WAEC SSCE - General Mathematics - 2004 (Essay)
Swali 1 Ripoti
The frequency distribution shows tha marks of 100 students in a Mathematics test.
| Marks | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| No. of Students |
2 | 4 | 9 | 13 | 18 | 32 | 13 | 5 | 3 | 1 |
(a) Draw cumulative frequency curve for the distribution .
(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.
(a) Less-than cumulative frequency table
| Marks | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 10.5 | 2 | 2 |
| 11–20 | 20.5 | 4 | 6 |
| 21–30 | 30.5 | 9 | 15 |
| 31–40 | 40.5 | 13 | 28 |
| 41–50 | 50.5 | 18 | 46 |
| 51–60 | 60.5 | 32 | 78 |
| 61–70 | 70.5 | 13 | 91 |
| 71–80 | 80.5 | 5 | 96 |
| 81–90 | 90.5 | 3 | 99 |
| 91–100 | 100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries, beginning with 0.5,0 . Draw a smooth increasing curve through the plotted points.
(b) Estimates from the ogive
The total frequency is 0.5+99.5=100 students.
- Median position = rac{100}{2}=50 . From cumulative frequency 50, the corresponding mark is approximately 52 marks.
- Lower quartile position = rac{100}{4}=25 . From cumulative frequency 25, the corresponding mark is approximately 39 marks.
- 60th percentile position = rac{60}{100} imes100=60 . From cumulative frequency 60, the corresponding mark is approximately 55 marks.
Maelezo ya Majibu
(a) Less-than cumulative frequency table
| Marks | Class boundary | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 10.5 | 2 | 2 |
| 11–20 | 20.5 | 4 | 6 |
| 21–30 | 30.5 | 9 | 15 |
| 31–40 | 40.5 | 13 | 28 |
| 41–50 | 50.5 | 18 | 46 |
| 51–60 | 60.5 | 32 | 78 |
| 61–70 | 70.5 | 13 | 91 |
| 71–80 | 80.5 | 5 | 96 |
| 81–90 | 90.5 | 3 | 99 |
| 91–100 | 100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries, beginning with 0.5,0 . Draw a smooth increasing curve through the plotted points.
(b) Estimates from the ogive
The total frequency is 0.5+99.5=100 students.
- Median position = rac{100}{2}=50 . From cumulative frequency 50, the corresponding mark is approximately 52 marks.
- Lower quartile position = rac{100}{4}=25 . From cumulative frequency 25, the corresponding mark is approximately 39 marks.
- 60th percentile position = rac{60}{100} imes100=60 . From cumulative frequency 60, the corresponding mark is approximately 55 marks.
Swali 2 Ripoti
(a) In the simultaneous equations : \(px + qy = 5 ; qx + py = -10\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.
(b) Solve : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
(a) Substitute \(x=1,\ y=-2\):
\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]
\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]
From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).
Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).
(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):
\[(4r-3)(3r+4)=(2r-1)(6r+1)\]
\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]
Maelezo ya Majibu
(a) Substitute \(x=1,\ y=-2\):
\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]
\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]
From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).
Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).
(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):
\[(4r-3)(3r+4)=(2r-1)(6r+1)\]
\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]
Swali 3 Ripoti
(a) Simplify : \(\sqrt{1001_{two}}\), leaving your answer in base two.
(b) 
In the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.
(c) 
In the diagram, P, Q, R and S are points of the circle centre O. \(\stackrel\frown{POQ} = 160°\), \(\stackrel\frown{QSR} = 45°\) and \(\stackrel\frown{PQS} = 40°\). Calculate, (i) < QPS ; (ii) < RQS.
(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.
First convert \(1001_{two}\) to base ten:
\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:
\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).
(b) Find \(z\) in terms of \(x\) and \(y\).
In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):
\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]Applying Pythagoras' theorem to right-angled triangle \(OKP\):
\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).
(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).
(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:
\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]In triangle \(PQS\), the three angles sum to \(180^\circ\):
\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\](ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.
- Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
- \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
- \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).
The four arcs around the circle sum to \(360^\circ\):
\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):
\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]Maelezo ya Majibu
(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.
First convert \(1001_{two}\) to base ten:
\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:
\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).
(b) Find \(z\) in terms of \(x\) and \(y\).
In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):
\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]Applying Pythagoras' theorem to right-angled triangle \(OKP\):
\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).
(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).
(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:
\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]In triangle \(PQS\), the three angles sum to \(180^\circ\):
\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\](ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.
- Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
- \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
- \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).
The four arcs around the circle sum to \(360^\circ\):
\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):
\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]Swali 4 Ripoti
The table shows the marks scored by a group of students in a class test.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 1 | 4 | 9 | 8 | 5 | 3 |
(a)(i) Calculate the mean mark ; (ii) Find the median.
(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?
(a) Prepare the frequency table:
| Mark, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | Total |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 4 | 9 | 8 | 5 | 3 | 30 |
| \(fx\) | 0 | 4 | 18 | 24 | 20 | 15 | 81 |
| Cumulative frequency | 1 | 5 | 14 | 22 | 27 | 30 |
(i) Mean mark
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{81}{30}=2.7\]
(ii) Median
There are \(30\) observations. The median lies between the 15th and 16th observations. From the cumulative frequencies, both the 15th and 16th observations have mark \(3\).
\[\text{Median}=3\]
(b) Pie chart
The sector angle for each mark is \(\dfrac{f}{30}\times360^\circ\). Thus, for mark \(2\):
\[\frac{9}{30}\times360^\circ=108^\circ\]
Therefore, the sectorial angle for mark \(2\) is \(108^\circ\).
The complete pie chart is shown below. Its sector angles are \(12^\circ,48^\circ,108^\circ,96^\circ,60^\circ\), and \(36^\circ\) for marks \(0,1,2,3,4\), and \(5\) respectively.
Maelezo ya Majibu
(a) Prepare the frequency table:
| Mark, \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | Total |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 4 | 9 | 8 | 5 | 3 | 30 |
| \(fx\) | 0 | 4 | 18 | 24 | 20 | 15 | 81 |
| Cumulative frequency | 1 | 5 | 14 | 22 | 27 | 30 |
(i) Mean mark
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{81}{30}=2.7\]
(ii) Median
There are \(30\) observations. The median lies between the 15th and 16th observations. From the cumulative frequencies, both the 15th and 16th observations have mark \(3\).
\[\text{Median}=3\]
(b) Pie chart
The sector angle for each mark is \(\dfrac{f}{30}\times360^\circ\). Thus, for mark \(2\):
\[\frac{9}{30}\times360^\circ=108^\circ\]
Therefore, the sectorial angle for mark \(2\) is \(108^\circ\).
The complete pie chart is shown below. Its sector angles are \(12^\circ,48^\circ,108^\circ,96^\circ,60^\circ\), and \(36^\circ\) for marks \(0,1,2,3,4\), and \(5\) respectively.
Swali 5 Ripoti
(a) 
In the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96° at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \(\pi = \frac{22}{7}\)).
(b) 
The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \(\pi = \frac{22}{7}\)].
(a) Area of the minor segment
The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).
Area of minor sector (with \(\pi = \tfrac{22}{7}\)):
\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]
Area of triangle OXY:
\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]
Area of minor segment:
\[20.95 - 12.43 = 8.52\text{ cm}^2\]
Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).
(b) Total area of the shaded portions
The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).
Shaded sector of the circle (angle \(80^\circ\)):
\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]
Shaded portion of the square (the four corners between the square and the circle):
\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]
\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]
\[196 - 154 = 42\text{ cm}^2\]
Total shaded area:
\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]
Maelezo ya Majibu
(a) Area of the minor segment
The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).
Area of minor sector (with \(\pi = \tfrac{22}{7}\)):
\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]
Area of triangle OXY:
\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]
Area of minor segment:
\[20.95 - 12.43 = 8.52\text{ cm}^2\]
Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).
(b) Total area of the shaded portions
The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).
Shaded sector of the circle (angle \(80^\circ\)):
\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]
Shaded portion of the square (the four corners between the square and the circle):
\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]
\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]
\[196 - 154 = 42\text{ cm}^2\]
Total shaded area:
\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]
Swali 6 Ripoti
(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \(\Delta\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.
(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?
(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):
\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]
By the cosine rule in \(\triangle AOB\):
\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]
\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]
(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:
\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]
Maelezo ya Majibu
(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):
\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]
By the cosine rule in \(\triangle AOB\):
\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]
\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]
(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:
\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]
Swali 7 Ripoti
(a) Simplify : \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.
(c) A man left N5,720 to be shared among his son and three daughters. Each daughter's share was \(\frac{3}{4}\) of the son's share. How much did the son receive?
(a) Factor the first bracket:
\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]
\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]
(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).
(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)
(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)
(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).
\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]
Maelezo ya Majibu
(a) Factor the first bracket:
\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]
\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]
(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).
(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)
(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)
(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).
\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]
Swali 8 Ripoti
(a) The sides of an isosceles triangle triangle are in the ratio \(7 : 5 : 7\). Calculate, correct to the nearest degree, the angle included between the equal sides.
(b) The sum of the interior angles of a regular polygon is 1440°. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.
(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:
\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]
\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]
(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):
\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]
(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)
Maelezo ya Majibu
(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:
\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]
\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]
(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):
\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]
(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)
Swali 9 Ripoti
(a) Copy and complete the following table of values for the relation \(y = 2x^{2} - 7x - 3\).
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 19 | -3 | -9 |
(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 7x - 3\) for \(-2 \leq x \leq 5\).
(c) From your graph, find the : (i) minimum value of y ;
(ii) gradient of the curve at x = 1.
(d) By drawing a suitable straight line, find the values of x for which \(2x^{2} - 7x - 5 = x + 4\).
(a) For \(y=2x^2-7x-3\), the completed table is:
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(19\) | \(6\) | \(-3\) | \(-8\) | \(-9\) | \(-6\) | \(1\) | \(12\) |
(b) The points were plotted using the stated scales, and joined with a smooth curve. The straight line \(y=x+6\) and the tangent at \(x=1\) are included for parts (c) and (d).
(c)
(i) The lowest point of the curve gives the minimum value
\[\boxed{y\approx -9}\]
(ii) The tangent at \(x=1\) passes through \((1,-8)\). Using two points on the tangent, for example \((0,-5)\) and \((1,-8)\),
\[\text{gradient}=\frac{-8-(-5)}{1-0}=\boxed{-3}.\]
(d) Rearrange the equation in terms of the curve already drawn:
\[2x^2-7x-5=x+4\]
\[2x^2-7x-3=x+6.\]
Thus, draw the straight line \(y=x+6\). From its points of intersection with the parabola on the graph,
\[\boxed{x\approx -0.9\ \text{ or }\ x\approx 4.9}.\]
Maelezo ya Majibu
(a) For \(y=2x^2-7x-3\), the completed table is:
| \(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(19\) | \(6\) | \(-3\) | \(-8\) | \(-9\) | \(-6\) | \(1\) | \(12\) |
(b) The points were plotted using the stated scales, and joined with a smooth curve. The straight line \(y=x+6\) and the tangent at \(x=1\) are included for parts (c) and (d).
(c)
(i) The lowest point of the curve gives the minimum value
\[\boxed{y\approx -9}\]
(ii) The tangent at \(x=1\) passes through \((1,-8)\). Using two points on the tangent, for example \((0,-5)\) and \((1,-8)\),
\[\text{gradient}=\frac{-8-(-5)}{1-0}=\boxed{-3}.\]
(d) Rearrange the equation in terms of the curve already drawn:
\[2x^2-7x-5=x+4\]
\[2x^2-7x-3=x+6.\]
Thus, draw the straight line \(y=x+6\). From its points of intersection with the parabola on the graph,
\[\boxed{x\approx -0.9\ \text{ or }\ x\approx 4.9}.\]
Swali 10 Ripoti
K(lat. 60°N, long. 50°W) is a point on the eart's surface. L is another point due East of K and the third point N is due North of K. The distance KL is 3520km and KN is 10951km.
(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \(\pi = \frac{22}{7}\) and the radius of the earth = 6400km).
(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.
\(R=6400\) km, \(\pi=\tfrac{22}{7}\). \(K(60^{\circ}\N,50^{\circ}\W)\).
(a)(i) Longitude of L (due east of \(K\), along latitude \(60^{\circ}\N\)). Radius of that parallel uses \(\cos60^{\circ}=\tfrac12\):
\[KL=\frac{\theta}{360}\times2\pi R\cos60^{\circ}\Rightarrow 3520=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\frac12.\]
Full parallel \(=\dfrac{44}{7}\times3200=20114.3\) km, so \(\theta=\dfrac{3520\times360}{20114.3}=63^{\circ}\) (eastward).
Starting at \(50^{\circ}\W\) and moving \(63^{\circ}\) east: \(-50^{\circ}+63^{\circ}=+13^{\circ}\). Longitude of \(L=13^{\circ}\E\).
(a)(ii) Latitude of N (due north of \(K\), along a meridian). Full meridian circle \(=2\pi R=\dfrac{44}{7}\times6400=40228.6\) km.
\[KN=\frac{\phi}{360}\times40228.6=10951\Rightarrow\phi=\frac{10951\times360}{40228.6}=98^{\circ}.\]
From \(60^{\circ}\N\), going \(98^{\circ}\) north first reaches the North Pole after \(30^{\circ}\), then continues \(68^{\circ}\) down the opposite meridian. Latitude of \(N=90^{\circ}-68^{\circ}=22^{\circ}\N\) (on the \(130^{\circ}\E\) meridian).
(b) Let total income be \(I\). Tax-free \(=20\%\), so taxable \(=0.8I\); tax \(=25\) kobo per naira \(=25\%\) of taxable:
\[0.25\times0.8I=1200\Rightarrow 0.2I=1200\Rightarrow I=\mathbf{N6000}.\]
Maelezo ya Majibu
\(R=6400\) km, \(\pi=\tfrac{22}{7}\). \(K(60^{\circ}\N,50^{\circ}\W)\).
(a)(i) Longitude of L (due east of \(K\), along latitude \(60^{\circ}\N\)). Radius of that parallel uses \(\cos60^{\circ}=\tfrac12\):
\[KL=\frac{\theta}{360}\times2\pi R\cos60^{\circ}\Rightarrow 3520=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\frac12.\]
Full parallel \(=\dfrac{44}{7}\times3200=20114.3\) km, so \(\theta=\dfrac{3520\times360}{20114.3}=63^{\circ}\) (eastward).
Starting at \(50^{\circ}\W\) and moving \(63^{\circ}\) east: \(-50^{\circ}+63^{\circ}=+13^{\circ}\). Longitude of \(L=13^{\circ}\E\).
(a)(ii) Latitude of N (due north of \(K\), along a meridian). Full meridian circle \(=2\pi R=\dfrac{44}{7}\times6400=40228.6\) km.
\[KN=\frac{\phi}{360}\times40228.6=10951\Rightarrow\phi=\frac{10951\times360}{40228.6}=98^{\circ}.\]
From \(60^{\circ}\N\), going \(98^{\circ}\) north first reaches the North Pole after \(30^{\circ}\), then continues \(68^{\circ}\) down the opposite meridian. Latitude of \(N=90^{\circ}-68^{\circ}=22^{\circ}\N\) (on the \(130^{\circ}\E\) meridian).
(b) Let total income be \(I\). Tax-free \(=20\%\), so taxable \(=0.8I\); tax \(=25\) kobo per naira \(=25\%\) of taxable:
\[0.25\times0.8I=1200\Rightarrow 0.2I=1200\Rightarrow I=\mathbf{N6000}.\]
Swali 11 Ripoti
(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.
(b) 
In the diagram, PR is a chord of the circle centre O and radius 30cm, < POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \(\pi = 3.142\)).
(a) Height of the building.
Let \(d\) be the horizontal distance between the tower and the building. The observer is at the top of the tower, height \(72\text{ m}\).
Bottom of the building (same level as the foot of the tower) has an angle of depression of \(62^{\circ}\), so the vertical drop is the full \(72\text{ m}\):
\[\tan 62^{\circ}=\frac{72}{d}\Rightarrow d=\frac{72}{\tan 62^{\circ}}=\frac{72}{1.8807}=38.28\text{ m}\]
Top of the building has an angle of depression of \(51^{\circ}\), so the drop from the tower top down to the building top is:
\[\text{drop}=d\tan 51^{\circ}=38.28\times 1.2349=47.27\text{ m}\]
The height of the building is the tower height minus this drop:
\[h=72-47.27=24.73\text{ m}\]
Correct to 2 significant figures, the height of the building is \(25\text{ m}\).
(b) Circle, centre O, radius 30 cm, \(\angle POR=120^{\circ}\), \(\pi=3.142\).
(i) Length of chord PR.
Using the cosine rule in triangle \(POR\) with \(|OP|=|OR|=30\text{ cm}\):
\[|PR|^{2}=30^{2}+30^{2}-2(30)(30)\cos 120^{\circ}\]
\[=900+900-1800(-0.5)=1800+900=2700\]
\[|PR|=\sqrt{2700}=51.96\text{ cm}\approx 52.0\text{ cm}\]
(ii) Length of arc PQR.
From the diagram, \(Q\) lies on the major arc, so arc \(PQR\) uses the reflex angle:
\[\text{Reflex }\angle POR=360^{\circ}-120^{\circ}=240^{\circ}\]
\[\text{Arc }PQR=\frac{240}{360}\times 2\pi r=\frac{240}{360}\times 2\times 3.142\times 30\]
\[=\frac{240}{360}\times 188.52=0.6667\times 188.52=125.7\text{ cm}\]
(iii) Perimeter of the shaded portion.
The shaded region is the major segment, bounded by the chord \(PR\) and the major arc \(PQR\):
\[\text{Perimeter}=|PR|+\text{arc }PQR=51.96+125.7=177.7\text{ cm}\]
Correct to 3 significant figures, the perimeter of the shaded portion is \(178\text{ cm}\).
Maelezo ya Majibu
(a) Height of the building.
Let \(d\) be the horizontal distance between the tower and the building. The observer is at the top of the tower, height \(72\text{ m}\).
Bottom of the building (same level as the foot of the tower) has an angle of depression of \(62^{\circ}\), so the vertical drop is the full \(72\text{ m}\):
\[\tan 62^{\circ}=\frac{72}{d}\Rightarrow d=\frac{72}{\tan 62^{\circ}}=\frac{72}{1.8807}=38.28\text{ m}\]
Top of the building has an angle of depression of \(51^{\circ}\), so the drop from the tower top down to the building top is:
\[\text{drop}=d\tan 51^{\circ}=38.28\times 1.2349=47.27\text{ m}\]
The height of the building is the tower height minus this drop:
\[h=72-47.27=24.73\text{ m}\]
Correct to 2 significant figures, the height of the building is \(25\text{ m}\).
(b) Circle, centre O, radius 30 cm, \(\angle POR=120^{\circ}\), \(\pi=3.142\).
(i) Length of chord PR.
Using the cosine rule in triangle \(POR\) with \(|OP|=|OR|=30\text{ cm}\):
\[|PR|^{2}=30^{2}+30^{2}-2(30)(30)\cos 120^{\circ}\]
\[=900+900-1800(-0.5)=1800+900=2700\]
\[|PR|=\sqrt{2700}=51.96\text{ cm}\approx 52.0\text{ cm}\]
(ii) Length of arc PQR.
From the diagram, \(Q\) lies on the major arc, so arc \(PQR\) uses the reflex angle:
\[\text{Reflex }\angle POR=360^{\circ}-120^{\circ}=240^{\circ}\]
\[\text{Arc }PQR=\frac{240}{360}\times 2\pi r=\frac{240}{360}\times 2\times 3.142\times 30\]
\[=\frac{240}{360}\times 188.52=0.6667\times 188.52=125.7\text{ cm}\]
(iii) Perimeter of the shaded portion.
The shaded region is the major segment, bounded by the chord \(PR\) and the major arc \(PQR\):
\[\text{Perimeter}=|PR|+\text{arc }PQR=51.96+125.7=177.7\text{ cm}\]
Correct to 3 significant figures, the perimeter of the shaded portion is \(178\text{ cm}\).
Swali 12 Ripoti
Using a ruler and a pair of compasses only,
(a) Construct : (i) \(\Delta PQR\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105°. (ii) locus \(L_{1}\) of points equidistant from P and Q. (iii) locus \(l_{2}\) of points equidistant Q and R.
(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \(l_{3}\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.
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Construction of Triangle PQR and the Required Loci
Given: Construct a triangle PQR such that
- PQ = 8 cm,
- PR = 7 cm,
- ∠QPR = 105°.
(a)(i) Construction of ΔPQR
- Using a ruler, draw a line segment PQ = 8 cm.
- At point P, construct an angle of 105°:
- Construct a 90° angle at P by drawing a perpendicular to PQ.
- Construct a 120° ray at P using equal arcs with the compasses.
- Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
- On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
- Join R to Q. This gives the required triangle ΔPQR.
(a)(ii) Locus L1 of points equidistant from P and Q
The locus of points equidistant from P and Q is the perpendicular bisector of PQ.
- With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
- Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
- Join the two arc-intersection points with a ruler.
This straight line is labelled L1.
(a)(iii) Locus l2 of points equidistant from Q and R
The locus of points equidistant from Q and R is the perpendicular bisector of QR.
- Draw equal arcs centred at Q and R, using a radius greater than half of QR.
- Join the two points where the arcs intersect.
This straight line is labelled l2.
(b)(i) Point of Intersection
Label the point where L1 and l2 intersect as T.
Since T lies on the perpendicular bisector of QR, TQ = TR.
Therefore, TP = TQ = TR.
Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.
```Maelezo ya Majibu
```html
Construction of Triangle PQR and the Required Loci
Given: Construct a triangle PQR such that
- PQ = 8 cm,
- PR = 7 cm,
- ∠QPR = 105°.
(a)(i) Construction of ΔPQR
- Using a ruler, draw a line segment PQ = 8 cm.
- At point P, construct an angle of 105°:
- Construct a 90° angle at P by drawing a perpendicular to PQ.
- Construct a 120° ray at P using equal arcs with the compasses.
- Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
- On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
- Join R to Q. This gives the required triangle ΔPQR.
(a)(ii) Locus L1 of points equidistant from P and Q
The locus of points equidistant from P and Q is the perpendicular bisector of PQ.
- With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
- Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
- Join the two arc-intersection points with a ruler.
This straight line is labelled L1.
(a)(iii) Locus l2 of points equidistant from Q and R
The locus of points equidistant from Q and R is the perpendicular bisector of QR.
- Draw equal arcs centred at Q and R, using a radius greater than half of QR.
- Join the two points where the arcs intersect.
This straight line is labelled l2.
(b)(i) Point of Intersection
Label the point where L1 and l2 intersect as T.
Since T lies on the perpendicular bisector of QR, TQ = TR.
Therefore, TP = TQ = TR.
Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.
```Swali 13 Ripoti
(a) Solve \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\)
(b) Simplify \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\), leaving your answer in surd form.
(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).
\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]
Equating indices:
\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]
(b) Rationalise each term:
\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]
\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]
Maelezo ya Majibu
(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).
\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]
Equating indices:
\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]
(b) Rationalise each term:
\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]
\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]
