WAEC SSCE - General Mathematics - 1990 (Essay)

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(a) The angles of a triangle sum to two right angles.

Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.

• \(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
• \(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).

The angles on the straight line XY at C add to two right angles:

\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]

Replacing the two alternate angles gives

\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]

so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)

(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).

\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:

\[ \angle CBD = A + \angle ACB = A + 2c. \]

Therefore

\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]

Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):

\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]

From (1) and (2),

\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]

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Rewrite the inequalities in terms of y:

Draw the three boundary lines as solid lines, since equality is included. Their pairwise intersections are:

The required solution set is the common region below \(y=2x+4\) and above both \(y=2-\frac{x}{3}\) and \(y=7x-9\). It is the shaded triangular region shown.

Hence, the solution region is the closed triangle with vertices

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Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.

Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with

\[ NT = 7 - 3 = 4\ \text{m}. \]

The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:

(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]

(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]

(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where

\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]

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(a) Multiply out the numerator and divide:

\[ \frac{0.016 \times 0.084}{0.48} = \frac{0.001344}{0.48} = 0.0028. \]

In standard form:

\[ 0.0028 = 2.8 \times 10^{-3}. \]

(b) Let the lengths form an A.P. with first term \(a\) and common difference \(d\).

\[ T_2 = a + d = 2, \qquad T_6 = a + 5d = 5. \]

Subtracting: \(4d = 3 \implies d = 0.75\ \text{m}\), and \(a = 2 - 0.75 = 1.25\ \text{m}\).

The eight poles, in order, are:

\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]

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(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:

(b) Since every student speaks at least one of the languages,

\[n(H\cup I\cup Y)=40.\]

Using the inclusion-exclusion principle,

\[40=25+16+21-8-11-6+x\]

\[40=37+x\]

\[x=3.\]

Therefore, 3 students speak all three languages.

Hence the numerical entries in the Venn diagram are:

\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]

\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]

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Take \(R = 6400\) km and \(\pi = 3.142\).

(a) Distance P to Q along the parallel of latitude \(40^\circ\)N. The longitude difference is \(38^\circ\text{E} + 22^\circ\text{W} = 60^\circ\). The radius of the parallel is \(R\cos 40^\circ\), so

\[ PQ = \frac{60}{360} \times 2\pi R \cos 40^\circ = \frac{1}{6} \times 2 \times 3.142 \times 6400 \times 0.7660 = 5134\ \text{km}. \]

(b) Distance Q to T along the meridian \(22^\circ\)W. The latitude difference is \(40^\circ - 28^\circ = 12^\circ\), and along a meridian the radius is \(R\):

\[ QT = \frac{12}{360} \times 2\pi R = \frac{1}{30} \times 2 \times 3.142 \times 6400 = 1340.6\ \text{km}. \]

(c) Average speed from P to T via Q over 12 hours. Total distance:

\[ PQ + QT = 5134 + 1340.6 = 6474.6\ \text{km}. \] \[ \text{Average speed} = \frac{6474.6}{12} = 539.6 \approx 540\ \text{km/h (to 3 s.f.).} \]

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(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):

\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]

\(V \approx 100.5\ \text{cm}^3\).

(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.

\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]

The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:

\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]

Time to fill:

\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]

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(a) The integers from 5 to 25 inclusive number \(25 - 5 + 1 = 21\).

• Multiples of 5: 5, 10, 15, 20, 25 \(\rightarrow\) 5 numbers.
• Multiples of 3: 6, 9, 12, 15, 18, 21, 24 \(\rightarrow\) 7 numbers.
• Multiples of both (i.e. of 15): 15 \(\rightarrow\) 1 number.

By inclusion and exclusion, the count of "multiple of 5 or 3" is \(5 + 7 - 1 = 11\).

\[ P(\text{multiple of 5 or 3}) = \frac{11}{21}. \]

(b) 10 balls: 4 blue, 6 red. Picking is with replacement, so each pick has \(P(\text{red}) = \tfrac{6}{10}\) and \(P(\text{blue}) = \tfrac{4}{10}\).

(i) Both red:

\[ \frac{6}{10} \times \frac{6}{10} = \frac{36}{100} = \frac{9}{25}. \]

(ii) Both the same colour (both red or both blue):

\[ \left(\frac{6}{10}\right)^2 + \left(\frac{4}{10}\right)^2 = \frac{36}{100} + \frac{16}{100} = \frac{52}{100} = \frac{13}{25}. \]

Antwoorddetails

There is an inconsistency in the supplied materials: the question asks for \(|AB|\), which is already given as \(8\text{ cm}\), whereas the reference answer gives \(|AC|=7\text{ cm}\). The reference answer therefore appears to intend \(|AC|\), not \(|AB|\).

Construction

1. Draw \(AB=8\text{ cm}\).
2. At \(B\), construct an angle of \(60^\circ\) with \(BA\). On this ray, mark \(C\) so that \(BC=5.5\text{ cm}\).
3. Construct a line through \(C\) parallel to \(AB\): construct a perpendicular to \(AB\) through \(C\), then construct a perpendicular to this new line through \(C\).
4. With centre \(B\) and radius \(8.3\text{ cm}\), draw an arc to meet the parallel through \(C\) at \(D\), choosing the intersection that gives the trapezium shown. Join \(A\) to \(D\) and \(C\) to \(D\).
5. Construct perpendiculars from \(D\) and \(C\) to \(AB\). Let their feet on \(AB\) be \(P\) and \(Q\), respectively. Since \(PQ\parallel DC\) and \(PD\parallel QC\), with right angles at \(P\) and \(Q\), \(PQCD\) is a rectangle.

Measurements

• \(|AB|=8.0\text{ cm}\), because this length was given and used to construct the figure.
• \(|QB|\approx2.8\text{ cm}\), which may be recorded as \(2.9\text{ cm}\) from a construction drawing.

The value of \(QB\) follows from the horizontal component of \(BC\): \(QB=5.5\cos60^\circ=2.75\text{ cm}\), so a ruler measurement should be close to \(2.8\text{ cm}\) or \(2.9\text{ cm}\).

If the intended measurement was \(|AC|\), as indicated by the reference answer, then \(|AC|\approx7.1\text{ cm}\), recorded as \(7\text{ cm}\).