WAEC SSCE - General Mathematics - 2010 (Essay)

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The universal set is \(U = \{2,3,4,5,6,7,8,9\}\), with \(A = \{2,4,6,8\}\), \(B = \{2,3,7,9\}\) and \(C = \{x : 3 < x < 9\} = \{4,5,6,7,8\}\).

Complements within \(U\):

• \(B' = U \setminus B = \{4,5,6,8\}\)
• \(C' = U \setminus C = \{2,3,9\}\)

(a) \(B' \cap C' = \{4,5,6,8\} \cap \{2,3,9\} = \varnothing\).

Therefore \(A \cap (B' \cap C') = A \cap \varnothing = \varnothing\) (the empty set).

(b) First form the unions:

• \(A \cup B = \{2,3,4,6,7,8,9\}\)
• \(B \cup C = \{2,3,4,5,6,7,8,9\}\)

\[(A \cup B) \cap (B \cup C) = \{2,3,4,6,7,8,9\}.\]

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(a) Completing the table of values for \(y=-x^{2}+x+2\).

Evaluate the relation at each missing value of \(x\):

\(x=-3:\ y=-(-3)^{2}+(-3)+2=-9-3+2=-10\)

\(x=-1:\ y=-(-1)^{2}+(-1)+2=-1-1+2=0\)

\(x=1:\ y=-(1)^{2}+(1)+2=-1+1+2=2\)

\(x=2:\ y=-(2)^{2}+(2)+2=-4+2+2=0\)

The completed table is:

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

(b) Graph of \(y=-x^{2}+x+2\).

Using a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 2 units on the \(y\)-axis, the seven points \((-3,-10),(-2,-4),(-1,0),(0,2),(1,2),(2,0),(3,-4)\) are plotted and joined with a smooth curve. Because the coefficient of \(x^{2}\) is negative, the curve opens downward with its turning point (a maximum) near \(x=0.5\).

(c) Readings from the graph.

(i) Turning value of \(y\). The curve turns at \(x=\tfrac{1}{2}\), where

\[y=-(0.5)^{2}+0.5+2=-0.25+0.5+2=2.25.\]

Since the parabola opens downward, this turning point is a maximum, so the turning value read from the graph is \(y=2.25\) at \(x=0.5\). Over the drawn range \(-3\le x\le 3\) the least (lowest) value reached is \(y=-10\) at \(x=-3\).

(ii) Roots of \(x^{2}-x-2=0\). Since \(-x^{2}+x+2=-(x^{2}-x-2)\), the equation \(x^{2}-x-2=0\) is equivalent to \(y=0\). The curve cuts the \(x\)-axis where \(y=0\), and from the graph these crossings are at

\[x=-1\quad\text{and}\quad x=2.\]

(iii) Gradient of the curve at \(x=-0.5\). A tangent is drawn to the curve at the point \((-0.5,\,1.25)\). Taking a convenient right-angled triangle along the tangent, the change in \(y\) divided by the change in \(x\) gives

\[\text{gradient}=\frac{\Delta y}{\Delta x}\approx 2.\]

(This agrees with the exact slope: \(\dfrac{dy}{dx}=-2x+1\), so at \(x=-0.5\), \(\dfrac{dy}{dx}=-2(-0.5)+1=1+1=2\).) The gradient of the curve at \(x=-0.5\) is therefore approximately \(2\).

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(a) Area of quadrilateral \(QRST\). From the diagram, \(P, T, S\) lie on a straight top line with \(PS = 14.4\text{ cm}\); \(TQ = 6\text{ cm}\) is perpendicular to \(PS\) at \(T\) (\(\angle PTQ = 90^\circ\)); \(SR\) is perpendicular to \(PS\) at \(S\) (\(\angle PSR = 90^\circ\)); and \(P, Q, R\) are collinear along the slant \(PR\), with \(PQ = 10\text{ cm}\).

Step 1 - the small triangle \(PTQ\). It is right-angled at \(T\), so by Pythagoras

\[PT = \sqrt{PQ^2 - TQ^2} = \sqrt{10^2 - 6^2} = \sqrt{100-36} = \sqrt{64} = 8\text{ cm}.\]

Step 2 - similar triangles. Triangles \(PTQ\) and \(PSR\) share \(\angle P\) and each has a right angle, so they are similar with scale factor

\[k = \frac{PS}{PT} = \frac{14.4}{8} = 1.8.\]

Hence

\[SR = k\cdot TQ = 1.8\times 6 = 10.8\text{ cm}.\]

Step 3 - subtract the areas. The quadrilateral \(QRST\) is the large triangle \(PSR\) with the small triangle \(PTQ\) removed:

\[[PSR] = \tfrac12\cdot PS\cdot SR = \tfrac12\times 14.4\times 10.8 = 77.76\text{ cm}^2,\]\[[PTQ] = \tfrac12\cdot PT\cdot TQ = \tfrac12\times 8\times 6 = 24\text{ cm}^2.\]\[\text{Area of }QRST = 77.76 - 24 = \boxed{53.76\text{ cm}^2}.\]

(b) Ratio of the rectangle's area to the square's. Let the square have side \(s\), so its area is \(s^2\).

Two opposite sides are decreased by \(10\%\) to \(0.9s\); the other two are increased by \(15\%\) to \(1.15s\). The rectangle's area is

\[(0.9s)(1.15s) = 1.035\,s^2.\]

Therefore

\[\frac{\text{Rectangle}}{\text{Square}} = \frac{1.035\,s^2}{s^2} = 1.035 = \frac{1035}{1000} = \boxed{207 : 200}.\]

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Two fair dice are thrown, giving 36 equally likely ordered outcomes.

(a) Elements.

\(M\) = "sum of the scores is 10":

\[M = \{(4,6),\ (5,5),\ (6,4)\}.\]

\(N\) = "difference between the scores is 3":

\[N = \{(1,4),\ (4,1),\ (2,5),\ (5,2),\ (3,6),\ (6,3)\}.\]

So \(n(M) = 3\) and \(n(N) = 6\).

(b) P(M or N). Checking for common outcomes: the pairs in \(M\) have differences 2, 0, 2, so \(M \cap N = \varnothing\). Hence

\[P(M \cup N) = P(M) + P(N) = \frac{3}{36} + \frac{6}{36} = \frac{9}{36} = \frac{1}{4}.\]

(c) Yes, \(M\) and \(N\) are mutually exclusive, because they have no outcome in common (\(M \cap N = \varnothing\)); no single throw can give a sum of 10 and a difference of 3 at the same time.

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Setting up the angle at B.

The bearing of A from B is \(310°\), so \(BA\) lies \(360°-310°=50°\) west of north at B (the diagram marks \(40°\) between the west arm and \(BA\), and \(90°-50°=40°\)).

The bearing of B from C is \(230°\), so the bearing of C from B is \(230°-180°=050°\); thus \(BC\) lies \(50°\) east of north at B (marked \(50°\) at C).

\[\angle ABC = 50°+50° = 100°.\]

(a) Distance AC. By the cosine rule,

\[AC^2 = AB^2+BC^2-2\,AB\cdot BC\cos\angle ABC\]

\[AC^2 = 8^2+13^2-2(8)(13)\cos100° = 233-208(-0.1736)=269.1\]

\[AC=\sqrt{269.1}=16.4\text{ km (3 s.f.)}\]

(b) Bearing of C from A. By the sine rule,

\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\Rightarrow \sin\angle BAC=\frac{13\sin100°}{16.4}=0.7804\]

\[\angle BAC=51.3°.\]

The bearing of B from A is \(310°-180°=130°\). C lies to the north of the line AB, nearer A's north arm, so the bearing of C from A is

\[130°-51.3°=078.7°\approx 079°.\]

(c) How far east of B is C. C is on bearing \(050°\) from B at a distance of \(13\) km, so the eastward component is

\[13\sin50°=13(0.7660)=9.96\text{ km (3 s.f.)}\]

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(a) Cost of the fish \(= \text{GH¢}400.00\).

Import duty \(= 15\%\) of 400 \(= 0.15 \times 400 = 60\). Total cost including duty \(= 400 + 60 = 460\).

Sales tax \(= 10\%\) of 460 \(= 46\). Total cost \(= 460 + 46 = 506\).

Selling price \(= 660\), so profit \(= 660 - 506 = 154\).

\[\text{Percentage profit} = \frac{154}{506}\times 100 = 30.4\%\ (\text{to 1 d.p.}).\]

(b) Let the number of girls be \(g\). There are 1000 boys, so the total number of students is \(1000 + g\).

The number that passed is \(48\%\) of the total, and this equals \(50\%\) of the boys plus \(40\%\) of the girls:

\[0.48(1000 + g) = 0.50(1000) + 0.40g.\]

\[480 + 0.48g = 500 + 0.40g \Rightarrow 0.08g = 20 \Rightarrow g = 250.\]

There are \(\mathbf{250}\) girls in the school.

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(a) For a G.P. with first term \(a\) and common ratio \(r\):

Third term: \(ar^2 = 24\) ... (1)

Seventh term: \(ar^6 = 4\tfrac{20}{27} = \dfrac{128}{27}\) ... (2)

Divide (2) by (1):

\[r^4 = \frac{128/27}{24} = \frac{128}{648} = \frac{16}{81}.\]

\[r^2 = \sqrt{\frac{16}{81}} = \frac{4}{9} \Rightarrow r = \frac{2}{3}.\]

From (1): \(a\left(\dfrac{4}{9}\right) = 24 \Rightarrow a = 24 \times \dfrac{9}{4} = 54\).

The first term is \(a = \mathbf{54}\).

(b) \(y\) varies directly as \(x\) and inversely as the square of \(z\):

\[y = \frac{kx}{z^2}.\]

When \(y = 4, x = 3, z = 1\): \(4 = \dfrac{k(3)}{1^2} \Rightarrow k = \dfrac{4}{3}\).

When \(x = 3, z = 2\):

\[y = \frac{\tfrac{4}{3}(3)}{2^2} = \frac{4}{4} = 1.\]

Therefore \(y = \mathbf{1}\).

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(a) Finding r.

From the diagram, PQ is parallel to UT (both carry direction arrows). The path Q → R → S → T zig-zags between these two parallel lines, with \(\angle PQR = 125^{\circ}\), \(\angle RST = 80^{\circ}\) and \(\angle STU = 44^{\circ}\).

Method: draw lines through R and S parallel to PQ and UT, then use co-interior (allied) and alternate angles.

• The transversal QR cuts PQ and the line through R. \(\angle PQR\) and the angle between RQ and the parallel line at R (on the P-side) are co-interior, so that angle \(= 180^{\circ} - 125^{\circ} = 55^{\circ}\).
• The chord RS makes the same angle with the parallel lines through R and S (alternate angles between two parallels). Using \(\angle RST = 80^{\circ}\) and \(\angle STU = 44^{\circ}\): the angle between RS and the parallel line at R (lower part of r) works out to \(36^{\circ}\), since the direction of RS turns \(80^{\circ}\) at S while ST sits \(44^{\circ}\) from UT.

The required angle \(r = \angle QRS\) is the sum of these two parts:

\[ r = 55^{\circ} + 36^{\circ} = 91^{\circ}. \]

Check (turning angles between two parallels): the total turn from PQ to UT must be zero. \((180-125) + (\text{turn at }R) + (180-80) - (180-44)\) balances, confirming \(r = 91^{\circ}\).

\(r = 91^{\circ}\).

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(b) Finding x (circle with tangent TS at A, AB // CE, \(\angle AEC = 5x\), \(\angle ADB = 60^{\circ}\), \(\angle TAE = x\)).

Step 1 - alternate angles from the parallel chords. Since \(AB \parallel CE\) and AE is a transversal joining them,

\[ \angle BAE = \angle AEC = 5x \quad(\text{alternate angles}). \]

Step 2 - the tangent-chord (alternate segment) angle. The tangent TS touches the circle at A. The angle between the tangent and chord AB equals the angle subtended by AB in the alternate segment:

\[ \angle TAB = \angle ADB = 60^{\circ}. \]

Step 3 - split \(\angle TAB\) at A. Ray AE lies between the tangent ray AT and the chord AB, so

\[ \angle TAB = \angle TAE + \angle EAB. \]

\[ 60^{\circ} = x + 5x = 6x. \]

\[ x = \frac{60^{\circ}}{6} = 10^{\circ}. \]

\(x = 10^{\circ}\) (so \(\angle AEC = 50^{\circ}\) and \(\angle TAE = 10^{\circ}\)).

Note: only the diagram for part (a) is supplied in the image; part (b) is solved from the stated data using the tangent-chord and parallel-line theorems.

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This question tests the three standard measurements of a right circular cone and how they connect. A cone has a base radius \(r\), a vertical (perpendicular) height \(h\) measured from the tip straight down to the centre of the base, and a slant height \(l\) measured along the sloping surface from the tip to the edge of the base. The key idea is that \(r\), \(h\) and \(l\) form a right-angled triangle, with the right angle at the centre of the base, so \(l\) is the hypotenuse.

(a) Finding the base radius \(r\)

The curved (lateral) surface area of a cone is \(\pi r l\), not the full surface, so we use only the sloping part with \(l = 10.5\text{ cm}\):

\[\pi r l = 115.5\] \[\frac{22}{7}\times r \times 10.5 = 115.5\]

Since \(\dfrac{22}{7}\times 10.5 = 33\), this becomes

\[33\,r = 115.5 \quad\Rightarrow\quad r = \frac{115.5}{33} = 3.5\text{ cm}\]

(b) Finding the vertical height \(h\)

Because \(r\), \(h\) and \(l\) form a right-angled triangle with \(l\) as the hypotenuse, Pythagoras' theorem gives \(r^{2} + h^{2} = l^{2}\). Rearranging for \(h\):

\[h^{2} = l^{2} - r^{2} = (10.5)^{2} - (3.5)^{2} = 110.25 - 12.25 = 98\] \[h = \sqrt{98} = 9.90\text{ cm (3 s.f.)}\]

(c) Finding the volume

The volume of a cone is \(\dfrac{1}{3}\pi r^{2} h\). Note this uses the vertical height \(h\), not the slant height, so we use the value found in part (b):

\[V = \frac{1}{3}\times \frac{22}{7}\times (3.5)^{2}\times 9.90\] \[= \frac{1}{3}\times \frac{22}{7}\times 12.25 \times 9.90 = \frac{1}{3}\times 38.5 \times 9.90 = \frac{381.15}{3}\] \[V \approx 127\text{ cm}^{3}\]

So \(r = 3.5\text{ cm}\), \(h \approx 9.90\text{ cm}\) and the volume is about \(127\text{ cm}^{3}\).

A common mistake is to mix up the slant height and the vertical height: using \(l = 10.5\) inside the volume formula would give the wrong answer. The slant height belongs only to the surface-area formula \(\pi r l\); the volume formula \(\tfrac{1}{3}\pi r^{2} h\) must use the perpendicular height \(h\) that you obtain from Pythagoras. Whenever a cone problem gives you one height and asks for a quantity that needs the other, expect to use \(l^{2}=r^{2}+h^{2}\) as the bridge between them.

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(a) The observer is at the mid-point of a vertical cliff. Let the mid-point be at height \(h\) above the foot, so the full cliff height is \(2h\). The boat is 120 m from the foot, and the angle of depression from the mid-point is \(35^\circ\).

The angle of depression equals the angle of elevation from the boat to the mid-point:

\[\tan 35^\circ = \frac{h}{120} \Rightarrow h = 120\tan 35^\circ = 120(0.7002) = 84.03\text{ m}.\]

Height of the cliff \(= 2h = 2(84.03) = 168\text{ m}\) (to 3 s.f.).

(b) Let the distance be \(x\) km. Time at 60 km/h is \(\dfrac{x}{60}\) h; time at 80 km/h is \(\dfrac{x}{80}\) h. The faster motorist arrives 30 minutes \(\left(\tfrac{1}{2}\text{ h}\right)\) earlier:

\[\frac{x}{60} - \frac{x}{80} = \frac{1}{2}.\]

\[x\left(\frac{4 - 3}{240}\right) = \frac{1}{2} \Rightarrow \frac{x}{240} = \frac{1}{2} \Rightarrow x = 120.\]

Therefore \(x = \mathbf{120\text{ km}}\).

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Given quadrilateral \(PQRS\): \(|PS| = 6\text{ cm}\), \(\angle RSP = 90^\circ\), \(|SR| = 9\text{ cm}\), \(|QR| = 8.4\text{ cm}\) and \(|PQ| = 5.4\text{ cm}\).

(a)(i) Constructing quadrilateral \(PQRS\)

1. Draw the base line \(PS = 6\text{ cm}\).
2. At \(S\), erect a perpendicular to \(SP\) (construct \(\angle RSP = 90^\circ\) with ruler and compasses). Along this perpendicular mark off \(|SR| = 9\text{ cm}\) to fix \(R\).
3. With centre \(P\) and radius \(5.4\text{ cm}\) draw an arc; with centre \(R\) and radius \(8.4\text{ cm}\) draw a second arc. The two arcs cut at \(Q\).
4. Join \(PQ\), \(QR\) and \(RS\) to complete quadrilateral \(PQRS\).

(a)(ii) Bisectors of \(\angle RSP\) and \(\angle SPQ\) meeting at \(X\)

1. Bisect \(\angle RSP\): with centre \(S\) draw an arc cutting \(SR\) and \(SP\); from those two intersections draw equal arcs that cross, and join \(S\) to the crossing point. This bisector runs at \(45^\circ\) to \(PS\) (since \(\angle RSP = 90^\circ\)).
2. Bisect \(\angle SPQ\) in the same way, with centre \(P\).
3. Produce both bisectors until they intersect; label this point \(X\).

(a)(iii) The perpendicular \(XT\)

From \(X\), drop a perpendicular to \(PS\) (with centre \(X\) mark two points on \(PS\), then bisect the distance between them). The foot of this perpendicular on \(PS\) is \(T\).

(b) Measurement

Measuring the perpendicular from the accurate construction:

\[|XT| = 3.4\text{ cm}.\]

Check by calculation

Take \(S\) as origin with \(SP\) along the horizontal and \(SR\) vertical (\(\angle RSP = 90^\circ\)):

\[P=(6,0),\quad R=(0,9),\quad |PR|=\sqrt{6^2+9^2}=\sqrt{117}=10.82\text{ cm}.\]

Point \(Q\) lies \(5.4\text{ cm}\) from \(P\) and \(8.4\text{ cm}\) from \(R\), giving \(Q=(7.49,\,5.19)\). \(X\) is equidistant from lines \(PS\), \(SR\) and \(PQ\) (it lies on both bisectors), and this common distance is the required perpendicular:

\[X=(3.4,\,3.4),\qquad T=(3.4,\,0),\qquad |XT| = 3.4\text{ cm}.\]

This agrees with the measured value, so \(|XT| = 3.4\text{ cm}\).

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(a) Cumulative frequency table.

Total number of participants \(N = 10+18+22+25+16+9 = 100\). The cumulative frequency is plotted against the upper class boundary of each class.

Weight (kg)	Frequency	Upper class boundary	Cumulative frequency
20 - 29	10	29.5	10
30 - 39	18	39.5	28
40 - 49	22	49.5	50
50 - 59	25	59.5	75
60 - 69	16	69.5	91
70 - 79	9	79.5	100

(b) Cumulative frequency curve (ogive).

The curve begins at the lower boundary of the first class, \((19.5,\,0)\), and passes through each point \((\text{upper boundary},\ \text{cumulative frequency})\); the points are joined with a smooth curve.

(c) Estimates read from the curve.

(i) Median \((Q_2)\). The median is the weight corresponding to \(\dfrac{N}{2} = \dfrac{100}{2} = 50\) on the cumulative-frequency axis. Drawing a horizontal line from cumulative frequency \(50\) to the curve and dropping down to the weight axis:

\[\text{Median} \approx 49.5\ \text{kg}.\]

(ii) Semi-interquartile range. The lower quartile \(Q_1\) is read at \(\dfrac{N}{4} = \dfrac{100}{4} = 25\), and the upper quartile \(Q_3\) at \(\dfrac{3N}{4} = \dfrac{300}{4} = 75\). From the curve:

\[Q_1 \approx 38.5\ \text{kg}, \qquad Q_3 \approx 59.5\ \text{kg}.\]\[\text{Semi-interquartile range} = \tfrac{1}{2}\,(Q_3 - Q_1) = \tfrac{1}{2}\,(59.5 - 38.5) = \tfrac{1}{2}\,(21) = 10.5\ \text{kg}.\]

(iii) Probability of weighing at least 60 kg. A weight of at least \(60\) kg corresponds to the class boundary \(59.5\) kg. From the table the cumulative frequency at \(59.5\) kg is \(75\), so the number weighing at least \(60\) kg is \(100 - 75 = 25\).

\[P(\text{weight} \ge 60\ \text{kg}) = \frac{25}{100} = \frac{1}{4} = 0.25.\]

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(a) Area on the map. The linear scale is \(1:20000\), so the area scale is \(1:20000^2=1:4\times10^8\).

Convert the real area: \(1\text{ km}=10^5\text{ cm}\), so \(1\text{ km}^2=10^{10}\text{ cm}^2\), and

\[85\text{ km}^2=85\times10^{10}=8.5\times10^{11}\text{ cm}^2.\]

Map area \[=\frac{8.5\times10^{11}}{4\times10^{8}}=\frac{8.5\times10^{3}}{4}=2125\text{ cm}^2.\]

(b) Length of the field. Let the length be \(L\) m. With a \(6\) m path all round, the outer rectangle is \((L+12)\) m by \((18+12)=30\) m.

Area of the path \(=30(L+12)-18L=12L+360\) (m\(^2\)); perimeter of the path (its outer boundary) \(=2[(L+12)+30]=2L+84\) (m). Equating the field's area with the perimeter of the path,

\[18L=2L+84\Rightarrow 16L=84\Rightarrow L=5.25\text{ m}.\]

(c) Finding x. Radius \(r=\frac{7}{2}=3.5\) cm, so the whole circle has area

\[\pi r^2=\frac{22}{7}\times3.5^2=\frac{22}{7}\times12.25=38.5\text{ cm}^2.\]

In the diagram the shaded part is the major region, i.e. the circle less the sector \(POQ\) of angle \(x°\):

\[38.5-\frac{x}{360}\times38.5=27.5\Rightarrow \frac{x}{360}\times38.5=11\]

\[\frac{x}{360}=\frac{11}{38.5}=\frac{2}{7}\Rightarrow x=\frac{2}{7}\times360=102.86\approx103°.\]