WAEC SSCE - General Mathematics - 1990 (Essay)
Swali 1 Ripoti
(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.
(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:
(i) both are red? (ii) both are the same colour?
(a) The integers from 5 to 25 inclusive number \(25 - 5 + 1 = 21\).
- Multiples of 5: 5, 10, 15, 20, 25 \(\rightarrow\) 5 numbers.
- Multiples of 3: 6, 9, 12, 15, 18, 21, 24 \(\rightarrow\) 7 numbers.
- Multiples of both (i.e. of 15): 15 \(\rightarrow\) 1 number.
By inclusion and exclusion, the count of "multiple of 5 or 3" is \(5 + 7 - 1 = 11\).
\[ P(\text{multiple of 5 or 3}) = \frac{11}{21}. \](b) 10 balls: 4 blue, 6 red. Picking is with replacement, so each pick has \(P(\text{red}) = \tfrac{6}{10}\) and \(P(\text{blue}) = \tfrac{4}{10}\).
(i) Both red:
\[ \frac{6}{10} \times \frac{6}{10} = \frac{36}{100} = \frac{9}{25}. \](ii) Both the same colour (both red or both blue):
\[ \left(\frac{6}{10}\right)^2 + \left(\frac{4}{10}\right)^2 = \frac{36}{100} + \frac{16}{100} = \frac{52}{100} = \frac{13}{25}. \]Maelezo ya Majibu
(a) The integers from 5 to 25 inclusive number \(25 - 5 + 1 = 21\).
- Multiples of 5: 5, 10, 15, 20, 25 \(\rightarrow\) 5 numbers.
- Multiples of 3: 6, 9, 12, 15, 18, 21, 24 \(\rightarrow\) 7 numbers.
- Multiples of both (i.e. of 15): 15 \(\rightarrow\) 1 number.
By inclusion and exclusion, the count of "multiple of 5 or 3" is \(5 + 7 - 1 = 11\).
\[ P(\text{multiple of 5 or 3}) = \frac{11}{21}. \](b) 10 balls: 4 blue, 6 red. Picking is with replacement, so each pick has \(P(\text{red}) = \tfrac{6}{10}\) and \(P(\text{blue}) = \tfrac{4}{10}\).
(i) Both red:
\[ \frac{6}{10} \times \frac{6}{10} = \frac{36}{100} = \frac{9}{25}. \](ii) Both the same colour (both red or both blue):
\[ \left(\frac{6}{10}\right)^2 + \left(\frac{4}{10}\right)^2 = \frac{36}{100} + \frac{16}{100} = \frac{52}{100} = \frac{13}{25}. \]Swali 2 Ripoti
An aeroplane flies from a town P(lat. 40°N, 38°E) to another town Q(lat. 40°N, 22°W). It later flies to a third town T(28°N, 22°W). Calculate the :
(a) distance between P and Q along their parallel of latitude ;
(b) distance between Q and T along their line of longitudes;
(c) average speed at which the aeroplane will fly from P to T via Q, if the journey takes 12 hours, correct to 3 significant figures. [Take the radius of the earth = 6400km ; \(\pi = 3.142\)]
Take \(R = 6400\) km and \(\pi = 3.142\).
(a) Distance P to Q along the parallel of latitude \(40^\circ\)N. The longitude difference is \(38^\circ\text{E} + 22^\circ\text{W} = 60^\circ\). The radius of the parallel is \(R\cos 40^\circ\), so
\[ PQ = \frac{60}{360} \times 2\pi R \cos 40^\circ = \frac{1}{6} \times 2 \times 3.142 \times 6400 \times 0.7660 = 5134\ \text{km}. \](b) Distance Q to T along the meridian \(22^\circ\)W. The latitude difference is \(40^\circ - 28^\circ = 12^\circ\), and along a meridian the radius is \(R\):
\[ QT = \frac{12}{360} \times 2\pi R = \frac{1}{30} \times 2 \times 3.142 \times 6400 = 1340.6\ \text{km}. \](c) Average speed from P to T via Q over 12 hours. Total distance:
\[ PQ + QT = 5134 + 1340.6 = 6474.6\ \text{km}. \] \[ \text{Average speed} = \frac{6474.6}{12} = 539.6 \approx 540\ \text{km/h (to 3 s.f.).} \]Maelezo ya Majibu
Take \(R = 6400\) km and \(\pi = 3.142\).
(a) Distance P to Q along the parallel of latitude \(40^\circ\)N. The longitude difference is \(38^\circ\text{E} + 22^\circ\text{W} = 60^\circ\). The radius of the parallel is \(R\cos 40^\circ\), so
\[ PQ = \frac{60}{360} \times 2\pi R \cos 40^\circ = \frac{1}{6} \times 2 \times 3.142 \times 6400 \times 0.7660 = 5134\ \text{km}. \](b) Distance Q to T along the meridian \(22^\circ\)W. The latitude difference is \(40^\circ - 28^\circ = 12^\circ\), and along a meridian the radius is \(R\):
\[ QT = \frac{12}{360} \times 2\pi R = \frac{1}{30} \times 2 \times 3.142 \times 6400 = 1340.6\ \text{km}. \](c) Average speed from P to T via Q over 12 hours. Total distance:
\[ PQ + QT = 5134 + 1340.6 = 6474.6\ \text{km}. \] \[ \text{Average speed} = \frac{6474.6}{12} = 539.6 \approx 540\ \text{km/h (to 3 s.f.).} \]Swali 3 Ripoti
The following is an incomplete table for the relation \(y = 2x^{2} - 5x + 1\)
| 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 \(x \approx -0.7\).
Maelezo ya Majibu
(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 \(x \approx -0.7\).
Swali 4 Ripoti
(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),
\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]Maelezo ya Majibu
(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),
\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]Swali 5 Ripoti
(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:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Maelezo ya Majibu
(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:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Swali 6 Ripoti
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
\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]Maelezo ya Majibu
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
\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]Swali 7 Ripoti
(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 \(7\text{ cm}\).
Maelezo ya Majibu
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 \(7\text{ cm}\).
Swali 8 Ripoti
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 |
