WAEC SSCE - General Mathematics - 1992 (Essay)
Question 1 Report
(a) Solve the following pair of simultaneous equations: \(2x + 5y = 6\frac{1}{2} ; 5x - 2y = 9\)
(b) If \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\), find x.
(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).
(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)
Answer Details
(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).
(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)
Question 2 Report
The table below shows the weekly profit in naira from a mini-market.
| Weekly profit (N) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |
| Freq | 6 | 6 | 12 | 11 | 10 | 5 |
(a) Draw the cumulative frequency curve of the data;
(b) From your graph, estimate the: (i) median; (ii) 80th percentile
(c) What is the modal weekly profit?
(a) Cumulative frequency table and curve (ogive)
The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.
| Weekly profit (N) | Upper class boundary | Frequency (f) | Cumulative frequency |
|---|---|---|---|
| 1 - 10 | 10.5 | 6 | 6 |
| 11 - 20 | 20.5 | 6 | 12 |
| 21 - 30 | 30.5 | 12 | 24 |
| 31 - 40 | 40.5 | 11 | 35 |
| 41 - 50 | 50.5 | 10 | 45 |
| 51 - 60 | 60.5 | 5 | 50 |
Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).
(b) Estimates from the graph
(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.
Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):
\[ \text{Median}=30.5+\frac{25-24}{35-24}\times 10 = 30.5+0.9 \approx N\,31 \](ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.
Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):
\[ P_{80}=40.5+\frac{40-35}{45-35}\times 10 = 40.5+5 = N\,45.5 \](c) Modal weekly profit
The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using
\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):
\[ \text{Mode}=20.5+\frac{(12-6)}{(12-6)+(12-11)}\times 10 = 20.5+\frac{60}{7} = 20.5+8.57 \]\[ \boxed{\text{Modal weekly profit}\approx N\,29.07} \]Answer Details
(a) Cumulative frequency table and curve (ogive)
The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.
| Weekly profit (N) | Upper class boundary | Frequency (f) | Cumulative frequency |
|---|---|---|---|
| 1 - 10 | 10.5 | 6 | 6 |
| 11 - 20 | 20.5 | 6 | 12 |
| 21 - 30 | 30.5 | 12 | 24 |
| 31 - 40 | 40.5 | 11 | 35 |
| 41 - 50 | 50.5 | 10 | 45 |
| 51 - 60 | 60.5 | 5 | 50 |
Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).
(b) Estimates from the graph
(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.
Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):
\[ \text{Median}=30.5+\frac{25-24}{35-24}\times 10 = 30.5+0.9 \approx N\,31 \](ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.
Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):
\[ P_{80}=40.5+\frac{40-35}{45-35}\times 10 = 40.5+5 = N\,45.5 \](c) Modal weekly profit
The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using
\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):
\[ \text{Mode}=20.5+\frac{(12-6)}{(12-6)+(12-11)}\times 10 = 20.5+\frac{60}{7} = 20.5+8.57 \]\[ \boxed{\text{Modal weekly profit}\approx N\,29.07} \]Question 3 Report
(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector. [Take \(\pi = \frac{22}{7}\)]
(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.
(a) Radius \(r=7\text{ cm}\), angle \(=108^{\circ}\). Arc length \[=\frac{108}{360}\times2\pi r=\frac{108}{360}\times2\times\frac{22}{7}\times7=\frac{3}{10}\times44=13.2\text{ cm}.\] Perimeter of sector \(=\) arc \(+2r=13.2+14=\) 27.2 cm.
(b) Angle of depression \(=30^{\circ}\) equals the angle of elevation of the cliff top from the boat. With the boat \(120\text{ m}\) from the foot: \[\tan30^{\circ}=\frac{h}{120}\Rightarrow h=120\tan30^{\circ}=120\times0.5774=69.28\text{ m}\approx69.3\text{ m}.\]
Answer Details
(a) Radius \(r=7\text{ cm}\), angle \(=108^{\circ}\). Arc length \[=\frac{108}{360}\times2\pi r=\frac{108}{360}\times2\times\frac{22}{7}\times7=\frac{3}{10}\times44=13.2\text{ cm}.\] Perimeter of sector \(=\) arc \(+2r=13.2+14=\) 27.2 cm.
(b) Angle of depression \(=30^{\circ}\) equals the angle of elevation of the cliff top from the boat. With the boat \(120\text{ m}\) from the foot: \[\tan30^{\circ}=\frac{h}{120}\Rightarrow h=120\tan30^{\circ}=120\times0.5774=69.28\text{ m}\approx69.3\text{ m}.\]
Question 4 Report
(a) The sides PQ and PR of \(\Delta\) PQR are produced to T and S respectively, such that TQR = 131° and < QRS = 98°. Find < QPR.
(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre. [Take \(\pi = \frac{22}{7}\)]
(a) \(PQ\) is produced to \(T\), so \(\angle PQR=180^{\circ}-131^{\circ}=49^{\circ}\). \(PR\) is produced to \(S\), so \(\angle PRQ=180^{\circ}-98^{\circ}=82^{\circ}\). The angles of \(\triangle PQR\) sum to \(180^{\circ}\): \[\angle QPR=180^{\circ}-49^{\circ}-82^{\circ}=49^{\circ}.\]
(b) Circumference \(=2\pi r=400\): \[r=\frac{400}{2\times\frac{22}{7}}=\frac{400\times7}{44}=\frac{2800}{44}=63.6\text{ m}\approx64\text{ m}.\]
Answer Details
(a) \(PQ\) is produced to \(T\), so \(\angle PQR=180^{\circ}-131^{\circ}=49^{\circ}\). \(PR\) is produced to \(S\), so \(\angle PRQ=180^{\circ}-98^{\circ}=82^{\circ}\). The angles of \(\triangle PQR\) sum to \(180^{\circ}\): \[\angle QPR=180^{\circ}-49^{\circ}-82^{\circ}=49^{\circ}.\]
(b) Circumference \(=2\pi r=400\): \[r=\frac{400}{2\times\frac{22}{7}}=\frac{400\times7}{44}=\frac{2800}{44}=63.6\text{ m}\approx64\text{ m}.\]
Question 5 Report
Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the :
(a) distance between Q and R ;
(b) baering of R from Q.
Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is
\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]
(a) Distance QR (cosine rule).
\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]
\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]
\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]
(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:
\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]
\[\angle PQR = 36.0^\circ.\]
The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is
\[255^\circ + 36^\circ = 291^\circ.\]
Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).
Answer Details
Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is
\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]
(a) Distance QR (cosine rule).
\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]
\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]
\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]
(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:
\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]
\[\angle PQR = 36.0^\circ.\]
The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is
\[255^\circ + 36^\circ = 291^\circ.\]
Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).
Question 6 Report
(a) If \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\), find x.
(b) Without using Mathematical Tables, evaluate: \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
(a) Write everything to base \(3\): \(9=3^{2}\), \(81=3^{4}\). \[9^{2x-1}=3^{4x-2},\qquad\frac{81^{x-2}}{3^{x}}=\frac{3^{4x-8}}{3^{x}}=3^{3x-8}.\] So \(4x-2=3x-8\Rightarrow x=-6\).
(b) \[\sqrt{\frac{0.81\times10^{-5}}{2.25\times10^{7}}}=\sqrt{\frac{0.81}{2.25}\times10^{-12}}=\sqrt{0.36\times10^{-12}}=0.6\times10^{-6}=6.0\times10^{-7}.\]
Answer Details
(a) Write everything to base \(3\): \(9=3^{2}\), \(81=3^{4}\). \[9^{2x-1}=3^{4x-2},\qquad\frac{81^{x-2}}{3^{x}}=\frac{3^{4x-8}}{3^{x}}=3^{3x-8}.\] So \(4x-2=3x-8\Rightarrow x=-6\).
(b) \[\sqrt{\frac{0.81\times10^{-5}}{2.25\times10^{7}}}=\sqrt{\frac{0.81}{2.25}\times10^{-12}}=\sqrt{0.36\times10^{-12}}=0.6\times10^{-6}=6.0\times10^{-7}.\]
Question 7 Report
(a) Using a ruler and a pair of compasses only, 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 triangle ABC .
(b) Label the point P where \(l_{1}\) and \(l_{2}\) intersect.
(c) Measure |CP|.
(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):
Steps.
- Draw a base line and mark \(|AB| = 5\text{ cm}\).
- Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
- With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
- Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
- Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).
(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.
(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).
This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence
\[|CP| = \sqrt{(2.5-(-3.75))^{2} + (4.33-6.50)^{2}} = \sqrt{6.25^{2} + 2.17^{2}} = \sqrt{43.75} \approx 6.7\text{ cm}.\]Answer Details
(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):
Steps.
- Draw a base line and mark \(|AB| = 5\text{ cm}\).
- Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
- With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
- Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
- Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).
(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.
(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).
This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence
\[|CP| = \sqrt{(2.5-(-3.75))^{2} + (4.33-6.50)^{2}} = \sqrt{6.25^{2} + 2.17^{2}} = \sqrt{43.75} \approx 6.7\text{ cm}.\]Question 8 Report
(a) If \(17x = 375^{2} - 356^{2}\), find the exact value of x.
(b) If \(4^{x} = 2^{\frac{1}{2}} \times 8\), find x.
(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.
(a) Using difference of two squares: \[375^{2}-356^{2}=(375-356)(375+356)=19\times731=13889.\] So \(17x=13889\Rightarrow x=\dfrac{13889}{17}=817\).
(b) \(4^{x}=2^{2x}\) and \(2^{\frac12}\times8=2^{\frac12}\times2^{3}=2^{\frac72}\). Thus \(2x=\tfrac72\Rightarrow x=\tfrac74\).
(c) \(S_{9}=72\Rightarrow\tfrac92(2a+8d)=72\Rightarrow a+4d=8\). Sum of next \(4\) terms \(=S_{13}-S_{9}=71\Rightarrow S_{13}=143\Rightarrow\tfrac{13}{2}(2a+12d)=143\Rightarrow a+6d=11\). Subtracting: \(2d=3\Rightarrow d=1.5\), then \(a=2\). The AP is \[2,\;3\tfrac12,\;5,\;6\tfrac12,\;8,\;\dots\]
Answer Details
(a) Using difference of two squares: \[375^{2}-356^{2}=(375-356)(375+356)=19\times731=13889.\] So \(17x=13889\Rightarrow x=\dfrac{13889}{17}=817\).
(b) \(4^{x}=2^{2x}\) and \(2^{\frac12}\times8=2^{\frac12}\times2^{3}=2^{\frac72}\). Thus \(2x=\tfrac72\Rightarrow x=\tfrac74\).
(c) \(S_{9}=72\Rightarrow\tfrac92(2a+8d)=72\Rightarrow a+4d=8\). Sum of next \(4\) terms \(=S_{13}-S_{9}=71\Rightarrow S_{13}=143\Rightarrow\tfrac{13}{2}(2a+12d)=143\Rightarrow a+6d=11\). Subtracting: \(2d=3\Rightarrow d=1.5\), then \(a=2\). The AP is \[2,\;3\tfrac12,\;5,\;6\tfrac12,\;8,\;\dots\]
Question 9 Report
(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.
(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.
| Marks | 4 | 5 | 6 | 7 | 8 | 9 |
| No of candidates | 5 | 8 | 5 | 6 | 4 | 2 |
Find the mean score to the nearest whole number.
(a) Pie chart of junior workers
First find the total number of workers:
\[ 78+36+44+52+30 = 240 \]The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):
\[ \text{Sector angle} = \frac{\text{Number in category}}{240}\times 360^\circ \]| Category | Number | Sector angle |
|---|---|---|
| Clerks | 78 | \(\frac{78}{240}\times360^\circ=117^\circ\) |
| Drivers | 36 | \(\frac{36}{240}\times360^\circ=54^\circ\) |
| Typists | 44 | \(\frac{44}{240}\times360^\circ=66^\circ\) |
| Messengers | 52 | \(\frac{52}{240}\times360^\circ=78^\circ\) |
| Others | 30 | \(\frac{30}{240}\times360^\circ=45^\circ\) |
| Total | 240 | \(360^\circ\) |
The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:
(b) Mean score of the aptitude test
Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):
| Marks \((x)\) | 4 | 5 | 6 | 7 | 8 | 9 | Total |
|---|---|---|---|---|---|---|---|
| No. of candidates \((f)\) | 5 | 8 | 5 | 6 | 4 | 2 | 30 |
| \(fx\) | 20 | 40 | 30 | 42 | 32 | 18 | 182 |
To the nearest whole number, the mean score is 6 marks.
Answer Details
(a) Pie chart of junior workers
First find the total number of workers:
\[ 78+36+44+52+30 = 240 \]The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):
\[ \text{Sector angle} = \frac{\text{Number in category}}{240}\times 360^\circ \]| Category | Number | Sector angle |
|---|---|---|
| Clerks | 78 | \(\frac{78}{240}\times360^\circ=117^\circ\) |
| Drivers | 36 | \(\frac{36}{240}\times360^\circ=54^\circ\) |
| Typists | 44 | \(\frac{44}{240}\times360^\circ=66^\circ\) |
| Messengers | 52 | \(\frac{52}{240}\times360^\circ=78^\circ\) |
| Others | 30 | \(\frac{30}{240}\times360^\circ=45^\circ\) |
| Total | 240 | \(360^\circ\) |
The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:
(b) Mean score of the aptitude test
Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):
| Marks \((x)\) | 4 | 5 | 6 | 7 | 8 | 9 | Total |
|---|---|---|---|---|---|---|---|
| No. of candidates \((f)\) | 5 | 8 | 5 | 6 | 4 | 2 | 30 |
| \(fx\) | 20 | 40 | 30 | 42 | 32 | 18 | 182 |
To the nearest whole number, the mean score is 6 marks.
Question 10 Report
(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?
(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:
(i) a red, then a black marble ; (ii) two black marbles.
(a) \(P(\text{rolling a }3)=\tfrac16\). For two coins, \(P(\text{at least one tail})=1-P(HH)=1-\tfrac14=\tfrac34\). These are independent, so \[P(3\text{ and a tail})=\frac16\times\frac34=\frac{3}{24}=\frac18.\]
(b) \(10\) marbles: \(7\) black, \(3\) red, drawn without replacement.
- (i) Red then black: \[\frac{3}{10}\times\frac{7}{9}=\frac{21}{90}=\frac{7}{30}.\]
- (ii) Two black: \[\frac{7}{10}\times\frac{6}{9}=\frac{42}{90}=\frac{7}{15}.\]
Answer Details
(a) \(P(\text{rolling a }3)=\tfrac16\). For two coins, \(P(\text{at least one tail})=1-P(HH)=1-\tfrac14=\tfrac34\). These are independent, so \[P(3\text{ and a tail})=\frac16\times\frac34=\frac{3}{24}=\frac18.\]
(b) \(10\) marbles: \(7\) black, \(3\) red, drawn without replacement.
- (i) Red then black: \[\frac{3}{10}\times\frac{7}{9}=\frac{21}{90}=\frac{7}{30}.\]
- (ii) Two black: \[\frac{7}{10}\times\frac{6}{9}=\frac{42}{90}=\frac{7}{15}.\]
Question 11 Report
(a) Copy and complete the following table for the relation \(y = \frac{5}{2} + x - 4x^{2}\)
| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | 1 | 2.5 |
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \(-2.0 \leq x \leq 2.0\).
(c) What is the maximum value of y?
(d) From your graph, obtain the roots of the equation \(8x^{2} - 2x - 5 = 0\)
(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).
Each missing value is found by substitution, e.g.
\[ x=-1.5:\; y=2.5+(-1.5)-4(-1.5)^2 = 2.5-1.5-9 = -8.0 \] \[ x=1:\; y=2.5+1-4(1)^2 = 2.5+1-4 = -0.5 \] \[ x=1.5:\; y=2.5+1.5-4(1.5)^2 = 2.5+1.5-9 = -5.0 \] \[ x=2.0:\; y=2.5+2-4(2)^2 = 2.5+2-16 = -11.5 \]| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | -8.0 | -2.5 | 1.0 | 2.5 | 2.0 | -0.5 | -5.0 | -11.5 |
(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).
(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:
\[ y_{\max}=2.5+0.125-4(0.125)^2 = 2.5+0.125-0.0625 \approx 2.6 \]So the maximum value of \(y\) is approximately 2.6.
(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):
\[ -4x^{2}+x+\tfrac{5}{2}=0 \quad\Longrightarrow\quad \tfrac{5}{2}+x-4x^{2}=0 \]This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:
\[ x \approx -0.7 \qquad\text{and}\qquad x \approx 0.9 \](Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)
Answer Details
(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).
Each missing value is found by substitution, e.g.
\[ x=-1.5:\; y=2.5+(-1.5)-4(-1.5)^2 = 2.5-1.5-9 = -8.0 \] \[ x=1:\; y=2.5+1-4(1)^2 = 2.5+1-4 = -0.5 \] \[ x=1.5:\; y=2.5+1.5-4(1.5)^2 = 2.5+1.5-9 = -5.0 \] \[ x=2.0:\; y=2.5+2-4(2)^2 = 2.5+2-16 = -11.5 \]| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | -8.0 | -2.5 | 1.0 | 2.5 | 2.0 | -0.5 | -5.0 | -11.5 |
(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).
(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:
\[ y_{\max}=2.5+0.125-4(0.125)^2 = 2.5+0.125-0.0625 \approx 2.6 \]So the maximum value of \(y\) is approximately 2.6.
(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):
\[ -4x^{2}+x+\tfrac{5}{2}=0 \quad\Longrightarrow\quad \tfrac{5}{2}+x-4x^{2}=0 \]This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:
\[ x \approx -0.7 \qquad\text{and}\qquad x \approx 0.9 \](Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)
Question 12 Report
(a) 
Calculate the area of the shaded segment of the circle shown in the diagram [Take \(\pi = \frac{22}{7}\)]
(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.
[Take \(\pi = \frac{22}{7}\)]
(a) Area of the shaded segment
The shaded segment is found by subtracting the area of the triangle formed by the two radii from the area of the sector:
\[\text{area of segment}=\text{area of sector}-\text{area of triangle}.\]
The angle at the centre is \(63^\circ\), not \(30^\circ\). Both sides of the triangle are radii of length \(10\text{ cm}\).
\[\text{Area of sector}=\frac{63}{360}\times\frac{22}{7}\times10^2=55\text{ cm}^2.\]
For the triangle, use \(\frac12 ab\sin C\):
\[\text{Area of triangle}=\frac12\times10\times10\times\sin63^\circ\]
\[=50\times0.891=44.55\text{ cm}^2.\]
Therefore,
\[\text{Area of shaded segment}=55-44.55=10.45\text{ cm}^2.\]
(b) Tin with radius \(3\text{ cm}\) and height \(6\text{ cm}\)
Total surface area includes the curved surface and both circular ends, so:
\[\text{TSA}=2\pi r(r+h)\]
\[=2\times\frac{22}{7}\times3\times(3+6)\]
\[=\frac{1188}{7}=169.71\text{ cm}^2.\]
Volume of the cylindrical tin is:
\[V=\pi r^2h\]
\[=\frac{22}{7}\times3^2\times6=\frac{1188}{7}=169.71\text{ cm}^3.\]
Since \(1000\text{ cm}^3=1\) litre:
\[169.71\div1000=0.16971\text{ litres}\]
Correct to two decimal places, the capacity is \(0.17\) litres.
Answers: shaded segment \(=10.45\text{ cm}^2\); total surface area \(=169.71\text{ cm}^2\); capacity \(=0.17\) litres.
Examination reminder: For a circular segment, use sector minus triangle, and take the central angle directly from the diagram. Here, using \(30^\circ\) instead of \(63^\circ\) changes both areas and gives the wrong segment area.
Answer Details
(a) Area of the shaded segment
The shaded segment is found by subtracting the area of the triangle formed by the two radii from the area of the sector:
\[\text{area of segment}=\text{area of sector}-\text{area of triangle}.\]
The angle at the centre is \(63^\circ\), not \(30^\circ\). Both sides of the triangle are radii of length \(10\text{ cm}\).
\[\text{Area of sector}=\frac{63}{360}\times\frac{22}{7}\times10^2=55\text{ cm}^2.\]
For the triangle, use \(\frac12 ab\sin C\):
\[\text{Area of triangle}=\frac12\times10\times10\times\sin63^\circ\]
\[=50\times0.891=44.55\text{ cm}^2.\]
Therefore,
\[\text{Area of shaded segment}=55-44.55=10.45\text{ cm}^2.\]
(b) Tin with radius \(3\text{ cm}\) and height \(6\text{ cm}\)
Total surface area includes the curved surface and both circular ends, so:
\[\text{TSA}=2\pi r(r+h)\]
\[=2\times\frac{22}{7}\times3\times(3+6)\]
\[=\frac{1188}{7}=169.71\text{ cm}^2.\]
Volume of the cylindrical tin is:
\[V=\pi r^2h\]
\[=\frac{22}{7}\times3^2\times6=\frac{1188}{7}=169.71\text{ cm}^3.\]
Since \(1000\text{ cm}^3=1\) litre:
\[169.71\div1000=0.16971\text{ litres}\]
Correct to two decimal places, the capacity is \(0.17\) litres.
Answers: shaded segment \(=10.45\text{ cm}^2\); total surface area \(=169.71\text{ cm}^2\); capacity \(=0.17\) litres.
Examination reminder: For a circular segment, use sector minus triangle, and take the central angle directly from the diagram. Here, using \(30^\circ\) instead of \(63^\circ\) changes both areas and gives the wrong segment area.
