WAEC SSCE - General Mathematics - 1997 (Essay)
Tambaya 1 Rahoto
The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.
| Marks (in %) | Freq |
| 44 - 46 | 2 |
| 47 - 49 | 5 |
| 50 - 52 | 11 |
| 53 - 55 | 20 |
| 56 - 61 | 42 |
| 62 - 64 | 46 |
| 65 - 67 | 36 |
| 68 - 70 | 9 |
| 71 - 73 | 3 |
(a) Make a cumulative frequency for the distribution
(b) Draw the cumulative frequency curve.
(c) From your graph, estimate the median mark.
(d) The cut-off mark was 63%. What percentage of the candidates was selected?
(a) Cumulative frequency table
Total frequency, \(N=174\).
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 44–46 | 43.5–46.5 | 2 | 2 |
| 47–49 | 46.5–49.5 | 5 | 7 |
| 50–52 | 49.5–52.5 | 11 | 18 |
| 53–55 | 52.5–55.5 | 20 | 38 |
| 56–61 | 55.5–61.5 | 42 | 80 |
| 62–64 | 61.5–64.5 | 46 | 126 |
| 65–67 | 64.5–67.5 | 36 | 162 |
| 68–70 | 67.5–70.5 | 9 | 171 |
| 71–73 | 70.5–73.5 | 3 | 174 |
(b) Cumulative frequency curve (ogive)
Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.
(c) Median mark
The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).
(d) Percentage selected
At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is
\[174-103=71.\]
\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]
Bayanin Amsa
(a) Cumulative frequency table
Total frequency, \(N=174\).
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 44–46 | 43.5–46.5 | 2 | 2 |
| 47–49 | 46.5–49.5 | 5 | 7 |
| 50–52 | 49.5–52.5 | 11 | 18 |
| 53–55 | 52.5–55.5 | 20 | 38 |
| 56–61 | 55.5–61.5 | 42 | 80 |
| 62–64 | 61.5–64.5 | 46 | 126 |
| 65–67 | 64.5–67.5 | 36 | 162 |
| 68–70 | 67.5–70.5 | 9 | 171 |
| 71–73 | 70.5–73.5 | 3 | 174 |
(b) Cumulative frequency curve (ogive)
Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.
(c) Median mark
The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).
(d) Percentage selected
At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is
\[174-103=71.\]
\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]
Tambaya 2 Rahoto
(a) Given that \(\sin x = \frac{5}{13}, 0° \leq x \leq 90°\), find \(\frac{\cos x - 2 \sin x }{2\tan x}\).
(b) 
The diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures :
(i) |QP| ; (ii) the height of the mountain.
(a) Given \(\sin x=\dfrac{5}{13}\) with \(0^\circ\le x\le 90^\circ\), evaluate \(\dfrac{\cos x-2\sin x}{2\tan x}\).
Using a \(5\)-\(12\)-\(13\) right triangle (adjacent \(=\sqrt{13^2-5^2}=\sqrt{144}=12\)):
\[\cos x=\frac{12}{13},\qquad \tan x=\frac{5}{12}.\]Substitute:
\[\frac{\cos x-2\sin x}{2\tan x}=\frac{\dfrac{12}{13}-2\cdot\dfrac{5}{13}}{2\cdot\dfrac{5}{12}}=\frac{\dfrac{12-10}{13}}{\dfrac{10}{12}}=\frac{\dfrac{2}{13}}{\dfrac{5}{6}}.\]\[=\frac{2}{13}\times\frac{6}{5}=\mathbf{\frac{12}{65}}.\](b) \(NQ\) is the vertical height of the mountain on horizontal ground \(PRN\) (with \(R\) between \(P\) and \(N\)). The angle of elevation of the top \(Q\) is \(30^\circ\) from \(P\) and \(70^\circ\) from \(R\), and \(PR=500\text{ m}\).
(i) \(|QP|\). In triangle \(PQR\), \(\angle QPR=30^\circ\). Since \(\angle QRN=70^\circ\) is exterior to the triangle at \(R\), the interior angle is
\[\angle QRP=180^\circ-70^\circ=110^\circ,\]\[\angle PQR=180^\circ-30^\circ-110^\circ=40^\circ.\]By the sine rule:
\[\frac{|QP|}{\sin\angle QRP}=\frac{|PR|}{\sin\angle PQR}\]\[|QP|=\frac{500\times\sin 110^\circ}{\sin 40^\circ}=\frac{500\times0.9397}{0.6428}\approx\mathbf{731\text{ m}}.\](ii) Height of the mountain \(NQ\). In right-angled triangle \(QNP\) (right angle at \(N\)), with \(\angle QPN=30^\circ\):
\[NQ=|QP|\sin 30^\circ=730.95\times0.5\approx\mathbf{365\text{ m}}.\](Check via triangle \(QRN\): \(QR=\dfrac{500\sin30^\circ}{\sin40^\circ}=388.9\text{ m}\), and \(NQ=QR\sin70^\circ=388.9\times0.9397\approx365\text{ m}\).)
Bayanin Amsa
(a) Given \(\sin x=\dfrac{5}{13}\) with \(0^\circ\le x\le 90^\circ\), evaluate \(\dfrac{\cos x-2\sin x}{2\tan x}\).
Using a \(5\)-\(12\)-\(13\) right triangle (adjacent \(=\sqrt{13^2-5^2}=\sqrt{144}=12\)):
\[\cos x=\frac{12}{13},\qquad \tan x=\frac{5}{12}.\]Substitute:
\[\frac{\cos x-2\sin x}{2\tan x}=\frac{\dfrac{12}{13}-2\cdot\dfrac{5}{13}}{2\cdot\dfrac{5}{12}}=\frac{\dfrac{12-10}{13}}{\dfrac{10}{12}}=\frac{\dfrac{2}{13}}{\dfrac{5}{6}}.\]\[=\frac{2}{13}\times\frac{6}{5}=\mathbf{\frac{12}{65}}.\](b) \(NQ\) is the vertical height of the mountain on horizontal ground \(PRN\) (with \(R\) between \(P\) and \(N\)). The angle of elevation of the top \(Q\) is \(30^\circ\) from \(P\) and \(70^\circ\) from \(R\), and \(PR=500\text{ m}\).
(i) \(|QP|\). In triangle \(PQR\), \(\angle QPR=30^\circ\). Since \(\angle QRN=70^\circ\) is exterior to the triangle at \(R\), the interior angle is
\[\angle QRP=180^\circ-70^\circ=110^\circ,\]\[\angle PQR=180^\circ-30^\circ-110^\circ=40^\circ.\]By the sine rule:
\[\frac{|QP|}{\sin\angle QRP}=\frac{|PR|}{\sin\angle PQR}\]\[|QP|=\frac{500\times\sin 110^\circ}{\sin 40^\circ}=\frac{500\times0.9397}{0.6428}\approx\mathbf{731\text{ m}}.\](ii) Height of the mountain \(NQ\). In right-angled triangle \(QNP\) (right angle at \(N\)), with \(\angle QPN=30^\circ\):
\[NQ=|QP|\sin 30^\circ=730.95\times0.5\approx\mathbf{365\text{ m}}.\](Check via triangle \(QRN\): \(QR=\dfrac{500\sin30^\circ}{\sin40^\circ}=388.9\text{ m}\), and \(NQ=QR\sin70^\circ=388.9\times0.9397\approx365\text{ m}\).)
Tambaya 3 Rahoto
(a) Using a ruler and a pair of compasses only, construst \(\Delta\) ABC in which |AB| = 7cm, |BC| = 5cm and < ABC = 75°. Measure |AC|.
(b) In (a) above, locate by construction, a point D such that CD is parallel to AB and D is equidistant from points A and C. Measure < BAD.
(a) Construction: Draw \(AB = 7\text{ cm}\). At \(B\), construct an angle of \(75^\circ\) by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On the \(75^\circ\) ray, mark \(C\) such that \(BC = 5\text{ cm}\). Join \(AC\).
\(|AC| = 7.5\text{ cm}\).
(b) Construction: Through \(C\), construct a line parallel to \(AB\). Construct the perpendicular bisector of \(AC\). Let it meet the parallel line at \(D\). Join \(AD\).
\(\angle BAD = 80^\circ\).
Bayanin Amsa
(a) Construction: Draw \(AB = 7\text{ cm}\). At \(B\), construct an angle of \(75^\circ\) by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On the \(75^\circ\) ray, mark \(C\) such that \(BC = 5\text{ cm}\). Join \(AC\).
\(|AC| = 7.5\text{ cm}\).
(b) Construction: Through \(C\), construct a line parallel to \(AB\). Construct the perpendicular bisector of \(AC\). Let it meet the parallel line at \(D\). Join \(AD\).
\(\angle BAD = 80^\circ\).
Tambaya 4 Rahoto
The solid is a cylinder surmounted by a hemispherical bowl. Calculate its
(a) total surface area ;
(b) volume (Take \(\pi = \frac{22}{7}\))
From the diagram the solid is a closed cylinder capped on top by a hemisphere. The hemisphere has height \(7\text{ cm}\); since a hemisphere's height equals its radius, the radius is \(r = 7\text{ cm}\), which is also the radius of the cylinder. The cylinder height is \(h = 10\text{ cm}\). Take \(\pi = \dfrac{22}{7}\).
(a) Total surface area.
The exposed surfaces are: the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the hemisphere. (The flat top of the cylinder is exactly covered by the flat face of the hemisphere, so it is not counted.)
Curved surface of cylinder:
\[2\pi r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{ cm}^2\]Base circle of cylinder:
\[\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2\]Curved surface of hemisphere:
\[2\pi r^2 = 2 \times 154 = 308 \text{ cm}^2\]Total surface area:
\[440 + 154 + 308 = 902 \text{ cm}^2\]Total surface area \(= 902\text{ cm}^2\).
(b) Volume.
Volume of cylinder:
\[\pi r^2 h = \frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ cm}^3\]Volume of hemisphere:
\[\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \approx 718.67 \text{ cm}^3\]Total volume:
\[1540 + 718.67 = 2258.67 \text{ cm}^3\]Total volume \(\approx 2258.67\text{ cm}^3\) (i.e. \(2258\tfrac{2}{3}\text{ cm}^3\)).
Bayanin Amsa
From the diagram the solid is a closed cylinder capped on top by a hemisphere. The hemisphere has height \(7\text{ cm}\); since a hemisphere's height equals its radius, the radius is \(r = 7\text{ cm}\), which is also the radius of the cylinder. The cylinder height is \(h = 10\text{ cm}\). Take \(\pi = \dfrac{22}{7}\).
(a) Total surface area.
The exposed surfaces are: the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the hemisphere. (The flat top of the cylinder is exactly covered by the flat face of the hemisphere, so it is not counted.)
Curved surface of cylinder:
\[2\pi r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{ cm}^2\]Base circle of cylinder:
\[\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2\]Curved surface of hemisphere:
\[2\pi r^2 = 2 \times 154 = 308 \text{ cm}^2\]Total surface area:
\[440 + 154 + 308 = 902 \text{ cm}^2\]Total surface area \(= 902\text{ cm}^2\).
(b) Volume.
Volume of cylinder:
\[\pi r^2 h = \frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ cm}^3\]Volume of hemisphere:
\[\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \approx 718.67 \text{ cm}^3\]Total volume:
\[1540 + 718.67 = 2258.67 \text{ cm}^3\]Total volume \(\approx 2258.67\text{ cm}^3\) (i.e. \(2258\tfrac{2}{3}\text{ cm}^3\)).
Tambaya 5 Rahoto
Above is the graph of the quadratic function \(y = ax^{2} + bx + c\) where a, b and c are constants. Using the graph, find :
(a)(i) the scales on both axes ; (ii) the equation of the line of symmetry of the curve ; (iii) the roots of the quadratic equation \(ax^{2} + bx + c = 0\)
(b) Use the coordinates of D, E and G to find the values of the constants a, b and c hence write down the quadratic function illustrated in the graph.
(c) Find the greatest value of y within the range \(-3 \leq x \leq 5\).
There is an inconsistency in the supplied reference answer. The coordinates \(D(0,1)\), \(E(1,-2)\), and \(G(3,4)\) give the quadratic \(y=2x^2-5x+1\). This equation has its vertex at \(x=1.25\), not at \(x=1\). Also, for \(-3\leq x\leq 5\), its greatest value is \(34\), not \(33.5\).
Graph consistent with the given coordinates and equation
(a)(i) Scales
- On the \(x\)-axis, \(2\text{ cm}\) represents \(1\) unit.
- On the \(y\)-axis, \(2\text{ cm}\) represents \(5\) units.
(a)(ii) Line of symmetry
The line of symmetry passes through the vertex. From the graph, the minimum is halfway between \(x=1\) and \(x=1.5\), so
\[x=1.25\]
Do not assume that \(E(1,-2)\) is the vertex. It is a point on the curve, but the curve continues slightly lower before turning upwards.
(a)(iii) Roots
The roots are the \(x\)-coordinates where the curve crosses the \(x\)-axis. Reading from the graph gives approximately
\[x\approx0.25\quad\text{and}\quad x\approx2.25.\]
Using the equation found in part (b), the more accurate values are
\[x=\frac{5-\sqrt{17}}{4}\approx0.219,\qquad x=\frac{5+\sqrt{17}}{4}\approx2.281.\]
(b) Finding \(a\), \(b\), and \(c\)
Use \(y=ax^2+bx+c\) and substitute the three stated points.
Using \(D(0,1)\):
\[1=a(0)^2+b(0)+c\]
\[c=1\]
Using \(E(1,-2)\):
\[-2=a+b+1\]
\[a+b=-3\qquad\text{(1)}\]
Using \(G(3,4)\):
\[4=9a+3b+1\]
\[9a+3b=3\]
\[3a+b=1\qquad\text{(2)}\]
Subtract equation (1) from equation (2):
\[(3a+b)-(a+b)=1-(-3)\]
\[2a=4\]
\[a=2\]
Then substitute \(a=2\) into \(a+b=-3\):
\[2+b=-3\]
\[b=-5\]
Therefore,
\[\boxed{y=2x^2-5x+1}\]
(c) Greatest value of \(y\) for \(-3\leq x\leq5\)
Since \(a=2>0\), the parabola opens upwards. Its vertex is a minimum, so the greatest value in the given closed interval must be at an endpoint.
At \(x=-3\):
\[y=2(-3)^2-5(-3)+1=18+15+1=34\]
At \(x=5\):
\[y=2(5)^2-5(5)+1=50-25+1=26\]
Since \(34>26\), the greatest value is
\[\boxed{34}\]
Examination reminder: For a quadratic opening upwards, check both endpoints when asked for the greatest value over a restricted interval. The vertex gives the minimum, not the maximum.
Bayanin Amsa
There is an inconsistency in the supplied reference answer. The coordinates \(D(0,1)\), \(E(1,-2)\), and \(G(3,4)\) give the quadratic \(y=2x^2-5x+1\). This equation has its vertex at \(x=1.25\), not at \(x=1\). Also, for \(-3\leq x\leq 5\), its greatest value is \(34\), not \(33.5\).
Graph consistent with the given coordinates and equation
(a)(i) Scales
- On the \(x\)-axis, \(2\text{ cm}\) represents \(1\) unit.
- On the \(y\)-axis, \(2\text{ cm}\) represents \(5\) units.
(a)(ii) Line of symmetry
The line of symmetry passes through the vertex. From the graph, the minimum is halfway between \(x=1\) and \(x=1.5\), so
\[x=1.25\]
Do not assume that \(E(1,-2)\) is the vertex. It is a point on the curve, but the curve continues slightly lower before turning upwards.
(a)(iii) Roots
The roots are the \(x\)-coordinates where the curve crosses the \(x\)-axis. Reading from the graph gives approximately
\[x\approx0.25\quad\text{and}\quad x\approx2.25.\]
Using the equation found in part (b), the more accurate values are
\[x=\frac{5-\sqrt{17}}{4}\approx0.219,\qquad x=\frac{5+\sqrt{17}}{4}\approx2.281.\]
(b) Finding \(a\), \(b\), and \(c\)
Use \(y=ax^2+bx+c\) and substitute the three stated points.
Using \(D(0,1)\):
\[1=a(0)^2+b(0)+c\]
\[c=1\]
Using \(E(1,-2)\):
\[-2=a+b+1\]
\[a+b=-3\qquad\text{(1)}\]
Using \(G(3,4)\):
\[4=9a+3b+1\]
\[9a+3b=3\]
\[3a+b=1\qquad\text{(2)}\]
Subtract equation (1) from equation (2):
\[(3a+b)-(a+b)=1-(-3)\]
\[2a=4\]
\[a=2\]
Then substitute \(a=2\) into \(a+b=-3\):
\[2+b=-3\]
\[b=-5\]
Therefore,
\[\boxed{y=2x^2-5x+1}\]
(c) Greatest value of \(y\) for \(-3\leq x\leq5\)
Since \(a=2>0\), the parabola opens upwards. Its vertex is a minimum, so the greatest value in the given closed interval must be at an endpoint.
At \(x=-3\):
\[y=2(-3)^2-5(-3)+1=18+15+1=34\]
At \(x=5\):
\[y=2(5)^2-5(5)+1=50-25+1=26\]
Since \(34>26\), the greatest value is
\[\boxed{34}\]
Examination reminder: For a quadratic opening upwards, check both endpoints when asked for the greatest value over a restricted interval. The vertex gives the minimum, not the maximum.
Tambaya 6 Rahoto
(a) The 6th term of an A.P is 35 and the 13th term is 77. Find the 20th term.
(b) 
The Venn diagram represents three subsets P, Q and R of the universal set U. Copy the Venn diagram. Shade and indicate the regions represented by (i) \(P \cap Q' \cap R\) ; (ii) \(P' \cap Q \cap R'\).
(a) The 20th term. For an A.P., \(T_n=a+(n-1)d\).
\[T_6=a+5d=35,\qquad T_{13}=a+12d=77.\]
Subtracting the first from the second: \(7d=42\Rightarrow d=6\). Then \(a=35-5(6)=5\).
\[T_{20}=a+19d=5+19(6)=5+114=119.\]
(b) Shading the Venn diagram. The three circles \(P\) (top left), \(Q\) (top right) and \(R\) (bottom) divide \(U\) into eight regions.
(i) \(P\cap Q'\cap R\): shade the region lying inside both \(P\) and \(R\) but outside \(Q\). This is the overlap (lens) of the \(P\) and \(R\) circles, with the central piece that also lies in \(Q\) left unshaded.
(ii) \(P'\cap Q\cap R'\): shade the region lying inside \(Q\) only, that is, the part of the \(Q\) circle that is outside both \(P\) and \(R\).
Bayanin Amsa
(a) The 20th term. For an A.P., \(T_n=a+(n-1)d\).
\[T_6=a+5d=35,\qquad T_{13}=a+12d=77.\]
Subtracting the first from the second: \(7d=42\Rightarrow d=6\). Then \(a=35-5(6)=5\).
\[T_{20}=a+19d=5+19(6)=5+114=119.\]
(b) Shading the Venn diagram. The three circles \(P\) (top left), \(Q\) (top right) and \(R\) (bottom) divide \(U\) into eight regions.
(i) \(P\cap Q'\cap R\): shade the region lying inside both \(P\) and \(R\) but outside \(Q\). This is the overlap (lens) of the \(P\) and \(R\) circles, with the central piece that also lies in \(Q\) left unshaded.
(ii) \(P'\cap Q\cap R'\): shade the region lying inside \(Q\) only, that is, the part of the \(Q\) circle that is outside both \(P\) and \(R\).
Tambaya 7 Rahoto
The table below shows how a company's sales manager spent his 1995 annual salary.
| Food | 30% |
| Rent | 18% |
| Car Maintenance | 25% |
| Savings | 12% |
| Taxes | 5% |
| Others | 10% |
(a) Represent this information on a pie chart.
(b) Find his savings at the end of the year if his annual salary was N60,000.00.
(a) Pie chart of the manager's annual expenditure
The angle for each sector is calculated using:
\[\text{Sector angle}=\frac{\text{Percentage}}{100}\times360^\circ\]
| Item | Percentage | Angle of sector |
|---|---|---|
| Food | 30% | \(108^\circ\) |
| Rent | 18% | \(64.8^\circ\) |
| Car Maintenance | 25% | \(90^\circ\) |
| Savings | 12% | \(43.2^\circ\) |
| Taxes | 5% | \(18^\circ\) |
| Others | 10% | \(36^\circ\) |
| Total | 100% | \(360^\circ\) |
(b)
\[\text{Savings}=12\%\text{ of }\text{₦}60,000\]
\[=\frac{12}{100}\times 60,000=\text{₦}7,200.00\]
His savings at the end of the year was \(\text{₦}7,200.00\).
Bayanin Amsa
(a) Pie chart of the manager's annual expenditure
The angle for each sector is calculated using:
\[\text{Sector angle}=\frac{\text{Percentage}}{100}\times360^\circ\]
| Item | Percentage | Angle of sector |
|---|---|---|
| Food | 30% | \(108^\circ\) |
| Rent | 18% | \(64.8^\circ\) |
| Car Maintenance | 25% | \(90^\circ\) |
| Savings | 12% | \(43.2^\circ\) |
| Taxes | 5% | \(18^\circ\) |
| Others | 10% | \(36^\circ\) |
| Total | 100% | \(360^\circ\) |
(b)
\[\text{Savings}=12\%\text{ of }\text{₦}60,000\]
\[=\frac{12}{100}\times 60,000=\text{₦}7,200.00\]
His savings at the end of the year was \(\text{₦}7,200.00\).
Tambaya 8 Rahoto
(a) Solve the simultaneous equation:
\[ \log_{10} x + \log_{10} y = 4 \]
\[ \log_{10} x + 2\log_{10} y = 3 \]
(b) The time, \(t\), taken to buy fuel at a petrol station varies directly as the number of vehicles \(V\) on queue and jointly varies inversely as the number of pumps \(P\) available in the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find:
(i) the relationship between \(t\), \(P\) and \(V\); (ii) the time it will take to fuel 50 vehicles in the station with 2 pumps; (iii) the number of pumps required to fuel 40 vehicles in 20 minutes.
(a) Let \(\log_{10}x = u\) and \(\log_{10}y = v\).
\[u + v = 4 \quad\text{...(1)}\qquad u + 2v = 3 \quad\text{...(2)}\]
Subtract (1) from (2): \(v = -1\). Then \(u = 4 - (-1) = 5\).
\[\log_{10}x = 5 \Rightarrow x = 10^5 = 100000, \qquad \log_{10}y = -1 \Rightarrow y = 10^{-1} = 0.1\]
(b) \(t\) varies directly as \(V\) and inversely as \(P\): \(t = \dfrac{kV}{P}\).
With \(P = 5,\, V = 20,\, t = 10\): \(10 = \dfrac{k(20)}{5} = 4k \Rightarrow k = 2.5\).
(i) \(t = \dfrac{2.5V}{P}\) (equivalently \(t = \dfrac{5V}{2P}\)).
(ii) \(V = 50,\, P = 2\): \(t = \dfrac{2.5(50)}{2} = 62.5\) minutes.
(iii) \(V = 40,\, t = 20\): \(20 = \dfrac{2.5(40)}{P} \Rightarrow 20 = \dfrac{100}{P} \Rightarrow P = 5\) pumps.
Bayanin Amsa
(a) Let \(\log_{10}x = u\) and \(\log_{10}y = v\).
\[u + v = 4 \quad\text{...(1)}\qquad u + 2v = 3 \quad\text{...(2)}\]
Subtract (1) from (2): \(v = -1\). Then \(u = 4 - (-1) = 5\).
\[\log_{10}x = 5 \Rightarrow x = 10^5 = 100000, \qquad \log_{10}y = -1 \Rightarrow y = 10^{-1} = 0.1\]
(b) \(t\) varies directly as \(V\) and inversely as \(P\): \(t = \dfrac{kV}{P}\).
With \(P = 5,\, V = 20,\, t = 10\): \(10 = \dfrac{k(20)}{5} = 4k \Rightarrow k = 2.5\).
(i) \(t = \dfrac{2.5V}{P}\) (equivalently \(t = \dfrac{5V}{2P}\)).
(ii) \(V = 50,\, P = 2\): \(t = \dfrac{2.5(50)}{2} = 62.5\) minutes.
(iii) \(V = 40,\, t = 20\): \(20 = \dfrac{2.5(40)}{P} \Rightarrow 20 = \dfrac{100}{P} \Rightarrow P = 5\) pumps.
Tambaya 9 Rahoto
(a)
PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT.
(b) 
In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.
(a) \(PQRST\) is a circle, centre \(C\), with \(PCS\) a straight line (so \(PS\) is a diameter), \(RS \parallel QT\), \(|QR| = |RS|\), and \(\angle QTS = 56^\circ\).
Work with arcs (an inscribed angle equals half its intercepted arc).
\(\angle QTS = 56^\circ\) stands on chord \(QS\), intercepting arc \(QRS\):
\[ \text{arc } QRS = 2 \times 56^\circ = 112^\circ \]
Equal chords \(|QR| = |RS|\) cut off equal arcs, so
\[ \text{arc } QR = \text{arc } RS = \tfrac{112^\circ}{2} = 56^\circ \]
(i) \(\angle SQT\). Because \(RS \parallel QT\), the arcs intercepted between these parallel chords are equal, so \(\text{arc } ST = \text{arc } QR = 56^\circ\). Then \(\angle SQT\) stands on chord \(ST\):
\[ \angle SQT = \tfrac{1}{2}\,\text{arc } ST = \tfrac{1}{2}\times 56^\circ = 28^\circ \]
(ii) \(\angle PQT\). Since \(PS\) is a diameter, each semicircle is \(180^\circ\). Taking the semicircle \(S \to T \to P\):
\[ \text{arc } ST + \text{arc } TP = 180^\circ \;\Rightarrow\; 56^\circ + \text{arc } TP = 180^\circ \;\Rightarrow\; \text{arc } TP = 124^\circ \]
\(\angle PQT\) stands on chord \(PT\), intercepting arc \(TP\):
\[ \angle PQT = \tfrac{1}{2}\,\text{arc } TP = \tfrac{1}{2}\times 124^\circ = 62^\circ \]
(b) \(B\) and \(C\) are on the horizontal plane with \(|BC| = 30\) cm; \(A\) is vertically above \(B\) with \(|AB| = 26\) cm, and \(D\) is vertically above \(C\) with \(|DC| = 40\) cm.
(i) Angle of depression of \(B\) from \(D\). The horizontal from \(D\) is level with \(C\). \(B\) lies \(40\) cm below that level and \(30\) cm across:
\[ \tan\theta = \frac{DC}{BC} = \frac{40}{30} = 1.3333 \]
\[ \theta = \tan^{-1}(1.3333) = 53.13^\circ \approx 53^\circ \]
(ii) Angle of depression of \(A\) from \(D\). \(A\) is \(26\) cm above \(B\), so its top is \(40 - 26 = 14\) cm below the level of \(D\); the horizontal separation is still \(30\) cm:
\[ \tan\alpha = \frac{40 - 26}{30} = \frac{14}{30} = 0.4667 \]
\[ \alpha = \tan^{-1}(0.4667) = 25.02^\circ \approx 25^\circ \]
Answers: (a)(i) \(28^\circ\); (a)(ii) \(62^\circ\); (b)(i) \(53^\circ\); (b)(ii) \(25^\circ\).
Bayanin Amsa
(a) \(PQRST\) is a circle, centre \(C\), with \(PCS\) a straight line (so \(PS\) is a diameter), \(RS \parallel QT\), \(|QR| = |RS|\), and \(\angle QTS = 56^\circ\).
Work with arcs (an inscribed angle equals half its intercepted arc).
\(\angle QTS = 56^\circ\) stands on chord \(QS\), intercepting arc \(QRS\):
\[ \text{arc } QRS = 2 \times 56^\circ = 112^\circ \]
Equal chords \(|QR| = |RS|\) cut off equal arcs, so
\[ \text{arc } QR = \text{arc } RS = \tfrac{112^\circ}{2} = 56^\circ \]
(i) \(\angle SQT\). Because \(RS \parallel QT\), the arcs intercepted between these parallel chords are equal, so \(\text{arc } ST = \text{arc } QR = 56^\circ\). Then \(\angle SQT\) stands on chord \(ST\):
\[ \angle SQT = \tfrac{1}{2}\,\text{arc } ST = \tfrac{1}{2}\times 56^\circ = 28^\circ \]
(ii) \(\angle PQT\). Since \(PS\) is a diameter, each semicircle is \(180^\circ\). Taking the semicircle \(S \to T \to P\):
\[ \text{arc } ST + \text{arc } TP = 180^\circ \;\Rightarrow\; 56^\circ + \text{arc } TP = 180^\circ \;\Rightarrow\; \text{arc } TP = 124^\circ \]
\(\angle PQT\) stands on chord \(PT\), intercepting arc \(TP\):
\[ \angle PQT = \tfrac{1}{2}\,\text{arc } TP = \tfrac{1}{2}\times 124^\circ = 62^\circ \]
(b) \(B\) and \(C\) are on the horizontal plane with \(|BC| = 30\) cm; \(A\) is vertically above \(B\) with \(|AB| = 26\) cm, and \(D\) is vertically above \(C\) with \(|DC| = 40\) cm.
(i) Angle of depression of \(B\) from \(D\). The horizontal from \(D\) is level with \(C\). \(B\) lies \(40\) cm below that level and \(30\) cm across:
\[ \tan\theta = \frac{DC}{BC} = \frac{40}{30} = 1.3333 \]
\[ \theta = \tan^{-1}(1.3333) = 53.13^\circ \approx 53^\circ \]
(ii) Angle of depression of \(A\) from \(D\). \(A\) is \(26\) cm above \(B\), so its top is \(40 - 26 = 14\) cm below the level of \(D\); the horizontal separation is still \(30\) cm:
\[ \tan\alpha = \frac{40 - 26}{30} = \frac{14}{30} = 0.4667 \]
\[ \alpha = \tan^{-1}(0.4667) = 25.02^\circ \approx 25^\circ \]
Answers: (a)(i) \(28^\circ\); (a)(ii) \(62^\circ\); (b)(i) \(53^\circ\); (b)(ii) \(25^\circ\).
Tambaya 10 Rahoto
(a) Use logarithm tables to evaluate \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\).
(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :
(i) both were boys ; (ii) the two were of different sexes.
(a) Evaluate \(\dfrac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\) with logarithms.
First \(\sqrt{0.00695}\): \(\log 0.00695 = \bar{3}.8420\), half of it \(= \bar{2}.9210\), so \(\sqrt{0.00695} = 0.08337\).
Numerator: \(\log 15.05 = 1.1775\); \(1.1775 + \bar{2}.9210 = 0.0985\), giving \(15.05 \times 0.08337 = 1.2547\).
Divide by \(695\): \(\log 695 = 2.8420\); \(0.0985 - 2.8420 = \bar{3}.2565\).
\[\text{answer} = 1.806 \times 10^{-3} \approx 0.00181\]
(b) Two boys and three girls; two are chosen at random from the five. Total ways \(= \binom{5}{2} = 10\).
(i) Both boys: \(\dfrac{\binom{2}{2}}{\binom{5}{2}} = \dfrac{1}{10}\).
(ii) Different sexes (one boy, one girl): \(\dfrac{2 \times 3}{10} = \dfrac{6}{10} = \dfrac{3}{5}\).
Bayanin Amsa
(a) Evaluate \(\dfrac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\) with logarithms.
First \(\sqrt{0.00695}\): \(\log 0.00695 = \bar{3}.8420\), half of it \(= \bar{2}.9210\), so \(\sqrt{0.00695} = 0.08337\).
Numerator: \(\log 15.05 = 1.1775\); \(1.1775 + \bar{2}.9210 = 0.0985\), giving \(15.05 \times 0.08337 = 1.2547\).
Divide by \(695\): \(\log 695 = 2.8420\); \(0.0985 - 2.8420 = \bar{3}.2565\).
\[\text{answer} = 1.806 \times 10^{-3} \approx 0.00181\]
(b) Two boys and three girls; two are chosen at random from the five. Total ways \(= \binom{5}{2} = 10\).
(i) Both boys: \(\dfrac{\binom{2}{2}}{\binom{5}{2}} = \dfrac{1}{10}\).
(ii) Different sexes (one boy, one girl): \(\dfrac{2 \times 3}{10} = \dfrac{6}{10} = \dfrac{3}{5}\).
Tambaya 11 Rahoto
(a) Copy and complete the binary multiplication table:
| x | 10 | 11 | 100 | 101 |
| 10 | 100 | 1000 | ||
| 11 | 110 | 1100 | ||
| 100 | 10000 | 10100 |
(b) Convert \(11.011_{two}\) to a number in base ten.
(c) Simplify \(\frac{9.6 \times 10^{18}}{0.24 \times 10^{5}}\) and express your answer in the form \(P \times 10^{m}\) where 1 < P < 10 and m is an integer.
(a) Binary multiplication table. Multiply in base ten, then convert back to base two (e.g. \(11\times 11 = 3\times 3 = 9 = 1001_{two}\), \(11\times 101 = 3\times 5 = 15 = 1111_{two}\)).
| \(\times\) | 10 | 11 | 100 | 101 |
|---|---|---|---|---|
| 10 | 100 | 110 | 1000 | 1010 |
| 11 | 110 | 1001 | 1100 | 1111 |
| 100 | 1000 | 1100 | 10000 | 10100 |
(b) Convert \(11.011_{two}\) to base ten.
\[1\times 2 + 1\times 1 + 0\times\tfrac{1}{2} + 1\times\tfrac{1}{4} + 1\times\tfrac{1}{8} = 2 + 1 + 0.25 + 0.125 = \mathbf{3.375}\](c) Simplify.
\[\frac{9.6\times 10^{18}}{0.24\times 10^{5}} = \frac{9.6}{0.24}\times 10^{18-5} = 40\times 10^{13} = 4.0\times 10^{14}\]So \(P = 4.0\) and \(m = 14\).
Bayanin Amsa
(a) Binary multiplication table. Multiply in base ten, then convert back to base two (e.g. \(11\times 11 = 3\times 3 = 9 = 1001_{two}\), \(11\times 101 = 3\times 5 = 15 = 1111_{two}\)).
| \(\times\) | 10 | 11 | 100 | 101 |
|---|---|---|---|---|
| 10 | 100 | 110 | 1000 | 1010 |
| 11 | 110 | 1001 | 1100 | 1111 |
| 100 | 1000 | 1100 | 10000 | 10100 |
(b) Convert \(11.011_{two}\) to base ten.
\[1\times 2 + 1\times 1 + 0\times\tfrac{1}{2} + 1\times\tfrac{1}{4} + 1\times\tfrac{1}{8} = 2 + 1 + 0.25 + 0.125 = \mathbf{3.375}\](c) Simplify.
\[\frac{9.6\times 10^{18}}{0.24\times 10^{5}} = \frac{9.6}{0.24}\times 10^{18-5} = 40\times 10^{13} = 4.0\times 10^{14}\]So \(P = 4.0\) and \(m = 14\).
Tambaya 12 Rahoto
(a) Given that \(\frac{5y - x}{8y + 3x} = \frac{1}{5}\), find the value of \(\frac{x}{y}\) to two decimal places.
(b) If 3 is a root of the quadratic equation \(x^{2} + bx - 15 = 0\), determine the value of b. Find the other root.
(a) \(\dfrac{5y - x}{8y + 3x} = \dfrac15\). Cross-multiplying:
\[5(5y - x) = 8y + 3x \;\Rightarrow\; 25y - 5x = 8y + 3x\]
\[17y = 8x \;\Rightarrow\; \frac{x}{y} = \frac{17}{8} = 2.13 \text{ (2 d.p.)}\]
(b) Since \(3\) is a root of \(x^2 + bx - 15 = 0\):
\[3^2 + 3b - 15 = 0 \;\Rightarrow\; 9 + 3b - 15 = 0 \;\Rightarrow\; 3b = 6 \;\Rightarrow\; b = 2\]
The equation becomes \(x^2 + 2x - 15 = 0 = (x + 5)(x - 3)\). Hence the other root is \(x = -5\).
(Check by product of roots: \(3 \times (-5) = -15\), matching the constant term.)
Bayanin Amsa
(a) \(\dfrac{5y - x}{8y + 3x} = \dfrac15\). Cross-multiplying:
\[5(5y - x) = 8y + 3x \;\Rightarrow\; 25y - 5x = 8y + 3x\]
\[17y = 8x \;\Rightarrow\; \frac{x}{y} = \frac{17}{8} = 2.13 \text{ (2 d.p.)}\]
(b) Since \(3\) is a root of \(x^2 + bx - 15 = 0\):
\[3^2 + 3b - 15 = 0 \;\Rightarrow\; 9 + 3b - 15 = 0 \;\Rightarrow\; 3b = 6 \;\Rightarrow\; b = 2\]
The equation becomes \(x^2 + 2x - 15 = 0 = (x + 5)(x - 3)\). Hence the other root is \(x = -5\).
(Check by product of roots: \(3 \times (-5) = -15\), matching the constant term.)
