WAEC SSCE - General Mathematics - 2009 (Essay)
Tambaya 1 Rahoto
(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44°. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman ; (ii) height of the flag pole.
(b)
In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.
(a) The woman is at a window \(30\text{ m}\) above the ground. The foot of the flag pole is \(25\text{ m}\) horizontally from the foot of the building, on the same level ground.
(i) Angle of depression of the foot of the pole. From the window, the horizontal distance to the pole is \(25\text{ m}\) and the vertical drop to the foot of the pole is the full window height \(30\text{ m}\):
\[\tan\theta=\frac{30}{25}=1.2\]\[\theta=\tan^{-1}(1.2)\approx 50.19^\circ\approx\mathbf{50^\circ}.\](ii) Height of the flag pole. The angle of depression of the top of the pole is \(44^\circ\). Over the horizontal distance \(25\text{ m}\), the vertical drop from window level down to the top of the pole is
\[h_1=25\tan 44^\circ=25\times0.9657=24.14\text{ m}.\]So the top of the pole is \(30-24.14=5.86\text{ m}\) above the ground. Since the pole stands on the same ground, its height is
\[|\text{pole}|=30-24.14\approx\mathbf{6\text{ m}}.\](b) \(O\) is the centre of the circle, \(\angle OQR=32^\circ\) and \(\angle TPQ=15^\circ\).
(i) \(\angle QPR\): \(OQ\) and \(OR\) are radii, so triangle \(OQR\) is isosceles and
\[\angle ORQ=\angle OQR=32^\circ.\]\[\angle QOR=180^\circ-32^\circ-32^\circ=116^\circ.\]\(\angle QOR\) is the angle subtended by chord \(QR\) at the centre, and \(\angle QPR\) is subtended by the same chord at the circumference, so it is half:
\[\angle QPR=\tfrac{1}{2}\times116^\circ=\mathbf{58^\circ}.\](ii) \(\angle TQO\): \(\angle TPQ=15^\circ\) stands on chord \(TQ\), so the central angle on the same chord is
\[\angle TOQ=2\times15^\circ=30^\circ.\]\(OT\) and \(OQ\) are radii, so triangle \(TOQ\) is isosceles:
\[\angle TQO=\angle OTQ=\frac{180^\circ-30^\circ}{2}=\mathbf{75^\circ}.\]Bayanin Amsa
(a) The woman is at a window \(30\text{ m}\) above the ground. The foot of the flag pole is \(25\text{ m}\) horizontally from the foot of the building, on the same level ground.
(i) Angle of depression of the foot of the pole. From the window, the horizontal distance to the pole is \(25\text{ m}\) and the vertical drop to the foot of the pole is the full window height \(30\text{ m}\):
\[\tan\theta=\frac{30}{25}=1.2\]\[\theta=\tan^{-1}(1.2)\approx 50.19^\circ\approx\mathbf{50^\circ}.\](ii) Height of the flag pole. The angle of depression of the top of the pole is \(44^\circ\). Over the horizontal distance \(25\text{ m}\), the vertical drop from window level down to the top of the pole is
\[h_1=25\tan 44^\circ=25\times0.9657=24.14\text{ m}.\]So the top of the pole is \(30-24.14=5.86\text{ m}\) above the ground. Since the pole stands on the same ground, its height is
\[|\text{pole}|=30-24.14\approx\mathbf{6\text{ m}}.\](b) \(O\) is the centre of the circle, \(\angle OQR=32^\circ\) and \(\angle TPQ=15^\circ\).
(i) \(\angle QPR\): \(OQ\) and \(OR\) are radii, so triangle \(OQR\) is isosceles and
\[\angle ORQ=\angle OQR=32^\circ.\]\[\angle QOR=180^\circ-32^\circ-32^\circ=116^\circ.\]\(\angle QOR\) is the angle subtended by chord \(QR\) at the centre, and \(\angle QPR\) is subtended by the same chord at the circumference, so it is half:
\[\angle QPR=\tfrac{1}{2}\times116^\circ=\mathbf{58^\circ}.\](ii) \(\angle TQO\): \(\angle TPQ=15^\circ\) stands on chord \(TQ\), so the central angle on the same chord is
\[\angle TOQ=2\times15^\circ=30^\circ.\]\(OT\) and \(OQ\) are radii, so triangle \(TOQ\) is isosceles:
\[\angle TQO=\angle OTQ=\frac{180^\circ-30^\circ}{2}=\mathbf{75^\circ}.\]Tambaya 2 Rahoto
(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both?
(b) In triangle PQR, M and N are points on the side PQ and PR respectively such that MN is parallel to QR. If < PRQ = 75°, PN = QN and < PNQ = 125°, determine :
(i) < NQR ; (ii) < NPM.
(a) \(n=30\); degrees \(=17\), diplomas \(=15\), neither \(=4\), so \(30-4=26\) have at least one.
\(|D\cup P|=|D|+|P|-|D\cap P| \Rightarrow 26=17+15-|D\cap P| \Rightarrow |D\cap P|=32-26=\mathbf{6}\).
Six candidates have both a degree and a diploma.
(b) In \(\triangle PQR\), \(M\) on \(PQ\), \(N\) on \(PR\), \(MN\parallel QR\); \(\angle PRQ=75^{\circ}\), \(PN=QN\), \(\angle PNQ=125^{\circ}\).
Triangle \(PNQ\) is isosceles (\(PN=QN\)), so \(\angle NPQ=\angle NQP\).
\(\angle NPQ+\angle NQP+\angle PNQ=180^{\circ}\Rightarrow 2\angle NPQ=180^{\circ}-125^{\circ}=55^{\circ}\Rightarrow \angle NPQ=\angle NQP=27.5^{\circ}\).
In \(\triangle PQR\): \(\angle P=27.5^{\circ}\), \(\angle R=75^{\circ}\), so \(\angle PQR=180^{\circ}-27.5^{\circ}-75^{\circ}=77.5^{\circ}\).
(i) \(\angle NQR=\angle PQR-\angle NQP=77.5^{\circ}-27.5^{\circ}=\mathbf{50^{\circ}}\).
(ii) \(\angle NPM\) is the angle at \(P\) between \(PM\) (on \(PQ\)) and \(PN\) (on \(PR\)), i.e. \(\angle QPR=\mathbf{27.5^{\circ}}\).
Bayanin Amsa
(a) \(n=30\); degrees \(=17\), diplomas \(=15\), neither \(=4\), so \(30-4=26\) have at least one.
\(|D\cup P|=|D|+|P|-|D\cap P| \Rightarrow 26=17+15-|D\cap P| \Rightarrow |D\cap P|=32-26=\mathbf{6}\).
Six candidates have both a degree and a diploma.
(b) In \(\triangle PQR\), \(M\) on \(PQ\), \(N\) on \(PR\), \(MN\parallel QR\); \(\angle PRQ=75^{\circ}\), \(PN=QN\), \(\angle PNQ=125^{\circ}\).
Triangle \(PNQ\) is isosceles (\(PN=QN\)), so \(\angle NPQ=\angle NQP\).
\(\angle NPQ+\angle NQP+\angle PNQ=180^{\circ}\Rightarrow 2\angle NPQ=180^{\circ}-125^{\circ}=55^{\circ}\Rightarrow \angle NPQ=\angle NQP=27.5^{\circ}\).
In \(\triangle PQR\): \(\angle P=27.5^{\circ}\), \(\angle R=75^{\circ}\), so \(\angle PQR=180^{\circ}-27.5^{\circ}-75^{\circ}=77.5^{\circ}\).
(i) \(\angle NQR=\angle PQR-\angle NQP=77.5^{\circ}-27.5^{\circ}=\mathbf{50^{\circ}}\).
(ii) \(\angle NPM\) is the angle at \(P\) between \(PM\) (on \(PQ\)) and \(PN\) (on \(PR\)), i.e. \(\angle QPR=\mathbf{27.5^{\circ}}\).
Tambaya 3 Rahoto
In the diagram, ABCDEF is a triangular prism. < ABC = < DEF = 90°, /AB/ = 24 cm, /BC/ = 7 cm and /CD/ = 40 cm. Calculate :
(a) /AC/ ;
(b) the total surface area of the prism.
The solid is a triangular prism with congruent right-angled triangular ends \(ABC\) and \(DEF\) (\(\angle ABC = \angle DEF = 90^\circ\)). The given lengths are \(|AB| = 24\) cm, \(|BC| = 7\) cm, and the prism length \(|CD| = 40\) cm.
(a) Length AC. Triangle \(ABC\) is right-angled at \(B\), so by Pythagoras:
\[ |AC| = \sqrt{|AB|^2 + |BC|^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\ \text{cm} \]
(b) Total surface area. The prism has two triangular ends and three rectangular faces. The prism length is \(40\) cm.
The two triangular ends:
\[ 2 \times \left(\tfrac{1}{2}\times 24 \times 7\right) = 2 \times 84 = 168\ \text{cm}^2 \]
The three rectangles have widths equal to the sides of the triangle \((24, 7, 25)\) and common length \(40\):
\[ (24 + 7 + 25)\times 40 = 56 \times 40 = 2240\ \text{cm}^2 \]
Total surface area:
\[ 168 + 2240 = 2408\ \text{cm}^2 \]
Answers: (a) \(|AC| = 25\) cm; (b) total surface area \(= 2408\ \text{cm}^2\).
Bayanin Amsa
The solid is a triangular prism with congruent right-angled triangular ends \(ABC\) and \(DEF\) (\(\angle ABC = \angle DEF = 90^\circ\)). The given lengths are \(|AB| = 24\) cm, \(|BC| = 7\) cm, and the prism length \(|CD| = 40\) cm.
(a) Length AC. Triangle \(ABC\) is right-angled at \(B\), so by Pythagoras:
\[ |AC| = \sqrt{|AB|^2 + |BC|^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\ \text{cm} \]
(b) Total surface area. The prism has two triangular ends and three rectangular faces. The prism length is \(40\) cm.
The two triangular ends:
\[ 2 \times \left(\tfrac{1}{2}\times 24 \times 7\right) = 2 \times 84 = 168\ \text{cm}^2 \]
The three rectangles have widths equal to the sides of the triangle \((24, 7, 25)\) and common length \(40\):
\[ (24 + 7 + 25)\times 40 = 56 \times 40 = 2240\ \text{cm}^2 \]
Total surface area:
\[ 168 + 2240 = 2408\ \text{cm}^2 \]
Answers: (a) \(|AC| = 25\) cm; (b) total surface area \(= 2408\ \text{cm}^2\).
Tambaya 4 Rahoto
(a) Copy and complete the table of values for \(y = \sin x + 2 \cos x\), correct to one decimal place.
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° |
| y | 2.2 | -1.2 | -2.0 | -1.9 |
(b) Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 units on the y- axis, draw the graph of \(y = \sin x + 2\cos x\) for \(0° \leq x \leq 240°\).
(c) Use your graph to solve the equation : (i) \(\sin x + 2 \cos x = 0\) ; (ii) \(\sin x = 2.1 - 2\cos x\).
(d) From the graph, find y when x = 171°.
(a) For each value of x, evaluate \(y=\sin x+2\cos x\), correct to one decimal place.
| \(x\) | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° |
|---|---|---|---|---|---|---|---|---|---|
| \(y\) | 2.0 | 2.2 | 1.9 | 1.0 | −0.1 | −1.2 | −2.0 | −2.2 | −1.9 |
(b) The plotted graph is shown below. The points are joined with a smooth curve.
(c)
(i) The x-intercept of the curve gives
\[x\approx117^\circ.\]
(ii) Draw the horizontal line \(y=2.1\). Its intersections with the curve give
\[x\approx9^\circ\quad\text{or}\quad x\approx45^\circ.\]
(d) Reading the ordinate at \(x=171^\circ\) from the graph gives
\[y\approx-1.75.\]
Bayanin Amsa
(a) For each value of x, evaluate \(y=\sin x+2\cos x\), correct to one decimal place.
| \(x\) | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° |
|---|---|---|---|---|---|---|---|---|---|
| \(y\) | 2.0 | 2.2 | 1.9 | 1.0 | −0.1 | −1.2 | −2.0 | −2.2 | −1.9 |
(b) The plotted graph is shown below. The points are joined with a smooth curve.
(c)
(i) The x-intercept of the curve gives
\[x\approx117^\circ.\]
(ii) Draw the horizontal line \(y=2.1\). Its intersections with the curve give
\[x\approx9^\circ\quad\text{or}\quad x\approx45^\circ.\]
(d) Reading the ordinate at \(x=171^\circ\) from the graph gives
\[y\approx-1.75.\]
Tambaya 5 Rahoto
(a) Given that \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = a + b\sqrt{6}\), find a and b.
(b) If \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\), find y.
(a) Expand \((\sqrt{3}-5\sqrt{2})(\sqrt{3}+\sqrt{2})\):
\(=\sqrt{3}\cdot\sqrt{3}+\sqrt{3}\cdot\sqrt{2}-5\sqrt{2}\cdot\sqrt{3}-5\sqrt{2}\cdot\sqrt{2}\)
\(=3+\sqrt{6}-5\sqrt{6}-10=-7-4\sqrt{6}\).
Comparing with \(a+b\sqrt{6}\): \(\mathbf{a=-7,\ b=-4}\).
(b) \(\dfrac{2^{1-y}\times 2^{y-1}}{2^{y+2}}=8^{2-3y}\).
Numerator: \(2^{(1-y)+(y-1)}=2^{0}=1\). Left side \(=2^{-(y+2)}\).
Right side: \(8^{2-3y}=2^{3(2-3y)}=2^{6-9y}\).
\(-(y+2)=6-9y \Rightarrow -y-2=6-9y \Rightarrow 8y=8 \Rightarrow y=\mathbf{1}\).
Bayanin Amsa
(a) Expand \((\sqrt{3}-5\sqrt{2})(\sqrt{3}+\sqrt{2})\):
\(=\sqrt{3}\cdot\sqrt{3}+\sqrt{3}\cdot\sqrt{2}-5\sqrt{2}\cdot\sqrt{3}-5\sqrt{2}\cdot\sqrt{2}\)
\(=3+\sqrt{6}-5\sqrt{6}-10=-7-4\sqrt{6}\).
Comparing with \(a+b\sqrt{6}\): \(\mathbf{a=-7,\ b=-4}\).
(b) \(\dfrac{2^{1-y}\times 2^{y-1}}{2^{y+2}}=8^{2-3y}\).
Numerator: \(2^{(1-y)+(y-1)}=2^{0}=1\). Left side \(=2^{-(y+2)}\).
Right side: \(8^{2-3y}=2^{3(2-3y)}=2^{6-9y}\).
\(-(y+2)=6-9y \Rightarrow -y-2=6-9y \Rightarrow 8y=8 \Rightarrow y=\mathbf{1}\).
Tambaya 6 Rahoto
The table shows the number of children per family in a community.
| No of children | 0 | 1 | 2 | 3 | 4 | 5 |
| No of families | 3 | 5 | 7 | 4 | 3 | 2 |
(a) Find the : (i) mode ; (ii) third quartile ; (iii) probability that a family has at least 2 children.
(b) If a pie chart were to be drawn for the data, what would be the sectorial angle representing families with one child?
Total number of families:
\[N=3+5+7+4+3+2=24.\]
| Number of children, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Number of families, \(f\) | 3 | 5 | 7 | 4 | 3 | 2 |
| Cumulative frequency | 3 | 8 | 15 | 19 | 22 | 24 |
(a)(i) Mode
The greatest frequency is \(7\), which corresponds to 2 children.
\[\boxed{\text{Mode}=2}\]
(a)(ii) Third quartile
\[Q_3=\frac{3}{4}\times24=18.\]
The 18th observation lies between cumulative frequencies 15 and 19, corresponding to 3 children.
\[\boxed{Q_3=3}\]
(a)(iii) Probability of at least 2 children
\[P(X\geq2)=\frac{7+4+3+2}{24}=\frac{16}{24}=\boxed{\frac{2}{3}}.\]
(b) Pie chart
The sector angle for each category is \(\dfrac{f}{24}\times360^\circ\). The completed pie chart is shown below.
For families with one child:
\[\frac{5}{24}\times360^\circ=\boxed{75^\circ}.\]
Bayanin Amsa
Total number of families:
\[N=3+5+7+4+3+2=24.\]
| Number of children, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Number of families, \(f\) | 3 | 5 | 7 | 4 | 3 | 2 |
| Cumulative frequency | 3 | 8 | 15 | 19 | 22 | 24 |
(a)(i) Mode
The greatest frequency is \(7\), which corresponds to 2 children.
\[\boxed{\text{Mode}=2}\]
(a)(ii) Third quartile
\[Q_3=\frac{3}{4}\times24=18.\]
The 18th observation lies between cumulative frequencies 15 and 19, corresponding to 3 children.
\[\boxed{Q_3=3}\]
(a)(iii) Probability of at least 2 children
\[P(X\geq2)=\frac{7+4+3+2}{24}=\frac{16}{24}=\boxed{\frac{2}{3}}.\]
(b) Pie chart
The sector angle for each category is \(\dfrac{f}{24}\times360^\circ\). The completed pie chart is shown below.
For families with one child:
\[\frac{5}{24}\times360^\circ=\boxed{75^\circ}.\]
Tambaya 7 Rahoto
(a) Using ruler and a pair of compasses only, construct : (i) quadrilateral PQRS such that /PQ/ = 10 cm, /QR/ = 8 cm, /PS/ = 6 cm, < PQR = 60° and < QPS = 75° ;
(ii) the locus \(l_{1}\) of points equidistant from QR and RS ; (iii) locus \(l_{2}\) of points equidistant from R and S ;
(b) Measure /RS/.
Construction and loci
- Draw \(PQ=10\text{ cm}\).
- At \(Q\), construct an angle of \(60^\circ\) with \(QP\). On this ray, mark \(R\) so that \(QR=8\text{ cm}\).
- At \(P\), construct an angle of \(75^\circ\) with \(PQ\). This can be made by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On this ray, mark \(S\) so that \(PS=6\text{ cm}\).
- Join \(R\) to \(S\) to complete quadrilateral \(PQRS\).
- The locus \(l_1\), equidistant from \(QR\) and \(RS\), is the internal angle bisector of \(\angle QRS\).
- The locus \(l_2\), equidistant from points \(R\) and \(S\), is the perpendicular bisector of \(RS\).
Measurement of \(RS\): \(RS\approx 4.6\text{ cm}\), measured to the nearest \(0.1\text{ cm}\).
A check from the given lengths and angles gives \[ RS=\sqrt{(10-8\cos60^\circ-6\cos75^\circ)^2+(8\sin60^\circ-6\sin75^\circ)^2} \approx 4.59\text{ cm}. \] Therefore, \(4.6\text{ cm}\) is the appropriate measured value. The stated value \(4.4\text{ cm}\) is not consistent with an accurate construction using the dimensions given.
Examination reminder: “Equidistant from two intersecting sides” indicates an angle bisector; “equidistant from two points” indicates a perpendicular bisector.
Bayanin Amsa
Construction and loci
- Draw \(PQ=10\text{ cm}\).
- At \(Q\), construct an angle of \(60^\circ\) with \(QP\). On this ray, mark \(R\) so that \(QR=8\text{ cm}\).
- At \(P\), construct an angle of \(75^\circ\) with \(PQ\). This can be made by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On this ray, mark \(S\) so that \(PS=6\text{ cm}\).
- Join \(R\) to \(S\) to complete quadrilateral \(PQRS\).
- The locus \(l_1\), equidistant from \(QR\) and \(RS\), is the internal angle bisector of \(\angle QRS\).
- The locus \(l_2\), equidistant from points \(R\) and \(S\), is the perpendicular bisector of \(RS\).
Measurement of \(RS\): \(RS\approx 4.6\text{ cm}\), measured to the nearest \(0.1\text{ cm}\).
A check from the given lengths and angles gives \[ RS=\sqrt{(10-8\cos60^\circ-6\cos75^\circ)^2+(8\sin60^\circ-6\sin75^\circ)^2} \approx 4.59\text{ cm}. \] Therefore, \(4.6\text{ cm}\) is the appropriate measured value. The stated value \(4.4\text{ cm}\) is not consistent with an accurate construction using the dimensions given.
Examination reminder: “Equidistant from two intersecting sides” indicates an angle bisector; “equidistant from two points” indicates a perpendicular bisector.
Tambaya 8 Rahoto
The marks scored by 50 students in a Geography examination are as follows :
60 54 40 67 53 73 37 55 62 43 44 69 39 32 45 58 48 67 39 51 46 59 40 52 61 48 23 60 59 47 65 58 74 47 40 59 68 51 50 50 71 51 26 36 38 70 46 40 51 42.
(a) Using class intervals 21 - 30, 31 - 40, ..., prepare a frequency distribution table.
(b) Calculate the mean mark of the distribution.
(c) What percentage of the students scored more than 60%?
(a) Frequency distribution table
| Class interval | Frequency \(f\) | Midpoint \(x\) | \(fx\) |
|---|---|---|---|
| 21 - 30 | 2 | 25.5 | 51.0 |
| 31 - 40 | 10 | 35.5 | 355.0 |
| 41 - 50 | 12 | 45.5 | 546.0 |
| 51 - 60 | 15 | 55.5 | 832.5 |
| 61 - 70 | 8 | 65.5 | 524.0 |
| 71 - 80 | 3 | 75.5 | 226.5 |
| Total | 50 | 2535.0 |
(b) Mean \[\bar{x} = \frac{\sum fx}{\sum f} = \frac{2535}{50} = \mathbf{50.7}.\]
(c) Marks more than \(60\): the values greater than \(60\) are \(61, 62, 65, 67, 67, 68, 69, 70, 71, 73, 74\) - that is \(11\) students. \[\text{percentage} = \frac{11}{50}\times 100 = \mathbf{22\%}.\]
Bayanin Amsa
(a) Frequency distribution table
| Class interval | Frequency \(f\) | Midpoint \(x\) | \(fx\) |
|---|---|---|---|
| 21 - 30 | 2 | 25.5 | 51.0 |
| 31 - 40 | 10 | 35.5 | 355.0 |
| 41 - 50 | 12 | 45.5 | 546.0 |
| 51 - 60 | 15 | 55.5 | 832.5 |
| 61 - 70 | 8 | 65.5 | 524.0 |
| 71 - 80 | 3 | 75.5 | 226.5 |
| Total | 50 | 2535.0 |
(b) Mean \[\bar{x} = \frac{\sum fx}{\sum f} = \frac{2535}{50} = \mathbf{50.7}.\]
(c) Marks more than \(60\): the values greater than \(60\) are \(61, 62, 65, 67, 67, 68, 69, 70, 71, 73, 74\) - that is \(11\) students. \[\text{percentage} = \frac{11}{50}\times 100 = \mathbf{22\%}.\]
Tambaya 9 Rahoto
(a) If \(9 \cos x - 7 = 1\) and \(0° \leq x \leq 90°\), find x.
(b) Given that x is an integer, find the three greatest values of x which satisfy the inequality \(7x < 2x - 13\).
(a) \(9\cos x-7=1\Rightarrow 9\cos x=8\Rightarrow \cos x=\dfrac{8}{9}=0.8889\).
For \(0^{\circ}\le x\le 90^{\circ}\): \(x=\cos^{-1}(0.8889)\approx \mathbf{27.3^{\circ}}\).
(b) \(7x<2x-13 \Rightarrow 7x-2x<-13 \Rightarrow 5x<-13 \Rightarrow x<-2.6\).
Since \(x\) is an integer, the values are \(\ldots,-5,-4,-3\). The three greatest are \(\mathbf{-3,\ -4,\ -5}\).
Bayanin Amsa
(a) \(9\cos x-7=1\Rightarrow 9\cos x=8\Rightarrow \cos x=\dfrac{8}{9}=0.8889\).
For \(0^{\circ}\le x\le 90^{\circ}\): \(x=\cos^{-1}(0.8889)\approx \mathbf{27.3^{\circ}}\).
(b) \(7x<2x-13 \Rightarrow 7x-2x<-13 \Rightarrow 5x<-13 \Rightarrow x<-2.6\).
Since \(x\) is an integer, the values are \(\ldots,-5,-4,-3\). The three greatest are \(\mathbf{-3,\ -4,\ -5}\).
Tambaya 10 Rahoto
(a) Simplify \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\)
(b) The graph of the equation \(y = Ax^{2} + Bx + C\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :
(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.
(a) Simplifying the algebraic fractions
Use the common denominator \((x-2)(x-1)\):
\[ \frac{x+2}{x-2}-\frac{x+3}{x-1} = \frac{(x+2)(x-1)-(x+3)(x-2)}{(x-2)(x-1)}. \] \[ = \frac{(x^2+x-2)-(x^2+x-6)}{(x-2)(x-1)} = \frac{x^2+x-2-x^2-x+6}{(x-2)(x-1)}. \] \[ \boxed{\frac{4}{(x-2)(x-1)}} \]This is valid for \(x\ne2\) and \(x\ne1\), since these values would make an original denominator zero.
(b) Finding the quadratic equation
The equation is \(y=Ax^2+Bx+C\). Since the graph passes through \((0,0)\), substitute \(x=0\) and \(y=0\):
\[ 0=A(0)^2+B(0)+C, \] \[ \boxed{C=0}. \]So the equation becomes:
\[ y=Ax^2+Bx. \]Using \((1,4)\):
\[ 4=A(1)^2+B(1), \] \[ A+B=4. \]Using \((2,10)\):
\[ 10=A(2)^2+B(2), \] \[ 4A+2B=10, \] \[ 2A+B=5. \]Subtract \(A+B=4\) from \(2A+B=5\):
\[ A=1. \]Then:
\[ 1+B=4 \quad\Rightarrow\quad B=3. \]Therefore:
\[ \boxed{A=1,\quad B=3} \]and the equation of the graph is:
\[ \boxed{y=x^2+3x}. \]Other x-intercept
At an x-intercept, \(y=0\):
\[ x^2+3x=0, \] \[ x(x+3)=0. \] \[ x=0 \quad\text{or}\quad x=-3. \]\((0,0)\) is the given intercept, so the other point where the graph cuts the x-axis is:
\[ \boxed{(-3,0)}. \]Examination reminder: \((1,4)\) and \((2,10)\) are points on the graph, not roots. A root is an x-value for which \(y=0\). Factoring \(x^2+3x\) gives \(x=0\) and \(x=-3\), not \(x=3\).
Bayanin Amsa
(a) Simplifying the algebraic fractions
Use the common denominator \((x-2)(x-1)\):
\[ \frac{x+2}{x-2}-\frac{x+3}{x-1} = \frac{(x+2)(x-1)-(x+3)(x-2)}{(x-2)(x-1)}. \] \[ = \frac{(x^2+x-2)-(x^2+x-6)}{(x-2)(x-1)} = \frac{x^2+x-2-x^2-x+6}{(x-2)(x-1)}. \] \[ \boxed{\frac{4}{(x-2)(x-1)}} \]This is valid for \(x\ne2\) and \(x\ne1\), since these values would make an original denominator zero.
(b) Finding the quadratic equation
The equation is \(y=Ax^2+Bx+C\). Since the graph passes through \((0,0)\), substitute \(x=0\) and \(y=0\):
\[ 0=A(0)^2+B(0)+C, \] \[ \boxed{C=0}. \]So the equation becomes:
\[ y=Ax^2+Bx. \]Using \((1,4)\):
\[ 4=A(1)^2+B(1), \] \[ A+B=4. \]Using \((2,10)\):
\[ 10=A(2)^2+B(2), \] \[ 4A+2B=10, \] \[ 2A+B=5. \]Subtract \(A+B=4\) from \(2A+B=5\):
\[ A=1. \]Then:
\[ 1+B=4 \quad\Rightarrow\quad B=3. \]Therefore:
\[ \boxed{A=1,\quad B=3} \]and the equation of the graph is:
\[ \boxed{y=x^2+3x}. \]Other x-intercept
At an x-intercept, \(y=0\):
\[ x^2+3x=0, \] \[ x(x+3)=0. \] \[ x=0 \quad\text{or}\quad x=-3. \]\((0,0)\) is the given intercept, so the other point where the graph cuts the x-axis is:
\[ \boxed{(-3,0)}. \]Examination reminder: \((1,4)\) and \((2,10)\) are points on the graph, not roots. A root is an x-value for which \(y=0\). Factoring \(x^2+3x\) gives \(x=0\) and \(x=-3\), not \(x=3\).
Tambaya 11 Rahoto
(a) If \(\log 5 = 0.6990, \log 7 = 0.8451\) and \(\log 8 = 0.9031\), evaluate \(\log (\frac{35 \times 49}{40 \div 56})\).
(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the
(i) value of x ; (ii) ratio of the number of children to the number of adults ; (iii) average amount paid per person ; (iv) percentage gain if the organisers spent D720 on the show.
(a) Simplify inside first. \(40 \div 56 = \dfrac{40}{56} = \dfrac{5}{7}\), so \[\frac{35\times 49}{40\div 56} = \frac{35\times 49}{5/7} = 35\times 49\times\frac{7}{5} = 7\times 49\times 7 = 7^4 = 2401.\] \[\log 2401 = \log 7^4 = 4\log 7 = 4(0.8451) = \mathbf{3.3804}.\]
(b) Let there be \(x\) children; adults \(= x + 60\). Adult pays \(D5\), child \(D2\), total \(D1280\): \[5(x + 60) + 2x = 1280 \Rightarrow 7x + 300 = 1280 \Rightarrow 7x = 980 \Rightarrow x = 140.\]
(i) \(x = \mathbf{140}\) children (adults \(= 200\)).
(ii) children : adults \(= 140 : 200 = \mathbf{7 : 10}.\)
(iii) Total people \(= 340\); average \(= \dfrac{1280}{340} = \mathbf{D3.76}\) per person.
(iv) Gain \(= 1280 - 720 = D560\); percentage gain \(= \dfrac{560}{720}\times 100 = \mathbf{77.8\%}.\)
Bayanin Amsa
(a) Simplify inside first. \(40 \div 56 = \dfrac{40}{56} = \dfrac{5}{7}\), so \[\frac{35\times 49}{40\div 56} = \frac{35\times 49}{5/7} = 35\times 49\times\frac{7}{5} = 7\times 49\times 7 = 7^4 = 2401.\] \[\log 2401 = \log 7^4 = 4\log 7 = 4(0.8451) = \mathbf{3.3804}.\]
(b) Let there be \(x\) children; adults \(= x + 60\). Adult pays \(D5\), child \(D2\), total \(D1280\): \[5(x + 60) + 2x = 1280 \Rightarrow 7x + 300 = 1280 \Rightarrow 7x = 980 \Rightarrow x = 140.\]
(i) \(x = \mathbf{140}\) children (adults \(= 200\)).
(ii) children : adults \(= 140 : 200 = \mathbf{7 : 10}.\)
(iii) Total people \(= 340\); average \(= \dfrac{1280}{340} = \mathbf{D3.76}\) per person.
(iv) Gain \(= 1280 - 720 = D560\); percentage gain \(= \dfrac{560}{720}\times 100 = \mathbf{77.8\%}.\)
Tambaya 12 Rahoto
(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take \(\pi = \frac{22}{7}\)].
(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the :
(i) length ; (ii) diagonal of the rectangle.
(a) A circle inscribed in a square has diameter equal to the side, so side \(= 2r.\) Then \[\text{perimeter of square} + \text{circumference} = 4(2r) + 2\pi r = 100.\] \[8r + 2\times\frac{22}{7}r = 100 \Rightarrow 8r + \frac{44}{7}r = 100 \Rightarrow \frac{56r + 44r}{7} = 100.\] \[\frac{100r}{7} = 100 \Rightarrow r = \mathbf{7\,\text{cm}}.\]
(b) Perimeter of rectangle \(= 60\,\text{cm}\), with length \(= 4\times\)breadth. \[2(l + b) = 60 \Rightarrow l + b = 30,\quad l = 4b \Rightarrow 5b = 30 \Rightarrow b = 6.\]
(i) Length \(l = 4(6) = \mathbf{24\,\text{cm}}.\)
(ii) Diagonal \(= \sqrt{l^2 + b^2} = \sqrt{24^2 + 6^2} = \sqrt{576 + 36} = \sqrt{612} \approx \mathbf{24.7\,\text{cm}}.\)
Bayanin Amsa
(a) A circle inscribed in a square has diameter equal to the side, so side \(= 2r.\) Then \[\text{perimeter of square} + \text{circumference} = 4(2r) + 2\pi r = 100.\] \[8r + 2\times\frac{22}{7}r = 100 \Rightarrow 8r + \frac{44}{7}r = 100 \Rightarrow \frac{56r + 44r}{7} = 100.\] \[\frac{100r}{7} = 100 \Rightarrow r = \mathbf{7\,\text{cm}}.\]
(b) Perimeter of rectangle \(= 60\,\text{cm}\), with length \(= 4\times\)breadth. \[2(l + b) = 60 \Rightarrow l + b = 30,\quad l = 4b \Rightarrow 5b = 30 \Rightarrow b = 6.\]
(i) Length \(l = 4(6) = \mathbf{24\,\text{cm}}.\)
(ii) Diagonal \(= \sqrt{l^2 + b^2} = \sqrt{24^2 + 6^2} = \sqrt{576 + 36} = \sqrt{612} \approx \mathbf{24.7\,\text{cm}}.\)
Tambaya 13 Rahoto
(a) How many numbers between 75 and 500 are divisible by 7?
(b) The 8th term of an Arithmetic Progression (A.P) is 5 times the 3rd term while the 7th term is 9 greater than the 4th term. Write the first 5 terms of the A.P.
(a) The multiples of \(7\) between \(75\) and \(500\): the first is \(77 = 7\times 11\) and the last is \(497 = 7\times 71.\) \[\text{count} = 71 - 11 + 1 = \mathbf{61}.\]
(b) Let the first term be \(a\) and common difference \(d.\)
8th term is \(5\) times the 3rd: \(a + 7d = 5(a + 2d) \Rightarrow a + 7d = 5a + 10d \Rightarrow 4a + 3d = 0.\)
7th term is \(9\) greater than the 4th: \((a + 6d) - (a + 3d) = 9 \Rightarrow 3d = 9 \Rightarrow d = 3.\)
Then \(4a + 3(3) = 0 \Rightarrow 4a = -9 \Rightarrow a = -\tfrac{9}{4} = -2.25.\)
First five terms (adding \(d = 3\) each time): \[\mathbf{-2\tfrac{1}{4},\ \tfrac{3}{4},\ 3\tfrac{3}{4},\ 6\tfrac{3}{4},\ 9\tfrac{3}{4}}.\]
Bayanin Amsa
(a) The multiples of \(7\) between \(75\) and \(500\): the first is \(77 = 7\times 11\) and the last is \(497 = 7\times 71.\) \[\text{count} = 71 - 11 + 1 = \mathbf{61}.\]
(b) Let the first term be \(a\) and common difference \(d.\)
8th term is \(5\) times the 3rd: \(a + 7d = 5(a + 2d) \Rightarrow a + 7d = 5a + 10d \Rightarrow 4a + 3d = 0.\)
7th term is \(9\) greater than the 4th: \((a + 6d) - (a + 3d) = 9 \Rightarrow 3d = 9 \Rightarrow d = 3.\)
Then \(4a + 3(3) = 0 \Rightarrow 4a = -9 \Rightarrow a = -\tfrac{9}{4} = -2.25.\)
First five terms (adding \(d = 3\) each time): \[\mathbf{-2\tfrac{1}{4},\ \tfrac{3}{4},\ 3\tfrac{3}{4},\ 6\tfrac{3}{4},\ 9\tfrac{3}{4}}.\]
