WAEC SSCE - General Mathematics - 2000 (Essay)

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(a) Completing the table for \(y=x^2-2x-5\). For example, when \(x=-3\), \(y=9+6-5=10\), and when \(x=2\), \(y=4-4-5=-5\).

x	-3	-2	-1	0	1	2	3	4	5
y	10	3	-2	-5	-6	-5	-2	3	10

(b) and (c) The plotted curve is the parabola \(y=x^2-2x-5\). The straight line is \(y=2x+3\), which passes through \((0,3)\) and \((3,9)\).

(d) At a point of intersection, both graphs have the same \(y\)-value:

\[ x^2-2x-5=2x+3 \] \[ x^2-4x-8=0 \]

(e) The roots are the \(x\)-coordinates of the intersections. From the graph, they are approximately \(x=-1.5\) and \(x=5.5\).

The more accurate solutions are \(x=2\pm2\sqrt{3}\), giving \(x\approx-1.46\) and \(x\approx5.46\). The supplied reference answer uses \(y=2x-3\), but this conflicts with the question stem, which states \(y=2x+3\). Therefore \(x^2-4x-8=0\) is the equation consistent with the stated line.

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This question tests accurate ruler-and-compasses construction and the ability to read a required measurement from the finished figure. The three given facts, \(|PQ| = 7\) cm, \(\angle PQR = 60^\circ\) and \(|PR| = 6\) cm, fix the triangle, and the interesting result is the size of \(\angle RMQ\).

Construction steps (ruler and compasses only):

1. Draw a base line and mark off \(|PQ| = 7\) cm.
2. At \(Q\), construct an angle of \(60^\circ\) using the equilateral-triangle arc method (an arc from \(Q\) cutting \(QP\), then the same radius stepped once round it), and draw the ray \(QR\).
3. With centre \(P\) and radius \(6\) cm, draw an arc to cut the \(60^\circ\) ray at \(R\). Join \(PR\) and \(QR\) to complete \(\Delta PQR\).
4. (b) Bisect \(PQ\) with equal arcs from \(P\) and \(Q\) above and below the line; the bisector meets \(PQ\) at its mid-point \(M\), so \(|PM| = |MQ| = 3.5\) cm. Join \(RM\).
5. (c) Place the protractor at \(M\) and read \(\angle RMQ\).

The accurately drawn figure:

(c) Measurement: \(\angle RMQ = 60^\circ\).

Why the answer is exactly \(60^\circ\). When the \(60^\circ\) ray at \(Q\) is drawn and an arc of radius \(6\) cm is swung from \(P\), the point \(R\) lands where \(PR\) is perpendicular to \(QR\), so \(\angle PRQ = 90^\circ\). In a right-angled triangle the mid-point of the hypotenuse is equidistant from all three vertices, so

\[|MP| = |MQ| = |MR| = 3.5\ \text{cm}.\]

Since \(|MQ| = |MR|\), triangle \(MQR\) is isosceles and \(\angle MRQ = \angle MQR = \angle PQR = 60^\circ\). The three angles of triangle \(MQR\) must sum to \(180^\circ\), so

\[\angle RMQ = 180^\circ - 60^\circ - 60^\circ = 60^\circ.\]

(In fact \(|QR| = 7\cos 60^\circ = 3.5\) cm as well, so \(MQR\) is equilateral, which confirms \(\angle RMQ = 60^\circ\).)

Examination reminder: the marks are awarded for the visible construction arcs (the \(60^\circ\) at \(Q\) and the perpendicular bisector of \(PQ\)) and for a measured value within about \(\pm 1^\circ\) of the true figure. A reading of \(90^\circ\) is a sign that \(RM\) was mistaken for the perpendicular bisector rather than the line to \(R\); measure the angle between \(MR\) and \(MQ\) at \(M\), and the correct value is \(60^\circ\).

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Reading the pie chart. The whole circle (\(360^\circ\)) represents all \(200\) pupils, so each degree stands for \(\frac{200}{360}=\frac{5}{9}\) of a pupil. The sectors read off the diagram are:

Class (marks)	Angle
31 - 40	\(54^\circ\)
41 - 50	\(90^\circ\)
51 - 60	\(108^\circ\)
61 - 70	\(54^\circ\)
71 - 80	\(36^\circ\)
81 - 90	\(18^\circ\)

Check: \(54+90+108+54+36+18=360^\circ\). Good.

(a)(i) Pupils scoring between 41 and 50. This is the \(90^\circ\) sector.

\[\frac{90}{360}\times 200 = 50 \text{ pupils.}\]

(a)(ii) Pupils scoring above 80. This is the \(81\text{-}90\) sector, \(18^\circ\).

\[\frac{18}{360}\times 200 = 10 \text{ pupils.}\]

(b) Fraction scoring at most 50 marks. “At most 50” means the \(31\text{-}40\) and \(41\text{-}50\) classes together: \(54^\circ+90^\circ=144^\circ\).

\[\text{Fraction}=\frac{144}{360}=\frac{2}{5}.\]

(c) Modal class. The modal class is the one with the greatest sector angle (largest frequency), which is \(108^\circ\).

Therefore the modal class is 51 - 60.

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(a) Since \(\cos x = 0.7431\) and \(0^\circ < x < 90^\circ\), from tables \(x = 42^\circ\).

(i) \(\sin42^\circ = 0.6691\), so

\[2\sin x = 2\times0.6691 = 1.338.\]

(ii) \(\dfrac{x}{2} = 21^\circ\), so

\[\tan\frac{x}{2} = \tan21^\circ = 0.3839.\]

(b) The five interior angles of a pentagon sum to \((5-2)\times180^\circ = 540^\circ\). With ratio \(2:3:4:4:5\), the total number of parts is \(2+3+4+4+5 = 18\).

\[\text{One part} = \frac{540^\circ}{18} = 30^\circ.\]

\[\text{Largest angle} = 5\times30^\circ = 150^\circ.\]

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(a) Tax payable. Taxable income \(= 150{,}000 - 40{,}000 = \text{N}110{,}000\). Tax at 25 kobo (\(\text{N}0.25\)) per naira:

\[\text{Tax} = 0.25\times110{,}000 = \text{N}27{,}500.\]

\[\text{Amount left} = 150{,}000 - 27{,}500 = \text{N}122{,}500.\]

(b) Percentage profit on the whole. The 650 copies are marked at N75 to yield 30% profit, so the total cost price is

\[\text{CP} = \frac{650\times75}{1.30} = \frac{48{,}750}{1.30} = \text{N}37{,}500.\]

Sales received: 300 copies at 5% discount \(= 300\times(75\times0.95) = 300\times71.25 = \text{N}21{,}375\); the other 350 copies at N75 \(= 350\times75 = \text{N}26{,}250\).

\[\text{Total sales} = 21{,}375 + 26{,}250 = \text{N}47{,}625.\]

\[\text{Profit} = 47{,}625 - 37{,}500 = \text{N}10{,}125.\]

\[\%\text{ profit} = \frac{10{,}125}{37{,}500}\times100 = 27\%.\]

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	Limes	Apples	Total
Good	10	8	18
Bad	6	6	12
Total	16	14	30

There are 30 fruits. Picking is without replacement, so the second denominator is 29.

(i) Both good limes (10 good limes):

\[ \frac{10}{30}\times\frac{9}{29}=\frac{90}{870}=\frac{3}{29}\approx 0.103 \]

(ii) Both bad fruits (12 bad fruits):

\[ \frac{12}{30}\times\frac{11}{29}=\frac{132}{870}=\frac{22}{145}\approx 0.152 \]

(iii) One good apple and one bad lime (either order; 8 good apples, 6 bad limes):

\[ \frac{8}{30}\times\frac{6}{29}+\frac{6}{30}\times\frac{8}{29}=\frac{48+48}{870}=\frac{96}{870}=\frac{16}{145}\approx 0.110 \]

(b) Solve \(\log_3(4x+1)-\log_3(3x-5)=2\). Combine the logs:

\[ \log_3\!\left(\frac{4x+1}{3x-5}\right)=2 \;\Rightarrow\; \frac{4x+1}{3x-5}=3^2=9 \] \[ 4x+1=9(3x-5)=27x-45 \;\Rightarrow\; 46=23x \;\Rightarrow\; x=2 \]

Check: \(3x-5=1>0\) and \(4x+1=9>0\), so \(x=2\) is valid.

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(a)

\[\frac{4.56\times3.6}{0.12} = \frac{16.416}{0.12} = 136.8 = 1.368\times10^{2}.\]

(b) Use the difference of two squares \(a^2 - b^2 = (a-b)(a+b)\):

\[(73.8)^2 - (26.2)^2 = (73.8 - 26.2)(73.8 + 26.2) = 47.6\times100 = 4760.\]

(c) Convert the mixed number to an improper fraction:

\[1\tfrac{19}{81} = \frac{81 + 19}{81} = \frac{100}{81},\qquad \sqrt{\frac{100}{81}} = \frac{10}{9}.\]

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(a) Find x and y.

The six numbers in ascending order are \(1,\;2,\;x,\;11,\;y,\;14\).

Use the mean. The mean of the six values is 8, so their sum is \(6\times 8=48\):

\[1+2+x+11+y+14=48\]

\[28+x+y=48\Rightarrow x+y=20\quad(1)\]

Use the median. For six values the median is the average of the 3rd and 4th values, which are \(x\) and \(11\):

\[\frac{x+11}{2}=9\Rightarrow x+11=18\Rightarrow x=7\]

Substitute into (1):

\[7+y=20\Rightarrow y=13\]

Check ascending order: \(1,2,7,11,13,14\) is valid. So \(x=7,\;y=13\).

(b) Area of quadrilateral MPQN.

Since \(MN\parallel PQ\), triangles \(LMN\) and \(LPQ\) are similar (equal angles at \(L\), and corresponding angles equal).

From the diagram, \(|LM|=3\text{ cm}\) and \(|MP|=1\text{ cm}\), so:

\[|LP|=|LM|+|MP|=3+1=4\text{ cm}\]

The ratio of corresponding sides is:

\[\frac{|LM|}{|LP|}=\frac{3}{4}\]

The ratio of areas of similar triangles is the square of the ratio of sides:

\[\frac{\text{Area }\Delta LMN}{\text{Area }\Delta LPQ}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\]

Given \(\text{Area }\Delta LMN=18\text{ cm}^{2}\):

\[18=\frac{9}{16}\times\text{Area }\Delta LPQ\]

\[\text{Area }\Delta LPQ=18\times\frac{16}{9}=32\text{ cm}^{2}\]

The quadrilateral \(MPQN\) is the region between the two parallel lines:

\[\text{Area }MPQN=\text{Area }\Delta LPQ-\text{Area }\Delta LMN=32-18=14\text{ cm}^{2}\]

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(a) Height risen up the slope

The 100 m is the distance along the slope (the hypotenuse), and the height risen is the vertical (opposite) side of a right triangle whose angle to the horizontal is \(24^\circ\).

Let \(h\) be the height.

\[\sin 24^\circ = \frac{h}{100}\]

\[h = 100 \times \sin 24^\circ = 100 \times 0.4067 = 40.67\text{ m}\]

Correct to the nearest metre, \(h \approx \mathbf{41\text{ m}}\).

(b) Angle BAC of the isosceles triangle

From the diagram, \(|AB| = |AC| = 5\text{ cm}\) and \(|BC| = 8\text{ cm}\). Drop a perpendicular from \(A\) to the midpoint \(M\) of \(BC\). Then \(|BM| = \tfrac{1}{2}\times 8 = 4\text{ cm}\), and \(AM\) bisects \(\angle BAC\).

In right triangle \(ABM\):

\[\sin(\angle BAM) = \frac{BM}{AB} = \frac{4}{5} = 0.8\]

\[\angle BAM = \sin^{-1}(0.8) = 53.13^\circ\]

Therefore

\[\angle BAC = 2 \times 53.13^\circ = 106.26^\circ \approx \mathbf{106^\circ}\]

(c) Height of the light-house

The two boats are on opposite sides of the light-house and in a straight line with its foot, 70 m apart. Let the foot of the light-house be \(F\), the height be \(h\), the horizontal distance to the boat with elevation \(71.6^\circ\) be \(d_1\), and to the boat with elevation \(45^\circ\) be \(d_2\).

\[d_1 + d_2 = 70\]

From the \(71.6^\circ\) boat: \(\tan 71.6^\circ = \dfrac{h}{d_1} = 3\), so \(d_1 = \dfrac{h}{3}\).

From the \(45^\circ\) boat: \(\tan 45^\circ = \dfrac{h}{d_2} = 1\), so \(d_2 = h\).

Substitute:

\[\frac{h}{3} + h = 70\]

\[\frac{h + 3h}{3} = 70 \quad\Rightarrow\quad \frac{4h}{3} = 70\]

\[h = \frac{70 \times 3}{4} = \frac{210}{4} = 52.5\text{ m}\]

The height of the light-house is \(\mathbf{52.5\text{ m}}\).

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(a) \(X\) is on longitude \(67^\circ E\) and \(Y\) on \(123^\circ E\); both lie on the equator. Take \(\pi=\frac{22}{7}\) and \(R=6400\text{ km}\).

(i) Distance between \(X\) and \(Y\) along the equator. The difference in longitude is

\[\theta=123^\circ-67^\circ=56^\circ.\]

Distance along the equator (a great circle):

\[d=\frac{\theta}{360^\circ}\times 2\pi R=\frac{56}{360}\times 2\times\frac{22}{7}\times 6400.\]\[d=\frac{56}{360}\times\frac{2\times22\times6400}{7}=\frac{56}{360}\times 40228.57\]\[d\approx 6257.78\text{ km}\approx\mathbf{6258\text{ km}}.\]

(ii) Distance of \(X\) from the North pole. The North pole is \(90^\circ\) of latitude from the equator, measured along a meridian (a great circle):

\[d=\frac{90}{360}\times 2\pi R=\frac{1}{4}\times 2\times\frac{22}{7}\times 6400\]\[d=\frac{1}{4}\times 40228.57\approx\mathbf{10057\text{ km}}.\]

(b) In circle centre \(O\), \(N\) is the mid-point of chord \(PQ\) with \(|PQ|=8\text{ cm}\), \(|ON|=3\text{ cm}\) and \(\angle ONR=20^\circ\). Find \(\angle ORN\).

Since \(N\) is the mid-point of the chord, \(ON\perp PQ\) and

\[|PN|=\tfrac{1}{2}|PQ|=4\text{ cm}.\]

The radius is \(|OP|\); by Pythagoras in right triangle \(ONP\):

\[|OP|=\sqrt{|ON|^2+|PN|^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ cm}.\]

\(R\) lies on the circle, so \(|OR|=5\text{ cm}\) too. Now apply the sine rule in triangle \(ONR\), where \(|ON|=3\), \(|OR|=5\) and \(\angle ONR=20^\circ\):

\[\frac{\sin\angle ORN}{|ON|}=\frac{\sin\angle ONR}{|OR|}\]\[\sin\angle ORN=\frac{|ON|\sin 20^\circ}{|OR|}=\frac{3\times0.3420}{5}=0.2052\]\[\angle ORN=\sin^{-1}(0.2052)\approx 11.8^\circ\approx\mathbf{12^\circ}.\]

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(a)(i) Volume of soil dug from the well.

The well is a cylinder of radius \(r=1\text{ m}\) and depth \(h=8\text{ m}\).

\[V=\pi r^2 h=\frac{22}{7}\times 1^2\times 8=\frac{176}{7}=25.14\text{ m}^3\]

The volume of soil dug out is \(25.14\text{ m}^3\) (to 2 d.p.).

(a)(ii) Thickness of the new layer of soil.

The soil is spread over a rectangular floor \(4\text{ m}\times 12\text{ m}\).

\[\text{Floor area}=4\times 12=48\text{ m}^2\]

Let the thickness be \(t\). Volume of the layer = floor area \(\times\) thickness, and this equals the volume of soil:

\[48\times t=\frac{176}{7}\]\[t=\frac{176}{7\times 48}=\frac{176}{336}=0.5238\text{ m}\]

Converting to centimetres: \(0.5238\times 100=52.38\text{ cm}\).

Correct to the nearest cm, the thickness is \(52\text{ cm}\).

(b) Quadrilateral \(ABCD\).

From the diagram, \(\angle DAB=90^{\circ}\), \(|AB|=3.3\text{ cm}\), \(|AD|=5.6\text{ cm}\), \(|BC|=3.9\text{ cm}\) and \(|DC|=5.2\text{ cm}\).

(i) Length of \(BD\).

Triangle \(ABD\) is right-angled at \(A\), so by Pythagoras:

\[BD^2=AB^2+AD^2=3.3^2+5.6^2=10.89+31.36=42.25\]\[BD=\sqrt{42.25}=6.5\text{ cm}\]

(ii) Show that \(\angle BCD=90^{\circ}\).

In triangle \(BCD\), \(|BC|=3.9\), \(|DC|=5.2\) and \(|BD|=6.5\). Test the converse of Pythagoras:

\[BC^2+DC^2=3.9^2+5.2^2=15.21+27.04=42.25\]\[BD^2=6.5^2=42.25\]

Since \(BC^2+DC^2=BD^2\), the converse of Pythagoras theorem holds, so the angle opposite \(BD\) is a right angle. Therefore \(\angle BCD=90^{\circ}\).

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(a) Common difference. With \(a = 8\), the 7th and 9th terms are \(8 + 6d\) and \(8 + 8d\), and their ratio is \(5 : 8\):

\[\frac{8 + 6d}{8 + 8d} = \frac{5}{8}\Rightarrow 8(8 + 6d) = 5(8 + 8d).\]

\[64 + 48d = 40 + 40d \Rightarrow 8d = -24 \Rightarrow d = -3.\]

(b)(i) Volume of the sphere (\(r = 2\) cm):

\[V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\times\tfrac{22}{7}\times8 = \frac{704}{21} = 33.52\text{ cm}^3.\]

(ii) Density:

\[\rho = \frac{\text{mass}}{\text{volume}} = \frac{11.2}{33.52} = 0.334\text{ g/cm}^3.\]

(iii) Mass of a sphere of radius 3 cm (same material). Mass varies as \(r^3\):

\[\text{mass} = 11.2\times\left(\frac{3}{2}\right)^3 = 11.2\times\frac{27}{8} = 37.8\text{ g}.\]

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(a) Substitute \(a=1,\ b=2,\ y=3\) into \(y = ax^2 - bx - 12\):

\[3 = x^2 - 2x - 12 \Rightarrow x^2 - 2x - 15 = 0.\]

Factorise: \((x - 5)(x + 3) = 0\), giving

\[x = 5\quad\text{or}\quad x = -3.\]

(b) Square both sides of \(\sqrt{x^2 + 1} = \tfrac{5}{4}\):

\[x^2 + 1 = \frac{25}{16} \Rightarrow x^2 = \frac{25}{16} - 1 = \frac{9}{16}.\]

\[x = \frac{3}{4}\quad(\text{positive value}).\]