WAEC SSCE - General Mathematics - 1998 (Essay)

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Answer Details

(a) Cumulative frequency table

Number of eggs per year	Class boundaries	Frequency	Cumulative frequency
45 - 49	44.5 - 49.5	10	10
50 - 54	49.5 - 54.5	36	46
55 - 59	54.5 - 59.5	64	110
60 - 64	59.5 - 64.5	52	162
65 - 69	64.5 - 69.5	28	190
70 - 74	69.5 - 74.5	10	200

The total number of chickens is \(N=200\). Plot the upper class boundaries against the cumulative frequencies, beginning with \((44.5,0)\), and join the points with a smooth curve.

(b) Interquartile range

\[Q_1=\frac{N}{4}=\frac{200}{4}=50,\qquad Q_3=\frac{3N}{4}=\frac{3(200)}{4}=150.\]

From the cumulative frequency curve, the corresponding egg numbers are approximately \(Q_1=54\) and \(Q_3=63\).

\[\text{Interquartile range}=Q_3-Q_1=63-54=9\text{ eggs}.\]

(c) Chickens laying at least \(60\) eggs per year:

\[52+28+10=90.\]

\[P(\text{at least }60\text{ eggs})=\frac{90}{200}=\frac{9}{20}=0.45.\]

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(a) Dimensions. Let the shorter side be \(x\) m; the longer side is \(x + 1.5\) m. The area is 13.5 m²:

\[x(x + 1.5) = 13.5 \Rightarrow x^2 + 1.5x - 13.5 = 0.\]

Multiply through by 2: \(2x^2 + 3x - 27 = 0\), which factorises as \((2x + 9)(x - 3) = 0\).

Taking the positive root, \(x = 3\). So the dimensions are

\[3\text{ m by }4.5\text{ m}.\]

(b) Area that can be carpeted. At N250.00 per square metre with N2,000.00 available:

\[\text{Area} = \frac{2000}{250} = 8\text{ m}^2.\]

So 8 m² of the floor can be covered (the floor is 13.5 m², so it cannot all be carpeted).

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(a) \(2Ax - Kx^{2}\)

When \(x = 1\),

\(2A(1) - K(1^{2}) = 7\)

\(2A - K = 7 \quad \ldots (1)\)

When \(x = 2\),

\(2A(2) - K(2^{2}) = 4\)

\(4A - 4K = 4\)

\(A - K = 1 \quad \ldots (2)\)

From (2),

\(A = 1 + K\)

Substitute into (1):

\(2(1 + K) - K = 7\)

\(2 + 2K - K = 7\)

\(K = 5\)

\(A = 1 + 5 = 6\)

\(\therefore A = 6,\quad K = 5\)

(b) \(x^{2} - 3x - 1 = 0\)

\(a = 1,\quad b = -3,\quad c = -1\)

Using the quadratic formula,

\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]

\[x = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(-1)}}{2(1)}\]

\[x = \frac{3 \pm \sqrt{9 + 4}}{2}\]

\[x = \frac{3 \pm \sqrt{13}}{2}\]

\[x = \frac{3 \pm 3.606}{2}\]

\[x = \frac{3 + 3.606}{2} = 3.303 \approx 3.3\]

or

\[x = \frac{3 - 3.606}{2} = -0.303 \approx -0.3\]

\(\therefore x = 3.3\) or \(x = -0.3\), correct to 1 decimal place.

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(a)(i) Final sales price. Marked price for a 30% profit:

\[\text{MP} = 6000\times1.30 = \text{N}7800.\]

Reduced by 10% in the sales:

\[\text{Sales price} = 7800\times0.90 = \text{N}7020.\]

(a)(ii) Percentage profit.

\[\%\text{ profit} = \frac{7020 - 6000}{6000}\times100 = \frac{1020}{6000}\times100 = 17\%.\]

(b) Let \(u = 2^x\). Then \(2^{(2x+1)} = 2\cdot2^{2x} = 2u^2\), so the equation becomes

\[2u^2 - 9u + 4 = 0 \Rightarrow (2u - 1)(u - 4) = 0.\]

\[u = 4\ \Rightarrow\ 2^x = 4\ \Rightarrow\ x = 2;\qquad u = \tfrac{1}{2}\ \Rightarrow\ 2^x = 2^{-1}\ \Rightarrow\ x = -1.\]

Answer: \(x = 2\) or \(x = -1\).

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Place \(X\) at the origin. \(Y\) is due east, and \(W\) is \(180\,\text{m}\) from \(X\) on bearing \(135^\circ\) (South-East), giving \(W = (180\sin135^\circ,\ 180\cos135^\circ) = (127.28,\ -127.28)\).

(a)(i) Finding \(|XY|\): \(Y = (y, 0)\). From \(W\) the bearing of \(Y\) is \(050^\circ\), so \[\tan50^\circ = \frac{y - 127.28}{0 - (-127.28)} = \frac{y - 127.28}{127.28}.\] \[y - 127.28 = 127.28\tan50^\circ = 151.68 \Rightarrow y = 278.96.\] Therefore \(|XY| \approx \mathbf{279\,\text{m}}\).

(a)(ii) Finding \(|ZW|\): \(Z\) lies on bearing \(046^\circ\) from \(X\): \(Z = t(\sin46^\circ,\cos46^\circ)\). From \(W\), \(Z\) is on bearing \(337^\circ\): \(Z = W + s(\sin337^\circ,\cos337^\circ)\). Solving \[0.7193t + 0.3907s = 127.28,\qquad 0.6947t - 0.9205s = -127.28,\] gives \(t = |XZ| = 72.2\) and \(s = |ZW| \approx \mathbf{193\,\text{m}}\), with \(Z = (51.95,\ 50.17)\).

(b) Bearing of \(Z\) from \(N\): \(N\) is on \(XY\) (the east line) with \(|XZ| = |ZN|\). Since \(Z = (51.95, 50.17)\) and \(|XZ| = 72.2\), \[(n - 51.95)^2 + 50.17^2 = 72.2^2 \Rightarrow n = 103.9\ (\text{taking } n \neq 0).\] So \(N = (103.9, 0)\) and \(\vec{NZ} = (-51.95,\ 50.17)\), which points North-West. \[\text{angle west of north} = \tan^{-1}\!\frac{51.95}{50.17} = 46^\circ \Rightarrow \text{bearing} = 360^\circ - 46^\circ = \mathbf{314^\circ}.\]

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(a) One number is chosen from \(\{2,3,4\}\) and one from \(\{1,3,5\}\). There are \(3\times 3 = 9\) equally likely pairs. We need the sum \(S\) to satisfy \(3 < S < 7\), i.e. \(S \in \{4,5,6\}\).

+	1	3	5
2	3	5	7
3	4	6	8
4	5	7	9

The sums strictly between \(3\) and \(7\) are: \((2,3)=5,\;(3,1)=4,\;(3,3)=6,\;(4,1)=5\) - that is \(4\) favourable outcomes.

\[P(3 < S < 7) = \frac{4}{9}.\]

(b) \(ABCD\) is a circle. \(DAE\) and \(CBE\) are straight lines meeting at the outer point \(E\) (with \(\angle E = m\)); \(ABF\) and \(DCF\) are straight lines meeting at the outer point \(F\) (with \(\angle F = y\)). Let the four arcs be \(AB = p,\; BC = q,\; CD = r,\; DA = s\), so \(p+q+r+s = 360^\circ\).

Angle at each external point = half the difference of the intercepted arcs.

\[m = \angle E = \tfrac12(\text{arc }DC - \text{arc }AB) = \tfrac12(r-p),\qquad y = \angle F = \tfrac12(\text{arc }AD - \text{arc }BC) = \tfrac12(s-q).\]

Given \(y + m = 90^\circ\):

\[\tfrac12(r-p) + \tfrac12(s-q) = 90^\circ \;\Rightarrow\; (r+s) - (p+q) = 180^\circ.\]

Combine with \(p+q+r+s = 360^\circ\). Adding the two equations gives \(2(r+s) = 540^\circ\), so

\[r+s = 270^\circ,\qquad p+q = 90^\circ.\]

At vertex \(B\) the straight lines \(CBE\) and \(ABF\) cross, and \(x\) is the angle \(\angle ABE = \angle FBC\) (vertically opposite), which is the supplement of the interior angle \(\angle ABC\). Now \(\angle ABC\) is inscribed on arc \(ADC\):

\[\angle ABC = \tfrac12(\text{arc }AD + \text{arc }DC) = \tfrac12(s+r) = \tfrac12(270^\circ) = 135^\circ.\]

Therefore

\[x = 180^\circ - \angle ABC = 180^\circ - 135^\circ = \boxed{45^\circ}.\]

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Answer Details

From the diagram, the pulley has centre O and radius \(r = 10\text{ cm}\). The string runs \(A \to S\) (tangent on the left), over the arc \(S \to R \to T\) at the top, then \(T \to B\) (tangent on the right). \(AMB\) is horizontal, tangent to the pulley at M, with \(\angle SAB = 90^\circ\) and \(\angle TBA = 60^\circ\).

(a)(i) Obtuse angle SOT

From A, the tangents \(AS\) and \(AM\) meet the radii at right angles, so in quadrilateral \(ASOM\):

\[\angle SOM = 360^\circ - 90^\circ - 90^\circ - \angle SAM = 360^\circ - 90^\circ - 90^\circ - 90^\circ = 90^\circ\]

From B, the tangents \(BT\) and \(BM\) give, in quadrilateral \(BTOM\):

\[\angle TOM = 360^\circ - 90^\circ - 90^\circ - \angle TBM = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ\]

Going from S round the bottom through M to T gives \(90^\circ + 120^\circ = 210^\circ\). The angle on the top (through R), the obtuse \(\angle SOT\), is

\[\angle SOT = 360^\circ - 210^\circ = 150^\circ\]

(a)(ii) Arc SRT

\[\text{arc }SRT = \frac{150}{360}\times 2\pi r = \frac{150}{360}\times 2\times\frac{22}{7}\times 10 = \frac{5}{12}\times\frac{440}{7}\]\[= \frac{2200}{84} = 26.19\text{ cm} \approx 26\text{ cm}\]

(a)(iii) |BT|

In right triangle \(OTB\), \(OT = 10\text{ cm}\), \(\angle OTB = 90^\circ\), and \(OB\) bisects \(\angle TBM\) so \(\angle TBO = 30^\circ\):

\[\tan 30^\circ = \frac{OT}{BT} \;\Rightarrow\; BT = \frac{10}{\tan 30^\circ} = 10\sqrt{3} = 17.32\text{ cm}\]

(b) Length of the string

First find \(AS\). In right triangle \(OSA\), \(OS = 10\text{ cm}\), \(\angle OSA = 90^\circ\), and \(OA\) bisects \(\angle SAM\) so \(\angle SAO = 45^\circ\):

\[\tan 45^\circ = \frac{OS}{AS} \;\Rightarrow\; AS = \frac{10}{\tan 45^\circ} = 10\text{ cm}\]

The string length is

\[AS + \text{arc }SRT + BT = 10 + 26.19 + 17.32 = 53.51\text{ cm}\]

Length of the string \(\approx 54\text{ cm}\).

Summary: \(\angle SOT = 150^\circ\), arc \(SRT \approx 26\text{ cm}\), \(|BT| = 10\sqrt{3}\approx 17.32\text{ cm}\), string \(\approx 54\text{ cm}\).

Answer Details

(a) Common ratio. For a GP, \(T_3 = ar^2 = 360\) and \(T_6 = ar^5 = 1215\). Dividing:

\[\frac{ar^5}{ar^2} = r^3 = \frac{1215}{360} = \frac{27}{8}\Rightarrow r = \sqrt[3]{\tfrac{27}{8}} = \tfrac{3}{2}.\]

(b) First term.

\[ar^2 = 360 \Rightarrow a\left(\tfrac{3}{2}\right)^2 = 360 \Rightarrow a\times\tfrac{9}{4} = 360 \Rightarrow a = 160.\]

(c) Sum of the first four terms. The terms are \(160, 240, 360, 540\):

\[S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{160\left(\left(\tfrac{3}{2}\right)^4 - 1\right)}{\tfrac{3}{2} - 1} = \frac{160(5.0625 - 1)}{0.5} = 1300.\]

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(a)(i) \(35 = 5\times7\), so by the addition law of logarithms:

\[\log_{10}35 = \log_{10}5 + \log_{10}7 = 0.6990 + 0.8451 = 1.5441.\]

(a)(ii) \(2.8 = \dfrac{28}{10} = \dfrac{4\times7}{10} = \dfrac{2^2\times7}{10}\):

\[\log_{10}2.8 = 2\log_{10}2 + \log_{10}7 - \log_{10}10 = 2(0.3010) + 0.8451 - 1 = 0.4471.\]

(b) Take logarithms of \(N^{0.8942} = 2.8\):

\[0.8942\log_{10}N = \log_{10}2.8 = 0.4471 \Rightarrow \log_{10}N = \frac{0.4471}{0.8942} = 0.5.\]

\[N = 10^{0.5} = \sqrt{10} = 3.16.\]

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