WAEC SSCE - General Mathematics - 1990 (Essay)
Ajụjụ 1 Ripọtì
(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:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Akọwa Nkọwa
(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:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Ajụjụ 2 Ripọtì
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 \(x \approx -0.7\).
Akọwa Nkọwa
(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 \(x \approx -0.7\).
Ajụjụ 3 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 4 Ripọtì
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:
\[ PQ + QT = 5134 + 1340.6 = 6474.6\ \text{km}. \] \[ \text{Average speed} = \frac{6474.6}{12} = 539.6 \approx 540\ \text{km/h (to 3 s.f.).} \]Akọwa Nkọwa
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:
\[ PQ + QT = 5134 + 1340.6 = 6474.6\ \text{km}. \] \[ \text{Average speed} = \frac{6474.6}{12} = 539.6 \approx 540\ \text{km/h (to 3 s.f.).} \]Ajụjụ 5 Ripọtì
(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:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Akọwa Nkọwa
(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:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Ajụjụ 6 Ripọtì
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:
\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]
\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]
Akọwa Nkọwa
(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:
\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]
\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]
Ajụjụ 7 Ripọtì
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):
\[ 20\% \times 12.5\% = 0.20 \times 0.125 = 0.025 = 2.5\%. \]Akọwa Nkọwa
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):
\[ 20\% \times 12.5\% = 0.20 \times 0.125 = 0.025 = 2.5\%. \]Ajụjụ 8 Ripọtì
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 |
