WAEC SSCE - General Mathematics - 1992 (Essay)

Maelezo ya Majibu

(a) Radius \(r=7\text{ cm}\), angle \(=108^{\circ}\). Arc length \[=\frac{108}{360}\times2\pi r=\frac{108}{360}\times2\times\frac{22}{7}\times7=\frac{3}{10}\times44=13.2\text{ cm}.\] Perimeter of sector \(=\) arc \(+2r=13.2+14=\) 27.2 cm.

(b) Angle of depression \(=30^{\circ}\) equals the angle of elevation of the cliff top from the boat. With the boat \(120\text{ m}\) from the foot: \[\tan30^{\circ}=\frac{h}{120}\Rightarrow h=120\tan30^{\circ}=120\times0.5774=69.28\text{ m}\approx69.3\text{ m}.\]

Maelezo ya Majibu

(a) Using difference of two squares: \[375^{2}-356^{2}=(375-356)(375+356)=19\times731=13889.\] So \(17x=13889\Rightarrow x=\dfrac{13889}{17}=817\).

(b) \(4^{x}=2^{2x}\) and \(2^{\frac12}\times8=2^{\frac12}\times2^{3}=2^{\frac72}\). Thus \(2x=\tfrac72\Rightarrow x=\tfrac74\).

(c) \(S_{9}=72\Rightarrow\tfrac92(2a+8d)=72\Rightarrow a+4d=8\). Sum of next \(4\) terms \(=S_{13}-S_{9}=71\Rightarrow S_{13}=143\Rightarrow\tfrac{13}{2}(2a+12d)=143\Rightarrow a+6d=11\). Subtracting: \(2d=3\Rightarrow d=1.5\), then \(a=2\). The AP is \[2,\;3\tfrac12,\;5,\;6\tfrac12,\;8,\;\dots\]

Maelezo ya Majibu

(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).

(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)

Maelezo ya Majibu

(a) Pie chart of junior workers

First find the total number of workers:

\[ 78+36+44+52+30 = 240 \]

The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):

\[ \text{Sector angle} = \frac{\text{Number in category}}{240}\times 360^\circ \]

Category	Number	Sector angle
Clerks	78	\(\frac{78}{240}\times360^\circ=117^\circ\)
Drivers	36	\(\frac{36}{240}\times360^\circ=54^\circ\)
Typists	44	\(\frac{44}{240}\times360^\circ=66^\circ\)
Messengers	52	\(\frac{52}{240}\times360^\circ=78^\circ\)
Others	30	\(\frac{30}{240}\times360^\circ=45^\circ\)
Total	240	\(360^\circ\)

The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:

(b) Mean score of the aptitude test

Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):

Marks \((x)\)	4	5	6	7	8	9	Total
No. of candidates \((f)\)	5	8	5	6	4	2	30
\(fx\)	20	40	30	42	32	18	182

\[ \sum f = 30, \qquad \sum fx = 20+40+30+42+32+18 = 182 \] \[ \text{Mean} = \frac{\sum fx}{\sum f} = \frac{182}{30} = 6.07 \]

To the nearest whole number, the mean score is 6 marks.

Maelezo ya Majibu

(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).

Each missing value is found by substitution, e.g.

\[ x=-1.5:\; y=2.5+(-1.5)-4(-1.5)^2 = 2.5-1.5-9 = -8.0 \] \[ x=1:\; y=2.5+1-4(1)^2 = 2.5+1-4 = -0.5 \] \[ x=1.5:\; y=2.5+1.5-4(1.5)^2 = 2.5+1.5-9 = -5.0 \] \[ x=2.0:\; y=2.5+2-4(2)^2 = 2.5+2-16 = -11.5 \]

x	-2.0	-1.5	-1.0	-0.5	0	0.5	1	1.5	2.0
y	-15.5	-8.0	-2.5	1.0	2.5	2.0	-0.5	-5.0	-11.5

(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).

(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:

\[ y_{\max}=2.5+0.125-4(0.125)^2 = 2.5+0.125-0.0625 \approx 2.6 \]

So the maximum value of \(y\) is approximately 2.6.

(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):

\[ -4x^{2}+x+\tfrac{5}{2}=0 \quad\Longrightarrow\quad \tfrac{5}{2}+x-4x^{2}=0 \]

This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:

\[ x \approx -0.7 \qquad\text{and}\qquad x \approx 0.9 \]

(Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)

Maelezo ya Majibu

(a) \(PQ\) is produced to \(T\), so \(\angle PQR=180^{\circ}-131^{\circ}=49^{\circ}\). \(PR\) is produced to \(S\), so \(\angle PRQ=180^{\circ}-98^{\circ}=82^{\circ}\). The angles of \(\triangle PQR\) sum to \(180^{\circ}\): \[\angle QPR=180^{\circ}-49^{\circ}-82^{\circ}=49^{\circ}.\]

(b) Circumference \(=2\pi r=400\): \[r=\frac{400}{2\times\frac{22}{7}}=\frac{400\times7}{44}=\frac{2800}{44}=63.6\text{ m}\approx64\text{ m}.\]

Maelezo ya Majibu

(a) Cumulative frequency table and curve (ogive)

The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.

Weekly profit (N)	Upper class boundary	Frequency (f)	Cumulative frequency
1 - 10	10.5	6	6
11 - 20	20.5	6	12
21 - 30	30.5	12	24
31 - 40	40.5	11	35
41 - 50	50.5	10	45
51 - 60	60.5	5	50

Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).

(b) Estimates from the graph

(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.

Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):

\[ \text{Median}=30.5+\frac{25-24}{35-24}\times 10 = 30.5+0.9 \approx N\,31 \]

(ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.

Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):

\[ P_{80}=40.5+\frac{40-35}{45-35}\times 10 = 40.5+5 = N\,45.5 \]

(c) Modal weekly profit

The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using

\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]

where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):

\[ \text{Mode}=20.5+\frac{(12-6)}{(12-6)+(12-11)}\times 10 = 20.5+\frac{60}{7} = 20.5+8.57 \]\[ \boxed{\text{Modal weekly profit}\approx N\,29.07} \]

Maelezo ya Majibu

(a) \(P(\text{rolling a }3)=\tfrac16\). For two coins, \(P(\text{at least one tail})=1-P(HH)=1-\tfrac14=\tfrac34\). These are independent, so \[P(3\text{ and a tail})=\frac16\times\frac34=\frac{3}{24}=\frac18.\]

(b) \(10\) marbles: \(7\) black, \(3\) red, drawn without replacement.

• (i) Red then black: \[\frac{3}{10}\times\frac{7}{9}=\frac{21}{90}=\frac{7}{30}.\]
• (ii) Two black: \[\frac{7}{10}\times\frac{6}{9}=\frac{42}{90}=\frac{7}{15}.\]

Maelezo ya Majibu

Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is

\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]

(a) Distance QR (cosine rule).

\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]

\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]

\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]

(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:

\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]

\[\angle PQR = 36.0^\circ.\]

The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is

\[255^\circ + 36^\circ = 291^\circ.\]

Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).

Maelezo ya Majibu

(a) Area of the shaded segment

The shaded segment is found by subtracting the area of the triangle formed by the two radii from the area of the sector:

\[\text{area of segment}=\text{area of sector}-\text{area of triangle}.\]

The angle at the centre is \(63^\circ\), not \(30^\circ\). Both sides of the triangle are radii of length \(10\text{ cm}\).

\[\text{Area of sector}=\frac{63}{360}\times\frac{22}{7}\times10^2=55\text{ cm}^2.\]

For the triangle, use \(\frac12 ab\sin C\):

\[\text{Area of triangle}=\frac12\times10\times10\times\sin63^\circ\]

\[=50\times0.891=44.55\text{ cm}^2.\]

Therefore,

\[\text{Area of shaded segment}=55-44.55=10.45\text{ cm}^2.\]

(b) Tin with radius \(3\text{ cm}\) and height \(6\text{ cm}\)

Total surface area includes the curved surface and both circular ends, so:

\[\text{TSA}=2\pi r(r+h)\]

\[=2\times\frac{22}{7}\times3\times(3+6)\]

\[=\frac{1188}{7}=169.71\text{ cm}^2.\]

Volume of the cylindrical tin is:

\[V=\pi r^2h\]

\[=\frac{22}{7}\times3^2\times6=\frac{1188}{7}=169.71\text{ cm}^3.\]

Since \(1000\text{ cm}^3=1\) litre:

\[169.71\div1000=0.16971\text{ litres}\]

Correct to two decimal places, the capacity is \(0.17\) litres.

Answers: shaded segment \(=10.45\text{ cm}^2\); total surface area \(=169.71\text{ cm}^2\); capacity \(=0.17\) litres.

Examination reminder: For a circular segment, use sector minus triangle, and take the central angle directly from the diagram. Here, using \(30^\circ\) instead of \(63^\circ\) changes both areas and gives the wrong segment area.

Maelezo ya Majibu

(a) Write everything to base \(3\): \(9=3^{2}\), \(81=3^{4}\). \[9^{2x-1}=3^{4x-2},\qquad\frac{81^{x-2}}{3^{x}}=\frac{3^{4x-8}}{3^{x}}=3^{3x-8}.\] So \(4x-2=3x-8\Rightarrow x=-6\).

(b) \[\sqrt{\frac{0.81\times10^{-5}}{2.25\times10^{7}}}=\sqrt{\frac{0.81}{2.25}\times10^{-12}}=\sqrt{0.36\times10^{-12}}=0.6\times10^{-6}=6.0\times10^{-7}.\]

Maelezo ya Majibu

(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):

Steps.

1. Draw a base line and mark \(|AB| = 5\text{ cm}\).
2. Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
3. With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
4. Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
5. Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).

(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.

(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).

This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence

\[|CP| = \sqrt{(2.5-(-3.75))^{2} + (4.33-6.50)^{2}} = \sqrt{6.25^{2} + 2.17^{2}} = \sqrt{43.75} \approx 6.7\text{ cm}.\]