times\frac38\times\frac14=\frac{3}{96}=\frac{1}{32}.\]

(c) Ade and Kujo only (Fati fails): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]

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(a) Frequency table. Sorting the 30 lengths into the given class intervals and tallying gives:

Length (cm)	Tally	Frequency \(f\)	Midpoint \(x\)	\(fx\)
45 - 49	|	1	47	47
50 - 54	|||	3	52	156
55 - 59	|||| |	6	57	342
60 - 64	|||| ||	7	62	434
65 - 69	|||| |	6	67	402
70 - 74	||||	4	72	288
75 - 79	||	2	77	154
80 - 84	|	1	82	82
Total		30		1905

(b) Histogram of the distribution. The bars are drawn against the class boundaries \(44.5, 49.5, 54.5, 59.5, 64.5, 69.5, 74.5, 79.5, 84.5\) on the horizontal axis, with heights equal to the class frequencies. Since all class widths are equal, the bar heights are simply the frequencies.

(c) Mean of the distribution.

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1905}{30}=63.5\text{ cm}.\]

(d) Probability of selecting a rod in the modal class. The modal class is the one with the highest frequency, \(60 - 64\) (frequency \(7\)). The number of rods in this class is \(7\) out of \(30\), so

\[P(\text{modal class})=\frac{7}{30}.\]

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(a) AP with \(a=3,\;d=4,\;n=28\). \[S_{28}=\frac{n}{2}\bigl(2a+(n-1)d\bigr)=14\bigl(6+27\times4\bigr)=14\times114=1596.\]

(b) \(x=\dfrac{2m}{1-m^{2}}=\dfrac{2m}{(1-m)(1+m)}\) and \(y=\dfrac{2m}{1+m}\). \[2x-y=\frac{4m}{(1-m)(1+m)}-\frac{2m}{1+m}=\frac{4m-2m(1-m)}{(1-m)(1+m)}=\frac{2m+2m^{2}}{(1-m)(1+m)}=\frac{2m(1+m)}{(1-m)(1+m)}=\frac{2m}{1-m}.\]

(c) Interior angles of a pentagon sum to \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]

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We are given the universal set \(\varepsilon = \{1, 2, 3, \dots, 15\}\) and the three subsets

\[A = \{1, 3, 5, 7, 9, 11\}, \quad B = \{2, 3, 5, 7, 11, 15\}, \quad C = \{3, 6, 9, 12, 15\}.\]

(a) Building the Venn diagram. Work out each region before drawing so that every element sits in exactly one place.

• In all three sets: \(A \cap B \cap C = \{3\}\).
• In \(A\) and \(B\) only: \(\{5, 7, 11\}\).
• In \(A\) and \(C\) only: \(\{9\}\).
• In \(B\) and \(C\) only: \(\{15\}\).
• In \(A\) only: \(\{1\}\).
• In \(B\) only: \(\{2\}\).
• In \(C\) only: \(\{6, 12\}\).
• Outside all three sets: \(\varepsilon \setminus (A \cup B \cup C) = \{4, 8, 10, 13, 14\}\).

These give the Venn diagram below.

(b) Reading answers from the diagram.

(i) \(A'\) is everything outside circle \(A\), so \(C \cap A'\) is the part of circle \(C\) that lies outside \(A\). Reading those regions of \(C\) gives the C-only part \(\{6, 12\}\) together with the \(B\cap C\)-only part \(\{15\}\):

\[C \cap A' = \{6, 12, 15\}.\]

(ii) First form \(B \cup C\) (all of circles \(B\) and \(C\)):

\[B \cup C = \{2, 3, 5, 6, 7, 9, 11, 12, 15\}.\]

Now intersect with \(A'\), keeping only the members of \(B \cup C\) that lie outside circle \(A\). Discarding \(3, 5, 7, 9, 11\) (which are inside \(A\)) leaves

\[A' \cap (B \cup C) = \{2, 6, 12, 15\}.\]

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(a) Evaluate \(\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3}-\dfrac{\sqrt{12}}{6}\right)\) without tables.

First simplify each surd. \(\dfrac{3}{\sqrt3}=\dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3.\)

Inside the bracket: \(\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) and \(\dfrac{\sqrt{12}}{6}=\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}.\)

\[\frac{2\sqrt3}{3}-\frac{\sqrt3}{3}=\frac{\sqrt3}{3}.\]

Therefore

\[\sqrt3\times\frac{\sqrt3}{3}=\frac{3}{3}=\boxed{1.}\]

(b) Circle, centre \(O\); \(AB\) produced to \(E\), \(\widehat{ACB}=49^\circ\), \(\widehat{CBE}=68^\circ\).

Since \(A,B,E\) are collinear, \(\widehat{ABC}\) and \(\widehat{CBE}\) are angles on a straight line:

\[\widehat{ABC}=180^\circ-68^\circ=112^\circ.\]

In \(\triangle ABC\):

\[\widehat{BAC}=180^\circ-\widehat{ABC}-\widehat{ACB}=180^\circ-112^\circ-49^\circ=19^\circ.\]

(i) Interior angle \(AOC\). \(\widehat{ABC}=112^\circ\) is the angle at the circumference standing on chord \(AC\); it subtends the major arc \(AC\), whose central angle (reflex \(AOC\)) is \(2\times112^\circ=224^\circ.\) Hence the interior (non-reflex) angle is

\[\widehat{AOC}=360^\circ-224^\circ=\boxed{136^\circ.}\]

(ii) Angle \(BOC\). \(\widehat{BAC}=19^\circ\) at the circumference subtends arc \(BC\); the angle at the centre on the same arc is twice as large:

\[\widehat{BOC}=2\times19^\circ=\boxed{38^\circ.}\]

(Check: \(\widehat{ACB}=49^\circ\Rightarrow\widehat{AOB}=98^\circ,\) and \(98^\circ+38^\circ+224^\circ=360^\circ.\))

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The relation illustrated is a downward parabola. Reading from the graph, the curve cuts the y-axis at \(y = 6\), cuts the x-axis at \(x = -1.5\) and \(x = 2\), and turns over near \(x = 0.25\). The annotated graph below shows these key features, which we read off to answer each part.

(a) Roots of \(ax^{2}+bx+c=0\)

The roots are where the curve crosses the x-axis (\(y=0\)). From the graph these are:

\[ x = -1.5 \quad \text{or} \quad x = 2. \]

(b) Values of a, b and c, and the relation

The curve cuts the positive y-axis at \(6\), so the intercept gives

\[ c = 6. \]

Substitute the coordinates of the two x-axis crossings, \(Q(2,\,0)\) and \(P(-1.5,\,0)\), into \(y = ax^{2}+bx+c\) with \(y = 0\) and \(c = 6\).

At \(x = 2\):

\[ a(2)^{2} + b(2) + 6 = 0 \implies 4a + 2b = -6 \quad \text{...(1)} \]

At \(x = -1.5\):

\[ a(-1.5)^{2} + b(-1.5) + 6 = 0 \implies \tfrac{9a}{4} - \tfrac{3b}{2} = -6 \implies 9a - 6b = -24 \quad \text{...(2)} \]

Multiply (1) by 3:

\[ 12a + 6b = -18 \quad \text{...(1a)} \]

Add (1a) and (2):

\[ 21a = -42 \implies a = -2. \]

Substitute \(a = -2\) into (1):

\[ 4(-2) + 2b = -6 \implies 2b = 2 \implies b = 1. \]

Therefore \((a,\,b,\,c) = (-2,\,1,\,6)\), and the relation illustrated by the graph is

\[ \boxed{\,y = -2x^{2} + x + 6\,}. \]

(c) Maximum value of y and the corresponding x

The maximum occurs at the turning point. Completing the derivative (or reading the vertex from the graph):

\[ \frac{dy}{dx} = -4x + 1 = 0 \implies x = \tfrac{1}{4} = 0.25. \]

Substituting \(x = 0.25\):

\[ y_{\max} = -2(0.25)^{2} + 0.25 + 6 = -0.125 + 0.25 + 6 = 6.125. \]

So the maximum value is \(y_{\max} \approx 6.13\) (about \(6.1\)), occurring at \(x = 0.25\), which agrees with the peak read from the graph.

(d) Values of x when \(y = 2\)

Draw the horizontal line \(y = 2\) and read where it meets the curve, or solve:

\[ -2x^{2} + x + 6 = 2 \implies -2x^{2} + x + 4 = 0 \implies 2x^{2} - x - 4 = 0. \]

Using the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(2)(-4)}}{2(2)} = \frac{1 \pm \sqrt{33}}{4} = \frac{1 \pm 5.745}{4}. \] \[ x = \frac{1 + 5.745}{4} = 1.69 \quad \text{or} \quad x = \frac{1 - 5.745}{4} = -1.19. \]

So when \(y = 2\), \(x \approx -1.2\) or \(x \approx 1.7\), matching the two points where the line \(y = 2\) cuts the curve on the graph.

Answer Details

(a) Simplify \((2a + b)^2 - (b - 2a)^2\). Note that \((b - 2a)^2 = (2a - b)^2\). This is a difference of two squares, \(P^2 - R^2 = (P + R)(P - R)\), with \(P = 2a + b\) and \(R = 2a - b\):

\[(2a + b)^2 - (2a - b)^2 = \big[(2a + b) + (2a - b)\big]\big[(2a + b) - (2a - b)\big].\]

\[= (4a)(2b) = 8ab.\]

(b) Given \(S = K\sqrt{m^2 + n^2}\).

(i) Make \(m\) the subject.

\[\frac{S}{K} = \sqrt{m^2 + n^2} \Rightarrow \left(\frac{S}{K}\right)^2 = m^2 + n^2.\]

\[m^2 = \left(\frac{S}{K}\right)^2 - n^2 \Rightarrow m = \sqrt{\left(\frac{S}{K}\right)^2 - n^2}.\]

(ii) With \(S = 12.2, K = 0.02, n = 1.1\):

\[\frac{S}{K} = \frac{12.2}{0.02} = 610, \qquad \left(\frac{S}{K}\right)^2 = 372100, \qquad n^2 = 1.21.\]

\[m = \sqrt{372100 - 1.21} = \sqrt{372098.79} = 609.999 \approx 610.\]

To the nearest whole number, the positive value of \(m = \mathbf{610}\).

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```html

Geometrical Construction

Given: Construct triangle XYZ such that

• |XY| = 8 cm,
• ∠YXZ = 60°,
• ∠XYZ = 30°.

(a) Construction steps (ruler and compasses only).

1. Draw a line segment XY = 8 cm.
2. At X, construct an angle of 60° with XY. This may be done by constructing an equilateral triangle on XY, since every angle in an equilateral triangle is 60°.
3. At Y, construct an angle of 30° with YX. First construct a 60° angle at Y, then bisect it using the standard compass construction for an angle bisector.
4. Let the 60° ray from X and the 30° ray from Y meet at Z. This gives triangle XYZ.
5. Construct the perpendicular from Z to XY. Let it meet XY at T. Hence, ZT ⟂ XY.
6. The locus of points equidistant from the lines ZY and XY is the angle bisector of ∠ZYX. Construct the internal angle bisector of the 30° angle at Y and call it l₁.
7. Let l₁ meet ZT at S.

(b) Finding |ST|.

Since ∠XYZ = 30° and l<sub>1</sub> bisects this angle,

∠SYT = 15°.

Also, triangle ZXY is right-angled at Z, because

∠XZY = 180° − 60° − 30° = 90°.

In the 30°–60°–90° triangle XYZ, with hypotenuse XY = 8 cm:

XZ = 4 cm,    YZ = 4√3 cm.

The foot of the perpendicular is T, and

YT = 6 cm.

In right triangle YST:

tan 15° = ST / YT.

ST = 6 tan 15° = 6(2 − √3).

On a ruler-and-compass construction, the measured value should be approximately 1.6 cm.

```

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(a) Let width \(=w\). Then length \(=2w\) and depth \(=w-4\). An open tank has a base plus four sides: \[\text{Area}=\underbrace{(2w)(w)}_{\text{base}}+\underbrace{2(2w)(w-4)+2(w)(w-4)}_{\text{four sides}}=2w^{2}+6w(w-4).\] Set equal to \(1440\): \[2w^{2}+6w^{2}-24w=1440\Rightarrow 8w^{2}-24w-1440=0\Rightarrow w^{2}-3w-180=0.\] \[(w-15)(w+12)=0\Rightarrow w=15.\] Length \(=2w=\) 30 m.

(b) First arrangement: \(\dfrac{3000x}{100}+\dfrac{2000y}{100}=640\Rightarrow 3x+2y=64.\) Swapped: \(\dfrac{2000x}{100}+\dfrac{3000y}{100}=660\Rightarrow 2x+3y=66.\) Solving: from \(9x+6y=192\) and \(4x+6y=132\), subtract \(\Rightarrow 5x=60\Rightarrow x=12\); then \(3(12)+2y=64\Rightarrow y=14\). So x = 12, y = 14.

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From the diagram: \(PQ=8\text{ m}\), \(QR=13\text{ m}\), the bearing of Q from P is \(050^\circ\), and the bearing of R from Q is \(130^\circ\).

(a)(i) Length PR. Find \(\angle PQR\). The bearing of P from Q is the back-bearing of \(050^\circ\):

\[050^\circ+180^\circ=230^\circ\]

The bearing of R from Q is \(130^\circ\), so

\[\angle PQR=230^\circ-130^\circ=100^\circ\]

Apply the cosine rule to triangle PQR:

\[PR^2=PQ^2+QR^2-2\,(PQ)(QR)\cos(\angle PQR)\]\[PR^2=8^2+13^2-2(8)(13)\cos100^\circ\]\[PR^2=64+169-208(-0.17365)=269.12\]\[PR=16.4\text{ m (3 s.f.)}\]

(a)(ii) Bearing of R from P. Use the sine rule to find \(\angle QPR\):

\[\frac{\sin\angle QPR}{QR}=\frac{\sin\angle PQR}{PR}\]\[\sin\angle QPR=\frac{13\sin100^\circ}{16.405}=\frac{12.803}{16.405}=0.78040\]\[\angle QPR=51.3^\circ\]

R lies clockwise of Q as seen from P, so the bearing of R from P is

\[050^\circ+51.3^\circ=101^\circ\ (\text{3 s.f.})\]

(b) Shortest distance from Q to PR, and the area. The area of triangle PQR is

\[\text{Area}=\tfrac12(PQ)(QR)\sin(\angle PQR)=\tfrac12(8)(13)\sin100^\circ\]\[\text{Area}=52(0.98481)=51.2\text{ m}^2\ (\text{3 s.f.})\]

The shortest distance from Q to PR is the perpendicular height h onto base PR. Using \(\text{Area}=\tfrac12(PR)(h)\):

\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]