WAEC SSCE - General Mathematics - 2010 (Essay)

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