WAEC SSCE - General Mathematics - 2018 (Essay)

(a) Finding \(x\).

Reading the diagram. \(Q\), \(R\) and \(S\) lie on a circle with centre \(O\). The straight line \(Q\,O\,S\) passes through the centre, so \(QS\) is a diameter, and it is produced beyond \(S\) to the external point \(T\). The angle at \(Q\) is \(x\), the right-angle mark is at \(R\), and the exterior angle at \(S\) (between \(SR\) and \(ST\)) is \((3x + 15)^\circ\).

Angle in a semicircle. Since \(QS\) is a diameter, the angle it subtends at \(R\) is a right angle:

Exterior-angle theorem in \(\triangle QRS\). The line \(Q\,S\,T\) is straight, so \((3x+15)^\circ\) is the exterior angle at \(S\); it equals the sum of the two remote interior angles \(\angle Q\) and \(\angle R\):

(The follow-up part (ii) is not legible in the source text, so only \(x\) is evaluated here.)

(b) Number-base equation: \(2N4_{\text{seven}} = 15N_{\text{nine}}\).

Expand each numeral in powers of its base, treating \(N\) as an unknown digit.

Left side (base 7):

Right side (base 9):

Equate:

Check: \(N = 4\) is a valid digit in both base 7 and base 9. Left \(= 102 + 28 = 130\); right \(= 126 + 4 = 130\). Both equal \(130_{\text{ten}}\). Correct.

In triangle PQR the three sides are marked as follows: \(|PQ| = 2y + x\), \(|QR| = 4y\) and \(|PR| = 6y - 2x + 1\). Since PQR is isosceles the two slant sides are equal, \(|QR| = |PR|\).

(a) Values of x and y

Perimeter equation:

\[ (2y + x) + 4y + (6y - 2x + 1) = 28 \]

\[ 12y - x + 1 = 28 \quad\Rightarrow\quad 12y - x = 27 \quad\text{...(1)} \]

Isosceles equation \((|QR| = |PR|)\):

\[ 4y = 6y - 2x + 1 \quad\Rightarrow\quad 2x - 2y = 1 \quad\text{...(2)} \]

From (1), \(x = 12y - 27\). Substituting into (2):

\[ 2(12y - 27) - 2y = 1 \Rightarrow 24y - 54 - 2y = 1 \Rightarrow 22y = 55 \]

\[ y = \frac{55}{22} = \frac{5}{2} = 2\tfrac{1}{2} \]

\[ x = 12\left(\tfrac{5}{2}\right) - 27 = 30 - 27 = 3 \]

Hence x = 3 and y = 2\(\tfrac{1}{2}\) (i.e. 2.5).

(b) Lengths of the sides

• \(|PQ| = 2y + x = 2\left(\tfrac{5}{2}\right) + 3 = 5 + 3 = \textbf{8 cm}\)
• \(|QR| = 4y = 4\left(\tfrac{5}{2}\right) = \textbf{10 cm}\)
• \(|PR| = 6y - 2x + 1 = 6\left(\tfrac{5}{2}\right) - 2(3) + 1 = 15 - 6 + 1 = \textbf{10 cm}\)

Check: \(8 + 10 + 10 = 28\) cm, which agrees with the given perimeter, and \(|QR| = |PR| = 10\) cm confirms the triangle is isosceles.

From the diagram the track is a rectangle capped by two semicircular ends. The two straight (parallel) sides are each 60 m long, and each semicircular end has diameter 120 m (the marked vertical distance between the two straights).

(a) Distance for one lap, then two laps.

The two semicircular ends together form one complete circle of diameter \(d = 120\) m. So one lap consists of the two straights plus one full circle:

\[ \text{One lap} = 2(60) + \pi d = 120 + \frac{22}{7}\times 120. \]

\[ \frac{22}{7}\times 120 = \frac{2640}{7} = 377.14\text{ m}. \]

\[ \text{One lap} = 120 + \frac{2640}{7} = \frac{840 + 2640}{7} = \frac{3480}{7} = 497.14\text{ m}. \]

Running round the track two times:

\[ D = 2 \times \frac{3480}{7} = \frac{6960}{7} = 994.29\text{ m} \;(\text{to 2 d.p.}). \]

Distance covered \(\approx 994.29\) m.

(b) Speed in km/h.

Time \(= 200\) s. First find the speed in metres per second:

\[ \text{Speed} = \frac{D}{t} = \frac{6960/7}{200} = \frac{6960}{1400} = 4.971\text{ m/s}. \]

Convert to km/h (multiply by \(\tfrac{3600}{1000} = 3.6\)):

\[ \text{Speed} = 4.971 \times 3.6 = 17.90\text{ km/h}. \]

Alternatively: \(D = \dfrac{6960}{7000}\) km \(= 0.99429\) km, \(t = \dfrac{200}{3600}\) h \(= \dfrac{1}{18}\) h, so speed \(= 0.99429 \times 18 = 17.90\) km/h.

Speed \(\approx 17.9\) km/h.

(a) The diagonals of a rhombus bisect each other at right angles. Half-diagonals are

A side of the rhombus is the hypotenuse of a right triangle with these legs:

Perimeter \(= 4s = 4 \times 6.9016 = 27.6\ \text{cm (to 1 d.p.).}\)

(b) Given \(\sin x = \tfrac{3}{5}\) with \(0^\circ < x < 90^\circ\), this is a 3-4-5 right triangle, so

Then

(a) The mean of five numbers is their sum divided by 5.

Since the mean of \(m, n, s, p, q\) is 12, the sum is \(5 \times 12 = 60\).

Adding the constants changes the sum by \(+4 - 3 + 6 - 2 + 8 = +13\).

New sum \(= 60 + 13 = 73\), so new mean \(= \dfrac{73}{5} = 14.6\).

(b) The 25th percentile (15 years) and the 75th percentile (65 years) enclose the middle 50% of the data.

Number of people between 15 and 65 years \(= 50\% \times 500 = 250\).

1. Length of fence within reach

The donkey can reach every point 17 m from the post, so its path forms a circle of radius 17 m. The fence is 15 m from the post. This makes a right-angled triangle from the post to the nearest point on the fence and to one end of the reachable section.

Using Pythagoras’ theorem on one half of the reachable fence length:

The total length of fence is twice this value:

2. Right square pyramid

The reference answer treats the given 32 cm as the sloping edge from the vertex \(V\) to a corner of the square base, for example \(VP\). In a square, the centre \(O\) is halfway along a diagonal, so \(OP\) is half the diagonal. The vertical height is \(VO\).

Height

First find the diagonal \(PR\) of the square base:

Since \(O\) is the centre of the square,

Triangle \(VOP\) is right-angled at \(O\), so:

Volume

The unit for volume must be \(\text{cm}^3\), not \(\text{cm}^2\). Also, distinguish between half a base edge, \(7.5\) cm, and the distance from the centre to a corner, \(OP=\frac{15\sqrt2}{2}\) cm. The latter is required here because the 32 cm line is \(VP\), ending at a corner of the base.

From the diagram, in triangle PSR the point T lies on PS and Q lies on PR, with TQ parallel to SR (shown by the arrows). Given \(|PT| = 4\text{ cm}\), \(|TS| = 6\text{ cm}\), \(|PQ| = 6\text{ cm}\) and \(\angle SPR = 30^\circ\).

Preliminary - use the parallel lines. Because \(TQ \parallel SR\), triangles PTQ and PSR are similar (equiangular). The full side PS is

\[|PS| = |PT| + |TS| = 4 + 6 = 10\text{ cm}.\]

The ratio of similarity is \(\dfrac{PT}{PS} = \dfrac{4}{10} = 0.4\), so

\[\frac{PQ}{PR} = 0.4 \;\Rightarrow\; |PR| = \frac{PQ}{0.4} = \frac{6}{0.4} = 15\text{ cm}.\]

(a) Finding |SR|. In triangle PSR use the Cosine Rule with \(|PS| = 10\), \(|PR| = 15\), \(\angle P = 30^\circ\):

\[|SR|^2 = |PS|^2 + |PR|^2 - 2|PS||PR|\cos 30^\circ.\]

\[|SR|^2 = 10^2 + 15^2 - 2(10)(15)\cos 30^\circ = 100 + 225 - 300(0.8660).\]

\[|SR|^2 = 325 - 259.8 = 65.2.\]

\[|SR| = \sqrt{65.2} = 8.07\ldots \approx 8\text{ cm}.\]

(b) Area of TQRS. TQRS is the trapezium left when the small triangle PTQ is removed from triangle PSR:

\[\text{Area of } TQRS = \text{Area of } \triangle PSR - \text{Area of } \triangle PTQ.\]

Using \(\text{Area} = \tfrac{1}{2}ab\sin C\) with the common angle \(30^\circ\):

\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]

\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]

\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]

\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]

(a) Lamin's cost is \(N300\). Selling to Bola at \(x\%\) profit means Bola pays \(300\left(1 + \frac{x}{100}\right)\).

Bola resells at another \(x\%\) profit, so James pays \(300\left(1 + \frac{x}{100}\right)^2\).

James pays \(N\left(6x + \frac{3}{4}\right)\) more than Lamin's \(N300\):

So \(x = 5\) (taking the positive value).

(b) Split \(3x - 2 < 10 + x < 2 + 5x\) into two inequalities.

Left: \(3x - 2 < 10 + x \Rightarrow 2x < 12 \Rightarrow x < 6\).

Right: \(10 + x < 2 + 5x \Rightarrow 8 < 4x \Rightarrow x > 2\).

Therefore \(2 < x < 6\).

(a) \(\tan x = \frac{5}{12}\) gives opposite 5, adjacent 12, hypotenuse \(\sqrt{5^2 + 12^2} = 13\).

So \(\sin x = \frac{5}{13}\) and \(\cos x = \frac{12}{13}\).

(b) Lawn is \(14 \times 11\); a path of width \(x\) all round gives an outer rectangle \((14 + 2x)\) by \((11 + 2x)\).

Discriminant \(= 25^2 + 4(2)(93) = 625 + 744 = 1369 = 37^2\).

The path is \(3\) m wide (the negative root is rejected).

(a) Diagram: from X, line XY runs at bearing \(078^{\circ}\) (285 m); from Y the man walks due south to Z; XZ \(= 307\) m closes the triangle.

Place X at the origin:

Z has the same easting as Y (a due-south move): \(Z = (278.8,\ z)\).

(b)(ii) Distance YZ \(= 59.3 - (-128.5) = 187.8 \approx 188\) m.

(b)(i) Bearing of X from Z. \(\vec{ZX} = (-278.8,\ 128.5)\), pointing north-west.

(a)(i) Modulo 5 tables on {2, 3, 4}

Addition \(\oplus\) (add, then take the remainder on division by 5):

Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):

(For example \(3\otimes 4 = 12 = 2\times 5 + 2 \equiv 2\), and \(4\oplus 3 = 7 = 5 + 2 \equiv 2\).)

(a)(ii)(a) Solve \(4\otimes e \oplus 2 \equiv 1 \ (\text{mod }5)\)

\(4\otimes e \equiv 1 \ominus 2 \equiv -1 \equiv 4 \ (\text{mod }5)\). So \(4e \equiv 4 \ (\text{mod }5)\). Multiplying both sides by the inverse of 4 (which is 4, since \(4\times 4 = 16 \equiv 1\)) gives \(e \equiv 16 \equiv 1 \ (\text{mod }5)\). Hence \(e = 1\).

(a)(ii)(b) Find n if \(4 \oplus n\otimes 2 \equiv 1 \ (\text{mod }5)\)

Multiplication first: \(n\otimes 2 = 2n\). Then \(4 \oplus 2n \equiv 1\), so \(2n \equiv 1 - 4 \equiv -3 \equiv 2 \ (\text{mod }5)\). Multiplying by the inverse of 2 (which is 3, since \(2\times 3 = 6 \equiv 1\)) gives \(n \equiv 6 \equiv 1 \ (\text{mod }5)\). Hence \(n = 1\).

(b) Validity of the statements (given \(p \Rightarrow q\))

Here p: Landi has cholera, q: Landi is in the hospital, and we are told \(p \Rightarrow q\) is true.

• (i) "If in hospital, then has cholera" is \(q \Rightarrow p\), the converse. It does not follow from \(p\Rightarrow q\), so it is not valid.
• (ii) "If not in hospital, then no cholera" is \(\lnot q \Rightarrow \lnot p\), the contrapositive of \(p\Rightarrow q\). It is logically equivalent, so it is valid.
• (iii) "If no cholera, then not in hospital" is \(\lnot p \Rightarrow \lnot q\), the inverse. It does not follow, so it is not valid.

(a) Construction (ruler and compasses only).

1. Draw \(|PQ| = 6.8\) cm. At Q construct \(\angle PQR = 120^{\circ}\) and draw the ray along which R lies.
2. Through P draw a line parallel to QR (since \(QR \parallel PS\)) and mark \(|PS| = 10.6\) cm. The diagonal \(|PR| = 9.3\) cm (arc from P) fixes R on its ray. Join R to S to complete trapezium PQRS.
3. \(l_1\): construct the perpendicular bisector of \(PR\) (points equidistant from P and R).
4. \(l_2\): construct the perpendicular bisector of \(QR\) (points equidistant from Q and R). Let \(l_1\) and \(l_2\) meet at Y.

(b) Expected measurements from an accurate drawing:

• \(|QR| \approx 3.8\) cm
• \(\angle PSR \approx 60^{\circ}\)
• \(|QY| \approx 5.4\) cm

(Accept small tolerances, since these are read off the construction.)

(a) Work part by part (reading \(\frac{2}{2} = 1\) as written).

(b) The vertical (north-south) distance is the northward component of the walk.

(a) and (b) The required coordinate graph, drawn to equal scales on both axes, is shown below. The vertices are joined in the order in which they are named.

The coordinates plotted are:

For a rotation of \(90^\circ\) anticlockwise about the origin, \((x,y)\mapsto(-y,x)\). Thus, for example, \(P(3,2)\mapsto P_1(-2,3)\).

Reflection in \(y=x\) maps \((x,y)\mapsto(y,x)\). Thus \(P_1(-2,3)\mapsto P_2(3,-2)\).

(c) Combining the two transformations gives

\[(x,y)\mapsto(-y,x)\mapsto(x,-y).\]

Hence \(T\) is a reflection in the x-axis.

(d) The line \(P_1Q_1\) has gradient

\[m=\frac{-1-3}{-5-(-2)}=\frac{-4}{-3}=\frac{4}{3}.\]

Its equation is

\[y-3=\frac{4}{3}(x+2).\]

At the x-axis, \(y=0\), so

\[-3=\frac{4}{3}(x+2),\qquad x=-\frac{17}{4}.\]

Therefore \(W\left(-\frac{17}{4},0\right)\). Since \(P_1S_1\) and \(R_1W\) are both horizontal, they are parallel. Thus \(P_1S_1R_1W\) is a trapezium.

(a)

At \(x = 2\): \(8 - 8 + 6 = 6\). At \(x = 1\): \(\tfrac{1}{2} - 2 + 3 = \tfrac{3}{2}\).

(b) \(P\) is the inverse of \(P^{-1}\). With \(P^{-1} = \begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}\), \(\det = (-1)(-3) - (1)(4) = -1\).

Check: \(P\,P^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

(a) Solving the inequalities simultaneously

First inequality:

\[ 5 - x > 1 \;\Rightarrow\; -x > 1 - 5 \;\Rightarrow\; -x > -4 \;\Rightarrow\; x < 4 \]

Second inequality:

\[ 9 + x \geq 8 \;\Rightarrow\; x \geq 8 - 9 \;\Rightarrow\; x \geq -1 \]

Combining both conditions:

\[ -1 \leq x < 4 \]

(b) Finding \(\angle QRS\)

From the diagram, \(O\) is the centre and \(PS\) passes through \(O\), so \(PS\) is a diameter. Also \(|PQ| = |QR|\) (equal chords, shown by the tick marks) and \(\angle PSR = 56^{\circ}\). The points \(P, Q, R\) lie on the arc on the same side of the diameter.

Step 1: Use equal chords. Equal chords subtend equal arcs, so \(\text{arc } PQ = \text{arc } QR\).

Step 2: Use the inscribed angle at S. \(\angle PSR = 56^{\circ}\) is the angle subtended at the circumference by the arc \(PQR\) (arc \(PR\) through \(Q\)). Hence

\[ \text{arc } PQR = 2 \times 56^{\circ} = 112^{\circ} \]

and since arc \(PQ = \) arc \(QR\), each equals \(56^{\circ}\).

Step 3: Find the remaining arc. Because \(PS\) is a diameter, the semicircle \(P\)-\(Q\)-\(R\)-\(S\) totals \(180^{\circ}\):

\[ \text{arc } RS = 180^{\circ} - \text{arc } PQ - \text{arc } QR = 180^{\circ} - 56^{\circ} - 56^{\circ} = 68^{\circ} \]

Step 4: Use the cyclic quadrilateral PQRS. Opposite angles of a cyclic quadrilateral are supplementary, so \(\angle QPS + \angle QRS = 180^{\circ}\).

\(\angle QPS\) subtends arc \(QRS = \text{arc } QR + \text{arc } RS = 56^{\circ} + 68^{\circ} = 124^{\circ}\), so

\[ \angle QPS = \tfrac{1}{2}\times 124^{\circ} = 62^{\circ} \]

Therefore

\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]

\(\angle QRS = 118^{\circ}\).

(a) Finding \(\angle XWZ\)

\(WXYZ\) is a quadrilateral inscribed in the circle (a cyclic quadrilateral) with vertices in order \(W, X, Y, Z\). Its diagonals are equal: \(|WY| = |XZ|\), and \(\angle WXY = 80^{\circ}\).

Step 1: Interpret the equal diagonals. A cyclic quadrilateral whose diagonals are equal is an isosceles trapezium. From the diagram, side \(WX\) is parallel to side \(ZY\) (\(WX \parallel ZY\)), and the legs \(WZ\) and \(XY\) are the equal sides.

Step 2: Apply the base-angle property. In an isosceles trapezium the two angles standing on the same parallel side are equal. Here \(\angle XWZ\) (at \(W\)) and \(\angle WXY\) (at \(X\)) both stand on the base \(WX\), so they are equal:

\[ \angle XWZ = \angle WXY = 80^{\circ} \]

Check (cyclic quadrilateral): The opposite angles are then \(\angle WZY = \angle XYZ = 180^{\circ} - 80^{\circ} = 100^{\circ}\), and \(80^{\circ} + 100^{\circ} = 180^{\circ}\), confirming that opposite angles are supplementary.

\(\angle XWZ = 80^{\circ}\).

(b) Rate of interest per annum

The man is charged \(2\) kobo per month for every \(N1.00\) borrowed. Since \(N1.00 = 100\) kobo, the monthly interest on \(N1.00\) is:

\[ \text{monthly rate} = \frac{2\text{ kobo}}{100\text{ kobo}} \times 100\% = 2\%\text{ per month} \]

There are \(12\) months in a year, so:

\[ \text{rate per annum} = 2\% \times 12 = 24\% \]

The interest was charged at 24% per annum.

Let the first term be \(a\) and the common difference \(d\).

Sum of 10 terms: \(S_{10} = \tfrac{10}{2}(2a + 9d) = 130 \Rightarrow 2a + 9d = 26\).

Fifth term is three times the first: \(a + 4d = 3a \Rightarrow 4d = 2a \Rightarrow a = 2d\).

Substitute: \(2(2d) + 9d = 26 \Rightarrow 13d = 26 \Rightarrow d = 2\), so \(a = 4\).

Common difference \(= 2\); first term \(= 4\).

Number of terms when the last term is 28:

(a)(i) With profit \(y = 20000\):

(a)(ii) Quadratic formula:

Taking the positive root: \(x = \dfrac{761.0}{2} \approx 380\) (in thousands), i.e. about \(380{,}000\) textbooks.

(b)(i) Diagram: vertical tree \(TQ\) with P on the ground so that \(\angle TPQ = 28^{\circ}\); R is the midpoint of \(TQ\).

(b)(ii) Let \(PQ = d\). Then \(TQ = d\tan 28^{\circ}\) and \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).

Let Manya's present age be \(m\). Then Musa \(= m + 3\).

Seven years ago Musa was twice Manya:

(1) Manya is \(10\) years and Musa is \(13\) years now.

(2) The present sum of ages is \(23\), and it rises by 2 each year. Let it take \(t\) years:

The sum of their ages will be 45 in 11 years.

(a) Number of shares bought \(= \dfrac{4800}{2.50} = 1920\) shares.

Dividend of 25k \(= N0.25\) per share:

(b) Let \(t = \dfrac{1 - x}{x}\); then \(\dfrac{x}{1 - x} = \dfrac{1}{t}\), and the equation becomes \(t + \dfrac{1}{t} = \dfrac{5}{2}\), i.e.

Completing the square:

\(\dfrac{1 - x}{x} = 2 \Rightarrow 1 - x = 2x \Rightarrow x = \tfrac{1}{3}\).

\(\dfrac{1 - x}{x} = \tfrac{1}{2} \Rightarrow 2 - 2x = x \Rightarrow x = \tfrac{2}{3}\).

So \(x = \dfrac{1}{3}\) or \(x = \dfrac{2}{3}\).

(a)(i) Diagram: a cone standing on its vertex (point downwards), base radius 21 cm at the top, water filling it to height \(h\).

(a)(ii) Height. \(4.158\) litres \(= 4158 \text{ cm}^3\).

(b) When the surface diameter drops to 28 cm, the surface radius is 14 cm. By similar cones the remaining water height is

Volume remaining:

Volume drawn out \(= 4158 - 1232 = 2926 \text{ cm}^3\).

(a) \(\sin X = \tfrac{3}{5} \Rightarrow \cos X = \tfrac{4}{5}\) (a 3-4-5 triangle).

(b) Volume of the pyramid \(= \tfrac{1}{3} \times (42 \times 11) \times 12 = \tfrac{1}{3} \times 462 \times 12 = 1848 \text{ cm}^3\).

The cone has the same height (12 cm) and the same volume (1848):

(a) Equation of the line

The gradient of the line through \((2,5)\) and \((-4,-7)\) is

\[ m=\frac{-7-5}{-4-2}=\frac{-12}{-6}=2. \]

Using the point \((2,5)\),

\[ y-5=2(x-2). \]

\[ y-5=2x-4 \]

\[ \boxed{y=2x+1}. \]

(b)(i) Illustration

As shown in the diagram above, \(Q\) is on a bearing of \(030^\circ\) from \(P\), while \(R\) is on a bearing of \(300^\circ\) from \(P\). Therefore, the angle between \(PQ\) and \(PR\) is

\[ \angle QPR=30^\circ+60^\circ=90^\circ. \]

Thus, \(\triangle PQR\) is right-angled at \(P\), with \(PQ=5\text{ km}\) and \(PR=8\text{ km}\).

(b)(ii)(1) Distance between \(Q\) and \(R\)

By Pythagoras' theorem,

\[ QR^2=PQ^2+PR^2 \]

\[ QR^2=5^2+8^2=25+64=89. \]

\[ QR=\sqrt{89}=9.433\ldots \]

\[ \boxed{QR=9.43\text{ km}} \]

correct to three significant figures.

(b)(ii)(2) Bearing of \(R\) from \(Q\)

First, find the angle at \(Q\):

\[ \tan \angle PQR=\frac{PR}{PQ}=\frac{8}{5}. \]

\[ \angle PQR=\tan^{-1}\left(\frac{8}{5}\right)=58.0^\circ. \]

The bearing of \(P\) from \(Q\) is the reverse bearing of \(030^\circ\):

\[ 030^\circ+180^\circ=210^\circ. \]

From the diagram, \(R\) lies clockwise from the direction \(QP\). Hence,

\[ \text{Bearing of }R\text{ from }Q=210^\circ+58^\circ=268^\circ. \]

\[ \boxed{268^\circ} \]

Price before the first sale. After a 40% cut the price is \(0.6\) of the original; after a further 30% cut it is \(0.7 \times 0.6 = 0.42\) of the original.

Article costing GH\unicode{x20B5}180.00. Final price \(= 0.42 \times 180 = \text{GH}\unicode{x20B5}75.60\).

Total reduction \(= 180 - 75.60 = \text{GH}\unicode{x20B5}104.40\).

Percentage reduction \(= \dfrac{104.40}{180} \times 100\% = 58\%\).

Purchase price \(= \text{N}900{,}000.00\) on 1 March 2011. The value on 28 February 2015 is after 4 full years.

Year 1 (to Feb 2012): depreciation 30%, so the value is multiplied by \(0.70\).

Years 2, 3, 4: depreciation 22% each year, so each year the value is multiplied by \(0.78\).

Now \((0.78)^3 = 0.474552\), and \(900000 \times 0.70 = 630000\):

Correct to the nearest hundred naira, the value of the car on 28 February 2015 is N299,000.00.

(a) Completed table for \(y=2\cos x-\sin x\):

For example, at \(x=30^\circ\):

\[y=2\cos30^\circ-\sin30^\circ=2(0.866)-0.5=1.232\ldots\approx1.23.\]

(b) Graph of \(y=2\cos x-\sin x\)

(c)(i) To solve \(2\cos x-\sin x=1\), draw or imagine the horizontal line \(y=1\). The required values are the \(x\)-coordinates where this line meets the curve:

\[x\approx36^\circ\quad\text{or}\quad x\approx270^\circ.\]

(c)(ii) Since

\[\tan x=2 \quad\Rightarrow\quad \frac{\sin x}{\cos x}=2 \quad\Rightarrow\quad \sin x=2\cos x,\]

this is equivalent to

\[2\cos x-\sin x=0.\]

Therefore, read the \(x\)-coordinates where the curve crosses the \(x\)-axis:

\[x\approx63^\circ\quad\text{or}\quad x\approx243^\circ.\]

Examination reminder: For an equation involving \(2\cos x-\sin x\), use the graph’s \(y\)-value. In particular, \(\tan x=2\) rearranges to \(2\cos x-\sin x=0\), so look for the \(x\)-intercepts of the curve.