WAEC SSCE - General Mathematics - 1993 (Essay)
Question 1 Report
(a) Copy and complete the following table of values for \(y = 3\sin 2\theta - \cos \theta\).
| \(\theta\) | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| y | -1.0 | 0 | 1.0 |
(b) Using a scale of 2cm to 30° on the \(\theta\) axis and 2cm to 1 unit on the y- axis, draw the graph of \(y = 3 \sin 2\theta - \cos \theta\) for \(0° \leq \theta \leq 180°\).
(c) Use your graph to find the : (i) solution of the equation \(3 \sin 2\theta - \cos \theta = 0\), correct to the nearest degree; (ii) maximum value of y, correct to one decimal place.
(a) Completing the table for \(y = 3\sin 2\theta - \cos\theta\). Substitute each value of \(\theta\), remembering that the angle inside the sine is \(2\theta\).
- \(\theta = 30^\circ:\ 3\sin 60^\circ - \cos 30^\circ = 2.598 - 0.866 = 1.732 \approx 1.7\)
- \(\theta = 60^\circ:\ 3\sin 120^\circ - \cos 60^\circ = 2.598 - 0.500 = 2.098 \approx 2.1\)
- \(\theta = 120^\circ:\ 3\sin 240^\circ - \cos 120^\circ = -2.598 + 0.500 = -2.098 \approx -2.1\)
- \(\theta = 150^\circ:\ 3\sin 300^\circ - \cos 150^\circ = -2.598 + 0.866 = -1.732 \approx -1.7\)
| \(\theta\) | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
|---|---|---|---|---|---|---|---|
| \(y\) | -1.0 | 1.7 | 2.1 | 0 | -2.1 | -1.7 | 1.0 |
(b) Graph of \(y = 3\sin 2\theta - \cos\theta\) for \(0^\circ \le \theta \le 180^\circ\). The seven table points are plotted and joined with a smooth curve; the curve rises above the tabulated points to a peak near \(\theta = 49^\circ\) and dips to a trough near \(\theta = 131^\circ\).
(c)(i) Solution of \(3\sin 2\theta - \cos\theta = 0\). These are the values of \(\theta\) where the curve crosses the \(\theta\)-axis. Reading the three crossings from the graph gives
\[\theta \approx 10^\circ,\ 90^\circ,\ 170^\circ.\]This is confirmed algebraically, since \(3\sin 2\theta - \cos\theta = \cos\theta(6\sin\theta - 1) = 0\) gives \(\cos\theta = 0\) (so \(\theta = 90^\circ\)) or \(\sin\theta = \tfrac{1}{6}\) (so \(\theta \approx 10^\circ\) and \(\theta \approx 170^\circ\)).
(c)(ii) Maximum value of \(y\). The highest point of the curve lies between the tabulated points, near \(\theta = 49^\circ\), not at \(\theta = 60^\circ\). Reading the peak gives
\[y_{\max} \approx 2.3\ \text{(to 1 decimal place)}.\]Examination note: the greatest value of \(y\) is read from the top of the smooth curve, which rises higher than any tabulated point. Taking the largest table value (\(2.1\) at \(\theta = 60^\circ\)) as the maximum is a common error; the true maximum of about \(2.3\) occurs near \(\theta = 49^\circ\).
Answer Details
(a) Completing the table for \(y = 3\sin 2\theta - \cos\theta\). Substitute each value of \(\theta\), remembering that the angle inside the sine is \(2\theta\).
- \(\theta = 30^\circ:\ 3\sin 60^\circ - \cos 30^\circ = 2.598 - 0.866 = 1.732 \approx 1.7\)
- \(\theta = 60^\circ:\ 3\sin 120^\circ - \cos 60^\circ = 2.598 - 0.500 = 2.098 \approx 2.1\)
- \(\theta = 120^\circ:\ 3\sin 240^\circ - \cos 120^\circ = -2.598 + 0.500 = -2.098 \approx -2.1\)
- \(\theta = 150^\circ:\ 3\sin 300^\circ - \cos 150^\circ = -2.598 + 0.866 = -1.732 \approx -1.7\)
| \(\theta\) | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
|---|---|---|---|---|---|---|---|
| \(y\) | -1.0 | 1.7 | 2.1 | 0 | -2.1 | -1.7 | 1.0 |
(b) Graph of \(y = 3\sin 2\theta - \cos\theta\) for \(0^\circ \le \theta \le 180^\circ\). The seven table points are plotted and joined with a smooth curve; the curve rises above the tabulated points to a peak near \(\theta = 49^\circ\) and dips to a trough near \(\theta = 131^\circ\).
(c)(i) Solution of \(3\sin 2\theta - \cos\theta = 0\). These are the values of \(\theta\) where the curve crosses the \(\theta\)-axis. Reading the three crossings from the graph gives
\[\theta \approx 10^\circ,\ 90^\circ,\ 170^\circ.\]This is confirmed algebraically, since \(3\sin 2\theta - \cos\theta = \cos\theta(6\sin\theta - 1) = 0\) gives \(\cos\theta = 0\) (so \(\theta = 90^\circ\)) or \(\sin\theta = \tfrac{1}{6}\) (so \(\theta \approx 10^\circ\) and \(\theta \approx 170^\circ\)).
(c)(ii) Maximum value of \(y\). The highest point of the curve lies between the tabulated points, near \(\theta = 49^\circ\), not at \(\theta = 60^\circ\). Reading the peak gives
\[y_{\max} \approx 2.3\ \text{(to 1 decimal place)}.\]Examination note: the greatest value of \(y\) is read from the top of the smooth curve, which rises higher than any tabulated point. Taking the largest table value (\(2.1\) at \(\theta = 60^\circ\)) as the maximum is a common error; the true maximum of about \(2.3\) occurs near \(\theta = 49^\circ\).
Question 2 Report
The table below shows the frequency distribution of the marks scored by fifty students in an examination.
| Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Freq | 2 | 3 | 4 | 6 | 13 | 10 | 5 | 3 | 2 | 2 |
(a) Draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the : (i) upper quartile; (ii) pass mark if 60% of the students passed.
(a) Cumulative frequency table
| Marks (%) | Upper class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 0–9 | 9.5 | 2 | 2 |
| 10–19 | 19.5 | 3 | 5 |
| 20–29 | 29.5 | 4 | 9 |
| 30–39 | 39.5 | 6 | 15 |
| 40–49 | 49.5 | 13 | 28 |
| 50–59 | 59.5 | 10 | 38 |
| 60–69 | 69.5 | 5 | 43 |
| 70–79 | 79.5 | 3 | 46 |
| 80–89 | 89.5 | 2 | 48 |
| 90–99 | 99.5 | 2 | 50 |
Plot the cumulative frequencies against the upper class boundaries, beginning with \\((-0.5,0)\\) , and join the points with a smooth increasing curve.
(b)(i) Upper quartile
\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]
From the curve, the mark corresponding to cumulative frequency \(37.5\) is approximately \(57\).
\[\boxed{Q_3\approx57\text{ marks}}\]
(b)(ii) Pass mark when 60% passed
Number who passed \(=0.60\times50=30\). Hence the number below the pass mark is \(50-30=20\).
From the curve, the mark corresponding to cumulative frequency \(20\) is approximately \(42\).
\[\boxed{\text{Pass mark}\approx42\text{ marks}}\]
Answer Details
(a) Cumulative frequency table
| Marks (%) | Upper class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 0–9 | 9.5 | 2 | 2 |
| 10–19 | 19.5 | 3 | 5 |
| 20–29 | 29.5 | 4 | 9 |
| 30–39 | 39.5 | 6 | 15 |
| 40–49 | 49.5 | 13 | 28 |
| 50–59 | 59.5 | 10 | 38 |
| 60–69 | 69.5 | 5 | 43 |
| 70–79 | 79.5 | 3 | 46 |
| 80–89 | 89.5 | 2 | 48 |
| 90–99 | 99.5 | 2 | 50 |
Plot the cumulative frequencies against the upper class boundaries, beginning with \\((-0.5,0)\\) , and join the points with a smooth increasing curve.
(b)(i) Upper quartile
\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]
From the curve, the mark corresponding to cumulative frequency \(37.5\) is approximately \(57\).
\[\boxed{Q_3\approx57\text{ marks}}\]
(b)(ii) Pass mark when 60% passed
Number who passed \(=0.60\times50=30\). Hence the number below the pass mark is \(50-30=20\).
From the curve, the mark corresponding to cumulative frequency \(20\) is approximately \(42\).
\[\boxed{\text{Pass mark}\approx42\text{ marks}}\]
Question 3 Report
The universal set \(\varepsilon\) is the set of all integers and the subset P, Q, R of \(\varepsilon\) are given by:
\(P = {x : x < 0} ; Q = {... , -5, -3, -1, 1, 3, 5} ; R = {x : -2 \leq x < 7}\)
(a) Find \(Q \cap R\).
(b) Find \(R'\) where R' is the complement of R with respect to \(\varepsilon\).
(c) Find \(P' \cup R'\)
(d) List the members of \((P \cap Q)'\).
The universal set \(\varepsilon\) is the set of all integers. Listing the given sets over the relevant range:
- \(P = \{x : x < 0\} = \{\dots, -3, -2, -1\}\) (negative integers)
- \(Q = \{\dots, -5, -3, -1, 1, 3, 5, \dots\}\) (odd integers)
- \(R = \{x : -2 \le x < 7\} = \{-2, -1, 0, 1, 2, 3, 4, 5, 6\}\)
(a) \(Q \cap R\) = odd integers lying in \(R\):
\[Q \cap R = \{-1,\, 1,\, 3,\, 5\}\]
(b) \(R'\) = all integers not in \(R\):
\[R' = \{\dots, -4, -3\} \cup \{7, 8, 9, \dots\} = \{x : x \le -3 \text{ or } x \ge 7\}\]
(c) \(P' = \{0, 1, 2, 3, \dots\}\) (non-negative integers). Then
\[P' \cup R' = \{0,1,2,\dots\} \cup \{\dots,-4,-3\}\cup\{7,8,\dots\} = \varepsilon \setminus \{-2,\,-1\}\]
i.e. every integer except \(-1\) and \(-2\).
(d) \(P \cap Q\) = negative odd integers \(= \{\dots, -5, -3, -1\}\). Its complement is every integer that is not a negative odd integer:
\[(P \cap Q)' = \{\dots, -6, -4, -2,\, 0, 1, 2, 3, 4, \dots\}\]
that is, all non-negative integers together with all negative even integers.
Answer Details
The universal set \(\varepsilon\) is the set of all integers. Listing the given sets over the relevant range:
- \(P = \{x : x < 0\} = \{\dots, -3, -2, -1\}\) (negative integers)
- \(Q = \{\dots, -5, -3, -1, 1, 3, 5, \dots\}\) (odd integers)
- \(R = \{x : -2 \le x < 7\} = \{-2, -1, 0, 1, 2, 3, 4, 5, 6\}\)
(a) \(Q \cap R\) = odd integers lying in \(R\):
\[Q \cap R = \{-1,\, 1,\, 3,\, 5\}\]
(b) \(R'\) = all integers not in \(R\):
\[R' = \{\dots, -4, -3\} \cup \{7, 8, 9, \dots\} = \{x : x \le -3 \text{ or } x \ge 7\}\]
(c) \(P' = \{0, 1, 2, 3, \dots\}\) (non-negative integers). Then
\[P' \cup R' = \{0,1,2,\dots\} \cup \{\dots,-4,-3\}\cup\{7,8,\dots\} = \varepsilon \setminus \{-2,\,-1\}\]
i.e. every integer except \(-1\) and \(-2\).
(d) \(P \cap Q\) = negative odd integers \(= \{\dots, -5, -3, -1\}\). Its complement is every integer that is not a negative odd integer:
\[(P \cap Q)' = \{\dots, -6, -4, -2,\, 0, 1, 2, 3, 4, \dots\}\]
that is, all non-negative integers together with all negative even integers.
Question 4 Report
(a) Simplify, without using Mathematical tables: \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)
(b) Without using Mathematical tables, calculate \(\sqrt{\frac{P}{Q}}\) where \(P = 3.6 \times 10^{-3}\) and \(Q = 2.25 \times 10^{6}\), leaving your answer in standard form.
(a) Use \(a\log b = \log b^a\) and combine:
\[\log_{10}\frac{30}{16} - 2\log_{10}\frac{5}{9} + \log_{10}\frac{400}{243} = \log_{10}\frac{30}{16} - \log_{10}\frac{25}{81} + \log_{10}\frac{400}{243}\]
\[= \log_{10}\!\left(\frac{30}{16} \times \frac{81}{25} \times \frac{400}{243}\right)\]
Simplify the product step by step: \(\dfrac{30}{16} = \dfrac{15}{8}\), and \(\dfrac{15}{8}\times\dfrac{81}{25} = \dfrac{243}{40}\), then \(\dfrac{243}{40}\times\dfrac{400}{243} = \dfrac{400}{40} = 10\).
\[= \log_{10} 10 = 1\]
(b) With \(P = 3.6\times10^{-3}\) and \(Q = 2.25\times10^{6}\):
\[\frac{P}{Q} = \frac{3.6\times10^{-3}}{2.25\times10^{6}} = 1.6\times10^{-9} = 16\times10^{-10}\]
\[\sqrt{\frac{P}{Q}} = \sqrt{16\times10^{-10}} = 4\times10^{-5}\]
In standard form, \(\sqrt{\dfrac{P}{Q}} = 4.0\times10^{-5}\).
Answer Details
(a) Use \(a\log b = \log b^a\) and combine:
\[\log_{10}\frac{30}{16} - 2\log_{10}\frac{5}{9} + \log_{10}\frac{400}{243} = \log_{10}\frac{30}{16} - \log_{10}\frac{25}{81} + \log_{10}\frac{400}{243}\]
\[= \log_{10}\!\left(\frac{30}{16} \times \frac{81}{25} \times \frac{400}{243}\right)\]
Simplify the product step by step: \(\dfrac{30}{16} = \dfrac{15}{8}\), and \(\dfrac{15}{8}\times\dfrac{81}{25} = \dfrac{243}{40}\), then \(\dfrac{243}{40}\times\dfrac{400}{243} = \dfrac{400}{40} = 10\).
\[= \log_{10} 10 = 1\]
(b) With \(P = 3.6\times10^{-3}\) and \(Q = 2.25\times10^{6}\):
\[\frac{P}{Q} = \frac{3.6\times10^{-3}}{2.25\times10^{6}} = 1.6\times10^{-9} = 16\times10^{-10}\]
\[\sqrt{\frac{P}{Q}} = \sqrt{16\times10^{-10}} = 4\times10^{-5}\]
In standard form, \(\sqrt{\dfrac{P}{Q}} = 4.0\times10^{-5}\).
Question 5 Report
(a) 
In the diagram, BA is parallel to DE. Find the value of x.
(b) Illustrate graphically and shade the region in which inequalities \(y - 2x < 5 ; 2y + x \geq 4 ; y + 2x \leq 10\) are satisfied.
(a)
Through C, draw the line RCE parallel to BA and DE.
Since BA is parallel to RC,
\[\angle BCR=52^\circ\]
Also, since RC is parallel to DE,
\[\angle RCD=48^\circ.\]
Hence, using angles about a point,
\[x=360^\circ-(52^\circ+48^\circ)=260^\circ.\]
Therefore, \(x=260^\circ\).
(b)
First draw the boundary lines:
\[y-2x=5\quad\Rightarrow\quad y=2x+5\]
\[2y+x=4\quad\Rightarrow\quad y=2-\frac{x}{2}\]
\[y+2x=10\quad\Rightarrow\quad y=10-2x.\]
| Boundary | Points used for plotting | Required side |
|---|---|---|
| \(y=2x+5\) | \((0,5),\ (2,9)\) | Below the line; boundary excluded |
| \(y=2-\frac{x}{2}\) | \((0,2),\ (4,0)\) | On or above the line |
| \(y=10-2x\) | \((0,10),\ (2,6)\) | On or below the line |
The shaded feasible region is shown below. The broken boundary \(y=2x+5\) is not included because the inequality is strict.
The corner points of the region are obtained from intersections of the boundary lines:
\[\left(-\frac65,\frac{13}{5}\right),\qquad \left(\frac54,\frac{15}{2}\right),\qquad \left(\frac{16}{3},-\frac23\right).\]
The points \(\left(-\frac65,\frac{13}{5}\right)\) and \(\left(\frac54,\frac{15}{2}\right)\) are excluded because they lie on \(y-2x=5\).
Answer Details
(a)
Through C, draw the line RCE parallel to BA and DE.
Since BA is parallel to RC,
\[\angle BCR=52^\circ\]
Also, since RC is parallel to DE,
\[\angle RCD=48^\circ.\]
Hence, using angles about a point,
\[x=360^\circ-(52^\circ+48^\circ)=260^\circ.\]
Therefore, \(x=260^\circ\).
(b)
First draw the boundary lines:
\[y-2x=5\quad\Rightarrow\quad y=2x+5\]
\[2y+x=4\quad\Rightarrow\quad y=2-\frac{x}{2}\]
\[y+2x=10\quad\Rightarrow\quad y=10-2x.\]
| Boundary | Points used for plotting | Required side |
|---|---|---|
| \(y=2x+5\) | \((0,5),\ (2,9)\) | Below the line; boundary excluded |
| \(y=2-\frac{x}{2}\) | \((0,2),\ (4,0)\) | On or above the line |
| \(y=10-2x\) | \((0,10),\ (2,6)\) | On or below the line |
The shaded feasible region is shown below. The broken boundary \(y=2x+5\) is not included because the inequality is strict.
The corner points of the region are obtained from intersections of the boundary lines:
\[\left(-\frac65,\frac{13}{5}\right),\qquad \left(\frac54,\frac{15}{2}\right),\qquad \left(\frac{16}{3},-\frac23\right).\]
The points \(\left(-\frac65,\frac{13}{5}\right)\) and \(\left(\frac54,\frac{15}{2}\right)\) are excluded because they lie on \(y-2x=5\).
Question 6 Report
(a)(i) Given that \(\log_{10} 5 = 0.699\) and \(\log_{10} 3 = 0.477\), find \(\log_{10} 45\), without using Mathematical tables.
(ii) Hence, solve \(x^{0.8265} = 45\).
(b) Use Mathematical tables to evaluate \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\)
(a)(i) Write \(45 = 9 \times 5 = 3^2 \times 5\):
\[\log_{10}45 = 2\log_{10}3 + \log_{10}5 = 2(0.477) + 0.699 = 0.954 + 0.699 = 1.653\]
(ii) Solve \(x^{0.8265} = 45\). Take logarithms of both sides:
\[0.8265\,\log_{10}x = \log_{10}45 = 1.653\]
\[\log_{10}x = \frac{1.653}{0.8265} = 2.000 \;\Rightarrow\; x = 10^{2} = 100\]
(b) Evaluate \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}}\) using logarithms.
| Number | Log |
|---|---|
| 2.067 | 0.3155 |
| 0.0348 | \(\bar{2}.5416\) |
| 0.538 | \(\bar{1}.7308\) |
Denominator log \(= \bar{2}.5416 + \bar{1}.7308 = \bar{2}.2724\). Then
\[\log(\text{fraction}) = 0.3155 - \bar{2}.2724 = 2.0431\]
\[\log(\text{answer}) = \tfrac12(2.0431) = 1.0216 \;\Rightarrow\; \text{answer} = 10.51\]
Hence \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}} \approx 10.51\).
Answer Details
(a)(i) Write \(45 = 9 \times 5 = 3^2 \times 5\):
\[\log_{10}45 = 2\log_{10}3 + \log_{10}5 = 2(0.477) + 0.699 = 0.954 + 0.699 = 1.653\]
(ii) Solve \(x^{0.8265} = 45\). Take logarithms of both sides:
\[0.8265\,\log_{10}x = \log_{10}45 = 1.653\]
\[\log_{10}x = \frac{1.653}{0.8265} = 2.000 \;\Rightarrow\; x = 10^{2} = 100\]
(b) Evaluate \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}}\) using logarithms.
| Number | Log |
|---|---|
| 2.067 | 0.3155 |
| 0.0348 | \(\bar{2}.5416\) |
| 0.538 | \(\bar{1}.7308\) |
Denominator log \(= \bar{2}.5416 + \bar{1}.7308 = \bar{2}.2724\). Then
\[\log(\text{fraction}) = 0.3155 - \bar{2}.2724 = 2.0431\]
\[\log(\text{answer}) = \tfrac12(2.0431) = 1.0216 \;\Rightarrow\; \text{answer} = 10.51\]
Hence \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}} \approx 10.51\).
Question 7 Report
(a) Simplify \(\frac{3}{m + 2n} - \frac{2}{m - 3n}\)
(b) A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.
(a) Common denominator \((m+2n)(m-3n)\):
\[\frac{3}{m+2n} - \frac{2}{m-3n} = \frac{3(m-3n) - 2(m+2n)}{(m+2n)(m-3n)}\]
\[= \frac{3m - 9n - 2m - 4n}{(m+2n)(m-3n)} = \frac{m - 13n}{(m+2n)(m-3n)}\]
(b) Let the tens digit be \(t\) and the units digit be \(u\). The number is \(10t + u\).
Sum of digits: \(t + u = 11\).
Interchanging the digits gives \(10u + t\), which exceeds the original by 9:
\[10u + t = (10t + u) + 9 \;\Rightarrow\; 9u - 9t = 9 \;\Rightarrow\; u - t = 1\]
Solving \(t + u = 11\) and \(u - t = 1\): \(u = 6,\; t = 5\).
The number is \(\mathbf{56}\).
Answer Details
(a) Common denominator \((m+2n)(m-3n)\):
\[\frac{3}{m+2n} - \frac{2}{m-3n} = \frac{3(m-3n) - 2(m+2n)}{(m+2n)(m-3n)}\]
\[= \frac{3m - 9n - 2m - 4n}{(m+2n)(m-3n)} = \frac{m - 13n}{(m+2n)(m-3n)}\]
(b) Let the tens digit be \(t\) and the units digit be \(u\). The number is \(10t + u\).
Sum of digits: \(t + u = 11\).
Interchanging the digits gives \(10u + t\), which exceeds the original by 9:
\[10u + t = (10t + u) + 9 \;\Rightarrow\; 9u - 9t = 9 \;\Rightarrow\; u - t = 1\]
Solving \(t + u = 11\) and \(u - t = 1\): \(u = 6,\; t = 5\).
The number is \(\mathbf{56}\).
Question 8 Report
(a) What is the 25th term of 5, 9, 13,... ?
(b) Find the 5th term of \(\frac{8}{9}, \frac{-4}{3}, 2, ...\).
(c) The 3rd and 6th terms of a G.P are \(48\) and \(14\frac{2}{9}\) respectively. Write down the first four terms of the G.P.
(a) A.P. \(5, 9, 13, \dots\) with \(a = 5,\; d = 4\).
\[T_{25} = a + 24d = 5 + 24(4) = 5 + 96 = 101\]
(b) G.P. \(\dfrac{8}{9},\, -\dfrac{4}{3},\, 2, \dots\) with \(a = \dfrac{8}{9}\) and common ratio
\[r = \frac{-4/3}{8/9} = -\frac{4}{3}\times\frac{9}{8} = -\frac{3}{2}\]
\[T_5 = ar^4 = \frac{8}{9}\left(-\frac{3}{2}\right)^4 = \frac{8}{9}\times\frac{81}{16} = \frac{9}{2} = 4\tfrac12\]
(c) For the G.P., \(T_3 = ar^2 = 48\) and \(T_6 = ar^5 = 14\tfrac29 = \dfrac{128}{9}\). Dividing:
\[r^3 = \frac{ar^5}{ar^2} = \frac{128/9}{48} = \frac{8}{27} \;\Rightarrow\; r = \frac{2}{3}\]
\[ar^2 = a\left(\tfrac49\right) = 48 \;\Rightarrow\; a = 108\]
The first four terms are \(108,\; 72,\; 48,\; 32\).
Answer Details
(a) A.P. \(5, 9, 13, \dots\) with \(a = 5,\; d = 4\).
\[T_{25} = a + 24d = 5 + 24(4) = 5 + 96 = 101\]
(b) G.P. \(\dfrac{8}{9},\, -\dfrac{4}{3},\, 2, \dots\) with \(a = \dfrac{8}{9}\) and common ratio
\[r = \frac{-4/3}{8/9} = -\frac{4}{3}\times\frac{9}{8} = -\frac{3}{2}\]
\[T_5 = ar^4 = \frac{8}{9}\left(-\frac{3}{2}\right)^4 = \frac{8}{9}\times\frac{81}{16} = \frac{9}{2} = 4\tfrac12\]
(c) For the G.P., \(T_3 = ar^2 = 48\) and \(T_6 = ar^5 = 14\tfrac29 = \dfrac{128}{9}\). Dividing:
\[r^3 = \frac{ar^5}{ar^2} = \frac{128/9}{48} = \frac{8}{27} \;\Rightarrow\; r = \frac{2}{3}\]
\[ar^2 = a\left(\tfrac49\right) = 48 \;\Rightarrow\; a = 108\]
The first four terms are \(108,\; 72,\; 48,\; 32\).
Question 9 Report
P and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along
(a) the parallel of latitude ;
(b) a great circle.
[Take \(\pi = \frac{22}{7}\) and radius of the earth = 6400km].
Question 10 Report
A simple measuring device is used at points X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to 5,200m above sea level, /XY/ = 4,000m and the measurements of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (Take \(\tan 15 = 0.3\) and \(\tan 35 = 0.7\))
Let the foot of the perpendicular from the peak \(P\) to the horizontal level of \(X\) and \(Y\) be \(N\), with \(Y\) nearer the mountain. Let \(|YN| = d\) and let \(h\) be the height of \(P\) above the level \(XY\).
From \(Y\) (angle of elevation \(35^\circ\)):
\[\tan 35^\circ = \frac{h}{d} \;\Rightarrow\; h = 0.7d\]
From \(X\) (angle of elevation \(15^\circ\), and \(|XN| = d + 4000\)):
\[\tan 15^\circ = \frac{h}{d + 4000} \;\Rightarrow\; h = 0.3(d + 4000)\]
Equating:
\[0.7d = 0.3d + 1200 \;\Rightarrow\; 0.4d = 1200 \;\Rightarrow\; d = 3000\text{ m}\]
\[h = 0.7(3000) = 2100\text{ m}\]
This \(h\) is the height of the peak above the level of \(X\) and \(Y\). Since \(X\) (and \(Y\)) are \(5200\) m above sea level, the height of the mountain above sea level is:
\[5200 + 2100 = 7300\text{ m}\]
Answer Details
Let the foot of the perpendicular from the peak \(P\) to the horizontal level of \(X\) and \(Y\) be \(N\), with \(Y\) nearer the mountain. Let \(|YN| = d\) and let \(h\) be the height of \(P\) above the level \(XY\).
From \(Y\) (angle of elevation \(35^\circ\)):
\[\tan 35^\circ = \frac{h}{d} \;\Rightarrow\; h = 0.7d\]
From \(X\) (angle of elevation \(15^\circ\), and \(|XN| = d + 4000\)):
\[\tan 15^\circ = \frac{h}{d + 4000} \;\Rightarrow\; h = 0.3(d + 4000)\]
Equating:
\[0.7d = 0.3d + 1200 \;\Rightarrow\; 0.4d = 1200 \;\Rightarrow\; d = 3000\text{ m}\]
\[h = 0.7(3000) = 2100\text{ m}\]
This \(h\) is the height of the peak above the level of \(X\) and \(Y\). Since \(X\) (and \(Y\)) are \(5200\) m above sea level, the height of the mountain above sea level is:
\[5200 + 2100 = 7300\text{ m}\]
Question 11 Report
A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :
(a) three are red;
(b) the first is blue and the other two are red;
(c) two are white and one is blue.
Question 12 Report
(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) 
In the diagram, O is the centre of the circle ACDB. If < CAO = 26° and < AOB = 130°. Calculate : (i) < OBC ; (ii) < COB.
(a) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.
Let \(O\) be the centre, and let arc \(AB\) subtend \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining circumference. Join \(CO\) and produce it to \(X\).
Triangle \(OAC\) is isosceles because \(OA = OC\) (radii), so \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the interior opposite angles:
\[\angle AOX = \angle OAC + \angle OCA = 2\,\angle OCA.\]Likewise, from isosceles triangle \(OBC\), \(\angle BOX = 2\,\angle OCB\). Adding,
\[\angle AOB = \angle AOX + \angle BOX = 2(\angle OCA + \angle OCB) = 2\,\angle ACB.\]Hence the central angle is twice the inscribed angle on the same arc. (Q.E.D.)
(b) \(O\) is the centre of circle \(ACDB\), with \(\angle CAO = 26^\circ\) and \(\angle AOB = 130^\circ\); \(AD\) is the diameter through \(O\) (the dotted line \(A\!-\!O\!-\!D\)).
First find \(\angle AOC\). In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles with base angles equal:
\[\angle OCA = \angle OAC = 26^\circ \;\Rightarrow\; \angle AOC = 180^\circ - 26^\circ - 26^\circ = 128^\circ.\]Since \(A,O,D\) are collinear (diameter), angles on the straight line at \(O\) give
\[\angle COD = 180^\circ - \angle AOC = 180^\circ - 128^\circ = 52^\circ,\qquad \angle BOD = 180^\circ - \angle AOB = 180^\circ - 130^\circ = 50^\circ.\](ii) \(\angle COB\). With \(C\) above and \(B\) below the diameter, \(\angle COB\) is made up of \(\angle COD\) and \(\angle DOB\):
\[\angle COB = \angle COD + \angle BOD = 52^\circ + 50^\circ = \boxed{102^\circ}.\](i) \(\angle OBC\). Triangle \(OBC\) is isosceles since \(OB = OC\) (radii), so its base angles are equal:
\[\angle OBC = \angle OCB = \frac{180^\circ - \angle COB}{2} = \frac{180^\circ - 102^\circ}{2} = \frac{78^\circ}{2} = \boxed{39^\circ}.\]Answer Details
(a) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.
Let \(O\) be the centre, and let arc \(AB\) subtend \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining circumference. Join \(CO\) and produce it to \(X\).
Triangle \(OAC\) is isosceles because \(OA = OC\) (radii), so \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the interior opposite angles:
\[\angle AOX = \angle OAC + \angle OCA = 2\,\angle OCA.\]Likewise, from isosceles triangle \(OBC\), \(\angle BOX = 2\,\angle OCB\). Adding,
\[\angle AOB = \angle AOX + \angle BOX = 2(\angle OCA + \angle OCB) = 2\,\angle ACB.\]Hence the central angle is twice the inscribed angle on the same arc. (Q.E.D.)
(b) \(O\) is the centre of circle \(ACDB\), with \(\angle CAO = 26^\circ\) and \(\angle AOB = 130^\circ\); \(AD\) is the diameter through \(O\) (the dotted line \(A\!-\!O\!-\!D\)).
First find \(\angle AOC\). In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles with base angles equal:
\[\angle OCA = \angle OAC = 26^\circ \;\Rightarrow\; \angle AOC = 180^\circ - 26^\circ - 26^\circ = 128^\circ.\]Since \(A,O,D\) are collinear (diameter), angles on the straight line at \(O\) give
\[\angle COD = 180^\circ - \angle AOC = 180^\circ - 128^\circ = 52^\circ,\qquad \angle BOD = 180^\circ - \angle AOB = 180^\circ - 130^\circ = 50^\circ.\](ii) \(\angle COB\). With \(C\) above and \(B\) below the diameter, \(\angle COB\) is made up of \(\angle COD\) and \(\angle DOB\):
\[\angle COB = \angle COD + \angle BOD = 52^\circ + 50^\circ = \boxed{102^\circ}.\](i) \(\angle OBC\). Triangle \(OBC\) is isosceles since \(OB = OC\) (radii), so its base angles are equal:
\[\angle OBC = \angle OCB = \frac{180^\circ - \angle COB}{2} = \frac{180^\circ - 102^\circ}{2} = \frac{78^\circ}{2} = \boxed{39^\circ}.\]
