WAEC SSCE - General Mathematics - 2013 (Essay)
Ibeere 1 Ìròyìn
(a) 
A segment of a circle is cut off from a rectangular board as shown in the diagram. If the radius of the circle is \(1\frac{1}{2}\) times the length of the chord; calculate, correct to 2 decimal places, the perimeter of the remaining portion. [Take \(\pi = \frac{22}{7}\)]
(b) Evaluate without using calculators or tables, \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\).
(a) Perimeter of the remaining board.
Reading the diagram. The board is a rectangle \(22\text{ cm}\) wide and \(12\text{ cm}\) tall. A circular segment (curved arch) is cut from the bottom edge, leaving flat pieces of \(5\text{ cm}\) on the left and \(3\text{ cm}\) on the right. Hence the chord (span of the cut) is
\[\text{chord} = 22 - 5 - 3 = 14\text{ cm}.\]The radius is \(1\tfrac{1}{2}\) times the chord:
\[r = \tfrac{3}{2}\times 14 = 21\text{ cm}.\]Central angle of the segment. Half the chord is \(7\text{ cm}\). If the chord subtends \(2\theta\) at the centre,
\[\sin\theta = \frac{7}{21} = \frac{1}{3} \;\Rightarrow\; \theta = 19.47^\circ,\qquad 2\theta = 38.94^\circ.\]Arc length of the cut. With \(\pi = \tfrac{22}{7}\):
\[\text{arc} = \frac{2\theta}{360}\times 2\pi r = \frac{38.94}{360}\times 2\times\frac{22}{7}\times 21 = \frac{38.94}{360}\times 132 = 14.28\text{ cm}.\]Perimeter of the remaining portion = top + two sides + two flat bottoms + arc:
\[P = 22 + 12 + 12 + 5 + 3 + 14.28 = 68.28\text{ cm}.\]Perimeter \(\approx \mathbf{68.28\text{ cm}}\) (2 d.p.).
(b) Surd evaluation.
\[\frac{3}{\sqrt{3}}\left(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\right).\]First, \(\dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{3\sqrt3}{3} = \sqrt{3}.\) Inside the bracket, \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}\) and \(\sqrt{12} = 2\sqrt3\), so \(\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt3}{6} = \dfrac{\sqrt3}{3}.\) Therefore
\[\frac{2}{\sqrt3} - \frac{\sqrt{12}}{6} = \frac{2\sqrt3}{3} - \frac{\sqrt3}{3} = \frac{\sqrt3}{3}.\]Multiplying:
\[\sqrt{3}\times\frac{\sqrt3}{3} = \frac{3}{3} = \mathbf{1}.\]Awọn alaye Idahun
(a) Perimeter of the remaining board.
Reading the diagram. The board is a rectangle \(22\text{ cm}\) wide and \(12\text{ cm}\) tall. A circular segment (curved arch) is cut from the bottom edge, leaving flat pieces of \(5\text{ cm}\) on the left and \(3\text{ cm}\) on the right. Hence the chord (span of the cut) is
\[\text{chord} = 22 - 5 - 3 = 14\text{ cm}.\]The radius is \(1\tfrac{1}{2}\) times the chord:
\[r = \tfrac{3}{2}\times 14 = 21\text{ cm}.\]Central angle of the segment. Half the chord is \(7\text{ cm}\). If the chord subtends \(2\theta\) at the centre,
\[\sin\theta = \frac{7}{21} = \frac{1}{3} \;\Rightarrow\; \theta = 19.47^\circ,\qquad 2\theta = 38.94^\circ.\]Arc length of the cut. With \(\pi = \tfrac{22}{7}\):
\[\text{arc} = \frac{2\theta}{360}\times 2\pi r = \frac{38.94}{360}\times 2\times\frac{22}{7}\times 21 = \frac{38.94}{360}\times 132 = 14.28\text{ cm}.\]Perimeter of the remaining portion = top + two sides + two flat bottoms + arc:
\[P = 22 + 12 + 12 + 5 + 3 + 14.28 = 68.28\text{ cm}.\]Perimeter \(\approx \mathbf{68.28\text{ cm}}\) (2 d.p.).
(b) Surd evaluation.
\[\frac{3}{\sqrt{3}}\left(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\right).\]First, \(\dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{3\sqrt3}{3} = \sqrt{3}.\) Inside the bracket, \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}\) and \(\sqrt{12} = 2\sqrt3\), so \(\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt3}{6} = \dfrac{\sqrt3}{3}.\) Therefore
\[\frac{2}{\sqrt3} - \frac{\sqrt{12}}{6} = \frac{2\sqrt3}{3} - \frac{\sqrt3}{3} = \frac{\sqrt3}{3}.\]Multiplying:
\[\sqrt{3}\times\frac{\sqrt3}{3} = \frac{3}{3} = \mathbf{1}.\]Ibeere 2 Ìròyìn
When one end of a ladder, LM, is placed against a vertical wall at a point 5 metres above the ground, the ladder makes an angle of 37° with the horizontal ground.
(a) Represent this information in a diagram ;
(b) Calculate, correct to 3 significant figures, the length of the ladder ;
(c) If the foot of the ladder is pushed towards the wall by 2 metres, calculate,correct to the nearest degree, the angle which the ladder nows makes with the ground.
(a) Diagram. A vertical wall meets the horizontal ground at a right angle. The ladder \(LM\) leans with its top \(L\) touching the wall \(5\) m above the ground and its foot \(M\) on the ground, making \(37^\circ\) with the ground. The height (5 m), the ground distance, and the ladder form a right-angled triangle.
(b) Length of the ladder. The 5 m height is opposite the \(37^\circ\) angle:
\[\sin 37^\circ = \frac{5}{|LM|} \;\Rightarrow\; |LM| = \frac{5}{\sin 37^\circ} = \frac{5}{0.6018} = 8.31\text{ m (3 s.f.)}\]
(c) New angle after moving the foot 2 m towards the wall. Original horizontal distance of the foot from the wall:
\[|LM|\cos 37^\circ = 8.308 \times 0.7986 = 6.635\text{ m}\]
Pushing the foot 2 m nearer gives a new base distance \(6.635 - 2 = 4.635\) m, with the ladder length unchanged (8.308 m). If \(\alpha\) is the new angle with the ground:
\[\cos\alpha = \frac{4.635}{8.308} = 0.5579 \;\Rightarrow\; \alpha = 56^\circ \text{ (nearest degree)}\]
Awọn alaye Idahun
(a) Diagram. A vertical wall meets the horizontal ground at a right angle. The ladder \(LM\) leans with its top \(L\) touching the wall \(5\) m above the ground and its foot \(M\) on the ground, making \(37^\circ\) with the ground. The height (5 m), the ground distance, and the ladder form a right-angled triangle.
(b) Length of the ladder. The 5 m height is opposite the \(37^\circ\) angle:
\[\sin 37^\circ = \frac{5}{|LM|} \;\Rightarrow\; |LM| = \frac{5}{\sin 37^\circ} = \frac{5}{0.6018} = 8.31\text{ m (3 s.f.)}\]
(c) New angle after moving the foot 2 m towards the wall. Original horizontal distance of the foot from the wall:
\[|LM|\cos 37^\circ = 8.308 \times 0.7986 = 6.635\text{ m}\]
Pushing the foot 2 m nearer gives a new base distance \(6.635 - 2 = 4.635\) m, with the ladder length unchanged (8.308 m). If \(\alpha\) is the new angle with the ground:
\[\cos\alpha = \frac{4.635}{8.308} = 0.5579 \;\Rightarrow\; \alpha = 56^\circ \text{ (nearest degree)}\]
Ibeere 3 Ìròyìn
(a) Copy and complete the table of values for the relation \(y = 3x^{2} - 5x - 7\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 35 | -7 | -9 | 5 |
(b) Using scales of 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = 3x^{2} - 5x - 7, -3 \leq x \leq 4\).
(c) From the graph : (i) find the roots of the equation \(3x^{2} - 5x - 7 = 0\) ; (ii) estimate the minimum value of y ; (iii) calculate the gradient of the curve at the point x = 2.
(a) For \(y=3x^{2}-5x-7\):
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 35 | 15 | 1 | -7 | -9 | -5 | 5 | 21 |
For example, \(y\) when \(x=-2\) is \(3(-2)^2-5(-2)-7=15\), while when \(x=2\), \(y=3(2)^2-5(2)-7=-5\).
(b) The plotted graph of \(y=3x^2-5x-7\), using the stated scales, is shown below.
(c)(i) The roots are the \(x\)-coordinates where the curve cuts the \(x\)-axis:
\[\boxed{x\approx-0.9\text{ and }x\approx2.6}\]
(c)(ii) The minimum ordinate, read at the turning point, is
\[\boxed{y\approx-9.1}\]
(c)(iii) Using two convenient points on the tangent at \(x=2\), approximately \((1.5,-8.5)\) and \((3.3,4.0)\),
\[\text{gradient}=\frac{4.0-(-8.5)}{3.3-1.5}=\frac{12.5}{1.8}=6.94\approx\boxed{6.9}.\]
Awọn alaye Idahun
(a) For \(y=3x^{2}-5x-7\):
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 35 | 15 | 1 | -7 | -9 | -5 | 5 | 21 |
For example, \(y\) when \(x=-2\) is \(3(-2)^2-5(-2)-7=15\), while when \(x=2\), \(y=3(2)^2-5(2)-7=-5\).
(b) The plotted graph of \(y=3x^2-5x-7\), using the stated scales, is shown below.
(c)(i) The roots are the \(x\)-coordinates where the curve cuts the \(x\)-axis:
\[\boxed{x\approx-0.9\text{ and }x\approx2.6}\]
(c)(ii) The minimum ordinate, read at the turning point, is
\[\boxed{y\approx-9.1}\]
(c)(iii) Using two convenient points on the tangent at \(x=2\), approximately \((1.5,-8.5)\) and \((3.3,4.0)\),
\[\text{gradient}=\frac{4.0-(-8.5)}{3.3-1.5}=\frac{12.5}{1.8}=6.94\approx\boxed{6.9}.\]
Ibeere 4 Ìròyìn
An aeroplane flies due North from a town T on the equator at a speed of 950km per hour for 4 hours to another town P. It then flies eastwards to town Q on longitude 65°E. If the longitude of T is 15°E,
(a) represent this information in a diagram ;
(b) calculate the : (i) latitude of P, correct to the nearest degree ; (ii) distance between P and Q, correct to four significant figures. [Take \(\pi = \frac{22}{7}\); Radius of the earth = 6400km].
(a) Draw a circle for the earth with the equator horizontal and the North pole on top. Mark \(T\) on the equator (longitude \(15^\circ E\)). The plane flies due North along the \(15^\circ E\) meridian to \(P\), then East along the parallel of latitude through \(P\) to \(Q\) on longitude \(65^\circ E\).
(b)(i) Latitude of P: Distance North \(= 950\times 4 = 3800\,\text{km}\), an arc along a meridian: \[3800 = \frac{\theta}{360}\times 2\pi R = \frac{\theta}{360}\times 2\times\frac{22}{7}\times 6400.\] \[2\pi R = 40228.6\,\text{km} \Rightarrow \theta = \frac{3800\times 360}{40228.6} = 34.0^\circ.\] Latitude of \(P \approx \mathbf{34^\circ N}.\)
(b)(ii) Distance P to Q: Along the parallel of latitude \(34^\circ N\), the difference in longitude is \(65^\circ - 15^\circ = 50^\circ\): \[PQ = \frac{50}{360}\times 2\pi R\cos34^\circ = \frac{50}{360}\times 40228.6\times 0.8290.\] \[PQ = 5587.3\times 0.8290 = 4632\,\text{km}.\] Correct to four significant figures, \(|PQ| \approx \mathbf{4632\,\text{km}}.\)
Awọn alaye Idahun
(a) Draw a circle for the earth with the equator horizontal and the North pole on top. Mark \(T\) on the equator (longitude \(15^\circ E\)). The plane flies due North along the \(15^\circ E\) meridian to \(P\), then East along the parallel of latitude through \(P\) to \(Q\) on longitude \(65^\circ E\).
(b)(i) Latitude of P: Distance North \(= 950\times 4 = 3800\,\text{km}\), an arc along a meridian: \[3800 = \frac{\theta}{360}\times 2\pi R = \frac{\theta}{360}\times 2\times\frac{22}{7}\times 6400.\] \[2\pi R = 40228.6\,\text{km} \Rightarrow \theta = \frac{3800\times 360}{40228.6} = 34.0^\circ.\] Latitude of \(P \approx \mathbf{34^\circ N}.\)
(b)(ii) Distance P to Q: Along the parallel of latitude \(34^\circ N\), the difference in longitude is \(65^\circ - 15^\circ = 50^\circ\): \[PQ = \frac{50}{360}\times 2\pi R\cos34^\circ = \frac{50}{360}\times 40228.6\times 0.8290.\] \[PQ = 5587.3\times 0.8290 = 4632\,\text{km}.\] Correct to four significant figures, \(|PQ| \approx \mathbf{4632\,\text{km}}.\)
Ibeere 5 Ìròyìn
(a) Two positive whole numbers p and q are such that p is greater than q and their sum is equal to three times their difference;
(i) Express p in terms of q ; (ii) Hence, evaluate \(\frac{p^{2} + q^{2}}{pq}\).
(b) A man sold 100 articles at 25 for N66.00 and made a gain of 32%. Calculate his gain or loss percent if he sold them at 20 for N50.00.
(a)(i) Sum equals three times the difference: \[p + q = 3(p - q) \Rightarrow p + q = 3p - 3q \Rightarrow 4q = 2p \Rightarrow \mathbf{p = 2q}.\]
(a)(ii) Substitute \(p = 2q\): \[\frac{p^2 + q^2}{pq} = \frac{(2q)^2 + q^2}{(2q)q} = \frac{4q^2 + q^2}{2q^2} = \frac{5q^2}{2q^2} = \frac{5}{2} = \mathbf{2.5}.\]
(b) Selling price of \(100\) articles at \(25\) for \(\text{N}66\): \(\dfrac{100}{25}\times 66 = 4\times 66 = \text{N}264.\)
This gives a \(32\%\) gain, so cost price \(= \dfrac{264}{1.32} = \text{N}200.\)
New selling price at \(20\) for \(\text{N}50\): \(\dfrac{100}{20}\times 50 = 5\times 50 = \text{N}250.\)
New gain \(= 250 - 200 = \text{N}50\), so \[\text{gain}\% = \frac{50}{200}\times 100 = \mathbf{25\%\ \text{gain}}.\]
Awọn alaye Idahun
(a)(i) Sum equals three times the difference: \[p + q = 3(p - q) \Rightarrow p + q = 3p - 3q \Rightarrow 4q = 2p \Rightarrow \mathbf{p = 2q}.\]
(a)(ii) Substitute \(p = 2q\): \[\frac{p^2 + q^2}{pq} = \frac{(2q)^2 + q^2}{(2q)q} = \frac{4q^2 + q^2}{2q^2} = \frac{5q^2}{2q^2} = \frac{5}{2} = \mathbf{2.5}.\]
(b) Selling price of \(100\) articles at \(25\) for \(\text{N}66\): \(\dfrac{100}{25}\times 66 = 4\times 66 = \text{N}264.\)
This gives a \(32\%\) gain, so cost price \(= \dfrac{264}{1.32} = \text{N}200.\)
New selling price at \(20\) for \(\text{N}50\): \(\dfrac{100}{20}\times 50 = 5\times 50 = \text{N}250.\)
New gain \(= 250 - 200 = \text{N}50\), so \[\text{gain}\% = \frac{50}{200}\times 100 = \mathbf{25\%\ \text{gain}}.\]
Ibeere 6 Ìròyìn
The frequency distribution table shows the marks obtained by 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 |
| Frequency | 2 | 3 | 5 | 13 | 19 | 31 | 13 | 9 | 4 | 1 |
(a) Draw the cumulative curve for the distribution.
(b) Use the graph to find the : (i) 60th percentile ; (ii) probability that a student passed the test if the pass mark was fixed at 35%.
(a) Cumulative frequency table
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 0.5–10.5 | 2 | 2 |
| 11–20 | 10.5–20.5 | 3 | 5 |
| 21–30 | 20.5–30.5 | 5 | 10 |
| 31–40 | 30.5–40.5 | 13 | 23 |
| 41–50 | 40.5–50.5 | 19 | 42 |
| 51–60 | 50.5–60.5 | 31 | 73 |
| 61–70 | 60.5–70.5 | 13 | 86 |
| 71–80 | 70.5–80.5 | 9 | 95 |
| 81–90 | 80.5–90.5 | 4 | 99 |
| 91–100 | 90.5–100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries and join the points with a smooth increasing curve.
(b)(i) 60th percentile
The 60th percentile corresponds to cumulative frequency 60. It lies in the class 51–60:
\[P_{60}=50.5+\left(\frac{60-42}{73-42}\right)\times 10=50.5+\frac{18}{31}\times10\approx56.3\%\]
Thus, the 60th percentile is approximately \(56.3\%\).
(b)(ii) Probability of passing at 35%
From the ogive, the cumulative frequency below 35% is approximately 16. Hence the number passing is approximately
\[100-16=84.\]
Therefore,
\[P(\text{student passes})=\frac{84}{100}=\boxed{0.84}.\]
Awọn alaye Idahun
(a) Cumulative frequency table
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 1–10 | 0.5–10.5 | 2 | 2 |
| 11–20 | 10.5–20.5 | 3 | 5 |
| 21–30 | 20.5–30.5 | 5 | 10 |
| 31–40 | 30.5–40.5 | 13 | 23 |
| 41–50 | 40.5–50.5 | 19 | 42 |
| 51–60 | 50.5–60.5 | 31 | 73 |
| 61–70 | 60.5–70.5 | 13 | 86 |
| 71–80 | 70.5–80.5 | 9 | 95 |
| 81–90 | 80.5–90.5 | 4 | 99 |
| 91–100 | 90.5–100.5 | 1 | 100 |
Plot cumulative frequency against the upper class boundaries and join the points with a smooth increasing curve.
(b)(i) 60th percentile
The 60th percentile corresponds to cumulative frequency 60. It lies in the class 51–60:
\[P_{60}=50.5+\left(\frac{60-42}{73-42}\right)\times 10=50.5+\frac{18}{31}\times10\approx56.3\%\]
Thus, the 60th percentile is approximately \(56.3\%\).
(b)(ii) Probability of passing at 35%
From the ogive, the cumulative frequency below 35% is approximately 16. Hence the number passing is approximately
\[100-16=84.\]
Therefore,
\[P(\text{student passes})=\frac{84}{100}=\boxed{0.84}.\]
Ibeere 7 Ìròyìn
(a) Using ruler and a pair of compasses only, construct : (i) a trapezium WXYZ such that |WX| = 10.2 cm, |XY| = 5.6 cm, |YZ| = 5.8 cm, < WXY = 60° and WX is parallel to YZ (ii) a perpendicular from Z to meet \(\overline{WX}\) at N.
(b) Measure : (i) |WZ| ; (ii) |ZN| .
(a) Construction
- Draw the line segment \(WX=10.2\text{ cm}\).
- At \(X\), construct \(\angle WXY=60^\circ\). On this ray, mark \(Y\) such that \(XY=5.6\text{ cm}\).
- Through \(Y\), construct a line parallel to \(WX\). On this parallel line, towards \(W\), mark \(Z\) such that \(YZ=5.8\text{ cm}\).
- Join \(Z\) to \(W\). From \(Z\), construct a perpendicular to \(WX\), meeting \(WX\) at \(N\).
(b) Measurements from the construction
\[ |WZ|=5.25\text{ cm} \]
\[ |ZN|=4.86\text{ cm} \]
Awọn alaye Idahun
(a) Construction
- Draw the line segment \(WX=10.2\text{ cm}\).
- At \(X\), construct \(\angle WXY=60^\circ\). On this ray, mark \(Y\) such that \(XY=5.6\text{ cm}\).
- Through \(Y\), construct a line parallel to \(WX\). On this parallel line, towards \(W\), mark \(Z\) such that \(YZ=5.8\text{ cm}\).
- Join \(Z\) to \(W\). From \(Z\), construct a perpendicular to \(WX\), meeting \(WX\) at \(N\).
(b) Measurements from the construction
\[ |WZ|=5.25\text{ cm} \]
\[ |ZN|=4.86\text{ cm} \]
Ibeere 8 Ìròyìn
(a) If (3 - x), 6, (7 - 5x) are consecutive terms of a geometric progression (GP) with constant ratio r > 0, find the :
(i) values of x ; (ii) constant ratio.
(b)
In the diagram, |AB| = 3 cm, |BC| = 4 cm, |CD| = 6 cm and |DA| = 7 cm. Calculate <ADC, correct to the nearest degree.
(a) Geometric progression
For three consecutive GP terms, the square of the middle term equals the product of the outer terms:
\[6^2 = (3 - x)(7 - 5x)\]\[36 = 21 - 15x - 7x + 5x^2\]\[36 = 5x^2 - 22x + 21\]\[5x^2 - 22x - 15 = 0\](i) Values of x
Using the quadratic formula with \(a = 5,\; b = -22,\; c = -15\):
\[x = \frac{22 \pm \sqrt{(-22)^2 - 4(5)(-15)}}{2(5)} = \frac{22 \pm \sqrt{484 + 300}}{10} = \frac{22 \pm \sqrt{784}}{10} = \frac{22 \pm 28}{10}\]\[x = 5 \quad\text{or}\quad x = -\frac{6}{10} = -\frac{3}{5}\](ii) Constant ratio
The ratio is \(r = \dfrac{6}{3 - x}\).
- If \(x = 5\): \(r = \dfrac{6}{3-5} = \dfrac{6}{-2} = -3\) (negative, rejected since \(r > 0\)).
- If \(x = -\dfrac{3}{5}\): \(r = \dfrac{6}{3 + \frac{3}{5}} = \dfrac{6}{\frac{18}{5}} = 6\times\dfrac{5}{18} = \dfrac{5}{3}\) (positive, accepted).
So the admissible value is \(x = -\dfrac{3}{5}\) with constant ratio \(r = \dfrac{5}{3}\). (Check: terms are \(3.6,\; 6,\; 10\), and \(\tfrac{6}{3.6} = \tfrac{10}{6} = \tfrac{5}{3}\).)
(b) Finding \(\angle ADC\)
In the quadrilateral, \(|AB| = 3\text{ cm}\), \(|BC| = 4\text{ cm}\), \(|CD| = 6\text{ cm}\), \(|DA| = 7\text{ cm}\), and the angle at B is \(90^\circ\). Draw the diagonal \(AC\).
In right triangle \(ABC\):
\[|AC|^2 = |AB|^2 + |BC|^2 = 3^2 + 4^2 = 9 + 16 = 25 \;\Rightarrow\; |AC| = 5\text{ cm}\]In triangle \(ACD\), apply the cosine rule with \(\angle ADC\) opposite \(AC\):
\[|AC|^2 = |DA|^2 + |DC|^2 - 2\,|DA|\,|DC|\cos(\angle ADC)\]\[25 = 7^2 + 6^2 - 2(7)(6)\cos(\angle ADC)\]\[25 = 49 + 36 - 84\cos(\angle ADC)\]\[84\cos(\angle ADC) = 85 - 25 = 60\]\[\cos(\angle ADC) = \frac{60}{84} = 0.7143\]\[\angle ADC = \cos^{-1}(0.7143) = 44.4^\circ \approx 44^\circ\]Therefore \(\angle ADC \approx 44^\circ\).
Awọn alaye Idahun
(a) Geometric progression
For three consecutive GP terms, the square of the middle term equals the product of the outer terms:
\[6^2 = (3 - x)(7 - 5x)\]\[36 = 21 - 15x - 7x + 5x^2\]\[36 = 5x^2 - 22x + 21\]\[5x^2 - 22x - 15 = 0\](i) Values of x
Using the quadratic formula with \(a = 5,\; b = -22,\; c = -15\):
\[x = \frac{22 \pm \sqrt{(-22)^2 - 4(5)(-15)}}{2(5)} = \frac{22 \pm \sqrt{484 + 300}}{10} = \frac{22 \pm \sqrt{784}}{10} = \frac{22 \pm 28}{10}\]\[x = 5 \quad\text{or}\quad x = -\frac{6}{10} = -\frac{3}{5}\](ii) Constant ratio
The ratio is \(r = \dfrac{6}{3 - x}\).
- If \(x = 5\): \(r = \dfrac{6}{3-5} = \dfrac{6}{-2} = -3\) (negative, rejected since \(r > 0\)).
- If \(x = -\dfrac{3}{5}\): \(r = \dfrac{6}{3 + \frac{3}{5}} = \dfrac{6}{\frac{18}{5}} = 6\times\dfrac{5}{18} = \dfrac{5}{3}\) (positive, accepted).
So the admissible value is \(x = -\dfrac{3}{5}\) with constant ratio \(r = \dfrac{5}{3}\). (Check: terms are \(3.6,\; 6,\; 10\), and \(\tfrac{6}{3.6} = \tfrac{10}{6} = \tfrac{5}{3}\).)
(b) Finding \(\angle ADC\)
In the quadrilateral, \(|AB| = 3\text{ cm}\), \(|BC| = 4\text{ cm}\), \(|CD| = 6\text{ cm}\), \(|DA| = 7\text{ cm}\), and the angle at B is \(90^\circ\). Draw the diagonal \(AC\).
In right triangle \(ABC\):
\[|AC|^2 = |AB|^2 + |BC|^2 = 3^2 + 4^2 = 9 + 16 = 25 \;\Rightarrow\; |AC| = 5\text{ cm}\]In triangle \(ACD\), apply the cosine rule with \(\angle ADC\) opposite \(AC\):
\[|AC|^2 = |DA|^2 + |DC|^2 - 2\,|DA|\,|DC|\cos(\angle ADC)\]\[25 = 7^2 + 6^2 - 2(7)(6)\cos(\angle ADC)\]\[25 = 49 + 36 - 84\cos(\angle ADC)\]\[84\cos(\angle ADC) = 85 - 25 = 60\]\[\cos(\angle ADC) = \frac{60}{84} = 0.7143\]\[\angle ADC = \cos^{-1}(0.7143) = 44.4^\circ \approx 44^\circ\]Therefore \(\angle ADC \approx 44^\circ\).
Ibeere 9 Ìròyìn
The area of a circle is \(154cm^{2}\). It is divided into three sectors such that two of the sectors are equal in size and the third sector is three times the size of the other two put together. Calculate the perimeter of the third sector. [Take \(\pi = \frac{22}{7}\)].
First find the radius from the area: \[\pi r^2 = 154 \Rightarrow \frac{22}{7}r^2 = 154 \Rightarrow r^2 = \frac{154\times 7}{22} = 49 \Rightarrow r = 7\,\text{cm}.\]
Let each of the two equal sectors have angle \(a\). The third sector is three times the two put together: \[\text{third} = 3(2a) = 6a.\] Total angle: \(2a + 6a = 8a = 360^\circ \Rightarrow a = 45^\circ.\)
So the third sector has angle \(6a = 270^\circ.\)
Arc length of third sector: \[\frac{270}{360}\times 2\pi r = \frac{3}{4}\times 2\times\frac{22}{7}\times 7 = \frac{3}{4}\times 44 = 33\,\text{cm}.\]
Perimeter of a sector = arc + two radii: \[33 + 2(7) = 33 + 14 = \mathbf{47\,\text{cm}}.\]
Awọn alaye Idahun
First find the radius from the area: \[\pi r^2 = 154 \Rightarrow \frac{22}{7}r^2 = 154 \Rightarrow r^2 = \frac{154\times 7}{22} = 49 \Rightarrow r = 7\,\text{cm}.\]
Let each of the two equal sectors have angle \(a\). The third sector is three times the two put together: \[\text{third} = 3(2a) = 6a.\] Total angle: \(2a + 6a = 8a = 360^\circ \Rightarrow a = 45^\circ.\)
So the third sector has angle \(6a = 270^\circ.\)
Arc length of third sector: \[\frac{270}{360}\times 2\pi r = \frac{3}{4}\times 2\times\frac{22}{7}\times 7 = \frac{3}{4}\times 44 = 33\,\text{cm}.\]
Perimeter of a sector = arc + two radii: \[33 + 2(7) = 33 + 14 = \mathbf{47\,\text{cm}}.\]
Ibeere 10 Ìròyìn
(a) The present ages of a father and his son are in the ratio 10 : 3. If the son is 15 years old now, in how many years will the ratio of their ages be 2 : 1?
(b) The arithmetic mean of x, y and z is 6 while that of x, y, z, l, u, v and w is 9. Calculate the arithmetic mean of l, u, v and w.
(a) Father : son \(= 10 : 3\), and the son is \(15\). So father's age \(= \dfrac{10}{3}\times 15 = 50\) years.
Let the ratio be \(2 : 1\) in \(x\) years: \[\frac{50 + x}{15 + x} = \frac{2}{1} \Rightarrow 50 + x = 2(15 + x) = 30 + 2x.\] \[50 - 30 = 2x - x \Rightarrow x = 20.\] The ratio will be \(2 : 1\) in \(\mathbf{20\ \text{years}}.\)
(b) Mean of \(x, y, z\) is \(6\): \(x + y + z = 18.\)
Mean of \(x, y, z, l, u, v, w\) is \(9\): \(x + y + z + l + u + v + w = 63.\)
So \(l + u + v + w = 63 - 18 = 45.\)
Mean of \(l, u, v, w = \dfrac{45}{4} = \mathbf{11.25}.\)
Awọn alaye Idahun
(a) Father : son \(= 10 : 3\), and the son is \(15\). So father's age \(= \dfrac{10}{3}\times 15 = 50\) years.
Let the ratio be \(2 : 1\) in \(x\) years: \[\frac{50 + x}{15 + x} = \frac{2}{1} \Rightarrow 50 + x = 2(15 + x) = 30 + 2x.\] \[50 - 30 = 2x - x \Rightarrow x = 20.\] The ratio will be \(2 : 1\) in \(\mathbf{20\ \text{years}}.\)
(b) Mean of \(x, y, z\) is \(6\): \(x + y + z = 18.\)
Mean of \(x, y, z, l, u, v, w\) is \(9\): \(x + y + z + l + u + v + w = 63.\)
So \(l + u + v + w = 63 - 18 = 45.\)
Mean of \(l, u, v, w = \dfrac{45}{4} = \mathbf{11.25}.\)
Ibeere 11 Ìròyìn
(a) Solve : \(7x + 4 < \frac{1}{2}(4x + 3)\).
(b) Salem, Sunday and Shaka shared a sum of N1,100.00. For every N2.00 that Salem gets, Sunday gets 50 kobo and for every N4.00 Sunday gets, Shaka gets N2.00. Find Shaka's share.
(a) \(7x + 4 < \tfrac{1}{2}(4x + 3)\).
Multiply the bracket: \(7x + 4 < 2x + \tfrac{3}{2}.\)
\(7x - 2x < \tfrac{3}{2} - 4 \Rightarrow 5x < -\tfrac{5}{2} \Rightarrow x < -\tfrac{1}{2}.\)
Solution: \(\mathbf{x < -0.5}.\)
(b) For every \(\text{N}2.00\) Salem gets, Sunday gets \(50\) kobo \((\text{N}0.50)\): so Salem : Sunday \(= 2 : 0.5 = 4 : 1.\)
For every \(\text{N}4.00\) Sunday gets, Shaka gets \(\text{N}2.00\): so Sunday : Shaka \(= 4 : 2 = 2 : 1.\)
Make Sunday common: Salem : Sunday \(= 8 : 2\) and Sunday : Shaka \(= 2 : 1\), giving \[\text{Salem : Sunday : Shaka} = 8 : 2 : 1.\] Total parts \(= 11\). One part \(= \dfrac{\text{N}1100}{11} = \text{N}100.\)
Shaka's share \(= 1 \times \text{N}100 = \mathbf{\text{N}100.00}.\)
Awọn alaye Idahun
(a) \(7x + 4 < \tfrac{1}{2}(4x + 3)\).
Multiply the bracket: \(7x + 4 < 2x + \tfrac{3}{2}.\)
\(7x - 2x < \tfrac{3}{2} - 4 \Rightarrow 5x < -\tfrac{5}{2} \Rightarrow x < -\tfrac{1}{2}.\)
Solution: \(\mathbf{x < -0.5}.\)
(b) For every \(\text{N}2.00\) Salem gets, Sunday gets \(50\) kobo \((\text{N}0.50)\): so Salem : Sunday \(= 2 : 0.5 = 4 : 1.\)
For every \(\text{N}4.00\) Sunday gets, Shaka gets \(\text{N}2.00\): so Sunday : Shaka \(= 4 : 2 = 2 : 1.\)
Make Sunday common: Salem : Sunday \(= 8 : 2\) and Sunday : Shaka \(= 2 : 1\), giving \[\text{Salem : Sunday : Shaka} = 8 : 2 : 1.\] Total parts \(= 11\). One part \(= \dfrac{\text{N}1100}{11} = \text{N}100.\)
Shaka's share \(= 1 \times \text{N}100 = \mathbf{\text{N}100.00}.\)
Ibeere 12 Ìròyìn
(a) Simplify, without using tables or calculator : \(\frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}}\).
(b) Given that \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\), evaluate, correct to 2 significant figures and without using tables or calculator, \(\log_{10} 1.125\).
(a) Numerator: \(\tfrac{3}{4}\left(3\tfrac{3}{8} + 1\tfrac{5}{8}\right) = \tfrac{3}{4}\left(\tfrac{27}{8} + \tfrac{13}{8}\right) = \tfrac{3}{4}\times\tfrac{40}{8} = \tfrac{3}{4}\times 5 = \tfrac{15}{4}.\)
Denominator: \(2\tfrac{1}{8} - 1\tfrac{1}{2} = \tfrac{17}{8} - \tfrac{12}{8} = \tfrac{5}{8}.\)
Therefore \(\dfrac{15/4}{5/8} = \dfrac{15}{4}\times\dfrac{8}{5} = \dfrac{120}{20} = \mathbf{6}.\)
(b) Write \(1.125 = \dfrac{9}{8} = \dfrac{3^2}{2^3}.\) Then \[\log_{10}1.125 = 2\log_{10}3 - 3\log_{10}2 = 2(0.4771) - 3(0.3010) = 0.9542 - 0.9030 = 0.0512.\] Correct to 2 significant figures, \(\log_{10}1.125 \approx \mathbf{0.051}.\)
Awọn alaye Idahun
(a) Numerator: \(\tfrac{3}{4}\left(3\tfrac{3}{8} + 1\tfrac{5}{8}\right) = \tfrac{3}{4}\left(\tfrac{27}{8} + \tfrac{13}{8}\right) = \tfrac{3}{4}\times\tfrac{40}{8} = \tfrac{3}{4}\times 5 = \tfrac{15}{4}.\)
Denominator: \(2\tfrac{1}{8} - 1\tfrac{1}{2} = \tfrac{17}{8} - \tfrac{12}{8} = \tfrac{5}{8}.\)
Therefore \(\dfrac{15/4}{5/8} = \dfrac{15}{4}\times\dfrac{8}{5} = \dfrac{120}{20} = \mathbf{6}.\)
(b) Write \(1.125 = \dfrac{9}{8} = \dfrac{3^2}{2^3}.\) Then \[\log_{10}1.125 = 2\log_{10}3 - 3\log_{10}2 = 2(0.4771) - 3(0.3010) = 0.9542 - 0.9030 = 0.0512.\] Correct to 2 significant figures, \(\log_{10}1.125 \approx \mathbf{0.051}.\)
Ibeere 13 Ìròyìn
A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,
(a) represent this information in a diagram ;
(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.
(a) Take the pole vertical with the lamp \(L\) at its top \((4.2\,\text{m})\). The boy \(BC\) is \(1.2\,\text{m}\) tall standing \(6\,\text{m}\) from the pole's foot; his shadow \(CS\) stretches away from the pole on the ground. The light ray from \(L\) passes over the boy's head to the shadow tip \(S\).
(b)(i) Let the shadow length be \(s\). By similar triangles (light from \(L\) grazing the boy's head): \[\frac{\text{lamp height}}{\text{pole-to-shadow-tip}} = \frac{\text{boy height}}{\text{boy-to-shadow-tip}} \Rightarrow \frac{4.2}{6 + s} = \frac{1.2}{s}.\] \[4.2 s = 1.2(6 + s) \Rightarrow 4.2s = 7.2 + 1.2s \Rightarrow 3s = 7.2 \Rightarrow s = 2.4.\] Length of the boy's shadow \(= \mathbf{2.4\,\text{m}}.\)
(b)(ii) Angle of elevation of the lamp from the boy (measured from the top of his head): the horizontal distance is \(6\,\text{m}\) and the lamp is \(4.2 - 1.2 = 3.0\,\text{m}\) above him. \[\tan\theta = \frac{3.0}{6} = 0.5 \Rightarrow \theta = 26.57^\circ \approx \mathbf{27^\circ}.\]
Awọn alaye Idahun
(a) Take the pole vertical with the lamp \(L\) at its top \((4.2\,\text{m})\). The boy \(BC\) is \(1.2\,\text{m}\) tall standing \(6\,\text{m}\) from the pole's foot; his shadow \(CS\) stretches away from the pole on the ground. The light ray from \(L\) passes over the boy's head to the shadow tip \(S\).
(b)(i) Let the shadow length be \(s\). By similar triangles (light from \(L\) grazing the boy's head): \[\frac{\text{lamp height}}{\text{pole-to-shadow-tip}} = \frac{\text{boy height}}{\text{boy-to-shadow-tip}} \Rightarrow \frac{4.2}{6 + s} = \frac{1.2}{s}.\] \[4.2 s = 1.2(6 + s) \Rightarrow 4.2s = 7.2 + 1.2s \Rightarrow 3s = 7.2 \Rightarrow s = 2.4.\] Length of the boy's shadow \(= \mathbf{2.4\,\text{m}}.\)
(b)(ii) Angle of elevation of the lamp from the boy (measured from the top of his head): the horizontal distance is \(6\,\text{m}\) and the lamp is \(4.2 - 1.2 = 3.0\,\text{m}\) above him. \[\tan\theta = \frac{3.0}{6} = 0.5 \Rightarrow \theta = 26.57^\circ \approx \mathbf{27^\circ}.\]
