WAEC SSCE - General Mathematics - 2015 (Essay)
Question 1 Report
(a) Make m the subject of the relations \(h = \frac{mt}{d(m + p)}\).
(b) 
In the diagram, WY and WZ are straight lines, O is the centre of circle WXM and < XWM = 48°. Calculate the value of < WYZ.
(c) An operation \(\star\) is defind on the set X = {1, 3, 5, 6} by \(m \star n = m + n + 2 (mod 7)\) where \(m, n \in X\).
(i) Draw a table for the operation.
(ii) Using the table, find the truth set of : (I) \(3 \star n = 3\) ; (II) \(n \star n = 3\).
(a)
\[ h=\frac{mt}{d(m+p)} \]
\[ dh(m+p)=mt \]
\[ dhm+dhp=mt \]
\[ dhp=mt-dhm=m(t-dh) \]
Hence,
\[ \boxed{m=\frac{dhp}{t-dh}} \]
(b)
Since \(WM\) is a diameter of the circle,
\[ \angle WXM=90^\circ \]
In \(\triangle WXM\),
\[ \angle WMX=180^\circ-(90^\circ+48^\circ)=42^\circ. \]
Since \(W, M\) and \(Z\) lie on a straight line,
\[ \angle XMZ=180^\circ-42^\circ=138^\circ. \]
Opposite angles of the cyclic quadrilateral are supplementary. Therefore,
\[ \angle WYZ=180^\circ-138^\circ=\boxed{42^\circ}. \]
(c)(i) Using \(m\star n=m+n+2\pmod 7\), the operation table is:
| \(\star\) | \(1\) | \(3\) | \(5\) | \(6\) |
|---|---|---|---|---|
| \(1\) | \(4\) | \(6\) | \(1\) | \(2\) |
| \(3\) | \(6\) | \(1\) | \(3\) | \(4\) |
| \(5\) | \(1\) | \(3\) | \(5\) | \(6\) |
| \(6\) | \(2\) | \(4\) | \(6\) | \(0\) |
(c)(ii)
(I) From the row headed \(3\), \(3\star n=3\) when \(n=5\).
\[ \boxed{\{5\}} \]
(II) The diagonal entries are \(4,1,5,0\). None is equal to \(3\). Thus,
\[ \boxed{\varnothing} \]
Answer Details
(a)
\[ h=\frac{mt}{d(m+p)} \]
\[ dh(m+p)=mt \]
\[ dhm+dhp=mt \]
\[ dhp=mt-dhm=m(t-dh) \]
Hence,
\[ \boxed{m=\frac{dhp}{t-dh}} \]
(b)
Since \(WM\) is a diameter of the circle,
\[ \angle WXM=90^\circ \]
In \(\triangle WXM\),
\[ \angle WMX=180^\circ-(90^\circ+48^\circ)=42^\circ. \]
Since \(W, M\) and \(Z\) lie on a straight line,
\[ \angle XMZ=180^\circ-42^\circ=138^\circ. \]
Opposite angles of the cyclic quadrilateral are supplementary. Therefore,
\[ \angle WYZ=180^\circ-138^\circ=\boxed{42^\circ}. \]
(c)(i) Using \(m\star n=m+n+2\pmod 7\), the operation table is:
| \(\star\) | \(1\) | \(3\) | \(5\) | \(6\) |
|---|---|---|---|---|
| \(1\) | \(4\) | \(6\) | \(1\) | \(2\) |
| \(3\) | \(6\) | \(1\) | \(3\) | \(4\) |
| \(5\) | \(1\) | \(3\) | \(5\) | \(6\) |
| \(6\) | \(2\) | \(4\) | \(6\) | \(0\) |
(c)(ii)
(I) From the row headed \(3\), \(3\star n=3\) when \(n=5\).
\[ \boxed{\{5\}} \]
(II) The diagonal entries are \(4,1,5,0\). None is equal to \(3\). Thus,
\[ \boxed{\varnothing} \]
Question 2 Report
(a) (i) Illustrate the following statements in a Venn diagram : All good Literature students in a school are in the General Arts class.
(ii) Use ths diagram to determine whether or not the following are valid conclusions from the given statement.
(1) Vivian is in the General Arts class therefore she is a good Literature student.
(2) Audu is not a good Literature student therefore he is not in the General Arts class;
(3) Kweku is not in the General Arts class therefore he is not a good Literature student.
(b) The cost (c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH¢950.00 to produce 600 bricks and GH¢ 1,030.00 to produce 1000 bricks,
(i) Find the relationship between c, h and n ; (ii) Calculate the cost of producing 500 bricks.
(a)(i) Venn diagram: Draw a rectangle for the school. Inside it draw a circle for the General Arts class \((A)\). Entirely within that circle draw a smaller circle for good Literature students \((L)\). This shows \(L \subset A\): every good Literature student is in the General Arts class, but not every General Arts student is a good Literature student.
(ii) Testing the conclusions:
- Vivian is in General Arts, therefore she is a good Literature student. NOT valid: being in \(A\) does not place her inside the smaller circle \(L\).
- Audu is not a good Literature student, therefore he is not in General Arts. NOT valid: he could be in \(A\) but outside \(L\).
- Kweku is not in General Arts, therefore he is not a good Literature student. VALID: since \(L \subset A\), anyone outside \(A\) must be outside \(L\).
(b) Cost \(c = h + y\) where \(y = kn\) (\(y\) varies directly as \(n\)), so \(c = h + kn\).
\(950 = h + 600k \quad(1)\)
\(1030 = h + 1000k \quad(2)\)
\((2) - (1):\ 80 = 400k \ \Rightarrow\ k = 0.2\). Then \(h = 950 - 600(0.2) = 830\).
(i) \(\mathbf{c = h + 0.2n}\), that is \(c = 830 + 0.2n\) (with \(h = \text{GH¢}830\)).
(ii) For \(n = 500\): \(c = 830 + 0.2(500) = 830 + 100 = \mathbf{\text{GH¢}930.00}\).
Answer Details
(a)(i) Venn diagram: Draw a rectangle for the school. Inside it draw a circle for the General Arts class \((A)\). Entirely within that circle draw a smaller circle for good Literature students \((L)\). This shows \(L \subset A\): every good Literature student is in the General Arts class, but not every General Arts student is a good Literature student.
(ii) Testing the conclusions:
- Vivian is in General Arts, therefore she is a good Literature student. NOT valid: being in \(A\) does not place her inside the smaller circle \(L\).
- Audu is not a good Literature student, therefore he is not in General Arts. NOT valid: he could be in \(A\) but outside \(L\).
- Kweku is not in General Arts, therefore he is not a good Literature student. VALID: since \(L \subset A\), anyone outside \(A\) must be outside \(L\).
(b) Cost \(c = h + y\) where \(y = kn\) (\(y\) varies directly as \(n\)), so \(c = h + kn\).
\(950 = h + 600k \quad(1)\)
\(1030 = h + 1000k \quad(2)\)
\((2) - (1):\ 80 = 400k \ \Rightarrow\ k = 0.2\). Then \(h = 950 - 600(0.2) = 830\).
(i) \(\mathbf{c = h + 0.2n}\), that is \(c = 830 + 0.2n\) (with \(h = \text{GH¢}830\)).
(ii) For \(n = 500\): \(c = 830 + 0.2(500) = 830 + 100 = \mathbf{\text{GH¢}930.00}\).
Question 3 Report
(a) By how much is the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) less than 7?
(b) The height, h m, of a dock above sea level is given by \(h = 6 + 4\cos (15p)°, 0 < p < 6\). Find :
(i) the value of h when p = 4 ; (ii) correct to two significant figures, the value of p when h = 9 m.
(a) Sum \(= 3\tfrac{2}{3} + 2\tfrac{1}{5} = \tfrac{11}{3} + \tfrac{11}{5} = \tfrac{55 + 33}{15} = \tfrac{88}{15}\).
It is less than 7 by \(7 - \tfrac{88}{15} = \tfrac{105 - 88}{15} = \tfrac{17}{15} = 1\tfrac{2}{15}\).
(b)(i) With \(p = 4\): \(15p = 60^{\circ}\) and \(\cos 60^{\circ} = 0.5\).
\[h = 6 + 4(0.5) = 8 \text{ m}\](b)(ii) With \(h = 9\):
\[9 = 6 + 4\cos(15p)^{\circ} \Rightarrow \cos(15p)^{\circ} = 0.75\] \[15p = \cos^{-1} 0.75 = 41.4^{\circ} \Rightarrow p = \frac{41.4}{15} = 2.8 \text{ (2 s.f.)}\]Answer Details
(a) Sum \(= 3\tfrac{2}{3} + 2\tfrac{1}{5} = \tfrac{11}{3} + \tfrac{11}{5} = \tfrac{55 + 33}{15} = \tfrac{88}{15}\).
It is less than 7 by \(7 - \tfrac{88}{15} = \tfrac{105 - 88}{15} = \tfrac{17}{15} = 1\tfrac{2}{15}\).
(b)(i) With \(p = 4\): \(15p = 60^{\circ}\) and \(\cos 60^{\circ} = 0.5\).
\[h = 6 + 4(0.5) = 8 \text{ m}\](b)(ii) With \(h = 9\):
\[9 = 6 + 4\cos(15p)^{\circ} \Rightarrow \cos(15p)^{\circ} = 0.75\] \[15p = \cos^{-1} 0.75 = 41.4^{\circ} \Rightarrow p = \frac{41.4}{15} = 2.8 \text{ (2 s.f.)}\]Question 4 Report
(a) Solve the inequality : \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)
(b) 
The diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\(cm^{2}\), find the values of x.
(a) Solving the inequality
\[4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1\]
Multiply every term by 8 (the LCM of the denominators) to clear fractions:
\[8(4) + 8\cdot\frac{3}{4}(x+2) \le 8\cdot\frac{3}{8}x + 8(1)\]
\[32 + 6(x+2) \le 3x + 8\]
\[32 + 6x + 12 \le 3x + 8\]
\[6x + 44 \le 3x + 8\]
Collect like terms:
\[6x - 3x \le 8 - 44\]
\[3x \le -36\]
\[x \le -12\]
(b) Finding x from the rectangle
From the diagram, rectangle PQRS has height \(|PS| = 20\ \text{cm}\). Along the base SR a square of side x cm is cut out, leaving 10 cm on the left and 10 cm on the right. So the full length of the rectangle is
\[|SR| = 10 + x + 10 = (20 + x)\ \text{cm}\]
Area of the whole rectangle:
\[(20 + x)\times 20 = 400 + 20x\]
Area of the square removed:
\[x \times x = x^2\]
The shaded (remaining) area is 484 cm\(^2\):
\[(400 + 20x) - x^2 = 484\]
\[-x^2 + 20x + 400 - 484 = 0\]
\[-x^2 + 20x - 84 = 0\]
\[x^2 - 20x + 84 = 0\]
Solve using factorisation (\(-6\) and \(-14\) multiply to 84 and add to \(-20\)):
\[(x - 6)(x - 14) = 0\]
\[x = 6 \quad\text{or}\quad x = 14\]
Both values are geometrically valid (the cut height stays within the 20 cm rectangle), so
\[x = 6\ \text{cm} \quad\text{or}\quad x = 14\ \text{cm}\]
Answer Details
(a) Solving the inequality
\[4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1\]
Multiply every term by 8 (the LCM of the denominators) to clear fractions:
\[8(4) + 8\cdot\frac{3}{4}(x+2) \le 8\cdot\frac{3}{8}x + 8(1)\]
\[32 + 6(x+2) \le 3x + 8\]
\[32 + 6x + 12 \le 3x + 8\]
\[6x + 44 \le 3x + 8\]
Collect like terms:
\[6x - 3x \le 8 - 44\]
\[3x \le -36\]
\[x \le -12\]
(b) Finding x from the rectangle
From the diagram, rectangle PQRS has height \(|PS| = 20\ \text{cm}\). Along the base SR a square of side x cm is cut out, leaving 10 cm on the left and 10 cm on the right. So the full length of the rectangle is
\[|SR| = 10 + x + 10 = (20 + x)\ \text{cm}\]
Area of the whole rectangle:
\[(20 + x)\times 20 = 400 + 20x\]
Area of the square removed:
\[x \times x = x^2\]
The shaded (remaining) area is 484 cm\(^2\):
\[(400 + 20x) - x^2 = 484\]
\[-x^2 + 20x + 400 - 484 = 0\]
\[-x^2 + 20x - 84 = 0\]
\[x^2 - 20x + 84 = 0\]
Solve using factorisation (\(-6\) and \(-14\) multiply to 84 and add to \(-20\)):
\[(x - 6)(x - 14) = 0\]
\[x = 6 \quad\text{or}\quad x = 14\]
Both values are geometrically valid (the cut height stays within the 20 cm rectangle), so
\[x = 6\ \text{cm} \quad\text{or}\quad x = 14\ \text{cm}\]
Question 5 Report
The table is for the relation \(y = px^{2} - 5x + q\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 21 | 6 | -12 | 0 | 13 |
(a)(i) Use the table to find the values of p and q.
(ii) Copy and complete the table.
(b) Using scales of 2cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of the relation for \(-3 \leq x \leq 5\).
(c) Use the graph to find :
(i) y when x = 1.8 ; (ii) x when y = -8.
(a)(i) Given that \(y=px^2-5x+q\).
When \(x=0\), \(y=-12\):
\[q=-12.\]
When \(x=4\), \(y=0\):
\[0=16p-5(4)-12\]
\[0=16p-32\]
\[p=2.\]
Therefore,
\[\boxed{y=2x^2-5x-12}.\]
(a)(ii) The completed table is:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|
| \(y\) | \(21\) | \(6\) | \(-5\) | \(-12\) | \(-15\) | \(-14\) | \(-9\) | \(0\) | \(13\) |
(b) The graph of \(y=2x^2-5x-12\), plotted using the stated scales, is shown below.
(c)
(i) From the graph, when \(x=1.8\),
\[\boxed{y\approx -14.5}.\]
(ii) Drawing the horizontal line \(y=-8\) and reading the two points of intersection with the curve gives
\[\boxed{x\approx -0.65\text{ or }x\approx 3.15}.\]
Answer Details
(a)(i) Given that \(y=px^2-5x+q\).
When \(x=0\), \(y=-12\):
\[q=-12.\]
When \(x=4\), \(y=0\):
\[0=16p-5(4)-12\]
\[0=16p-32\]
\[p=2.\]
Therefore,
\[\boxed{y=2x^2-5x-12}.\]
(a)(ii) The completed table is:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|
| \(y\) | \(21\) | \(6\) | \(-5\) | \(-12\) | \(-15\) | \(-14\) | \(-9\) | \(0\) | \(13\) |
(b) The graph of \(y=2x^2-5x-12\), plotted using the stated scales, is shown below.
(c)
(i) From the graph, when \(x=1.8\),
\[\boxed{y\approx -14.5}.\]
(ii) Drawing the horizontal line \(y=-8\) and reading the two points of intersection with the curve gives
\[\boxed{x\approx -0.65\text{ or }x\approx 3.15}.\]
Question 6 Report
The table shows the marks scored by some candidates in an examination.
| Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.
(b) Use the curve to estimate, correct to one decimal place, the :
(i) Lowest mark for distinction if 5% of the candidates passed with distinction ; (ii) probability of selecting a candidate who scored at most 45%.
(a) Cumulative frequency table
| Marks (%) | Frequency, \(f\) | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 0–9 | 7 | 9.5 | 7 |
| 10–19 | 11 | 19.5 | 18 |
| 20–29 | 17 | 29.5 | 35 |
| 30–39 | 20 | 39.5 | 55 |
| 40–49 | 29 | 49.5 | 84 |
| 50–59 | 34 | 59.5 | 118 |
| 60–69 | 30 | 69.5 | 148 |
| 70–79 | 25 | 79.5 | 173 |
| 80–89 | 21 | 89.5 | 194 |
| 90–99 | 6 | 99.5 | 200 |
Total number of candidates \(=200\).
Plot the upper class boundaries against the cumulative frequencies, including \((-0.5,0)\), and draw a smooth cumulative frequency curve.
(b)(i) The number who passed with distinction is
\[5\%\text{ of }200=10.\]
Hence the lowest distinction mark has cumulative frequency
\[200-10=190.\]
From the cumulative frequency curve, a cumulative frequency of \(190\) gives a mark of approximately \(87.5\%\).
Lowest mark for distinction \(=87.5\%\).
(b)(ii) From the curve, the cumulative frequency at \(45\%\) is approximately \(69\).
\[P(\text{score at most }45\%)=\frac{69}{200}=0.345\approx 0.3\text{ (to 1 d.p.)}.\]
Answer Details
(a) Cumulative frequency table
| Marks (%) | Frequency, \(f\) | Upper class boundary | Cumulative frequency |
|---|---|---|---|
| 0–9 | 7 | 9.5 | 7 |
| 10–19 | 11 | 19.5 | 18 |
| 20–29 | 17 | 29.5 | 35 |
| 30–39 | 20 | 39.5 | 55 |
| 40–49 | 29 | 49.5 | 84 |
| 50–59 | 34 | 59.5 | 118 |
| 60–69 | 30 | 69.5 | 148 |
| 70–79 | 25 | 79.5 | 173 |
| 80–89 | 21 | 89.5 | 194 |
| 90–99 | 6 | 99.5 | 200 |
Total number of candidates \(=200\).
Plot the upper class boundaries against the cumulative frequencies, including \((-0.5,0)\), and draw a smooth cumulative frequency curve.
(b)(i) The number who passed with distinction is
\[5\%\text{ of }200=10.\]
Hence the lowest distinction mark has cumulative frequency
\[200-10=190.\]
From the cumulative frequency curve, a cumulative frequency of \(190\) gives a mark of approximately \(87.5\%\).
Lowest mark for distinction \(=87.5\%\).
(b)(ii) From the curve, the cumulative frequency at \(45\%\) is approximately \(69\).
\[P(\text{score at most }45\%)=\frac{69}{200}=0.345\approx 0.3\text{ (to 1 d.p.)}.\]
Question 7 Report
A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.
(a) Illustrate this information in a diagram.
(b) If the volume of the reservoir is \(333\frac{1}{3}\pi m^{3}\), calculate, correct to the nearest whole number, the :
(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take \(\pi = \frac{22}{7}\)].
(a) Diagram: A cone sitting on top of a hemisphere, sharing the same circular base of radius \(r\). The flat face of the hemisphere coincides with the base of the cone, and the cone height is \(h = 6r\).
(b) Let the common radius be \(r\), so \(h = 6r\).
\(V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi r^2(6r) = 2\pi r^3\)
\(V_{\text{hemisphere}} = \tfrac{2}{3}\pi r^3\)
Total: \(2\pi r^3 + \tfrac{2}{3}\pi r^3 = \tfrac{8}{3}\pi r^3 = 333\tfrac{1}{3}\pi = \tfrac{1000}{3}\pi\).
\(8r^3 = 1000 \ \Rightarrow\ r^3 = 125 \ \Rightarrow\ r = 5\text{ m}\) (so \(h = 30\text{ m}\)).
(I) Volume of hemisphere: \(\tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\times\tfrac{22}{7}\times 125 = \dfrac{5500}{21} = 261.9 \approx \mathbf{262\text{ m}^3}\).
(II) Total surface area \(=\) curved surface of cone \(+\) curved surface of hemisphere (the joining faces are internal).
Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{25 + 900} = \sqrt{925} = 30.41\text{ m}\).
\(\pi r l = \tfrac{22}{7}\times 5\times 30.41 = 477.9\text{ m}^2\); \(2\pi r^2 = 2\times\tfrac{22}{7}\times 25 = 157.1\text{ m}^2\).
Total \(= 477.9 + 157.1 = 635.0 \approx \mathbf{635\text{ m}^2}\).
Answer Details
(a) Diagram: A cone sitting on top of a hemisphere, sharing the same circular base of radius \(r\). The flat face of the hemisphere coincides with the base of the cone, and the cone height is \(h = 6r\).
(b) Let the common radius be \(r\), so \(h = 6r\).
\(V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi r^2(6r) = 2\pi r^3\)
\(V_{\text{hemisphere}} = \tfrac{2}{3}\pi r^3\)
Total: \(2\pi r^3 + \tfrac{2}{3}\pi r^3 = \tfrac{8}{3}\pi r^3 = 333\tfrac{1}{3}\pi = \tfrac{1000}{3}\pi\).
\(8r^3 = 1000 \ \Rightarrow\ r^3 = 125 \ \Rightarrow\ r = 5\text{ m}\) (so \(h = 30\text{ m}\)).
(I) Volume of hemisphere: \(\tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\times\tfrac{22}{7}\times 125 = \dfrac{5500}{21} = 261.9 \approx \mathbf{262\text{ m}^3}\).
(II) Total surface area \(=\) curved surface of cone \(+\) curved surface of hemisphere (the joining faces are internal).
Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{25 + 900} = \sqrt{925} = 30.41\text{ m}\).
\(\pi r l = \tfrac{22}{7}\times 5\times 30.41 = 477.9\text{ m}^2\); \(2\pi r^2 = 2\times\tfrac{22}{7}\times 25 = 157.1\text{ m}^2\).
Total \(= 477.9 + 157.1 = 635.0 \approx \mathbf{635\text{ m}^2}\).
Question 8 Report
(a) Using ruler and a pair of compasses only, construct a :
(i) Trapezium WXYZ such that |WX| = 8 cm, |XY| = 5.5 cm, |YZ| = 8.3 cm, < WXY = 60° and WX // ZY;
(ii) rectangle PQYZ where P and Q are on WX
(b) Measure : (i) |QX| ; (ii) < XWZ.
(a) Construction
- Draw the straight line WX = 8.0 cm.
- At X, construct an angle of 60° with XW. On this ray, mark Y such that XY = 5.5 cm.
- Through Y, construct a line parallel to WX. On it, mark Z such that YZ = 8.3 cm, on the side shown. Join Z to W to complete trapezium WXYZ.
- From Y and Z, construct perpendiculars to the line WX, meeting it at Q and P respectively. Join PQ. Thus, PQYZ is a rectangle.
(b) Measurements from the construction
(i) \(|QX| = 2.7\text{ cm}\).
(ii) \(\angle XWZ = 75^\circ\).
Answer Details
(a) Construction
- Draw the straight line WX = 8.0 cm.
- At X, construct an angle of 60° with XW. On this ray, mark Y such that XY = 5.5 cm.
- Through Y, construct a line parallel to WX. On it, mark Z such that YZ = 8.3 cm, on the side shown. Join Z to W to complete trapezium WXYZ.
- From Y and Z, construct perpendiculars to the line WX, meeting it at Q and P respectively. Join PQ. Thus, PQYZ is a rectangle.
(b) Measurements from the construction
(i) \(|QX| = 2.7\text{ cm}\).
(ii) \(\angle XWZ = 75^\circ\).
Question 9 Report
A trapezium PQRS is such that PQ // RS and the perpendicular P to RS is 40 cm. If |PQ| = 20 cm, |SP| = 50 cm and |SR| = 60 cm. Calculate, correct to 2 significant figures, the
(a) Area of the trapezium ; (b) < QRS.
(a) Area. The perpendicular distance between the parallel sides is 40 cm.
\[A = \tfrac{1}{2}(PQ + SR) \times h = \tfrac{1}{2}(20 + 60) \times 40 = 1600 \text{ cm}^2\](b) Angle QRS. Place S at \((0, 0)\) and R at \((60, 0)\). P is 40 cm above RS; since \(SP = 50\), its horizontal offset is \(\sqrt{50^2 - 40^2} = 30\), so P is at \((30, 40)\). As \(PQ \parallel SR\) with \(PQ = 20\), Q is at \((50, 40)\).
At R, \(\vec{RQ} = (-10, 40)\) and \(\vec{RS} = (-60, 0)\):
\[\cos(\angle QRS) = \frac{(-10)(-60) + 0}{\sqrt{10^2 + 40^2} \times 60} = \frac{600}{2474} = 0.2425\] \[\angle QRS = 76^{\circ} \text{ (2 s.f.)}\]Answer Details
(a) Area. The perpendicular distance between the parallel sides is 40 cm.
\[A = \tfrac{1}{2}(PQ + SR) \times h = \tfrac{1}{2}(20 + 60) \times 40 = 1600 \text{ cm}^2\](b) Angle QRS. Place S at \((0, 0)\) and R at \((60, 0)\). P is 40 cm above RS; since \(SP = 50\), its horizontal offset is \(\sqrt{50^2 - 40^2} = 30\), so P is at \((30, 40)\). As \(PQ \parallel SR\) with \(PQ = 20\), Q is at \((50, 40)\).
At R, \(\vec{RQ} = (-10, 40)\) and \(\vec{RS} = (-60, 0)\):
\[\cos(\angle QRS) = \frac{(-10)(-60) + 0}{\sqrt{10^2 + 40^2} \times 60} = \frac{600}{2474} = 0.2425\] \[\angle QRS = 76^{\circ} \text{ (2 s.f.)}\]Question 10 Report
(a) Without using Mathematical tables or calculators, simplify : \(\frac{2\tan 60° + \cos 30°}{\sin 60°}\)
(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :
(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.
(a) Use exact surd values: \(\tan 60° = \sqrt{3}\), \(\cos 30° = \dfrac{\sqrt{3}}{2}\), \(\sin 60° = \dfrac{\sqrt{3}}{2}\).
Numerator: \(2\sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{4\sqrt{3} + \sqrt{3}}{2} = \dfrac{5\sqrt{3}}{2}\).
\(\dfrac{2\tan 60° + \cos 30°}{\sin 60°} = \dfrac{5\sqrt{3}/2}{\sqrt{3}/2} = \mathbf{5}\)
(b) Horizontal distance to the tower \(= 1050\text{ m}\). The top has depression \(36°\) and the base \(41°\). The vertical drop from the aeroplane's level to each point is \(1050\tan(\text{angle})\).
(i) Height of tower \(=\) (drop to base) \(-\) (drop to top):
\(= 1050\tan 41° - 1050\tan 36° = 1050(0.8693 - 0.7265)\)
\(= 1050 \times 0.1428 = 149.9 \approx \mathbf{150\text{ m}}\).
(ii) Shortest distance (line of sight to the base):
\(= \dfrac{1050}{\cos 41°} = \dfrac{1050}{0.7547} = 1391.3 \approx \mathbf{1391\text{ m}}\).
Answer Details
(a) Use exact surd values: \(\tan 60° = \sqrt{3}\), \(\cos 30° = \dfrac{\sqrt{3}}{2}\), \(\sin 60° = \dfrac{\sqrt{3}}{2}\).
Numerator: \(2\sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{4\sqrt{3} + \sqrt{3}}{2} = \dfrac{5\sqrt{3}}{2}\).
\(\dfrac{2\tan 60° + \cos 30°}{\sin 60°} = \dfrac{5\sqrt{3}/2}{\sqrt{3}/2} = \mathbf{5}\)
(b) Horizontal distance to the tower \(= 1050\text{ m}\). The top has depression \(36°\) and the base \(41°\). The vertical drop from the aeroplane's level to each point is \(1050\tan(\text{angle})\).
(i) Height of tower \(=\) (drop to base) \(-\) (drop to top):
\(= 1050\tan 41° - 1050\tan 36° = 1050(0.8693 - 0.7265)\)
\(= 1050 \times 0.1428 = 149.9 \approx \mathbf{150\text{ m}}\).
(ii) Shortest distance (line of sight to the base):
\(= \dfrac{1050}{\cos 41°} = \dfrac{1050}{0.7547} = 1391.3 \approx \mathbf{1391\text{ m}}\).
Question 11 Report
(a) Without using Mathematical tables or calculators, simplify:
\(3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}\)
(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.
(a) Brackets first:
\[5\tfrac{1}{3} - 2\tfrac{3}{4} = \tfrac{16}{3} - \tfrac{11}{4} = \tfrac{64 - 33}{12} = \tfrac{31}{12}\] \[3\tfrac{4}{9} \div \tfrac{31}{12} = \tfrac{31}{9} \times \tfrac{12}{31} = \tfrac{12}{9} = \tfrac{4}{3}\] \[\tfrac{4}{3} + 5\tfrac{9}{10} = \tfrac{4}{3} + \tfrac{59}{10} = \tfrac{40 + 177}{30} = \tfrac{217}{30} = 7\tfrac{7}{30}\](b) One number from \(\{2, 3, 4\}\) and one from \(\{1, 3, 5\}\): \(3 \times 3 = 9\) equally likely pairs.
Pairs whose sum satisfies \(3 < \text{sum} < 7\): \((2,3) = 5,\ (3,1) = 4,\ (3,3) = 6,\ (4,1) = 5\), giving 4 outcomes.
\[P = \frac{4}{9}\]Answer Details
(a) Brackets first:
\[5\tfrac{1}{3} - 2\tfrac{3}{4} = \tfrac{16}{3} - \tfrac{11}{4} = \tfrac{64 - 33}{12} = \tfrac{31}{12}\] \[3\tfrac{4}{9} \div \tfrac{31}{12} = \tfrac{31}{9} \times \tfrac{12}{31} = \tfrac{12}{9} = \tfrac{4}{3}\] \[\tfrac{4}{3} + 5\tfrac{9}{10} = \tfrac{4}{3} + \tfrac{59}{10} = \tfrac{40 + 177}{30} = \tfrac{217}{30} = 7\tfrac{7}{30}\](b) One number from \(\{2, 3, 4\}\) and one from \(\{1, 3, 5\}\): \(3 \times 3 = 9\) equally likely pairs.
Pairs whose sum satisfies \(3 < \text{sum} < 7\): \((2,3) = 5,\ (3,1) = 4,\ (3,3) = 6,\ (4,1) = 5\), giving 4 outcomes.
\[P = \frac{4}{9}\]Question 12 Report
(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2, Find the number of sides of the polygon.
(b) 
The diagram shows a circle PQRS with centre O, < UQR = 68°, < TPS = 74° and < QSR = 40°. Calculate the value of < PRS.
(a) Number of sides of the polygon
At each vertex the interior and exterior angles are supplementary, so they add up to \(180^\circ\). The ratio interior : exterior is \(5:2\), giving \(5+2 = 7\) equal parts.
\[\text{Exterior angle} = \frac{2}{7}\times 180^\circ = \frac{360^\circ}{7}\]For any regular polygon the exterior angles sum to \(360^\circ\), so the number of sides is
\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{\tfrac{360^\circ}{7}} = 7\]The polygon has 7 sides (a regular heptagon).
(b) Finding \(\angle PRS\)
From the diagram, \(TPQU\) is a straight line through the two points \(P\) and \(Q\) on the circle, and \(PQRS\) is a cyclic quadrilateral. The given angles are \(\angle TPS = 74^\circ\), \(\angle UQR = 68^\circ\) and \(\angle QSR = 40^\circ\).
Step 1: Interior angles of the cyclic quadrilateral. Since \(TPQ\) is a straight line,
\[\angle SPQ = 180^\circ - \angle TPS = 180^\circ - 74^\circ = 106^\circ\]Since \(PQU\) is a straight line,
\[\angle PQR = 180^\circ - \angle UQR = 180^\circ - 68^\circ = 112^\circ\]Step 2: Use the opposite-angles property of a cyclic quadrilateral.
\[\angle SRQ = 180^\circ - \angle SPQ = 180^\circ - 106^\circ = 74^\circ\]Step 3: Split \(\angle SRQ\) using angles in the same segment. The angle \(\angle PRQ\) stands on chord \(PQ\), as does \(\angle PSQ\); angles in the same segment are equal, so \(\angle PRQ = \angle PSQ\).
Now find \(\angle PSQ\). We have \(\angle PSR = 180^\circ - \angle PQR = 180^\circ - 112^\circ = 68^\circ\), and this splits as \(\angle PSR = \angle PSQ + \angle QSR\). Therefore
\[\angle PSQ = 68^\circ - 40^\circ = 28^\circ \;\Rightarrow\; \angle PRQ = 28^\circ\]Step 4: Obtain \(\angle PRS\). Since \(\angle SRQ = \angle PRS + \angle PRQ\),
\[\angle PRS = 74^\circ - 28^\circ = \boxed{46^\circ}\]Hence \(\angle PRS = 46^\circ\).
Answer Details
(a) Number of sides of the polygon
At each vertex the interior and exterior angles are supplementary, so they add up to \(180^\circ\). The ratio interior : exterior is \(5:2\), giving \(5+2 = 7\) equal parts.
\[\text{Exterior angle} = \frac{2}{7}\times 180^\circ = \frac{360^\circ}{7}\]For any regular polygon the exterior angles sum to \(360^\circ\), so the number of sides is
\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{\tfrac{360^\circ}{7}} = 7\]The polygon has 7 sides (a regular heptagon).
(b) Finding \(\angle PRS\)
From the diagram, \(TPQU\) is a straight line through the two points \(P\) and \(Q\) on the circle, and \(PQRS\) is a cyclic quadrilateral. The given angles are \(\angle TPS = 74^\circ\), \(\angle UQR = 68^\circ\) and \(\angle QSR = 40^\circ\).
Step 1: Interior angles of the cyclic quadrilateral. Since \(TPQ\) is a straight line,
\[\angle SPQ = 180^\circ - \angle TPS = 180^\circ - 74^\circ = 106^\circ\]Since \(PQU\) is a straight line,
\[\angle PQR = 180^\circ - \angle UQR = 180^\circ - 68^\circ = 112^\circ\]Step 2: Use the opposite-angles property of a cyclic quadrilateral.
\[\angle SRQ = 180^\circ - \angle SPQ = 180^\circ - 106^\circ = 74^\circ\]Step 3: Split \(\angle SRQ\) using angles in the same segment. The angle \(\angle PRQ\) stands on chord \(PQ\), as does \(\angle PSQ\); angles in the same segment are equal, so \(\angle PRQ = \angle PSQ\).
Now find \(\angle PSQ\). We have \(\angle PSR = 180^\circ - \angle PQR = 180^\circ - 112^\circ = 68^\circ\), and this splits as \(\angle PSR = \angle PSQ + \angle QSR\). Therefore
\[\angle PSQ = 68^\circ - 40^\circ = 28^\circ \;\Rightarrow\; \angle PRQ = 28^\circ\]Step 4: Obtain \(\angle PRS\). Since \(\angle SRQ = \angle PRS + \angle PRQ\),
\[\angle PRS = 74^\circ - 28^\circ = \boxed{46^\circ}\]Hence \(\angle PRS = 46^\circ\).
Question 13 Report
(a) The first term of an Arithmetic Progression (AP) is 8, the ratio of the 7th term to the 9th term is 5 : 8, find the common difference of the AP.
(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.
