WAEC SSCE - General Mathematics - 2005 (Essay)

Answer Details

(a) Completing the table. For each value of \(x\) we evaluate \(y = x^{2} - 2x - 2\) by building it up in rows:

\(x\)	-4	-3	-2	-1	0	1	2	3	4
\(x^{2}\)	16	9	4	1	0	1	4	9	16
\(-2x\)	8	6	4	2	0	-2	-4	-6	-8
\(-2\)	-2	-2	-2	-2	-2	-2	-2	-2	-2
\(y\)	22	13	6	1	-2	-3	-2	1	6

The three missing entries are \(y=13\) (at \(x=-3\)), \(y=1\) (at \(x=-1\)) and \(y=-3\) (at \(x=1\)).

(b) Graph of \(y = x^{2} - 2x - 2\). Plotting the nine points \((x,y)\) and joining them with a smooth curve gives the parabola below. The dashed horizontal line \(y = 2.5\) is used in part (c)(ii) and the vertical dashed line \(x = 1\) is the line of symmetry from part (c)(iii).

(c) Using the graph.

(i) Roots of \(x^{2} - 2x - 2 = 0\). These are the values of \(x\) where the curve cuts the \(x\)-axis, i.e. where \(y = 0\). The curve crosses the \(x\)-axis at \[ x \approx -0.7 \quad \text{and} \quad x \approx 2.7. \]

(ii) Values of \(x\) for which \(x^{2} - 2x - 4\tfrac{1}{2} = 0\). Rearranging so that the left side becomes our plotted expression: \[ x^{2} - 2x - 4\tfrac{1}{2} = 0 \;\Rightarrow\; x^{2} - 2x - 2 = 2\tfrac{1}{2} = 2.5. \] So we draw the horizontal line \(y = 2.5\) and read off where it meets the curve: \[ x \approx -1.3 \quad \text{and} \quad x \approx 3.3. \]

(iii) Equation of the line of symmetry. The lowest point (vertex) of the curve occurs at \(x = \dfrac{-(-2)}{2(1)} = 1\), where \(y = -3\). The curve is symmetrical about the vertical line through this point, so the line of symmetry is \[ x = 1. \]

Answer Details

(a) Distance A to B along the parallel of latitude 53°N, through a longitude difference of \(85^\circ - 25^\circ = 60^\circ\):

\[AB = \frac{60}{360}\times 2\pi R\cos53^\circ = \frac{60}{360}\times 2\times\frac{22}{7}\times6400\times\cos53^\circ.\]

\[AB = \frac{1}{6}\times 40228.6\times0.6018 = 4035\text{ km (to the nearest km)}.\]

(b) Time to reach B at 400 km/h:

\[t = \frac{4035}{400} = 10.09 \approx 10\text{ hours}.\]

(c) Latitude of C. Flying south along a meridian, 1° corresponds to

\[\frac{2\pi R}{360} = \frac{1}{360}\times2\times\frac{22}{7}\times6400 = 111.75\text{ km}.\]

The 2000 km southward change in latitude is

\[\frac{2000}{111.75} = 17.9^\circ.\]

\[\text{Latitude of C} = 53^\circ - 17.9^\circ = 35.1^\circ \approx 35^\circ\text{N}.\]

Answer Details

(a) Mean age

Take the class mid-value (class-mark) \(x\) of each interval and form \(fx\).

Age (years)	Mid-value \(x\)	Frequency \(f\)	\(fx\)
10 - 14	12	7	84
15 - 19	17	18	306
20 - 24	22	25	550
25 - 29	27	17	459
30 - 34	32	9	288
35 - 39	37	4	148
Total		80	1835

\[ \bar{x}=\frac{\sum fx}{\sum f}=\frac{1835}{80}=22.9375 \]

Mean age \( \approx \mathbf{22.9\ years}\) (to 1 d.p.).

(b)(i) Histogram

First convert each class to its continuous class boundaries, then draw bars whose heights equal the frequencies (the bars touch, since the boundaries are continuous).

Age (years)	Class boundaries	Frequency \(f\)
10 - 14	9.5 - 14.5	7
15 - 19	14.5 - 19.5	18
20 - 24	19.5 - 24.5	25
25 - 29	24.5 - 29.5	17
30 - 34	29.5 - 34.5	9
35 - 39	34.5 - 39.5	4

(b)(ii) Modal age from the histogram

The tallest bar is the modal class \(20\text{-}24\) (boundaries \(19.5\text{-}24.5\)). To read the mode, join the top-left corner of the modal bar to the top-left corner of the bar after it, and the top-right corner of the modal bar to the top-right corner of the bar before it. From the point where these two lines cross, drop a vertical line to the age axis.

The vertical line meets the axis at approximately \(\mathbf{21.8\ years}\).

This agrees with the calculation \[ \text{Mode}=L+\frac{\Delta_1}{\Delta_1+\Delta_2}\times c =19.5+\frac{25-18}{(25-18)+(25-17)}\times 5 =19.5+\frac{7}{15}\times 5 =19.5+2.3=21.8\ \text{years}. \]

(c) Probability of being in the modal class

The modal class \(20\text{-}24\) contains \(25\) members out of a total of \(80\):

\[ P(\text{modal class})=\frac{25}{80}=\frac{5}{16}=0.3125 \]

Answer Details

(a) Circle through A, B, C, D with tangent XY at A.

From the diagram: \(\angle ADB=20^\circ\) (at D), and the tangent-secant angle at Y (between tangent \(YA\) and the secant through \(B\) and \(C\)) is \(\angle AYC=69^\circ\), with \(C,\,B,\,Y\) in a straight line.

(i) \(\angle CAX\).

\(\angle ADB=20^\circ\) is an inscribed angle on chord \(AB\), so

\[\text{arc } AB=2\times20^\circ=40^\circ.\]

The tangent-chord angle \(\angle BAY\) equals the angle in the alternate segment \(\angle ADB\):

\[\angle BAY=20^\circ.\]

In triangle \(ABY\), \(\angle AYB=69^\circ\), so

\[\angle ABY=180^\circ-69^\circ-20^\circ=91^\circ,\]

and since \(C,B,Y\) are collinear,

\[\angle ABC=180^\circ-\angle ABY=180^\circ-91^\circ=89^\circ.\]

By the alternate segment theorem, the tangent-chord angle \(\angle CAX\) equals the inscribed angle \(\angle ABC\) in the alternate segment:

\[\angle CAX=\angle ABC=\boxed{89^\circ}.\]

(ii) \(\angle ABY\).

From the working above,

\[\angle ABY=180^\circ-69^\circ-20^\circ=\boxed{91^\circ}.\]

(b) Factors of \(m^{2}-km+c\).

If \((m+1)\) and \((m-3)\) are factors, then

\[m^{2}-km+c=(m+1)(m-3)=m^{2}-3m+m-3=m^{2}-2m-3.\]

Comparing coefficients:

\[-k=-2\ \Rightarrow\ \boxed{k=2},\qquad c=\boxed{-3}.\]

Answer Details

(a) Two dice give \(6\times6 = 36\) equally likely outcomes.

(i) Same digit: the pairs are (1,1),(2,2),...,(6,6), that is 6 outcomes.

\[P(\text{same digit}) = \tfrac{6}{36} = \tfrac{1}{6}.\]

(ii) Total greater than 5: count the totals that are 5 or less: total 2 (1 way), 3 (2), 4 (3), 5 (4), giving \(1+2+3+4 = 10\) outcomes.

\[P(\text{total} > 5) = \tfrac{36-10}{36} = \tfrac{26}{36} = \tfrac{13}{18}.\]

(b) \(x = \cos30^\circ = \tfrac{\sqrt3}{2}\Rightarrow x^2 = \tfrac{3}{4};\quad y = \sin30^\circ = \tfrac{1}{2}\Rightarrow y^2 = \tfrac{1}{4}.\)

\[\frac{x^2 + y^2}{y^2 - x^2} = \frac{\tfrac{3}{4}+\tfrac{1}{4}}{\tfrac{1}{4}-\tfrac{3}{4}} = \frac{1}{-\tfrac{1}{2}} = -2.\]

Answer Details

(a) Smallest integer satisfying the inequality

\[x + 8 < 4x - 15\]

Collect like terms:

\[8 + 15 < 4x - x\]

\[23 < 3x\]

\[x > \frac{23}{3} = 7\tfrac{2}{3} \approx 7.67\]

The smallest integer greater than \(7.67\) is \(\mathbf{8}\).

(b) Monthly income of the sales girl

Fixed salary \(= \text{N}2{,}500\).

Commission \(= 5\text{ kobo in the naira} = \dfrac{5}{100}\text{ naira per naira} = \text{N}0.05\) per naira of sales.

Commission on N200,000 sales:

\[0.05 \times 200{,}000 = \text{N}10{,}000\]

Total income:

\[2{,}500 + 10{,}000 = \text{N}12{,}500\]

Her income for the month is \(\mathbf{\text{N}12{,}500.00}\).

(c) Area of the window

The window is a rectangle with a semicircle on top. From the diagram, the semicircular part has radius \(35\text{ cm}\), so the width of the rectangle equals the diameter:

\[\text{width} = 2 \times 35 = 70\text{ cm}, \qquad \text{height} = 50\text{ cm}\]

Area of rectangle:

\[A_1 = 70 \times 50 = 3{,}500\text{ cm}^2\]

Area of semicircle: (with \(\pi = \tfrac{22}{7}\))

\[A_2 = \frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7} \times 35^2 = \frac{1}{2}\times \frac{22}{7} \times 1225\]

\[A_2 = \frac{1}{2}\times 22 \times 175 = \frac{1}{2}\times 3{,}850 = 1{,}925\text{ cm}^2\]

Total area of window:

\[A = A_1 + A_2 = 3{,}500 + 1{,}925 = \mathbf{5{,}425\text{ cm}^2}\]

Answer Details

(a) Value of x

In the diagram \(\triangle ABD\) is right-angled at \(B\), with \(|AB| = 3\ \text{cm}\) and hypotenuse \(|AD| = 5\ \text{cm}\). Point \(C\) lies on \(BD\), \(\angle ACB = 61^\circ\) and \(\angle DAC = x^\circ\).

Step 1 - the whole angle at A. In right triangle \(ABD\), \(AB\) is adjacent to \(\angle BAD\) and \(AD\) is the hypotenuse, so

\[\cos(\angle BAD)=\frac{|AB|}{|AD|}=\frac{3}{5}=0.6,\qquad \angle BAD=\cos^{-1}(0.6)=53.13^\circ.\]

Step 2 - the lower part of that angle. Triangle \(ACB\) is also right-angled at \(B\), so its three angles give

\[\angle BAC=180^\circ-90^\circ-61^\circ=29^\circ.\]

Step 3 - subtract. Since \(C\) lies on \(BD\), the angle \(\angle DAC\) is the difference between the whole angle at \(A\) and \(\angle BAC\):

\[x=\angle BAD-\angle BAC=53.13^\circ-29^\circ=24.13^\circ.\]

Correct to one decimal place, \(x = \mathbf{24.1}\).

(b) The pyramid

\(OABCD\) is a pyramid on a square base of side \(2\ \text{cm}\) with a slant height of \(4\ \text{cm}\). In a pyramid the slant height is the distance from the apex \(O\) down the middle of a triangular face to the midpoint of a base edge (call that midpoint \(M\)). It is not the sloping corner edge \(OA\); mixing the two up is the usual error here. The foot \(N\) of the vertical height is the centre of the square base.

(i) Vertical height

\(N\) is the centre of the base and \(M\) is the midpoint of a base edge, so \(NM\) is half the side:

\[NM=\tfrac{1}{2}\times 2=1\ \text{cm}.\]

The vertical height \(h=ON\), the slant height \(l=OM=4\ \text{cm}\) and \(NM=1\ \text{cm}\) form a right triangle at \(N\). By Pythagoras' theorem,

\[l^{2}=h^{2}+NM^{2}\ \Rightarrow\ h^{2}=4^{2}-1^{2}=16-1=15,\]

\[h=\sqrt{15}=3.873\ \text{cm}.\]

Correct to three significant figures, \(h = \mathbf{3.87\ \text{cm}}\).

(ii) Volume

\[V=\frac{1}{3}\times(\text{base area})\times h=\frac{1}{3}\times(2\times 2)\times\sqrt{15}=\frac{4\sqrt{15}}{3}.\]

\[V=\frac{4\times 3.873}{3}=\frac{15.49}{3}=5.164\ \text{cm}^{3}.\]

Correct to three significant figures, \(V = \mathbf{5.16\ \text{cm}^{3}}\).

Answer Details

(a) Simplify inside the root first:

\[\frac{0.18\times12.5}{0.05\times0.2} = \frac{2.25}{0.01} = 225,\qquad \sqrt{225} = 15.\]

(b) Write each surd in simplest form: \(\sqrt{18} = 3\sqrt2\) and \(\sqrt{50} = 5\sqrt2\).

\[\frac{8 - 4\sqrt{18}}{\sqrt{50}} = \frac{8 - 12\sqrt2}{5\sqrt2} = \frac{8}{5\sqrt2} - \frac{12\sqrt2}{5\sqrt2} = \frac{4\sqrt2}{5} - \frac{12}{5} = \frac{4\sqrt2 - 12}{5}.\]

(c) The relation is \(x = \dfrac{k y^3}{z^2}\). Use \(x=108,\,y=3,\,z=4\):

\[108 = \frac{k(27)}{16}\Rightarrow k = \frac{108\times16}{27} = 64.\]

Now find z when \(x=4000,\,y=10\):

\[4000 = \frac{64(1000)}{z^2}\Rightarrow z^2 = \frac{64000}{4000} = 16\Rightarrow z = 4.\]

Answer Details

(a) Evaluate \(\log_{10}\sqrt{30} - \log_{10}\sqrt{6} + \log_{10}\sqrt{2}\).

Write each surd as a power and combine using the laws of logarithms. Since \(\sqrt{n}=n^{1/2}\),

\[\tfrac{1}{2}\log_{10}30 - \tfrac{1}{2}\log_{10}6 + \tfrac{1}{2}\log_{10}2 = \tfrac{1}{2}\left(\log_{10}30 - \log_{10}6 + \log_{10}2\right).\]

Using \(\log a - \log b + \log c = \log\dfrac{ac}{b}\):

\[\tfrac{1}{2}\log_{10}\!\left(\frac{30\times 2}{6}\right) = \tfrac{1}{2}\log_{10}\!\left(\frac{60}{6}\right) = \tfrac{1}{2}\log_{10}10.\]

Since \(\log_{10}10 = 1\),

\[= \tfrac{1}{2}\times 1 = \boxed{\tfrac{1}{2}}.\]

(b) \(U=\{1,2,3,\dots,10\}\), \(A=\{1,2,3,4,5\}\), \(B=\{2,3,5\}\), \(C=\{6,8,10\}\).

(i) Filling the Venn diagram. Every element of \(B\) is also in \(A\) (\(2,3,5\in A\)), so the circle \(B\) lies completely inside circle \(A\), exactly as shown. \(C\) has no element in common with \(A\), so it is drawn separately. The regions are filled as:

• Inside \(B\) (which is inside \(A\)): \(2,\;3,\;5\)
• Inside \(A\) but outside \(B\): \(1,\;4\)
• Inside \(C\): \(6,\;8,\;10\)
• Outside every circle: \(7,\;9\)

(The diagram already shows \(1\) in the \(A\)-only region, \(5\) in \(B\), \(6,10\) in \(C\) and \(9\) outside; the remaining elements \(2,3\) join \(B\), \(4\) joins the \(A\)-only region, \(8\) joins \(C\) and \(7\) joins the outside.)

(ii) \(A\cap C\). \(A=\{1,2,3,4,5\}\) and \(C=\{6,8,10\}\) share no element, so

\[A\cap C = \varnothing\;\;(\text{the empty set}).\]

(iii) \(A\cap B'\). First \(B' = U - B = \{1,4,6,7,8,9,10\}\). Then take the elements common to \(A\):

\[A\cap B' = \{1,2,3,4,5\}\cap\{1,4,6,7,8,9,10\} = \{1,\;4\}.\]

Answer Details

Solution

The completed scale construction is shown below. Everything that follows is read directly from it.

(a) Accurate scale drawing (1 cm to 10 m). Each real length is divided by 10 to get the drawing length:

\(|AB| = 85\text{ m} = \dfrac{85}{10} = 8.5\text{ cm}, \qquad |CA| = |CB| = 111\text{ m} = \dfrac{111}{10} = 11.1\text{ cm}.\)

Points A, B and C are plotted with ruler and compasses to these lengths, giving the triangular plot above.

(b) The two loci.

(i) \(l_1\), the locus of points equidistant from the sides \(AC\) and \(BC\), is the bisector of angle \(ACB\). It is constructed from C (shown by the two equal angle marks) and drawn as the blue dashed line.

(ii) \(l_2\), the locus of points 60 m from A, is a circle centred at A with radius \(\dfrac{60}{10} = 6\text{ cm}\). The relevant part of this arc is drawn in red.

(c) Locating T and finding \(|TC|\). The tree T lies on both loci, so it is the point where \(l_1\) meets \(l_2\) inside the plot. Measuring the drawing length of \(CT\):

\(CT = 6.0\text{ cm}.\)

Converting back to the real distance using the scale (multiply by 10):

\(|TC| = 6.0 \times 10 = 60\text{ m}.\)

(d) Region for the flagpole P. P must be:

• nearer AC than BC – this is the side of the bisector \(l_1\) lying toward \(AC\) (the upper part of the plot), and
• more than 60 m from A – this is the region outside the arc \(l_2\), i.e. on the C-side of the arc.

The set of points satisfying both conditions is the green hatched region above (bounded by side \(CA\), the bisector \(l_1\) and the arc \(l_2\)). Any point P chosen inside that shaded region is a valid position for the flagpole.

Answer Details

(a) The regular polygon

(i) Value of n. At any vertex the interior and exterior angles are supplementary, and here the interior angle is 120° greater than the exterior angle. Let the exterior angle be \(x\):

\[x + (x + 120^\circ) = 180^\circ\] \[2x + 120^\circ = 180^\circ \implies 2x = 60^\circ \implies x = 30^\circ.\]

For any polygon the exterior angles sum to \(360^\circ\), so

\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{30^\circ} = 12.\]

The polygon has 12 sides.

(ii) Sum of all the interior angles.

\[\text{Sum} = (n-2)\times 180^\circ = (12-2)\times 180^\circ = 10 \times 180^\circ = 1800^\circ.\]

The interior angles sum to 1800°.

(b) The boy's journey

(i) Sketch of the movement. He walks 6 km from \(P\) to \(Q\) on a bearing of 065°, then 13 km from \(Q\) to \(R\) on a bearing of 146°. A North line is drawn at each point so the bearings can be measured clockwise from North.

(ii) Distance PR. First find the interior angle of the triangle at \(Q\). The bearing of \(P\) from \(Q\) is \(065^\circ + 180^\circ = 245^\circ\), and the bearing of \(R\) from \(Q\) is \(146^\circ\). Working from the North line at \(Q\), the angle between \(QP\) and the North line is \(180^\circ - 65^\circ = 115^\circ\) on the western side, giving the angle inside the triangle:

\[\angle PQR = (180^\circ - 146^\circ) + 65^\circ = 34^\circ + 65^\circ = 99^\circ.\]

Apply the cosine rule to triangle \(PQR\):

\[|PR|^{2} = |PQ|^{2} + |QR|^{2} - 2\,|PQ|\,|QR|\cos(\angle PQR)\] \[|PR|^{2} = 6^{2} + 13^{2} - 2(6)(13)\cos 99^\circ\] \[|PR|^{2} = 36 + 169 - 156\cos 99^\circ\]

Since \(\cos 99^\circ = -0.1564\):

\[|PR|^{2} = 205 - 156(-0.1564) = 205 + 24.40 = 229.40\] \[|PR| = \sqrt{229.40} = 15.15\text{ km}.\]

Correct to the nearest kilometre, \(|PR| \approx \mathbf{15\text{ km}}\).

Answer Details

(a) A negative index inverts the base:

\[\left(\tfrac{64}{125}\right)^{-\frac{2}{3}} = \left(\tfrac{125}{64}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\tfrac{125}{64}}\right)^{2} = \left(\tfrac{5}{4}\right)^{2} = \tfrac{25}{16}.\]

\[2 \div \tfrac{25}{16} = 2 \times \tfrac{16}{25} = \tfrac{32}{25} = 1\tfrac{7}{25}.\]

(b) At the point of intersection k the two y-values are equal:

\[3x + 5 = -4x - 1 \Rightarrow 7x = -6 \Rightarrow x = -\tfrac{6}{7}.\]

\[y = 3\left(-\tfrac{6}{7}\right) + 5 = -\tfrac{18}{7} + \tfrac{35}{7} = \tfrac{17}{7}.\]

Coordinates of k: \(\left(-\tfrac{6}{7},\ \tfrac{17}{7}\right)\).

Answer Details

(a) Work "of" and \(\div\) before subtraction:

\[\tfrac{5}{8}\text{ of }2\tfrac{1}{2} = \tfrac{5}{8}\times\tfrac{5}{2} = \tfrac{25}{16},\qquad \tfrac{3}{4}\div\tfrac{3}{5} = \tfrac{3}{4}\times\tfrac{5}{3} = \tfrac{5}{4} = \tfrac{20}{16}.\]

\[\tfrac{25}{16} - \tfrac{20}{16} = \tfrac{5}{16}.\]

(b) The cone and the pyramid have equal heights \(h\) and equal volumes.

\[V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h,\qquad V_{\text{pyramid}} = \tfrac{1}{3}(\text{base area})h.\]

Since the volumes and heights are equal, the base areas are equal:

\[\pi r^2 = 154 \Rightarrow \tfrac{22}{7}r^2 = 154 \Rightarrow r^2 = 154\times\tfrac{7}{22} = 49.\]

\[r = 7\text{ cm}.\]