WAEC SSCE - General Mathematics - 1990 (Essay)

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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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(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) 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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Intended dimensions: 50 cm by 40 cm. Actual dimensions made: 60 cm by 35 cm.

Percentage error \(= \dfrac{|\text{actual} - \text{intended}|}{\text{intended}} \times 100\%\).

(a)(i) Length:

\[ \frac{|60 - 50|}{50} \times 100\% = \frac{10}{50} \times 100\% = 20\%. \]

Breadth:

\[ \frac{|35 - 40|}{40} \times 100\% = \frac{5}{40} \times 100\% = 12.5\%. \]

(a)(ii) Area: intended area \(= 50 \times 40 = 2000\ \text{cm}^2\); actual area \(= 60 \times 35 = 2100\ \text{cm}^2\).

\[ \frac{|2100 - 2000|}{2000} \times 100\% = \frac{100}{2000} \times 100\% = 5\%. \]

(b) Product of the two errors in a(i):

\[ 20\% \times 12.5\% = 0.20 \times 0.125 = 0.025 = 2.5\%. \]

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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 \(7\text{ cm}\).

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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),

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