WAEC SSCE - General Mathematics - 2013 (Essay)

(a) Father : son \(= 10 : 3\), and the son is \(15\). So father's age \(= \dfrac{10}{3}\times 15 = 50\) years.

Let the ratio be \(2 : 1\) in \(x\) years: \[\frac{50 + x}{15 + x} = \frac{2}{1} \Rightarrow 50 + x = 2(15 + x) = 30 + 2x.\] \[50 - 30 = 2x - x \Rightarrow x = 20.\] The ratio will be \(2 : 1\) in \(\mathbf{20\ \text{years}}.\)

(b) Mean of \(x, y, z\) is \(6\): \(x + y + z = 18.\)

Mean of \(x, y, z, l, u, v, w\) is \(9\): \(x + y + z + l + u + v + w = 63.\)

So \(l + u + v + w = 63 - 18 = 45.\)

Mean of \(l, u, v, w = \dfrac{45}{4} = \mathbf{11.25}.\)

(a) Diagram. A vertical wall meets the horizontal ground at a right angle. The ladder \(LM\) leans with its top \(L\) touching the wall \(5\) m above the ground and its foot \(M\) on the ground, making \(37^\circ\) with the ground. The height (5 m), the ground distance, and the ladder form a right-angled triangle.

(b) Length of the ladder. The 5 m height is opposite the \(37^\circ\) angle:

\[\sin 37^\circ = \frac{5}{|LM|} \;\Rightarrow\; |LM| = \frac{5}{\sin 37^\circ} = \frac{5}{0.6018} = 8.31\text{ m (3 s.f.)}\]

(c) New angle after moving the foot 2 m towards the wall. Original horizontal distance of the foot from the wall:

\[|LM|\cos 37^\circ = 8.308 \times 0.7986 = 6.635\text{ m}\]

Pushing the foot 2 m nearer gives a new base distance \(6.635 - 2 = 4.635\) m, with the ladder length unchanged (8.308 m). If \(\alpha\) is the new angle with the ground:

\[\cos\alpha = \frac{4.635}{8.308} = 0.5579 \;\Rightarrow\; \alpha = 56^\circ \text{ (nearest degree)}\]

(a) Draw a circle for the earth with the equator horizontal and the North pole on top. Mark \(T\) on the equator (longitude \(15^\circ E\)). The plane flies due North along the \(15^\circ E\) meridian to \(P\), then East along the parallel of latitude through \(P\) to \(Q\) on longitude \(65^\circ E\).

(b)(i) Latitude of P: Distance North \(= 950\times 4 = 3800\,\text{km}\), an arc along a meridian: \[3800 = \frac{\theta}{360}\times 2\pi R = \frac{\theta}{360}\times 2\times\frac{22}{7}\times 6400.\] \[2\pi R = 40228.6\,\text{km} \Rightarrow \theta = \frac{3800\times 360}{40228.6} = 34.0^\circ.\] Latitude of \(P \approx \mathbf{34^\circ N}.\)

(b)(ii) Distance P to Q: Along the parallel of latitude \(34^\circ N\), the difference in longitude is \(65^\circ - 15^\circ = 50^\circ\): \[PQ = \frac{50}{360}\times 2\pi R\cos34^\circ = \frac{50}{360}\times 40228.6\times 0.8290.\] \[PQ = 5587.3\times 0.8290 = 4632\,\text{km}.\] Correct to four significant figures, \(|PQ| \approx \mathbf{4632\,\text{km}}.\)

(a) Perimeter of the remaining board.

Reading the diagram. The board is a rectangle \(22\text{ cm}\) wide and \(12\text{ cm}\) tall. A circular segment (curved arch) is cut from the bottom edge, leaving flat pieces of \(5\text{ cm}\) on the left and \(3\text{ cm}\) on the right. Hence the chord (span of the cut) is

\[\text{chord} = 22 - 5 - 3 = 14\text{ cm}.\]

The radius is \(1\tfrac{1}{2}\) times the chord:

\[r = \tfrac{3}{2}\times 14 = 21\text{ cm}.\]

Central angle of the segment. Half the chord is \(7\text{ cm}\). If the chord subtends \(2\theta\) at the centre,

\[\sin\theta = \frac{7}{21} = \frac{1}{3} \;\Rightarrow\; \theta = 19.47^\circ,\qquad 2\theta = 38.94^\circ.\]

Arc length of the cut. With \(\pi = \tfrac{22}{7}\):

\[\text{arc} = \frac{2\theta}{360}\times 2\pi r = \frac{38.94}{360}\times 2\times\frac{22}{7}\times 21 = \frac{38.94}{360}\times 132 = 14.28\text{ cm}.\]

Perimeter of the remaining portion = top + two sides + two flat bottoms + arc:

\[P = 22 + 12 + 12 + 5 + 3 + 14.28 = 68.28\text{ cm}.\]

Perimeter \(\approx \mathbf{68.28\text{ cm}}\) (2 d.p.).

(b) Surd evaluation.

\[\frac{3}{\sqrt{3}}\left(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\right).\]

First, \(\dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{3\sqrt3}{3} = \sqrt{3}.\) Inside the bracket, \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}\) and \(\sqrt{12} = 2\sqrt3\), so \(\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt3}{6} = \dfrac{\sqrt3}{3}.\) Therefore

\[\frac{2}{\sqrt3} - \frac{\sqrt{12}}{6} = \frac{2\sqrt3}{3} - \frac{\sqrt3}{3} = \frac{\sqrt3}{3}.\]

Multiplying:

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

(a) Numerator: \(\tfrac{3}{4}\left(3\tfrac{3}{8} + 1\tfrac{5}{8}\right) = \tfrac{3}{4}\left(\tfrac{27}{8} + \tfrac{13}{8}\right) = \tfrac{3}{4}\times\tfrac{40}{8} = \tfrac{3}{4}\times 5 = \tfrac{15}{4}.\)

Denominator: \(2\tfrac{1}{8} - 1\tfrac{1}{2} = \tfrac{17}{8} - \tfrac{12}{8} = \tfrac{5}{8}.\)

Therefore \(\dfrac{15/4}{5/8} = \dfrac{15}{4}\times\dfrac{8}{5} = \dfrac{120}{20} = \mathbf{6}.\)

(b) Write \(1.125 = \dfrac{9}{8} = \dfrac{3^2}{2^3}.\) Then \[\log_{10}1.125 = 2\log_{10}3 - 3\log_{10}2 = 2(0.4771) - 3(0.3010) = 0.9542 - 0.9030 = 0.0512.\] Correct to 2 significant figures, \(\log_{10}1.125 \approx \mathbf{0.051}.\)

(a) Take the pole vertical with the lamp \(L\) at its top \((4.2\,\text{m})\). The boy \(BC\) is \(1.2\,\text{m}\) tall standing \(6\,\text{m}\) from the pole's foot; his shadow \(CS\) stretches away from the pole on the ground. The light ray from \(L\) passes over the boy's head to the shadow tip \(S\).

(b)(i) Let the shadow length be \(s\). By similar triangles (light from \(L\) grazing the boy's head): \[\frac{\text{lamp height}}{\text{pole-to-shadow-tip}} = \frac{\text{boy height}}{\text{boy-to-shadow-tip}} \Rightarrow \frac{4.2}{6 + s} = \frac{1.2}{s}.\] \[4.2 s = 1.2(6 + s) \Rightarrow 4.2s = 7.2 + 1.2s \Rightarrow 3s = 7.2 \Rightarrow s = 2.4.\] Length of the boy's shadow \(= \mathbf{2.4\,\text{m}}.\)

(b)(ii) Angle of elevation of the lamp from the boy (measured from the top of his head): the horizontal distance is \(6\,\text{m}\) and the lamp is \(4.2 - 1.2 = 3.0\,\text{m}\) above him. \[\tan\theta = \frac{3.0}{6} = 0.5 \Rightarrow \theta = 26.57^\circ \approx \mathbf{27^\circ}.\]

(a) \(7x + 4 < \tfrac{1}{2}(4x + 3)\).

Multiply the bracket: \(7x + 4 < 2x + \tfrac{3}{2}.\)

\(7x - 2x < \tfrac{3}{2} - 4 \Rightarrow 5x < -\tfrac{5}{2} \Rightarrow x < -\tfrac{1}{2}.\)

Solution: \(\mathbf{x < -0.5}.\)

(b) For every \(\text{N}2.00\) Salem gets, Sunday gets \(50\) kobo \((\text{N}0.50)\): so Salem : Sunday \(= 2 : 0.5 = 4 : 1.\)

For every \(\text{N}4.00\) Sunday gets, Shaka gets \(\text{N}2.00\): so Sunday : Shaka \(= 4 : 2 = 2 : 1.\)

Make Sunday common: Salem : Sunday \(= 8 : 2\) and Sunday : Shaka \(= 2 : 1\), giving \[\text{Salem : Sunday : Shaka} = 8 : 2 : 1.\] Total parts \(= 11\). One part \(= \dfrac{\text{N}1100}{11} = \text{N}100.\)

Shaka's share \(= 1 \times \text{N}100 = \mathbf{\text{N}100.00}.\)

(a) Geometric progression

For three consecutive GP terms, the square of the middle term equals the product of the outer terms:

\[6^2 = (3 - x)(7 - 5x)\]\[36 = 21 - 15x - 7x + 5x^2\]\[36 = 5x^2 - 22x + 21\]\[5x^2 - 22x - 15 = 0\]

(i) Values of x

Using the quadratic formula with \(a = 5,\; b = -22,\; c = -15\):

\[x = \frac{22 \pm \sqrt{(-22)^2 - 4(5)(-15)}}{2(5)} = \frac{22 \pm \sqrt{484 + 300}}{10} = \frac{22 \pm \sqrt{784}}{10} = \frac{22 \pm 28}{10}\]\[x = 5 \quad\text{or}\quad x = -\frac{6}{10} = -\frac{3}{5}\]

(ii) Constant ratio

The ratio is \(r = \dfrac{6}{3 - x}\).

• If \(x = 5\): \(r = \dfrac{6}{3-5} = \dfrac{6}{-2} = -3\) (negative, rejected since \(r > 0\)).
• If \(x = -\dfrac{3}{5}\): \(r = \dfrac{6}{3 + \frac{3}{5}} = \dfrac{6}{\frac{18}{5}} = 6\times\dfrac{5}{18} = \dfrac{5}{3}\) (positive, accepted).

So the admissible value is \(x = -\dfrac{3}{5}\) with constant ratio \(r = \dfrac{5}{3}\). (Check: terms are \(3.6,\; 6,\; 10\), and \(\tfrac{6}{3.6} = \tfrac{10}{6} = \tfrac{5}{3}\).)

(b) Finding \(\angle ADC\)

In the quadrilateral, \(|AB| = 3\text{ cm}\), \(|BC| = 4\text{ cm}\), \(|CD| = 6\text{ cm}\), \(|DA| = 7\text{ cm}\), and the angle at B is \(90^\circ\). Draw the diagonal \(AC\).

In right triangle \(ABC\):

\[|AC|^2 = |AB|^2 + |BC|^2 = 3^2 + 4^2 = 9 + 16 = 25 \;\Rightarrow\; |AC| = 5\text{ cm}\]

In triangle \(ACD\), apply the cosine rule with \(\angle ADC\) opposite \(AC\):

\[|AC|^2 = |DA|^2 + |DC|^2 - 2\,|DA|\,|DC|\cos(\angle ADC)\]\[25 = 7^2 + 6^2 - 2(7)(6)\cos(\angle ADC)\]\[25 = 49 + 36 - 84\cos(\angle ADC)\]\[84\cos(\angle ADC) = 85 - 25 = 60\]\[\cos(\angle ADC) = \frac{60}{84} = 0.7143\]\[\angle ADC = \cos^{-1}(0.7143) = 44.4^\circ \approx 44^\circ\]

Therefore \(\angle ADC \approx 44^\circ\).

First find the radius from the area: \[\pi r^2 = 154 \Rightarrow \frac{22}{7}r^2 = 154 \Rightarrow r^2 = \frac{154\times 7}{22} = 49 \Rightarrow r = 7\,\text{cm}.\]

Let each of the two equal sectors have angle \(a\). The third sector is three times the two put together: \[\text{third} = 3(2a) = 6a.\] Total angle: \(2a + 6a = 8a = 360^\circ \Rightarrow a = 45^\circ.\)

So the third sector has angle \(6a = 270^\circ.\)

Arc length of third sector: \[\frac{270}{360}\times 2\pi r = \frac{3}{4}\times 2\times\frac{22}{7}\times 7 = \frac{3}{4}\times 44 = 33\,\text{cm}.\]

Perimeter of a sector = arc + two radii: \[33 + 2(7) = 33 + 14 = \mathbf{47\,\text{cm}}.\]

(a)(i) Sum equals three times the difference: \[p + q = 3(p - q) \Rightarrow p + q = 3p - 3q \Rightarrow 4q = 2p \Rightarrow \mathbf{p = 2q}.\]

(a)(ii) Substitute \(p = 2q\): \[\frac{p^2 + q^2}{pq} = \frac{(2q)^2 + q^2}{(2q)q} = \frac{4q^2 + q^2}{2q^2} = \frac{5q^2}{2q^2} = \frac{5}{2} = \mathbf{2.5}.\]

(b) Selling price of \(100\) articles at \(25\) for \(\text{N}66\): \(\dfrac{100}{25}\times 66 = 4\times 66 = \text{N}264.\)

This gives a \(32\%\) gain, so cost price \(= \dfrac{264}{1.32} = \text{N}200.\)

New selling price at \(20\) for \(\text{N}50\): \(\dfrac{100}{20}\times 50 = 5\times 50 = \text{N}250.\)

New gain \(= 250 - 200 = \text{N}50\), so \[\text{gain}\% = \frac{50}{200}\times 100 = \mathbf{25\%\ \text{gain}}.\]