WAEC SSCE - General Mathematics - 1999 (Essay)

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(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]

(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.

(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]

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(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).

(i) Volume of metal used.

The metal is the hollow shell between the outer and inner cylinders:

\[V_{\text{metal}}=\pi(R^2-r^2)\times L\]\[R^2-r^2=5^2-3.5^2=25-12.25=12.75\text{ cm}^2\]\[V_{\text{metal}}=\frac{22}{7}\times 12.75\times 2800\]\[=\frac{22}{7}\times 35700=22\times 5100=112200\text{ cm}^3\]

The volume of metal is \(112200\text{ cm}^3\).

(ii) Volume of water the pipe holds when full.

This is the inner cylinder's volume:

\[V_{\text{water}}=\pi r^2 L=\frac{22}{7}\times 3.5^2\times 2800\]\[=\frac{22}{7}\times 12.25\times 2800=\frac{22}{7}\times 34300=22\times 4900=107800\text{ cm}^3\]

Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):

\[\frac{107800}{1000}=107.8\text{ litres}\]

The pipe holds \(107.8\) litres (to 1 d.p.).

(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.

By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):

\[\angle PMN=\angle MLN \quad\text{...(1)}\]

Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:

\[\angle PMN=\angle MNL \quad\text{...(2)}\]

From (1) and (2):

\[\angle MLN=\angle MNL\]

In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).

Therefore triangle \(LMN\) is isosceles. (Q.E.D.)

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(a) Completing the table of values

For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):

\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)

x	0°	30°	60°	90°	120°	150°	180°	210°
y	9.0	10.3	8.8	5.0	-0.2	-5.3	-9.0	-10.3

(b) Graph of \(y = 9\cos x + 5\sin x\)

Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).

(c) Solving from the graph

(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:

\[ x \approx 119^\circ \]

(Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))

(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:

\[ x \approx 99^\circ \]

(Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))

(d) Maximum value of y

The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):

\[ y_{\max}=\sqrt{81+25}=\sqrt{106}\approx 10.3 \]

The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).

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(a) \[\left(\frac{4}{25}\right)^{-\frac12}=\left(\frac{25}{4}\right)^{\frac12}=\frac52,\qquad 2^{4}=16,\qquad \left(\frac{15}{2}\right)^{-2}=\left(\frac{2}{15}\right)^{2}=\frac{4}{225}.\] So \[\frac52\times16\div\frac{4}{225}=40\times\frac{225}{4}=2250.\]

(b) \[\log_{5}\!\frac35+3\log_{5}\!\frac52-\log_{5}\!\frac{81}{8}=\log_{5}\!\left(\frac35\cdot\left(\frac52\right)^{3}\cdot\frac{8}{81}\right)=\log_{5}\!\left(\frac35\cdot\frac{125}{8}\cdot\frac{8}{81}\right)=\log_{5}\!\frac{25}{27}.\] Hence the value is \[\log_{5}\frac{25}{27}=2-3\log_{5}3\approx-0.048.\]

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Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).

(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):

\[|AB|^2 = |BD|^2 - |AD|^2 = 17^2 - 15^2 = 289 - 225 = 64,\]\[|AB| = \sqrt{64} = \mathbf{8\text{ cm}}.\]

(b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):

\[\text{Area} = \tfrac{1}{2}\times |BC| \times |AB| = \tfrac{1}{2}\times 9 \times 8 = \mathbf{36\text{ cm}^2}.\]

(c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then

\[|CD| = \sqrt{(15-9)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = \mathbf{10\text{ cm}}.\]

(d) Perimeter of the trapezium.

\[P = |AB| + |BC| + |CD| + |DA| = 8 + 9 + 10 + 15 = \mathbf{42\text{ cm}}.\]

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In \(\triangle PBC\), the right angle is at \(C\), \(|BC|=5\text{ cm}\), \(\angle PBC=37^{\circ}\), and \(P\) is the midpoint of \(AC\).

(a) \[\tan37^{\circ}=\frac{PC}{BC}\Rightarrow PC=5\tan37^{\circ}=5\times0.7536=3.768\text{ cm}.\] Since \(P\) is the midpoint, \(|AC|=2\times PC=7.536\approx\) 7.54 cm (3 s.f.).

(b) In \(\triangle ABC\), \[\tan(\angle ABC)=\frac{AC}{BC}=\frac{7.536}{5}=1.5072\Rightarrow\angle ABC=56.44^{\circ}.\] Therefore \[\angle PBA=\angle ABC-\angle PBC=56.44^{\circ}-37^{\circ}\approx19.4^{\circ}.\]

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Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.

(a) Height reached up the wall. In right triangle \(TQF\):

\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\]

(b)(i) Distance the top slides down. First the original foot distance:

\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]

New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):

\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]

Distance slid down:

\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\]

(b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):

\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]

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The scale gives \(1\) coordinate unit \(=1\text{ cm}\). Therefore \(PQ=7\text{ cm}\), since \(P(-5,3)\) and \(Q(2,3)\) have the same \(y\)-coordinate.

Locus \(L\): Points equidistant from \(P\) and \(Q\) lie on the perpendicular bisector of \(PQ\). The midpoint of \(PQ\) is

\[ \left(\frac{-5+2}{2},\frac{3+3}{2}\right)=(-1.5,3). \]

Since \(PQ\) is horizontal, its perpendicular bisector is the vertical line \(x=-1.5\).

Construction of \(R\) and \(S\): Draw arcs of radius \(5\text{ cm}\), centred at \(P\) and at \(Q\). Their two intersections are \(R\) and \(S\). Each intersection is \(5\text{ cm}\) from both \(P\) and \(Q\), so \(PR=RQ=QS=SP=5\text{ cm}\), giving a rhombus.

Half of \(PQ\) is \(3.5\text{ cm}\). Using Pythagoras in the right triangle from the midpoint of \(PQ\) to \(R\):

\[ \text{vertical distance}=\sqrt{5^2-3.5^2} =\sqrt{25-12.25} =\sqrt{12.75} =\frac{\sqrt{51}}{2}\approx3.57. \]

Therefore, the coordinates are

\[ R=\left(-1.5,\;3+\frac{\sqrt{51}}{2}\right)\approx(-1.5,6.57), \]

\[ S=\left(-1.5,\;3-\frac{\sqrt{51}}{2}\right)\approx(-1.5,-0.57). \]

The diagonal \(PQ=7\text{ cm}\), and the other diagonal is

\[ RS=2\left(\frac{\sqrt{51}}{2}\right)=\sqrt{51}\text{ cm}. \]

Hence the area is

\[ \text{Area}=\frac{1}{2}\times PQ\times RS =\frac{1}{2}\times7\times\sqrt{51} \approx24.74\text{ cm}^2. \]

The supplied reference coordinates \((-1.6,6.5)\) and \((-1.6,0.6)\) are not consistent with the perpendicular bisector \(x=-1.5\), and the positive \(y\)-coordinate for \(S\) is incorrect. A construction read from a graph may reasonably give approximately \(R(-1.5,6.6)\) and \(S(-1.5,-0.6)\).

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(a) With \(\varepsilon=\{1,2,\dots,20\}\):

• (i) \(A=\{\text{multiples of }5\}=\{5,10,15,20\}.\)
• (ii) \(B=\{\text{multiples of }4\}=\{4,8,12,16,20\}.\)
• (iii) \(C=\{\text{multiples of }3\}=\{3,6,9,12,15,18\}.\)

(b)

• (i) \(A\cap B=\{20\}.\)
• (ii) \(A\cap C=\{15\}.\)
• (iii) \(B\cup C=\{3,4,6,8,9,12,15,16,18,20\}.\)

(c) \[(A\cap B)\cup(A\cap C)=\{20\}\cup\{15\}=\{15,20\}.\] Also \(A\cap(B\cup C)\): from \(A=\{5,10,15,20\}\), those in \(B\cup C\) are \(\{15,20\}\). Since both sides equal \(\{15,20\}\), the identity \((A\cap B)\cup(A\cap C)=A\cap(B\cup C)\) is verified.

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(a) \(3y-2x=21\) and \(4y+5x=5\). Multiply the first by \(5\) and the second by \(2\): \[15y-10x=105,\qquad 8y+10x=10.\] Adding: \(23y=115\Rightarrow y=5\). Then \(3(5)-2x=21\Rightarrow-2x=6\Rightarrow x=-3\). So \(x=-3,\;y=5\).

(b) Card \(6\) wins \(\$60\); cards \(1\text{–}5\) lose \(\$10\times(\text{number})\).

• (i) A loss occurs whenever cards \(1,2,3,4,5\) are picked: \[P(\text{losing})=\frac{5}{6}.\]
• (ii) Losing exactly \(\$20\) in two picks: the outcomes per card are \(-10,-20,-30,-40,-50,+60\). The only pair of picks totalling \(-\$20\) is card \(1\) followed by card \(1\) \((-10-10=-20)\). \[P=\frac16\times\frac16=\frac{1}{36}.\]

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Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.

\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]

(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]

(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.

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Marks	3	4	5	6	7	8
Frequency	5	x-1	x	9	4	1

(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):

\[ \Sigma fx = 15+4(x-1)+5x+54+28+8 = 101+9x \]

Since mean \(=5\):

\[ \frac{101+9x}{18+2x}=5 \Rightarrow 101+9x=90+10x \Rightarrow x=11 \]

So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).

(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.

(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.

(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):

\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]

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(a) Let the cost price be \(C\). Marked to gain \(20\%\): marked \(=1.2C\). Sold at \(10\%\) reduction: \[0.9\times1.2C=1.08C=864\Rightarrow C=\frac{864}{1.08}=\text{N}800.\] The original cost was N800.

(b) Length \((x+5)\), width \((x-2)\), diagonal \((x+6)\). By Pythagoras: \[(x+5)^{2}+(x-2)^{2}=(x+6)^{2}.\] \[x^{2}+10x+25+x^{2}-4x+4=x^{2}+12x+36\Rightarrow x^{2}-6x-7=0\Rightarrow(x-7)(x+1)=0.\] Taking the positive value, \(x=7\).

(ii) Area \(=(x+5)(x-2)=12\times5=60\text{ m}^{2}.\)