WAEC SSCE - General Mathematics - 1993 (Essay)

Bayanin Amsa

(a) Common denominator \((m+2n)(m-3n)\):

\[\frac{3}{m+2n} - \frac{2}{m-3n} = \frac{3(m-3n) - 2(m+2n)}{(m+2n)(m-3n)}\]

\[= \frac{3m - 9n - 2m - 4n}{(m+2n)(m-3n)} = \frac{m - 13n}{(m+2n)(m-3n)}\]

(b) Let the tens digit be \(t\) and the units digit be \(u\). The number is \(10t + u\).

Sum of digits: \(t + u = 11\).

Interchanging the digits gives \(10u + t\), which exceeds the original by 9:

\[10u + t = (10t + u) + 9 \;\Rightarrow\; 9u - 9t = 9 \;\Rightarrow\; u - t = 1\]

Solving \(t + u = 11\) and \(u - t = 1\): \(u = 6,\; t = 5\).

The number is \(\mathbf{56}\).

Bayanin Amsa

The supplied school reference is inconsistent with the question. The question gives longitudes \(33^\circ\text{W}\) and \(20^\circ\text{E}\), so the difference in longitude is:

\[ 33^\circ+20^\circ=53^\circ \]

The use of \(23^\circ\text{E}\), giving \(56^\circ\), does not match the stated data. Therefore the distances must be calculated using \(53^\circ\), not \(56^\circ\).

(a) Distance along the parallel of latitude

A parallel is a circle smaller than the Equator. At latitude \(55^\circ\), its radius is:

\[ r=R\cos55^\circ \] \[ r=6400\cos55^\circ \]

The required arc is \(\frac{53}{360}\) of this parallel’s circumference:

\[ \begin{aligned} d &=\frac{53}{360}\times 2\pi R\cos55^\circ\\ &=\frac{53}{360}\times2\times\frac{22}{7}\times6400\cos55^\circ\\ &\approx\frac{53}{360}\times23075.1\\ &\approx 3397\text{ km} \end{aligned} \]

Distance along the parallel of latitude: \(\boxed{3397\text{ km}}\)

(b) Distance along a great circle

The shortest route over the Earth’s surface is an arc of a great circle. Let \(\theta\) be the angle at the Earth’s centre subtended by \(P\) and \(Q\). For two points at the same latitude:

\[ \cos\theta = \sin^2 55^\circ+\cos^2 55^\circ\cos53^\circ \] \[ \begin{aligned} \cos\theta &\approx (0.8192)^2+(0.5736)^2(0.6018)\\ &\approx0.8690 \end{aligned} \] \[ \theta\approx\cos^{-1}(0.8690)\approx29.66^\circ \]

The great-circle distance is therefore:

\[ \begin{aligned} d &=\frac{\theta}{360}\times2\pi R\\ &=\frac{29.66}{360}\times2\times\frac{22}{7}\times6400\\ &\approx3314\text{ km} \end{aligned} \]

Distance along the great circle: \(\boxed{3314\text{ km (approximately)}}\)

The great-circle distance is slightly shorter than the distance along the parallel because a parallel other than the Equator is not a great circle. In examination questions, first find the longitude difference carefully: longitudes on opposite sides of the Greenwich meridian are added.

Bayanin Amsa

The universal set \(\varepsilon\) is the set of all integers. Listing the given sets over the relevant range:

• \(P = \{x : x < 0\} = \{\dots, -3, -2, -1\}\) (negative integers)
• \(Q = \{\dots, -5, -3, -1, 1, 3, 5, \dots\}\) (odd integers)
• \(R = \{x : -2 \le x < 7\} = \{-2, -1, 0, 1, 2, 3, 4, 5, 6\}\)

(a) \(Q \cap R\) = odd integers lying in \(R\):

\[Q \cap R = \{-1,\, 1,\, 3,\, 5\}\]

(b) \(R'\) = all integers not in \(R\):

\[R' = \{\dots, -4, -3\} \cup \{7, 8, 9, \dots\} = \{x : x \le -3 \text{ or } x \ge 7\}\]

(c) \(P' = \{0, 1, 2, 3, \dots\}\) (non-negative integers). Then

\[P' \cup R' = \{0,1,2,\dots\} \cup \{\dots,-4,-3\}\cup\{7,8,\dots\} = \varepsilon \setminus \{-2,\,-1\}\]

i.e. every integer except \(-1\) and \(-2\).

(d) \(P \cap Q\) = negative odd integers \(= \{\dots, -5, -3, -1\}\). Its complement is every integer that is not a negative odd integer:

\[(P \cap Q)' = \{\dots, -6, -4, -2,\, 0, 1, 2, 3, 4, \dots\}\]

that is, all non-negative integers together with all negative even integers.

Bayanin Amsa

(a) A.P. \(5, 9, 13, \dots\) with \(a = 5,\; d = 4\).

\[T_{25} = a + 24d = 5 + 24(4) = 5 + 96 = 101\]

(b) G.P. \(\dfrac{8}{9},\, -\dfrac{4}{3},\, 2, \dots\) with \(a = \dfrac{8}{9}\) and common ratio

\[r = \frac{-4/3}{8/9} = -\frac{4}{3}\times\frac{9}{8} = -\frac{3}{2}\]

\[T_5 = ar^4 = \frac{8}{9}\left(-\frac{3}{2}\right)^4 = \frac{8}{9}\times\frac{81}{16} = \frac{9}{2} = 4\tfrac12\]

(c) For the G.P., \(T_3 = ar^2 = 48\) and \(T_6 = ar^5 = 14\tfrac29 = \dfrac{128}{9}\). Dividing:

\[r^3 = \frac{ar^5}{ar^2} = \frac{128/9}{48} = \frac{8}{27} \;\Rightarrow\; r = \frac{2}{3}\]

\[ar^2 = a\left(\tfrac49\right) = 48 \;\Rightarrow\; a = 108\]

The first four terms are \(108,\; 72,\; 48,\; 32\).

Bayanin Amsa

(a)(i) Write \(45 = 9 \times 5 = 3^2 \times 5\):

\[\log_{10}45 = 2\log_{10}3 + \log_{10}5 = 2(0.477) + 0.699 = 0.954 + 0.699 = 1.653\]

(ii) Solve \(x^{0.8265} = 45\). Take logarithms of both sides:

\[0.8265\,\log_{10}x = \log_{10}45 = 1.653\]

\[\log_{10}x = \frac{1.653}{0.8265} = 2.000 \;\Rightarrow\; x = 10^{2} = 100\]

(b) Evaluate \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}}\) using logarithms.

Number	Log
2.067	0.3155
0.0348	\(\bar{2}.5416\)
0.538	\(\bar{1}.7308\)

Denominator log \(= \bar{2}.5416 + \bar{1}.7308 = \bar{2}.2724\). Then

\[\log(\text{fraction}) = 0.3155 - \bar{2}.2724 = 2.0431\]

\[\log(\text{answer}) = \tfrac12(2.0431) = 1.0216 \;\Rightarrow\; \text{answer} = 10.51\]

Hence \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}} \approx 10.51\).

Bayanin Amsa

(a) Completing the table for \(y = 3\sin 2\theta - \cos\theta\). Substitute each value of \(\theta\), remembering that the angle inside the sine is \(2\theta\).

• \(\theta = 30^\circ:\ 3\sin 60^\circ - \cos 30^\circ = 2.598 - 0.866 = 1.732 \approx 1.7\)
• \(\theta = 60^\circ:\ 3\sin 120^\circ - \cos 60^\circ = 2.598 - 0.500 = 2.098 \approx 2.1\)
• \(\theta = 120^\circ:\ 3\sin 240^\circ - \cos 120^\circ = -2.598 + 0.500 = -2.098 \approx -2.1\)
• \(\theta = 150^\circ:\ 3\sin 300^\circ - \cos 150^\circ = -2.598 + 0.866 = -1.732 \approx -1.7\)

\(\theta\)	0°	30°	60°	90°	120°	150°	180°
\(y\)	-1.0	1.7	2.1	0	-2.1	-1.7	1.0

(b) Graph of \(y = 3\sin 2\theta - \cos\theta\) for \(0^\circ \le \theta \le 180^\circ\). The seven table points are plotted and joined with a smooth curve; the curve rises above the tabulated points to a peak near \(\theta = 49^\circ\) and dips to a trough near \(\theta = 131^\circ\).

(c)(i) Solution of \(3\sin 2\theta - \cos\theta = 0\). These are the values of \(\theta\) where the curve crosses the \(\theta\)-axis. Reading the three crossings from the graph gives

\[\theta \approx 10^\circ,\ 90^\circ,\ 170^\circ.\]

This is confirmed algebraically, since \(3\sin 2\theta - \cos\theta = \cos\theta(6\sin\theta - 1) = 0\) gives \(\cos\theta = 0\) (so \(\theta = 90^\circ\)) or \(\sin\theta = \tfrac{1}{6}\) (so \(\theta \approx 10^\circ\) and \(\theta \approx 170^\circ\)).

(c)(ii) Maximum value of \(y\). The highest point of the curve lies between the tabulated points, near \(\theta = 49^\circ\), not at \(\theta = 60^\circ\). Reading the peak gives

\[y_{\max} \approx 2.3\ \text{(to 1 decimal place)}.\]

Examination note: the greatest value of \(y\) is read from the top of the smooth curve, which rises higher than any tabulated point. Taking the largest table value (\(2.1\) at \(\theta = 60^\circ\)) as the maximum is a common error; the true maximum of about \(2.3\) occurs near \(\theta = 49^\circ\).

Bayanin Amsa

Bayanin Amsa

(a) Use \(a\log b = \log b^a\) and combine:

\[\log_{10}\frac{30}{16} - 2\log_{10}\frac{5}{9} + \log_{10}\frac{400}{243} = \log_{10}\frac{30}{16} - \log_{10}\frac{25}{81} + \log_{10}\frac{400}{243}\]

\[= \log_{10}\!\left(\frac{30}{16} \times \frac{81}{25} \times \frac{400}{243}\right)\]

Simplify the product step by step: \(\dfrac{30}{16} = \dfrac{15}{8}\), and \(\dfrac{15}{8}\times\dfrac{81}{25} = \dfrac{243}{40}\), then \(\dfrac{243}{40}\times\dfrac{400}{243} = \dfrac{400}{40} = 10\).

\[= \log_{10} 10 = 1\]

(b) With \(P = 3.6\times10^{-3}\) and \(Q = 2.25\times10^{6}\):

\[\frac{P}{Q} = \frac{3.6\times10^{-3}}{2.25\times10^{6}} = 1.6\times10^{-9} = 16\times10^{-10}\]

\[\sqrt{\frac{P}{Q}} = \sqrt{16\times10^{-10}} = 4\times10^{-5}\]

In standard form, \(\sqrt{\dfrac{P}{Q}} = 4.0\times10^{-5}\).

Bayanin Amsa

(a) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.

Let \(O\) be the centre, and let arc \(AB\) subtend \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining circumference. Join \(CO\) and produce it to \(X\).

Triangle \(OAC\) is isosceles because \(OA = OC\) (radii), so \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the interior opposite angles:

\[\angle AOX = \angle OAC + \angle OCA = 2\,\angle OCA.\]

Likewise, from isosceles triangle \(OBC\), \(\angle BOX = 2\,\angle OCB\). Adding,

\[\angle AOB = \angle AOX + \angle BOX = 2(\angle OCA + \angle OCB) = 2\,\angle ACB.\]

Hence the central angle is twice the inscribed angle on the same arc. (Q.E.D.)

(b) \(O\) is the centre of circle \(ACDB\), with \(\angle CAO = 26^\circ\) and \(\angle AOB = 130^\circ\); \(AD\) is the diameter through \(O\) (the dotted line \(A\!-\!O\!-\!D\)).

First find \(\angle AOC\). In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles with base angles equal:

\[\angle OCA = \angle OAC = 26^\circ \;\Rightarrow\; \angle AOC = 180^\circ - 26^\circ - 26^\circ = 128^\circ.\]

Since \(A,O,D\) are collinear (diameter), angles on the straight line at \(O\) give

\[\angle COD = 180^\circ - \angle AOC = 180^\circ - 128^\circ = 52^\circ,\qquad \angle BOD = 180^\circ - \angle AOB = 180^\circ - 130^\circ = 50^\circ.\]

(ii) \(\angle COB\). With \(C\) above and \(B\) below the diameter, \(\angle COB\) is made up of \(\angle COD\) and \(\angle DOB\):

\[\angle COB = \angle COD + \angle BOD = 52^\circ + 50^\circ = \boxed{102^\circ}.\]

(i) \(\angle OBC\). Triangle \(OBC\) is isosceles since \(OB = OC\) (radii), so its base angles are equal:

\[\angle OBC = \angle OCB = \frac{180^\circ - \angle COB}{2} = \frac{180^\circ - 102^\circ}{2} = \frac{78^\circ}{2} = \boxed{39^\circ}.\]

Bayanin Amsa

(a) Finding x.

From the diagram: BA is parallel to DE, the angle at B (angle ABC) is \(52^\circ\), and the angle at D is a reflex angle of \(312^\circ\), while x is the angle at C (angle BCD) formed by the broken line B-C-D.

First find the true (non-reflex) angle at D:

\[\angle CDE = 360^\circ - 312^\circ = 48^\circ\]

Method: draw an auxiliary line through C parallel to both BA and DE. Because \(BA \parallel DE\), a line drawn through C parallel to them is parallel to each, and it splits the angle at C into two parts.

• BC is a transversal cutting \(BA\) and the line through C. The alternate angles are equal, so the part of x adjacent to CB equals \(\angle ABC = 52^\circ\).
• CD is a transversal cutting \(DE\) and the line through C. The alternate angles are equal, so the part of x adjacent to CD equals \(\angle CDE = 48^\circ\).

Adding the two parts:

\[x = 52^\circ + 48^\circ = 100^\circ\]

x = 100\(^\circ\).

(b) Graph of the inequalities.

The three inequalities are \(y - 2x < 5\), \(2y + x \geq 4\) and \(y + 2x \leq 10\). Draw the three boundary lines, decide each shaded side with a test point, and the required region is where all three overlap.

Boundary line 1: \(y - 2x = 5\), i.e. \(y = 2x + 5\). Passes through \((0,5)\) and \((-2.5,0)\). Broken line (strict \(<\)). Test \((0,0)\): \(0-0=0<5\) true, so shade the side containing the origin (below/right of this line).

Boundary line 2: \(2y + x = 4\), i.e. \(y = \dfrac{4-x}{2}\). Passes through \((0,2)\) and \((4,0)\). Solid line (\(\geq\)). Test \((0,0)\): \(0+0=0\), not \(\geq 4\), so shade the side NOT containing the origin (above this line).

Boundary line 3: \(y + 2x = 10\), i.e. \(y = 10 - 2x\). Passes through \((0,10)\) and \((5,0)\). Solid line (\(\leq\)). Test \((0,0)\): \(0+0=0\leq 10\) true, so shade the side containing the origin (below/left of this line).

The required region is the closed polygon common to all three shaded areas: on or above \(2y+x=4\), on or below \(y+2x=10\), and below \(y=2x+5\). A convenient interior check point such as \((2,2)\) satisfies all three: \(2-4=-2<5\), \(4+2=6\geq4\), \(2+4=6\leq10\). Shade the overlapping region.

Bayanin Amsa

Let the foot of the perpendicular from the peak \(P\) to the horizontal level of \(X\) and \(Y\) be \(N\), with \(Y\) nearer the mountain. Let \(|YN| = d\) and let \(h\) be the height of \(P\) above the level \(XY\).

From \(Y\) (angle of elevation \(35^\circ\)):

\[\tan 35^\circ = \frac{h}{d} \;\Rightarrow\; h = 0.7d\]

From \(X\) (angle of elevation \(15^\circ\), and \(|XN| = d + 4000\)):

\[\tan 15^\circ = \frac{h}{d + 4000} \;\Rightarrow\; h = 0.3(d + 4000)\]

Equating:

\[0.7d = 0.3d + 1200 \;\Rightarrow\; 0.4d = 1200 \;\Rightarrow\; d = 3000\text{ m}\]

\[h = 0.7(3000) = 2100\text{ m}\]

This \(h\) is the height of the peak above the level of \(X\) and \(Y\). Since \(X\) (and \(Y\)) are \(5200\) m above sea level, the height of the mountain above sea level is:

\[5200 + 2100 = 7300\text{ m}\]

Bayanin Amsa

Total balls \(= 12 + 16 + 8 = 36\). Three are drawn without replacement.

(a) All three red:

\[P = \frac{12}{36}\times\frac{11}{35}\times\frac{10}{34} = \frac{1320}{42840} = \frac{11}{357} \approx 0.031\]

(b) First blue, then two reds (in that order):

\[P = \frac{8}{36}\times\frac{12}{35}\times\frac{11}{34} = \frac{1056}{42840} = \frac{44}{1785} \approx 0.025\]

(c) Two white and one blue (in any order): the favourable arrangements are WWB, WBW, BWW. Using selections,

\[P = \frac{\binom{16}{2}\binom{8}{1}}{\binom{36}{3}} = \frac{120 \times 8}{7140} = \frac{960}{7140} = \frac{16}{119} \approx 0.134\]