Akọwa Nkọwa

\(Y\) is on bearing \(135^\circ\), \(Z\) on bearing \(225^\circ\) from \(X\). The angle between them at \(X\) is \[\angle YXZ = 225^\circ - 135^\circ = 90^\circ,\] so triangle \(YXZ\) is right-angled at \(X\).

(a) By Pythagoras: \[|YZ| = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = \mathbf{100\,\text{km}}.\]

(b) Using coordinates: \(Y = (60\sin135^\circ, 60\cos135^\circ) = (42.43, -42.43)\) and \(Z = (80\sin225^\circ, 80\cos225^\circ) = (-56.57, -56.57).\) \[\vec{YZ} = (-98.99,\ -14.14),\] which points South-West. The bearing satisfies \(\sin B = \dfrac{-98.99}{100},\ \cos B = \dfrac{-14.14}{100}\) (third quadrant), so \[B = 180^\circ + \tan^{-1}\!\frac{98.99}{14.14} = 180^\circ + 81.9^\circ \approx \mathbf{262^\circ}.\] Bearing of \(Z\) from \(Y\) is about \(262^\circ.\)

Akọwa Nkọwa

(a) From the diagram, AOB is a straight line, so the angles on the upper side of the line sum to \(180^\circ\), and the angles on the lower side also sum to \(180^\circ\).

Upper side (angles \(AOC\) and \(COB\)):

\[3(x+y) + 45 = 180\]\[3(x+y) = 135 \Rightarrow x + y = 45 \quad (1)\]

Lower side (angles \(AOD\) and \(DOB\)):

\[(5x + y) + y = 180\]\[5x + 2y = 180 \quad (2)\]

From (1), \(y = 45 - x\). Substitute into (2):

\[5x + 2(45 - x) = 180\]\[5x + 90 - 2x = 180\]\[3x = 90 \Rightarrow x = 30\]

Then \(y = 45 - 30 = 15\).

\(x = 30,\quad y = 15\)

Check: \(3(x+y)=135^\circ\), \(135+45=180^\circ\) (upper). \(5x+y=165^\circ\), \(165+15=180^\circ\) (lower). Correct.

(b) Let the base of the pole be \(B\), and let the two points be \(P\) and \(Q\) on opposite sides. The pole \(BT = 33\ \text{m}\) is vertical.

For the point with elevation \(53^\circ\):

\[\tan 53^\circ = \frac{33}{PB} \Rightarrow PB = \frac{33}{\tan 53^\circ} = \frac{33}{1.3270} = 24.867\ \text{m}\]

For the point with elevation \(67^\circ\):

\[\tan 67^\circ = \frac{33}{QB} \Rightarrow QB = \frac{33}{\tan 67^\circ} = \frac{33}{2.3559} = 14.008\ \text{m}\]

Since the points are on opposite sides of the pole, the distance between them is:

\[PQ = PB + QB = 24.867 + 14.008 = 38.875\ \text{m}\]

Distance \(\approx 38.9\ \text{m}\) (to 3 significant figures).

Akọwa Nkọwa

(a) Using four-figure logarithm tables: \[\log(0.004592) = \bar{3}.6621 = -3 + 0.6621 = -2.3379.\] \[(0.004592)^{1/3}:\quad \frac{1}{3}(-2.3379) = -0.7793 = \bar{1}.2207.\] Antilog of \(0.2207\) is \(1.662\), so the value is \(1.662\times 10^{-1} \approx \mathbf{0.1662}.\)

(b) \(\log_{10} y + 3\log_{10} x = 2 \Rightarrow \log_{10}(yx^3) = 2 \Rightarrow yx^3 = 10^2 = 100.\) \[\mathbf{y = \frac{100}{x^3}}.\]

(c) \(3x - 2y = 21\ \ (1)\), \(4x + 5y = 5\ \ (2).\)

\((1)\times 5:\ 15x - 10y = 105.\quad (2)\times 2:\ 8x + 10y = 10.\) Adding: \(23x = 115 \Rightarrow x = 5.\)

From \((1)\): \(3(5) - 2y = 21 \Rightarrow -2y = 6 \Rightarrow y = -3.\)

Solution: \(\mathbf{x = 5,\ y = -3}.\)

Akọwa Nkọwa

(a) Height of the closed cylinder

For a cylinder closed at both ends, the total surface area is

\[TSA = 2\pi r(r+h)\]

With \(r = 3.5\text{ cm}\), \(TSA = 209\text{ cm}^2\) and \(\pi = \tfrac{22}{7}\):

\[2\times\frac{22}{7}\times 3.5\,(3.5+h) = 209\]

Now \(2\times\dfrac{22}{7}\times 3.5 = 22\), so

\[22(3.5+h) = 209 \;\Rightarrow\; 3.5+h = \frac{209}{22} = 9.5\]\[h = 9.5 - 3.5 = 6\text{ cm}\]

The height of the cylinder is \(6\text{ cm}\).

(b) Circle with tangent ABC at B

From the diagram, \(EB\) is a diameter (it passes through the centre O). The given angles are the inscribed angle \(\angle BDF = 66^\circ\) at D and the tangent-chord angle \(\angle DBC = 57^\circ\) at B. G is the point where chord \(FD\) crosses the diameter \(EB\).

Finding the arcs

The inscribed angle \(\angle BDF\) stands on arc \(BF\) (not containing D), so

\[\text{arc }BF = 2\times 66^\circ = 132^\circ\]

The tangent-chord angle \(\angle DBC\) equals half the intercepted arc \(BD\):

\[\text{arc }BD = 2\times 57^\circ = 114^\circ\]

Since \(EB\) is a diameter, each semicircle is \(180^\circ\):

\[\text{arc }EF = 180^\circ - \text{arc }BF = 180^\circ - 132^\circ = 48^\circ\]\[\text{arc }ED = 180^\circ - \text{arc }BD = 180^\circ - 114^\circ = 66^\circ\]

(Check: \(114+66+48+132 = 360^\circ\). Correct.)

(i) \(\angle EBF\)

\(\angle EBF\) is the inscribed angle at B standing on arc \(EF\):

\[\angle EBF = \tfrac{1}{2}\,\text{arc }EF = \tfrac{1}{2}\times 48^\circ = 24^\circ\]

(ii) \(\angle BGF\)

G is the intersection of chords \(EB\) and \(FD\). The angle between two chords equals half the sum of the two arcs it intercepts (arc \(BF\) and the vertically opposite arc \(ED\)):

\[\angle BGF = \tfrac{1}{2}\big(\text{arc }BF + \text{arc }ED\big) = \tfrac{1}{2}(132^\circ + 66^\circ) = \tfrac{1}{2}\times 198^\circ = 99^\circ\]

Therefore \(\angle EBF = 24^\circ\) and \(\angle BGF = 99^\circ\).

Akọwa Nkọwa

(a) \[\frac{3.69\times 10^{5}}{1.64\times 10^{-3}} = \frac{3.69}{1.64}\times 10^{5-(-3)} = 2.25\times 10^{8}.\] Answer in standard form: \(\mathbf{2.25\times 10^{8}}.\)

(b) Simple interest for one year. Bank A: \(\dfrac{20000\times y\times 1}{100} = 200y.\) Bank B: \(\dfrac{25000\times 1.5y\times 1}{100} = 375y.\)

Total interest: \[200y + 375y = 4600 \Rightarrow 575y = 4600 \Rightarrow y = \frac{4600}{575} = \mathbf{8}.\]

Akọwa Nkọwa

(a) Arithmetic progression.

Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).

Given \(T_3 = -9\) and \(T_8 = 26\):

\[a + 2d = -9 \quad\text{...(1)}\]\[a + 7d = 26 \quad\text{...(2)}\]

(i) Common difference. Subtract (1) from (2):

\[5d = 35 \quad\Rightarrow\quad d = 7\]

(ii) First term. Substitute \(d = 7\) into (1):

\[a + 2(7) = -9 \quad\Rightarrow\quad a + 14 = -9 \quad\Rightarrow\quad a = -23\]

Common difference \(= 7\); first term \(= -23\).

(b) Area of the trapezium PQZY.

From the diagram, \(PQ \parallel YZ\), with \(|XP| = 2\text{ cm}\), \(|PY| = 3\text{ cm}\), \(|PQ| = 6\text{ cm}\) and area of \(\triangle XPQ = 24\text{ cm}^2\). P lies on XY and Q lies on XZ.

Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\) (equal corresponding angles). The ratio of corresponding sides along XY is:

\[\frac{XP}{XY} = \frac{XP}{XP + PY} = \frac{2}{2 + 3} = \frac{2}{5}\]

For similar triangles the ratio of areas is the square of the ratio of sides:

\[\frac{\text{Area }\triangle XPQ}{\text{Area }\triangle XYZ} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\]

Therefore:

\[\text{Area }\triangle XYZ = 24 \times \frac{25}{4} = 150 \text{ cm}^2\]

The trapezium PQZY is what remains when \(\triangle XPQ\) is removed from \(\triangle XYZ\):

\[\text{Area of trapezium PQZY} = 150 - 24 = 126 \text{ cm}^2\]

Area of trapezium PQZY \(= 126\text{ cm}^2\).

Akọwa Nkọwa

Take the class mid-values \(x\) and build a working table (\(N=1+5+10+9+5=30\)):

Class	Mid-value x	f	fx	fx²
0 - 4	2	1	2	4
5 - 9	7	5	35	245
10 - 14	12	10	120	1440
15 - 19	17	9	153	2601
20 - 24	22	5	110	2420
Total		30	420	6710

(a) Mean.

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{420}{30}=14.\]

(b) Standard deviation.

\[s=\sqrt{\frac{\sum fx^{2}}{N}-\bar{x}^{2}}=\sqrt{\frac{6710}{30}-14^{2}}=\sqrt{223.67-196}=\sqrt{27.67}\approx 5.26.\]

The mean number of absentees is 14 and the standard deviation is 5.26 (to 2 d.p.).

Akọwa Nkọwa

(a) Simplify \(\dfrac{x^{2}-y^{2}}{3x+3y}\).

Factorise the numerator as a difference of two squares and take out the common factor in the denominator:

\[\frac{x^{2}-y^{2}}{3x+3y} = \frac{(x-y)(x+y)}{3(x+y)} = \frac{x-y}{3}.\]

(b) Reading the diagram. \(PQRS\) is a rectangle with \(P\) top-left, \(Q\) top-right, \(R\) bottom-right and \(S\) bottom-left. \(K\) lies on the base \(SR\) with \(|SK| = |KR|\) (so \(K\) is the midpoint of \(SR\)). The line \(PK = 15\text{ cm}\) is drawn, and \(\angle PKS = 30^\circ\). Triangle \(PSK\) has its right angle at the rectangle corner \(S\).

(i) Find \(|PS|\). In right-angled triangle \(PSK\), \(PS\) is opposite the \(30^\circ\) angle at \(K\) and \(PK\) is the hypotenuse:

\[|PS| = PK\sin 30^\circ = 15\times 0.5 = 7.50\text{ cm}.\]

(ii) Find \(|SK|\). \(SK\) is adjacent to the \(30^\circ\) angle:

\[|SK| = PK\cos 30^\circ = 15\times 0.8660 = 12.99 \approx 13.0\text{ cm}.\]

(iii) Area of the shaded portion. The unshaded region is triangle \(PSK\); the shaded portion is the rest of the rectangle. Since \(K\) is the midpoint of \(SR\), the full base is \(SR = 2|SK| = 2\times 12.99 = 25.98\text{ cm}\), and the height is \(|PS| = 7.50\text{ cm}\).

\[\text{Area of rectangle} = 25.98\times 7.50 = 194.85\text{ cm}^2,\]

\[\text{Area of triangle } PSK = \tfrac{1}{2}\times |SK|\times |PS| = \tfrac{1}{2}\times 12.99\times 7.50 = 48.71\text{ cm}^2.\]

\[\text{Shaded area} = 194.85 - 48.71 = 146.14 \approx 146\text{ cm}^2.\]

Answers: (a) \(\dfrac{x-y}{3}\); (b)(i) \(|PS| = 7.50\text{ cm}\); (ii) \(|SK| = 13.0\text{ cm}\); (iii) shaded area \(= 146\text{ cm}^2\) (3 s.f.).

Akọwa Nkọwa

Out of \(24\) apples, \(6\) are bad, so \(18\) are good. Because the drawing is with replacement, each draw is independent with \[P(\text{good}) = \frac{18}{24} = \frac{3}{4},\qquad P(\text{bad}) = \frac{6}{24} = \frac{1}{4}.\]

(a) First two good, third bad: \[\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4} = \mathbf{\frac{9}{64}}.\]

(b) All three bad: \[\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4} = \mathbf{\frac{1}{64}}.\]

(c) All three good: \[\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4} = \mathbf{\frac{27}{64}}.\]

Akọwa Nkọwa

(a) Numerator: \(3\tfrac{1}{12} + \tfrac{7}{8} = \tfrac{37}{12} + \tfrac{7}{8} = \tfrac{74}{24} + \tfrac{21}{24} = \tfrac{95}{24}.\)

Denominator: \(2\tfrac{1}{4} - \tfrac{1}{6} = \tfrac{9}{4} - \tfrac{1}{6} = \tfrac{27}{12} - \tfrac{2}{12} = \tfrac{25}{12}.\)

\[\frac{95/24}{25/12} = \frac{95}{24}\times\frac{12}{25} = \frac{95}{2\times 25} = \frac{95}{50} = \frac{19}{10} = \mathbf{1\tfrac{9}{10}}.\]

(b)(i) \(p = \dfrac{m}{2} - \dfrac{n^2}{5m}.\) Multiply through by \(5m\): \[5mp = \frac{5m^2}{2} - n^2 \Rightarrow n^2 = \frac{5m^2}{2} - 5mp \Rightarrow \mathbf{n = \sqrt{\frac{5m^2}{2} - 5mp}}.\]

(b)(ii) With \(p = 14,\ m = -8\): \[n^2 = \frac{5(-8)^2}{2} - 5(-8)(14) = \frac{320}{2} + 560 = 160 + 560 = 720.\] \[n = \sqrt{720} = \mathbf{26.8}\ (\text{to 3 s.f.}).\]

Akọwa Nkọwa

(a) Factorise numerator and denominator: \[x^2 - 8x + 16 = (x-4)^2,\qquad x^2 - 7x + 12 = (x-3)(x-4).\] \[\frac{(x-4)^2}{(x-3)(x-4)} = \frac{x-4}{x-3}\quad (x\neq 4).\]

(b) If \(\tfrac{1}{2}, \tfrac{1}{x}, \tfrac{1}{3}\) are successive terms of an A.P., the common difference is constant: \[\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}.\] The left side is \(\dfrac{2 - x}{2x}\) and the right side is \(\dfrac{x - 3}{3x}\), so \[\frac{2 - x}{2x} = \frac{x - 3}{3x}.\] Multiplying both sides by \(3x\) and dividing by \((x-3)\): \[\frac{2 - x}{x - 3} = \frac{2x}{3x} = \frac{2}{3}.\] Hence \(\dfrac{2 - x}{x - 3} = \dfrac{2}{3}\), as required.

Akọwa Nkọwa

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Construction of Triangle PQR and the Required Loci

(a)(i) Construct triangle PQR

1. Using a ruler, draw a line segment PQ = 10 cm.
2. At point Q, construct a perpendicular to PQ using a ruler and compasses. This gives a line making an angle of 90° with PQ.
3. Using the compasses set to 7 cm, measure from Q along the perpendicular line and mark the point R.
4. Join P to R. The required triangle PQR is formed.

Therefore, PQ = 10 cm, QR = 7 cm, and ∠PQR = 90°.

(a)(ii) Construct locus l<sub>1</sub>

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. With centres Q and R, draw arcs of equal radius greater than half of QR.
2. Let the arcs intersect at two points.
3. Join the two intersection points with a straight line.

This line is labelled l<sub>1</sub>.

(a)(iii) Construct locus l<sub>2</sub>

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With centres P and Q, draw arcs of equal radius greater than half of PQ.
2. Join the two points where the arcs intersect.

This line is labelled l<sub>2</sub>.

(b) Locate point O

Point O, which is equidistant from P, Q, and R, is the point of intersection of l<sub>1</sub> and l<sub>2</sub>.

Since triangle PQR is right-angled at Q, the point O is also the midpoint of the hypotenuse PR. It is the centre of the circle passing through P, Q, and R.

```

Akọwa Nkọwa

Perimeter of the tunnel cross-section.

The cross-section is made up of the straight base \(AB=100\text{ m}\) together with the arc that domes over the top. The arc has radius \(r=56\text{ m}\), and the diagram shows the arc is the major arc (it rises above and beyond a semicircle).

Step 1: Angle subtended by the chord at the centre.

Let \(O\) be the centre and let the perpendicular from \(O\) bisect the chord \(AB\). Half the chord is \(50\text{ m}\). If \(\theta\) is half the central angle of the minor arc:

\[\sin\theta=\frac{50}{56}=0.8929\Rightarrow\theta=63.26^{\circ}\]

So the minor-arc central angle is:

\[2\theta=126.52^{\circ}\]

Step 2: Reflex angle for the major (tunnel) arc.

\[\text{Reflex angle}=360^{\circ}-126.52^{\circ}=233.48^{\circ}\]

Step 3: Length of the major arc.

\[\text{Arc}=\frac{233.48}{360}\times 2\pi r=\frac{233.48}{360}\times 2\times\frac{22}{7}\times 56\]

\[=\frac{233.48}{360}\times 352=0.6486\times 352=228.3\text{ m}\]

Step 4: Perimeter.

\[\text{Perimeter}=AB+\text{arc}=100+228.3=328.3\text{ m}\]

Correct to the nearest metre, the perimeter of the cross-section is \(328\text{ m}\).