WAEC SSCE - General Mathematics - 1990 (Essay)
Question 1 Report
The weights to the nearest kilogram, of a group of 50 students in a College of Technology are given below:
65, 70, 60, 46, 51, 55, 59, 63, 68, 53, 47, 53, 72, 53, 67, 62, 64, 70, 57, 56, 73, 56, 48, 51, 58, 63, 65, 62, 49, 64, 53, 59, 63, 50, 48, 72, 67, 56, 61, 64, 66, 52, 49, 62, 71, 58, 53, 69, 63, 59.
(a) Prepare a grouped fraquency table with class intervals 45 - 49, 50 - 54, 55 - 59 etc.
(b) Using an assumed mean of 62 or otherwise, calculate the mean and standard deviation of the grouped data, correct to one decimal place.
(a) Grouped frequency table
| Column 1 | Column 2 | Column 3 |
|---|---|---|
| Data | Data | Data |
| Data | Data | Data |
| Data | Data | Data |
| Class Interval | Tally | Frequency, \(f\) |
|---|---|---|
| 45 - 49 | ||||| | | 6 |
| 50 - 54 | ||||| |||| | 9 |
| 55 - 59 | ||||| ||||| | 10 |
| 60 - 64 | ||||| ||||| || | 12 |
| 65 - 69 | ||||| || | 7 |
| 70 - 74 | ||||| | | 6 |
| Total | 50 |
(b) Let the assumed mean, \(A = 62\).
| Class Interval | Mid-value, \(x\) | \(d=x-62\) | \(d^2\) | \(f\) | \(fd\) | \(fd^2\) | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 45 - 49 | 47 | -15 | 225 | 6 | -90 | 1350 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 50 - 54 | 52 | -10 | 100 | 9 | -90 | 900 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| Column 1 | Column 2 | Column 3 |
|---|---|---|
| Data | Data | Data |
| Data | Data | Data |
| Data | Data | Data |
| Class Interval | Tally | Frequency, \(f\) |
|---|---|---|
| 45 - 49 | ||||| | | 6 |
| 50 - 54 | ||||| |||| | 9 |
| 55 - 59 | ||||| ||||| | 10 |
| 60 - 64 | ||||| ||||| || | 12 |
| 65 - 69 | ||||| || | 7 |
| 70 - 74 | ||||| | | 6 |
| Total | 50 |
(b) Let the assumed mean, \(A = 62\).
| Class Interval | Mid-value, \(x\) | \(d=x-62\) | \(d^2\) | \(f\) | \(fd\) | \(fd^2\) | |||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 45 - 49 | 47 | -15 | 225 | 6 | -90 | 1350 | |||||||||||||||||||||||||||||||||||||||||||||||||||||
| 50 - 54 | 52 | -10 | 100 | 9 | -90 | 900 | |||||||||||||||||||||||||||||||||||||||||||||||||||||
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 8 | 1 | -1 | 26 |
(a) Copy and complete the table.
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 10 units on the y- axis, draw the graph of the relation \(y = 2x^{2} - 5x + 1\) for \(-3 \leq x \leq 5\).
(c) Using the same scale and axes, draw the graph of \(y = x + 6\).
(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \(2x^{2} - 5x + 1 = x + 6\).
(a) Completing the table for \(y = 2x^{2} - 5x + 1\)
Substitute each x-value, for example \(x=-3:\; 2(9)-5(-3)+1 = 18+15+1 = 34\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|---|
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) The graph
Plot the nine points with 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, and join them with a smooth U-shaped parabola.
(c) The line \(y = x + 6\)
Use two points: at \(x=-3,\;y=3\) and at \(x=5,\;y=11\). Draw the straight line through \((-3,3)\) and \((5,11)\).
(d)(i) Least value of y
The lowest point of the parabola occurs at the vertex, \(x = \dfrac{5}{2\times 2} = 1.25\). Then
\(y = 2(1.25)^{2} - 5(1.25) + 1 = 3.125 - 6.25 + 1 = -2.125\).
From the graph, the least value of y is \(\approx -2.1\), occurring at \(x \approx 1.3\).
(d)(ii) Solution of \(2x^{2} - 5x + 1 = x + 6\)
The solutions are the x-coordinates where the curve meets the line. Algebraically:
\(2x^{2} - 5x + 1 = x + 6 \Rightarrow 2x^{2} - 6x - 5 = 0\).
\(x = \dfrac{6 \pm \sqrt{36+40}}{4} = \dfrac{6 \pm \sqrt{76}}{4}\), giving \(x \approx 3.7\) and \(x \approx -0.7\).
Answer Details
(a) Completing the table for \(y = 2x^{2} - 5x + 1\)
Substitute each x-value, for example \(x=-3:\; 2(9)-5(-3)+1 = 18+15+1 = 34\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|---|
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) The graph
Plot the nine points with 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, and join them with a smooth U-shaped parabola.
(c) The line \(y = x + 6\)
Use two points: at \(x=-3,\;y=3\) and at \(x=5,\;y=11\). Draw the straight line through \((-3,3)\) and \((5,11)\).
(d)(i) Least value of y
The lowest point of the parabola occurs at the vertex, \(x = \dfrac{5}{2\times 2} = 1.25\). Then
\(y = 2(1.25)^{2} - 5(1.25) + 1 = 3.125 - 6.25 + 1 = -2.125\).
From the graph, the least value of y is \(\approx -2.1\), occurring at \(x \approx 1.3\).
(d)(ii) Solution of \(2x^{2} - 5x + 1 = x + 6\)
The solutions are the x-coordinates where the curve meets the line. Algebraically:
\(2x^{2} - 5x + 1 = x + 6 \Rightarrow 2x^{2} - 6x - 5 = 0\).
\(x = \dfrac{6 \pm \sqrt{36+40}}{4} = \dfrac{6 \pm \sqrt{76}}{4}\), giving \(x \approx 3.7\) and \(x \approx -0.7\).
Question 5 Report
Show on a graph, the area which gives the solution set of the inequalities: \(y - 2x \leq 4 ; 3y + x \geq 6 ; y \geq 7x - 9\).
Rewrite the inequalities in terms of y:
Draw the three boundary lines as solid lines, since equality is included. Their pairwise intersections are:
The required solution set is the common region below \(y=2x+4\) and above both \(y=2-\frac{x}{3}\) and \(y=7x-9\). It is the shaded triangular region shown.
Hence, the solution region is the closed triangle with vertices
Answer Details
Rewrite the inequalities in terms of y:
Draw the three boundary lines as solid lines, since equality is included. Their pairwise intersections are:
The required solution set is the common region below \(y=2x+4\) and above both \(y=2-\frac{x}{3}\) and \(y=7x-9\). It is the shaded triangular region shown.
Hence, the solution region is the closed triangle with vertices
Question 6 Report
(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\(\pi = 3.142\)]
(b) A hemispherical tank of diameter which is 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.
(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):
\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]\(V \approx 100.5\ \text{cm}^3\).
(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.
\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:
\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]Time to fill:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Answer Details
(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):
\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]\(V \approx 100.5\ \text{cm}^3\).
(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.
\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:
\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]Time to fill:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Question 7 Report
(a) Simplify \(\frac{0.016 \times 0.084}{0.48}\) [Leave your answer in standard form].
(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.
(a) Multiply out the numerator and divide:
\[ \frac{0.016 \times 0.084}{0.48} = \frac{0.001344}{0.48} = 0.0028. \]In standard form:
\[ 0.0028 = 2.8 \times 10^{-3}. \](b) Let the lengths form an A.P. with first term \(a\) and common difference \(d\).
\[ T_2 = a + d = 2, \qquad T_6 = a + 5d = 5. \]Subtracting: \(4d = 3 \implies d = 0.75\ \text{m}\), and \(a = 2 - 0.75 = 1.25\ \text{m}\).
The eight poles, in order, are:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Answer Details
(a) Multiply out the numerator and divide:
\[ \frac{0.016 \times 0.084}{0.48} = \frac{0.001344}{0.48} = 0.0028. \]In standard form:
\[ 0.0028 = 2.8 \times 10^{-3}. \](b) Let the lengths form an A.P. with first term \(a\) and common difference \(d\).
\[ T_2 = a + d = 2, \qquad T_6 = a + 5d = 5. \]Subtracting: \(4d = 3 \implies d = 0.75\ \text{m}\), and \(a = 2 - 0.75 = 1.25\ \text{m}\).
The eight poles, in order, are:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Question 8 Report
The feet of two vertical poles of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30°. Calculate the :
(i) distance RT ; (ii) distance of the foot of the taller pole from P, correct to three significant figures ; (iii) angle of elevation of T from P, correct to one decimal place.
Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.
Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with
\[ NT = 7 - 3 = 4\ \text{m}. \]The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:
(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]
(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]
(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where
\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]Answer Details
Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.
Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with
\[ NT = 7 - 3 = 4\ \text{m}. \]The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:
(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]
(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]
(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where
\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]Question 9 Report
(a) ABCD is a trapezium in which AB // DC, |AB| = 8cm, < ABC = 60°, |BC| = 5.5cm and |BD| = 8.3cm. Using a ruler and a pair of compasses only, construct:
(i) the trapezium ABCD ; (ii) a rectangle PQCD, where P, Q are two points AB;
(b) Measure |AB| and |QB|.
There is an inconsistency in the supplied materials: the question asks for \(|AB|\), which is already given as \(8\text{ cm}\), whereas the reference answer gives \(|AC|=7\text{ cm}\). The reference answer therefore appears to intend \(|AC|\), not \(|AB|\).
Construction
- Draw \(AB=8\text{ cm}\).
- At \(B\), construct an angle of \(60^\circ\) with \(BA\). On this ray, mark \(C\) so that \(BC=5.5\text{ cm}\).
- Construct a line through \(C\) parallel to \(AB\): construct a perpendicular to \(AB\) through \(C\), then construct a perpendicular to this new line through \(C\).
- With centre \(B\) and radius \(8.3\text{ cm}\), draw an arc to meet the parallel through \(C\) at \(D\), choosing the intersection that gives the trapezium shown. Join \(A\) to \(D\) and \(C\) to \(D\).
- Construct perpendiculars from \(D\) and \(C\) to \(AB\). Let their feet on \(AB\) be \(P\) and \(Q\), respectively. Since \(PQ\parallel DC\) and \(PD\parallel QC\), with right angles at \(P\) and \(Q\), \(PQCD\) is a rectangle.
Measurements
- \(|AB|=8.0\text{ cm}\), because this length was given and used to construct the figure.
- \(|QB|\approx2.8\text{ cm}\), which may be recorded as \(2.9\text{ cm}\) from a construction drawing.
The value of \(QB\) follows from the horizontal component of \(BC\): \(QB=5.5\cos60^\circ=2.75\text{ cm}\), so a ruler measurement should be close to \(2.8\text{ cm}\) or \(2.9\text{ cm}\).
If the intended measurement was \(|AC|\), as indicated by the reference answer, then \(|AC|\approx7.1\text{ cm}\), recorded as \(7\text{ cm}\).
Answer Details
There is an inconsistency in the supplied materials: the question asks for \(|AB|\), which is already given as \(8\text{ cm}\), whereas the reference answer gives \(|AC|=7\text{ cm}\). The reference answer therefore appears to intend \(|AC|\), not \(|AB|\).
Construction
- Draw \(AB=8\text{ cm}\).
- At \(B\), construct an angle of \(60^\circ\) with \(BA\). On this ray, mark \(C\) so that \(BC=5.5\text{ cm}\).
- Construct a line through \(C\) parallel to \(AB\): construct a perpendicular to \(AB\) through \(C\), then construct a perpendicular to this new line through \(C\).
- With centre \(B\) and radius \(8.3\text{ cm}\), draw an arc to meet the parallel through \(C\) at \(D\), choosing the intersection that gives the trapezium shown. Join \(A\) to \(D\) and \(C\) to \(D\).
- Construct perpendiculars from \(D\) and \(C\) to \(AB\). Let their feet on \(AB\) be \(P\) and \(Q\), respectively. Since \(PQ\parallel DC\) and \(PD\parallel QC\), with right angles at \(P\) and \(Q\), \(PQCD\) is a rectangle.
Measurements
- \(|AB|=8.0\text{ cm}\), because this length was given and used to construct the figure.
- \(|QB|\approx2.8\text{ cm}\), which may be recorded as \(2.9\text{ cm}\) from a construction drawing.
The value of \(QB\) follows from the horizontal component of \(BC\): \(QB=5.5\cos60^\circ=2.75\text{ cm}\), so a ruler measurement should be close to \(2.8\text{ cm}\) or \(2.9\text{ cm}\).
If the intended measurement was \(|AC|\), as indicated by the reference answer, then \(|AC|\approx7.1\text{ cm}\), recorded as \(7\text{ cm}\).
Question 10 Report
A carpenter was told to make a rectangular desk with top of dimension 50cm by 40cm. The carpenter actually made the desk 60cm by 35cm.
(a) Calculate the percentage error in the (i) length and the breadth ; (ii) area of the table top.
(b) Find the product of the two errors in a(i).
Intended dimensions: 50 cm by 40 cm. Actual dimensions made: 60 cm by 35 cm.
Percentage error \(= \dfrac{|\text{actual} - \text{intended}|}{\text{intended}} \times 100\%\).
(a)(i) Length:
\[ \frac{|60 - 50|}{50} \times 100\% = \frac{10}{50} \times 100\% = 20\%. \]Breadth:
\[ \frac{|35 - 40|}{40} \times 100\% = \frac{5}{40} \times 100\% = 12.5\%. \](a)(ii) Area: intended area \(= 50 \times 40 = 2000\ \text{cm}^2\); actual area \(= 60 \times 35 = 2100\ \text{cm}^2\).
\[ \frac{|2100 - 2000|}{2000} \times 100\% = \frac{100}{2000} \times 100\% = 5\%. \](b) Product of the two errors in a(i):
\[ 20\% \times 12.5\% = 0.20 \times 0.125 = 0.025 = 2.5\%. \]Answer Details
Intended dimensions: 50 cm by 40 cm. Actual dimensions made: 60 cm by 35 cm.
Percentage error \(= \dfrac{|\text{actual} - \text{intended}|}{\text{intended}} \times 100\%\).
(a)(i) Length:
\[ \frac{|60 - 50|}{50} \times 100\% = \frac{10}{50} \times 100\% = 20\%. \]Breadth:
\[ \frac{|35 - 40|}{40} \times 100\% = \frac{5}{40} \times 100\% = 12.5\%. \](a)(ii) Area: intended area \(= 50 \times 40 = 2000\ \text{cm}^2\); actual area \(= 60 \times 35 = 2100\ \text{cm}^2\).
\[ \frac{|2100 - 2000|}{2000} \times 100\% = \frac{100}{2000} \times 100\% = 5\%. \](b) Product of the two errors in a(i):
\[ 20\% \times 12.5\% = 0.20 \times 0.125 = 0.025 = 2.5\%. \]Question 11 Report
(a) Prove that the sum of the angles in a triangle is 2 right angles.
(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that < CAE + < CBD = 2 < CEB.
(a) The angles of a triangle sum to two right angles.
Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.
- \(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
- \(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).
The angles on the straight line XY at C add to two right angles:
\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]Replacing the two alternate angles gives
\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)
(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).
\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:
\[ \angle CBD = A + \angle ACB = A + 2c. \]Therefore
\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):
\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]From (1) and (2),
\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]Answer Details
(a) The angles of a triangle sum to two right angles.
Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.
- \(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
- \(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).
The angles on the straight line XY at C add to two right angles:
\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]Replacing the two alternate angles gives
\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)
(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).
\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:
\[ \angle CBD = A + \angle ACB = A + 2c. \]Therefore
\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):
\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]From (1) and (2),
\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]Question 12 Report
In a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speak at least one of the these three languages. If 8 speak Hausa and Igbo, 11 speak Hausa and Yoruba and 6 speak Igbo and Yoruba.
(a) Draw a Venn diagram to illustrate the information, using x to represent the number of students that speak all three languages.
(b) calculate the value of x.
(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:
(b) Since every student speaks at least one of the languages,
\[n(H\cup I\cup Y)=40.\]
Using the inclusion-exclusion principle,
\[40=25+16+21-8-11-6+x\]
\[40=37+x\]
\[x=3.\]
Therefore, 3 students speak all three languages.
Hence the numerical entries in the Venn diagram are:
\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]
\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]
Answer Details
(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:
(b) Since every student speaks at least one of the languages,
\[n(H\cup I\cup Y)=40.\]
Using the inclusion-exclusion principle,
\[40=25+16+21-8-11-6+x\]
\[40=37+x\]
\[x=3.\]
Therefore, 3 students speak all three languages.
Hence the numerical entries in the Venn diagram are:
\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]
\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]
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