WAEC SSCE - General Mathematics - 1988 (Essay)
Tambaya 1 Rahoto
A sector of a circle of radius 7cm subtending an angle of 270° at the centre of the circle is used to form a cone.
(a) Find the base radius of the cone.
(b) Calculate the area of the base of the cone to the nearest square centimetre.
[Take \(\pi = \frac{22}{7}\)]
When a sector is rolled into a cone, the arc length of the sector becomes the circumference of the cone's base, and the sector radius becomes the slant height.
(a) Arc length \(=\dfrac{270}{360}\times2\pi\times7=\dfrac{3}{4}\times2\times\dfrac{22}{7}\times7=\dfrac{3}{4}\times44=33\text{ cm}.\)
Base circumference \(=2\pi r=33\):
\[2\times\frac{22}{7}\times r=33\Rightarrow\frac{44}{7}r=33\Rightarrow r=\frac{33\times7}{44}=5.25\text{ cm}.\]
Base radius \(=5.25\) cm.
(b) Base area \(=\pi r^{2}=\dfrac{22}{7}\times(5.25)^{2}=\dfrac{22}{7}\times27.5625=86.625\text{ cm}^{2}\).
\(\approx 87\text{ cm}^{2}\) (nearest square centimetre).
Bayanin Amsa
When a sector is rolled into a cone, the arc length of the sector becomes the circumference of the cone's base, and the sector radius becomes the slant height.
(a) Arc length \(=\dfrac{270}{360}\times2\pi\times7=\dfrac{3}{4}\times2\times\dfrac{22}{7}\times7=\dfrac{3}{4}\times44=33\text{ cm}.\)
Base circumference \(=2\pi r=33\):
\[2\times\frac{22}{7}\times r=33\Rightarrow\frac{44}{7}r=33\Rightarrow r=\frac{33\times7}{44}=5.25\text{ cm}.\]
Base radius \(=5.25\) cm.
(b) Base area \(=\pi r^{2}=\dfrac{22}{7}\times(5.25)^{2}=\dfrac{22}{7}\times27.5625=86.625\text{ cm}^{2}\).
\(\approx 87\text{ cm}^{2}\) (nearest square centimetre).
Tambaya 2 Rahoto
The number of items produced by a company over a five- year period is given below:
| Year | 1978 | 1979 | 1980 | 1981 | 1982 |
| No produced | 4100 | 2500 | 1500 | 1800 | 9200 |
(i) Plot a bar chart for this information; (ii) What is the average production for the five- year period.
(i) Bar chart of company production
The bar chart below represents the number of items produced in each year. The horizontal axis shows the years and the vertical axis shows the number of items produced.
(ii) Average production
| Year | 1978 | 1979 | 1980 | 1981 | 1982 |
|---|---|---|---|---|---|
| Items produced | 4100 | 2500 | 1500 | 1800 | 9200 |
\[\text{Average production}=\frac{4100+2500+1500+1800+9200}{5}=\frac{19100}{5}=3820\]
Therefore, the average production for the five-year period is 3,820 items.
Bayanin Amsa
(i) Bar chart of company production
The bar chart below represents the number of items produced in each year. The horizontal axis shows the years and the vertical axis shows the number of items produced.
(ii) Average production
| Year | 1978 | 1979 | 1980 | 1981 | 1982 |
|---|---|---|---|---|---|
| Items produced | 4100 | 2500 | 1500 | 1800 | 9200 |
\[\text{Average production}=\frac{4100+2500+1500+1800+9200}{5}=\frac{19100}{5}=3820\]
Therefore, the average production for the five-year period is 3,820 items.
Tambaya 3 Rahoto
The table below shows the frequency distribution of the marks of 800 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 |
| Freq | 10 | 40 | 80 | 140 | 170 | 130 | 100 | 70 | 40 | 20 |
(a) (i) Construct a cumulative frequency table ; (ii) Draw the Ogive ; (iii) Use your ogive to determine the 50th percentile.
(b) The candidates that scored less than 25% are to be withdrawn from the institution, while those that scored than 75% are to be awarded scholarship. Estimate the number of students that will be retained, but will not enjoy the award.
(a)(i) Cumulative frequency table. \(N=800\).
| Marks | Frequency | Upper boundary | Cumulative frequency |
|---|---|---|---|
| 0 - 9 | 10 | 9.5 | 10 |
| 10 - 19 | 40 | 19.5 | 50 |
| 20 - 29 | 80 | 29.5 | 130 |
| 30 - 39 | 140 | 39.5 | 270 |
| 40 - 49 | 170 | 49.5 | 440 |
| 50 - 59 | 130 | 59.5 | 570 |
| 60 - 69 | 100 | 69.5 | 670 |
| 70 - 79 | 70 | 79.5 | 740 |
| 80 - 89 | 40 | 89.5 | 780 |
| 90 - 99 | 20 | 99.5 | 800 |
(a)(ii) Ogive. Plot cumulative frequency against upper class boundaries and join with a smooth curve.
(a)(iii) 50th percentile (median). At \(\frac{N}{2}=400\), interpolating in the 40 - 49 class:
\[P_{50}=39.5+\frac{400-270}{170}\times10 \approx 47.\]
(b) Students retained but not awarded. Withdraw those scoring under 25 marks; award scholarships to those scoring over 75 marks. From the ogive:
\[\text{below }25:\ 50+\frac{25-19.5}{10}\times80 \approx 94,\]
\[\text{below }75:\ 670+\frac{75-69.5}{10}\times70 \approx 708.5.\]
Retained but not awarded \(=708.5-94 \approx 615\) students. So about 615 candidates are kept without a scholarship.
Bayanin Amsa
(a)(i) Cumulative frequency table. \(N=800\).
| Marks | Frequency | Upper boundary | Cumulative frequency |
|---|---|---|---|
| 0 - 9 | 10 | 9.5 | 10 |
| 10 - 19 | 40 | 19.5 | 50 |
| 20 - 29 | 80 | 29.5 | 130 |
| 30 - 39 | 140 | 39.5 | 270 |
| 40 - 49 | 170 | 49.5 | 440 |
| 50 - 59 | 130 | 59.5 | 570 |
| 60 - 69 | 100 | 69.5 | 670 |
| 70 - 79 | 70 | 79.5 | 740 |
| 80 - 89 | 40 | 89.5 | 780 |
| 90 - 99 | 20 | 99.5 | 800 |
(a)(ii) Ogive. Plot cumulative frequency against upper class boundaries and join with a smooth curve.
(a)(iii) 50th percentile (median). At \(\frac{N}{2}=400\), interpolating in the 40 - 49 class:
\[P_{50}=39.5+\frac{400-270}{170}\times10 \approx 47.\]
(b) Students retained but not awarded. Withdraw those scoring under 25 marks; award scholarships to those scoring over 75 marks. From the ogive:
\[\text{below }25:\ 50+\frac{25-19.5}{10}\times80 \approx 94,\]
\[\text{below }75:\ 670+\frac{75-69.5}{10}\times70 \approx 708.5.\]
Retained but not awarded \(=708.5-94 \approx 615\) students. So about 615 candidates are kept without a scholarship.
Tambaya 4 Rahoto
The table below is for the relation \(y = 2 + x - x^{2}\)
| x | -2 | -1.5 | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
| y | -4 | -1.75 | 0 | 1.25 | 2 | 2.25 | 2 | 1.25 | 0 | -1.75 | -4 |
(a) Using a scale of 2cm to 1 unit on each axis, draw the graph of the relation in the interval \(-2 \leq x \leq 3\).
(b) From your graph, find the greatest value of y and the value of x for which this occurs.
(c) Using the same scale and axes, draw the graph of \(y = 1 - x\)
(d) Use your graphs to solve the equation \(1 + 2x - x^{2} = 0\)
(a) Using the scale 2 cm to 1 unit on both axes, plot the given points and draw a smooth curve through them. The completed graph is shown below.
| \(x\) | -2 | -1.5 | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=2+x-x^2\) | -4 | -1.75 | 0 | 1.25 | 2 | 2.25 | 2 | 1.25 | 0 | -1.75 | -4 |
(b) The greatest value of \(y\) is \(2.25\). It occurs when \(x=0.5\).
(c) For \(y=1-x\), use, for example, the points \((-2,3)\), \((0,1)\), \((1,0)\), \((2,-1)\) and \((3,-2)\). Join them with a straight line on the same axes, as shown.
(d) The required equation may be written as
\[1+2x-x^2=0\iff 2+x-x^2=1-x.\]
Hence its solutions are the \(x\)-coordinates of the intersections of \(y=2+x-x^2\) and \(y=1-x\). Reading from the graph gives
\[\boxed{x\approx -0.4\quad\text{or}\quad x\approx 2.4}.\]
(The exact values are \(x=1\pm\sqrt2\).)
Bayanin Amsa
(a) Using the scale 2 cm to 1 unit on both axes, plot the given points and draw a smooth curve through them. The completed graph is shown below.
| \(x\) | -2 | -1.5 | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=2+x-x^2\) | -4 | -1.75 | 0 | 1.25 | 2 | 2.25 | 2 | 1.25 | 0 | -1.75 | -4 |
(b) The greatest value of \(y\) is \(2.25\). It occurs when \(x=0.5\).
(c) For \(y=1-x\), use, for example, the points \((-2,3)\), \((0,1)\), \((1,0)\), \((2,-1)\) and \((3,-2)\). Join them with a straight line on the same axes, as shown.
(d) The required equation may be written as
\[1+2x-x^2=0\iff 2+x-x^2=1-x.\]
Hence its solutions are the \(x\)-coordinates of the intersections of \(y=2+x-x^2\) and \(y=1-x\). Reading from the graph gives
\[\boxed{x\approx -0.4\quad\text{or}\quad x\approx 2.4}.\]
(The exact values are \(x=1\pm\sqrt2\).)
Tambaya 5 Rahoto
Using a ruler and a pair of compasses only,
(a) construct (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120° ; (ii) the locus \(L_{1}\) of points equidistant from A and B ; (iii) the locus \(L_{2}\) of points equidistant from Ab and AC, which passes through the triangle ABC.
(b) Label the point P where \(L_{1}\) and \(L_{2}\) intersect;
(c) Measure |CP|.
Construction
- Draw the line segment \(AB=5\text{ cm}\).
- At \(A\), construct an angle of \(120^\circ\). With centre \(A\) and radius \(7.5\text{ cm}\), cut this ray at \(C\). Join \(BC\).
- With centres \(A\) and \(B\), and with the same compass radius greater than \(2.5\text{ cm}\), draw arcs above and below \(AB\) to intersect. Join the arc-intersection points. This is \(L_1\), the perpendicular bisector of \(AB\).
- Construct the internal angle bisector of \(\angle BAC\). This is \(L_2\), the locus of points equidistant from the lines \(AB\) and \(AC\).
- Label the intersection of \(L_1\) and \(L_2\) as \(P\).
Measurement:
Measuring the constructed line gives
\[|CP|\approx 6.6\text{ cm}.\]Bayanin Amsa
Construction
- Draw the line segment \(AB=5\text{ cm}\).
- At \(A\), construct an angle of \(120^\circ\). With centre \(A\) and radius \(7.5\text{ cm}\), cut this ray at \(C\). Join \(BC\).
- With centres \(A\) and \(B\), and with the same compass radius greater than \(2.5\text{ cm}\), draw arcs above and below \(AB\) to intersect. Join the arc-intersection points. This is \(L_1\), the perpendicular bisector of \(AB\).
- Construct the internal angle bisector of \(\angle BAC\). This is \(L_2\), the locus of points equidistant from the lines \(AB\) and \(AC\).
- Label the intersection of \(L_1\) and \(L_2\) as \(P\).
Measurement:
Measuring the constructed line gives
\[|CP|\approx 6.6\text{ cm}.\]Tambaya 6 Rahoto
(a) Prove that the sum of the angles in a triangle is two right angles.
(b) In a triangle LMN, the side NM is produced to P and the bisector of < LNP meets ML produced at Q. If < LMN = 46°, and < MLN = 80°, calculate < LQN, stating clearly your reasins.
(a) The angle sum of a triangle is two right angles.
Let \(\triangle ABC\) have interior angles \(a,b,c\) at \(A,B,C\). Through \(A\) draw a line \(XY\) parallel to \(BC\).
- \(\angle XAB=\angle ABC=b\) (alternate angles, \(XY\parallel BC\)).
- \(\angle YAC=\angle ACB=c\) (alternate angles, \(XY\parallel BC\)).
Angles on the straight line \(XY\) at \(A\): \(\angle XAB+\angle BAC+\angle YAC=180^{\circ}\), i.e. \(b+a+c=180^{\circ}\). Hence \(a+b+c=180^{\circ}=\) two right angles. \(\blacksquare\)
(b) In \(\triangle LMN\): \(\angle LMN=46^{\circ}\), \(\angle MLN=80^{\circ}\), so \(\angle LNM=180^{\circ}-46^{\circ}-80^{\circ}=54^{\circ}\).
\(NM\) is produced to \(P\), so \(\angle LNP=180^{\circ}-\angle LNM=126^{\circ}\) (angles on a straight line). Its bisector \(NQ\) gives \(\angle LNQ=\tfrac12(126^{\circ})=63^{\circ}\).
Since \(M,L,Q\) are collinear (\(Q\) on \(ML\) produced), \(\angle NLQ=180^{\circ}-\angle NLM=180^{\circ}-80^{\circ}=100^{\circ}\).
In \(\triangle LNQ\): \(\angle LQN=180^{\circ}-\angle LNQ-\angle NLQ=180^{\circ}-63^{\circ}-100^{\circ}=\mathbf{17^{\circ}}.\)
Bayanin Amsa
(a) The angle sum of a triangle is two right angles.
Let \(\triangle ABC\) have interior angles \(a,b,c\) at \(A,B,C\). Through \(A\) draw a line \(XY\) parallel to \(BC\).
- \(\angle XAB=\angle ABC=b\) (alternate angles, \(XY\parallel BC\)).
- \(\angle YAC=\angle ACB=c\) (alternate angles, \(XY\parallel BC\)).
Angles on the straight line \(XY\) at \(A\): \(\angle XAB+\angle BAC+\angle YAC=180^{\circ}\), i.e. \(b+a+c=180^{\circ}\). Hence \(a+b+c=180^{\circ}=\) two right angles. \(\blacksquare\)
(b) In \(\triangle LMN\): \(\angle LMN=46^{\circ}\), \(\angle MLN=80^{\circ}\), so \(\angle LNM=180^{\circ}-46^{\circ}-80^{\circ}=54^{\circ}\).
\(NM\) is produced to \(P\), so \(\angle LNP=180^{\circ}-\angle LNM=126^{\circ}\) (angles on a straight line). Its bisector \(NQ\) gives \(\angle LNQ=\tfrac12(126^{\circ})=63^{\circ}\).
Since \(M,L,Q\) are collinear (\(Q\) on \(ML\) produced), \(\angle NLQ=180^{\circ}-\angle NLM=180^{\circ}-80^{\circ}=100^{\circ}\).
In \(\triangle LNQ\): \(\angle LQN=180^{\circ}-\angle LNQ-\angle NLQ=180^{\circ}-63^{\circ}-100^{\circ}=\mathbf{17^{\circ}}.\)
Tambaya 7 Rahoto
In the diagram, < PQR = < PSQ = 90°, |PS| = 9 cm, |SR| = 16 cm and |SQ| = x cm.
(a) Find the value of x using a trigonometric ratio.
(b) Calculate : (i) the size of < QRS to the nearest degree; (ii) |PQ|.
From the diagram, S lies on \(PR\) with \(\angle PQR = 90^\circ\) (right angle at Q) and \(\angle PSQ = 90^\circ\), so \(QS\) is the altitude from Q onto the hypotenuse \(PR\). The given lengths are \(|PS| = 9\text{ cm}\), \(|SR| = 16\text{ cm}\) and \(|SQ| = x\).
(a) Value of x by a trigonometric ratio
Let \(\angle QPS = \alpha\). In right triangle \(PSQ\):
\[\tan\alpha = \frac{QS}{PS} = \frac{x}{9}\]In right triangle \(QSR\), the angle \(\angle SQR\) equals \(\alpha\) (both equal \(90^\circ - \angle R\)), so
\[\tan\alpha = \frac{SR}{QS} = \frac{16}{x}\]Equating the two expressions:
\[\frac{x}{9} = \frac{16}{x} \;\Rightarrow\; x^2 = 9\times 16 = 144\]\[x = 12\text{ cm}\](b)(i) Size of \(\angle QRS\) to the nearest degree
In right triangle \(QSR\):
\[\tan(\angle QRS) = \frac{QS}{SR} = \frac{12}{16} = 0.75\]\[\angle QRS = \tan^{-1}(0.75) = 36.87^\circ \approx 37^\circ\](b)(ii) \(|PQ|\)
In right triangle \(PSQ\), by Pythagoras:
\[|PQ|^2 = |PS|^2 + |QS|^2 = 9^2 + 12^2 = 81 + 144 = 225\]\[|PQ| = \sqrt{225} = 15\text{ cm}\]Therefore \(x = 12\text{ cm}\), \(\angle QRS \approx 37^\circ\) and \(|PQ| = 15\text{ cm}\).
Bayanin Amsa
From the diagram, S lies on \(PR\) with \(\angle PQR = 90^\circ\) (right angle at Q) and \(\angle PSQ = 90^\circ\), so \(QS\) is the altitude from Q onto the hypotenuse \(PR\). The given lengths are \(|PS| = 9\text{ cm}\), \(|SR| = 16\text{ cm}\) and \(|SQ| = x\).
(a) Value of x by a trigonometric ratio
Let \(\angle QPS = \alpha\). In right triangle \(PSQ\):
\[\tan\alpha = \frac{QS}{PS} = \frac{x}{9}\]In right triangle \(QSR\), the angle \(\angle SQR\) equals \(\alpha\) (both equal \(90^\circ - \angle R\)), so
\[\tan\alpha = \frac{SR}{QS} = \frac{16}{x}\]Equating the two expressions:
\[\frac{x}{9} = \frac{16}{x} \;\Rightarrow\; x^2 = 9\times 16 = 144\]\[x = 12\text{ cm}\](b)(i) Size of \(\angle QRS\) to the nearest degree
In right triangle \(QSR\):
\[\tan(\angle QRS) = \frac{QS}{SR} = \frac{12}{16} = 0.75\]\[\angle QRS = \tan^{-1}(0.75) = 36.87^\circ \approx 37^\circ\](b)(ii) \(|PQ|\)
In right triangle \(PSQ\), by Pythagoras:
\[|PQ|^2 = |PS|^2 + |QS|^2 = 9^2 + 12^2 = 81 + 144 = 225\]\[|PQ| = \sqrt{225} = 15\text{ cm}\]Therefore \(x = 12\text{ cm}\), \(\angle QRS \approx 37^\circ\) and \(|PQ| = 15\text{ cm}\).
Tambaya 8 Rahoto
Two towns K and Q are on the parallel of latitude 46°N. The longitude of town K is 130°W and that of town Q is 103°W. A third town P also on latitude 46°N is on longitude 23°E, Calculate:
(i) the length of the parallel of latitude 46°N, to the nearest 100km;
(ii) the distance between K and Q, correct to the nearest 100km;
(iii) the distance between Q and P measured along the parallel of latitude, to the nearest 10km.
[Take \(\pi = 3.142\); Radius of the earth = 6400km]
Radius \(R=6400\) km, \(\pi=3.142\), latitude \(46^{\circ}\N\), \(\cos46^{\circ}=0.6947\).
(i) Length of the parallel of latitude \(46^{\circ}\)N
\[L=2\pi R\cos46^{\circ}=2(3.142)(6400)(0.6947)=27{,}940\text{ km}\approx\mathbf{27{,}900\text{ km}}\ (\text{nearest }100).\]
(ii) Distance K to Q (same latitude). \(K\) at \(130^{\circ}\W\), \(Q\) at \(103^{\circ}\W\): difference \(=27^{\circ}\).
\[d_{KQ}=\frac{27}{360}\times27{,}940=2095\text{ km}\approx\mathbf{2100\text{ km}}\ (\text{nearest }100).\]
(iii) Distance Q to P along the parallel. \(Q\) at \(103^{\circ}\W\), \(P\) at \(23^{\circ}\E\): difference \(=103^{\circ}+23^{\circ}=126^{\circ}\).
\[d_{QP}=\frac{126}{360}\times27{,}940=9779\text{ km}\approx\mathbf{9780\text{ km}}\ (\text{nearest }10).\]
Bayanin Amsa
Radius \(R=6400\) km, \(\pi=3.142\), latitude \(46^{\circ}\N\), \(\cos46^{\circ}=0.6947\).
(i) Length of the parallel of latitude \(46^{\circ}\)N
\[L=2\pi R\cos46^{\circ}=2(3.142)(6400)(0.6947)=27{,}940\text{ km}\approx\mathbf{27{,}900\text{ km}}\ (\text{nearest }100).\]
(ii) Distance K to Q (same latitude). \(K\) at \(130^{\circ}\W\), \(Q\) at \(103^{\circ}\W\): difference \(=27^{\circ}\).
\[d_{KQ}=\frac{27}{360}\times27{,}940=2095\text{ km}\approx\mathbf{2100\text{ km}}\ (\text{nearest }100).\]
(iii) Distance Q to P along the parallel. \(Q\) at \(103^{\circ}\W\), \(P\) at \(23^{\circ}\E\): difference \(=103^{\circ}+23^{\circ}=126^{\circ}\).
\[d_{QP}=\frac{126}{360}\times27{,}940=9779\text{ km}\approx\mathbf{9780\text{ km}}\ (\text{nearest }10).\]
Tambaya 9 Rahoto
(a) Given that \(\sin \alpha = 0.3907\), use tables to find the value of : (i) \(\tan \alpha\) ; (ii) \(\cos \alpha\).
(b) A ladder of length 4.5m leans against a vertical wall making an angle of 50° with the horizontal ground. If the bottom of a window is 4m above the ground, what is the distance between the top of the ladder and the bottom of the window? (Answer correct to the nearest cm)
(a) \(\sin\alpha=0.3907\Rightarrow\alpha=23^{\circ}\).
- (i) \(\tan\alpha=\tan23^{\circ}\approx0.4245.\)
- (ii) \(\cos\alpha=\cos23^{\circ}\approx0.9205\) (also \(\cos\alpha=\sqrt{1-0.3907^{2}}=\sqrt{0.8474}=0.9205\)).
(b) The ladder reaches a height up the wall of \[h=4.5\sin50^{\circ}=4.5\times0.7660=3.447\text{ m}.\] The bottom of the window is \(4\text{ m}\) above the ground, so the gap between the top of the ladder and the window bottom is \[4-3.447=0.553\text{ m}\approx55\text{ cm}.\]
Bayanin Amsa
(a) \(\sin\alpha=0.3907\Rightarrow\alpha=23^{\circ}\).
- (i) \(\tan\alpha=\tan23^{\circ}\approx0.4245.\)
- (ii) \(\cos\alpha=\cos23^{\circ}\approx0.9205\) (also \(\cos\alpha=\sqrt{1-0.3907^{2}}=\sqrt{0.8474}=0.9205\)).
(b) The ladder reaches a height up the wall of \[h=4.5\sin50^{\circ}=4.5\times0.7660=3.447\text{ m}.\] The bottom of the window is \(4\text{ m}\) above the ground, so the gap between the top of the ladder and the window bottom is \[4-3.447=0.553\text{ m}\approx55\text{ cm}.\]
Tambaya 10 Rahoto
Illustrate the following on graph paper and shade the region which satisfies all the three inequalities at the same time :
\(- x + 5y \leq 10 ; 3x - 4y \leq 8\) and \(x > -1\).
Rewrite the boundary equations as
\[ -x+5y=10\quad\Rightarrow\quad y=2+\frac{x}{5}, \]
\[ 3x-4y=8\quad\Rightarrow\quad y=-2+\frac{3x}{4}, \]
together with the vertical boundary \(x=-1\).
Use the following values to plot the two straight lines.
| For \(-x+5y=10\) | \(x=-4\) | \(x=0\) | \(x=4\) |
|---|---|---|---|
| \(y=2+\frac{x}{5}\) | \(1.2\) | \(2\) | \(2.8\) |
| For \(3x-4y=8\) | \(x=-4\) | \(x=0\) | \(x=4\) |
|---|---|---|---|
| \(y=-2+\frac{3x}{4}\) | \(-5\) | \(-2\) | \(1\) |
Plot the graph below. Draw the first two boundary lines solid, since their inequalities include equality. Draw \(x=-1\) as a broken line, since \(x>-1\) does not include the boundary.
Using \((0,0)\) as a test point:
\[0\leq10,\qquad 0\leq8,\qquad 0>-1.\]
Hence select the side containing \((0,0)\) for each inequality. Therefore the required region is
\[\boxed{\;x>-1,\qquad -2+\frac{3x}{4}\leq y\leq2+\frac{x}{5}\;}\]
The two sloping lines meet at
\[2+\frac{x}{5}=-2+\frac{3x}{4}\Rightarrow x=\frac{80}{11},\quad y=\frac{38}{11}.\]
Thus shade the triangular region between the two solid lines, to the right of the broken line \(x=-1\). The broken left-hand edge is not part of the region.
Bayanin Amsa
Rewrite the boundary equations as
\[ -x+5y=10\quad\Rightarrow\quad y=2+\frac{x}{5}, \]
\[ 3x-4y=8\quad\Rightarrow\quad y=-2+\frac{3x}{4}, \]
together with the vertical boundary \(x=-1\).
Use the following values to plot the two straight lines.
| For \(-x+5y=10\) | \(x=-4\) | \(x=0\) | \(x=4\) |
|---|---|---|---|
| \(y=2+\frac{x}{5}\) | \(1.2\) | \(2\) | \(2.8\) |
| For \(3x-4y=8\) | \(x=-4\) | \(x=0\) | \(x=4\) |
|---|---|---|---|
| \(y=-2+\frac{3x}{4}\) | \(-5\) | \(-2\) | \(1\) |
Plot the graph below. Draw the first two boundary lines solid, since their inequalities include equality. Draw \(x=-1\) as a broken line, since \(x>-1\) does not include the boundary.
Using \((0,0)\) as a test point:
\[0\leq10,\qquad 0\leq8,\qquad 0>-1.\]
Hence select the side containing \((0,0)\) for each inequality. Therefore the required region is
\[\boxed{\;x>-1,\qquad -2+\frac{3x}{4}\leq y\leq2+\frac{x}{5}\;}\]
The two sloping lines meet at
\[2+\frac{x}{5}=-2+\frac{3x}{4}\Rightarrow x=\frac{80}{11},\quad y=\frac{38}{11}.\]
Thus shade the triangular region between the two solid lines, to the right of the broken line \(x=-1\). The broken left-hand edge is not part of the region.
Tambaya 11 Rahoto
In a certain school, the lesson periods for each week are as itemised below: English 10, Mathematics 7, Biology 3, Statistics 4, Ibo 3, others 9. Draw a pie chart to illustrate this information.
Total number of lesson periods:
\[10+7+3+4+3+9=36\]
Hence, the angle for each lesson period is:
\[\frac{360^\circ}{36}=10^\circ\]
| Subject | Number of periods | Angle of sector |
|---|---|---|
| English | 10 | \(10\times10^\circ=100^\circ\) |
| Mathematics | 7 | \(7\times10^\circ=70^\circ\) |
| Biology | 3 | \(3\times10^\circ=30^\circ\) |
| Statistics | 4 | \(4\times10^\circ=40^\circ\) |
| Ibo | 3 | \(3\times10^\circ=30^\circ\) |
| Others | 9 | \(9\times10^\circ=90^\circ\) |
| Total | 36 | \(360^\circ\) |
The required pie chart is:
Bayanin Amsa
Total number of lesson periods:
\[10+7+3+4+3+9=36\]
Hence, the angle for each lesson period is:
\[\frac{360^\circ}{36}=10^\circ\]
| Subject | Number of periods | Angle of sector |
|---|---|---|
| English | 10 | \(10\times10^\circ=100^\circ\) |
| Mathematics | 7 | \(7\times10^\circ=70^\circ\) |
| Biology | 3 | \(3\times10^\circ=30^\circ\) |
| Statistics | 4 | \(4\times10^\circ=40^\circ\) |
| Ibo | 3 | \(3\times10^\circ=30^\circ\) |
| Others | 9 | \(9\times10^\circ=90^\circ\) |
| Total | 36 | \(360^\circ\) |
The required pie chart is:
Tambaya 12 Rahoto
(a) The subsets A, B and C of a universal set are defined as follows :
A = {m, a, p, e} ; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u}. List the elements of the following sets.
(i) \(A \cup B\) ; (ii) \(A \cup C\) ; (iii) \(A \cup (B \cap C)\).
(b) Out of the 400 students in the final year in a Senior Secondary School, 300 are offering Biology and 190 are offering Chemistry.
(i) How many students are offering both Biology and Chemistry, if only 70 students are offering neither Biology nor Chemistry? (ii) How many students are offering at least one of Biology or Chemistry?
(a) \(A=\{m,a,p,e\}\), \(B=\{a,e,i,o,u\}\), \(C=\{l,m,n,o,p,q,r,s,t,u\}\).
- (i) \(A\cup B=\{a,e,i,m,o,p,u\}\).
- (ii) \(A\cup C=\{a,e,l,m,n,o,p,q,r,s,t,u\}\).
- (iii) \(B\cap C=\{o,u\}\), so \(A\cup(B\cap C)=\{a,e,m,o,p,u\}\).
(b) Total \(=400\); Biology \(n(B)=300\); Chemistry \(n(C)=190\); neither \(=70\).
(ii) At least one \(=400-70=\mathbf{330}\) students.
(i) By inclusion-exclusion, both \(=n(B)+n(C)-n(B\cup C)=300+190-330=\mathbf{160}\) students.
Bayanin Amsa
(a) \(A=\{m,a,p,e\}\), \(B=\{a,e,i,o,u\}\), \(C=\{l,m,n,o,p,q,r,s,t,u\}\).
- (i) \(A\cup B=\{a,e,i,m,o,p,u\}\).
- (ii) \(A\cup C=\{a,e,l,m,n,o,p,q,r,s,t,u\}\).
- (iii) \(B\cap C=\{o,u\}\), so \(A\cup(B\cap C)=\{a,e,m,o,p,u\}\).
(b) Total \(=400\); Biology \(n(B)=300\); Chemistry \(n(C)=190\); neither \(=70\).
(ii) At least one \(=400-70=\mathbf{330}\) students.
(i) By inclusion-exclusion, both \(=n(B)+n(C)-n(B\cup C)=300+190-330=\mathbf{160}\) students.
