WAEC SSCE - General Mathematics - 2017 (Essay)
Tambaya 1 Rahoto
(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34° and < TSQ = 65°.
(i) Illustrate the information in a diagram; (ii) find the value of : (a) < RTS ; (b) < SRP.
(b) 
In the diagram, /VZ/ = /YZ/, < YXZ = 20° and < ZVY = 52°. Calculate the size of < WYZ.
(a) Tangent \(PQ\) touches the circle \(RST\) at \(S\), with \(PRT\) a straight secant, \(\angle TPS=34^\circ\) and \(\angle TSQ=65^\circ\).
(i) Diagram.
(ii)(a) Finding \(\angle RTS\). Since \(P\), \(S\) and \(Q\) lie on the tangent line, \(\angle TSQ=65^\circ\) is the exterior angle of triangle \(PST\) at \(S\). An exterior angle equals the sum of the two interior opposite angles, \(\angle TPS\) and \(\angle PTS\). Because \(R\) lies on \(PT\), the angle \(\angle PTS\) is the same as \(\angle RTS\):
\[ \angle TSQ=\angle TPS+\angle RTS \]
\[ 65^\circ=34^\circ+\angle RTS\ \Rightarrow\ \angle RTS=31^\circ. \]
(ii)(b) Finding \(\angle SRP\). By the alternate segment theorem, the tangent-chord angle \(\angle TSQ\) equals the inscribed angle in the alternate segment standing on chord \(ST\), which is \(\angle SRT\):
\[ \angle SRT=\angle TSQ=65^\circ. \]
Since \(PRT\) is a straight line, \(\angle SRP\) and \(\angle SRT\) are angles on a straight line:
\[ \angle SRP=180^\circ-\angle SRT=180^\circ-65^\circ=115^\circ. \]
(b) In the given diagram \(VZ=YZ\), \(\angle YXZ=20^\circ\) and \(\angle ZVY=52^\circ\); find \(\angle WYZ\).
Triangle \(VZY\) has \(VZ=YZ\), so it is isosceles and its base angles at \(V\) and \(Y\) are equal:
\[ \angle ZYV=\angle ZVY=52^\circ. \]
The inscribed angle \(\angle ZVY=52^\circ\) stands on chord \(ZY\), so the arc \(ZY\) it cuts off is
\[ \text{arc } ZY=2\times52^\circ=104^\circ. \]
\(X\) is an external point with the two secants \(XVY\) and \(XWZ\). The angle between two secants from an external point equals half the difference of the two intercepted arcs:
\[ \angle YXZ=\tfrac{1}{2}\big(\text{arc } YZ-\text{arc } VW\big) \]
\[ 20^\circ=\tfrac{1}{2}\big(104^\circ-\text{arc } VW\big)\ \Rightarrow\ \text{arc } VW=64^\circ. \]
Equal chords cut off equal arcs, so \(VZ=YZ\) gives \(\text{arc } VWZ=\text{arc } YZ=104^\circ\). Since \(W\) lies on the arc between \(V\) and \(Z\),
\[ \text{arc } WZ=\text{arc } VWZ-\text{arc } VW=104^\circ-64^\circ=40^\circ. \]
Finally, \(\angle WYZ\) is the inscribed angle at \(Y\) standing on chord \(WZ\), so it is half of arc \(WZ\):
\[ \angle WYZ=\tfrac{1}{2}\times40^\circ=20^\circ. \]
Bayanin Amsa
(a) Tangent \(PQ\) touches the circle \(RST\) at \(S\), with \(PRT\) a straight secant, \(\angle TPS=34^\circ\) and \(\angle TSQ=65^\circ\).
(i) Diagram.
(ii)(a) Finding \(\angle RTS\). Since \(P\), \(S\) and \(Q\) lie on the tangent line, \(\angle TSQ=65^\circ\) is the exterior angle of triangle \(PST\) at \(S\). An exterior angle equals the sum of the two interior opposite angles, \(\angle TPS\) and \(\angle PTS\). Because \(R\) lies on \(PT\), the angle \(\angle PTS\) is the same as \(\angle RTS\):
\[ \angle TSQ=\angle TPS+\angle RTS \]
\[ 65^\circ=34^\circ+\angle RTS\ \Rightarrow\ \angle RTS=31^\circ. \]
(ii)(b) Finding \(\angle SRP\). By the alternate segment theorem, the tangent-chord angle \(\angle TSQ\) equals the inscribed angle in the alternate segment standing on chord \(ST\), which is \(\angle SRT\):
\[ \angle SRT=\angle TSQ=65^\circ. \]
Since \(PRT\) is a straight line, \(\angle SRP\) and \(\angle SRT\) are angles on a straight line:
\[ \angle SRP=180^\circ-\angle SRT=180^\circ-65^\circ=115^\circ. \]
(b) In the given diagram \(VZ=YZ\), \(\angle YXZ=20^\circ\) and \(\angle ZVY=52^\circ\); find \(\angle WYZ\).
Triangle \(VZY\) has \(VZ=YZ\), so it is isosceles and its base angles at \(V\) and \(Y\) are equal:
\[ \angle ZYV=\angle ZVY=52^\circ. \]
The inscribed angle \(\angle ZVY=52^\circ\) stands on chord \(ZY\), so the arc \(ZY\) it cuts off is
\[ \text{arc } ZY=2\times52^\circ=104^\circ. \]
\(X\) is an external point with the two secants \(XVY\) and \(XWZ\). The angle between two secants from an external point equals half the difference of the two intercepted arcs:
\[ \angle YXZ=\tfrac{1}{2}\big(\text{arc } YZ-\text{arc } VW\big) \]
\[ 20^\circ=\tfrac{1}{2}\big(104^\circ-\text{arc } VW\big)\ \Rightarrow\ \text{arc } VW=64^\circ. \]
Equal chords cut off equal arcs, so \(VZ=YZ\) gives \(\text{arc } VWZ=\text{arc } YZ=104^\circ\). Since \(W\) lies on the arc between \(V\) and \(Z\),
\[ \text{arc } WZ=\text{arc } VWZ-\text{arc } VW=104^\circ-64^\circ=40^\circ. \]
Finally, \(\angle WYZ\) is the inscribed angle at \(Y\) standing on chord \(WZ\), so it is half of arc \(WZ\):
\[ \angle WYZ=\tfrac{1}{2}\times40^\circ=20^\circ. \]
Tambaya 2 Rahoto
(a) It takes 8 students two- thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one- third of an hour at the same rate?
(b) A chord, 20 cm long, is 12 cm from the centre of the circle. Calculate, correct to one decimal place, the :
(i) angle subtended by the chord at the centre of the circle;
(ii) perimeter of the minor segment cut off by the chord. [Take \(\pi = 3.142\)].
(a) The number of tanks filled varies jointly with the number of students and the time worked: tanks \(= k \times (\text{students})\times(\text{time})\).
From the given data, 8 students in \(\tfrac{2}{3}\) h fill 12 tanks:
\[12 = k \times 8 \times \frac{2}{3} = \frac{16k}{3} \Rightarrow k = \frac{36}{16} = 2.25.\]
For 4 students in \(\tfrac{1}{3}\) h:
\[\text{tanks} = 2.25 \times 4 \times \frac{1}{3} = 2.25 \times \frac{4}{3} = 3.\]
So \(\mathbf{3}\) tanks are filled.
(b) A chord of length 20 cm is 12 cm from the centre. The perpendicular from the centre bisects the chord, giving a right triangle with legs 12 and 10.
Radius: \(r = \sqrt{12^2 + 10^2} = \sqrt{244} = 15.62\text{ cm}\).
(i) If \(\theta\) is the angle subtended at the centre, then in the right triangle \(\tan\left(\tfrac{\theta}{2}\right) = \dfrac{10}{12}\):
\[\frac{\theta}{2} = 39.8^\circ \Rightarrow \theta = 79.6^\circ.\]
(ii) The perimeter of the minor segment \(=\) chord \(+\) arc length. Arc length \(= \dfrac{\theta}{360^\circ}\times 2\pi r\):
\[\text{arc} = \frac{79.6}{360}\times 2(3.142)(15.62) = 0.2211 \times 98.16 = 21.7\text{ cm}.\]
\[\text{Perimeter} = 20 + 21.7 = 41.7\text{ cm}.\]
Bayanin Amsa
(a) The number of tanks filled varies jointly with the number of students and the time worked: tanks \(= k \times (\text{students})\times(\text{time})\).
From the given data, 8 students in \(\tfrac{2}{3}\) h fill 12 tanks:
\[12 = k \times 8 \times \frac{2}{3} = \frac{16k}{3} \Rightarrow k = \frac{36}{16} = 2.25.\]
For 4 students in \(\tfrac{1}{3}\) h:
\[\text{tanks} = 2.25 \times 4 \times \frac{1}{3} = 2.25 \times \frac{4}{3} = 3.\]
So \(\mathbf{3}\) tanks are filled.
(b) A chord of length 20 cm is 12 cm from the centre. The perpendicular from the centre bisects the chord, giving a right triangle with legs 12 and 10.
Radius: \(r = \sqrt{12^2 + 10^2} = \sqrt{244} = 15.62\text{ cm}\).
(i) If \(\theta\) is the angle subtended at the centre, then in the right triangle \(\tan\left(\tfrac{\theta}{2}\right) = \dfrac{10}{12}\):
\[\frac{\theta}{2} = 39.8^\circ \Rightarrow \theta = 79.6^\circ.\]
(ii) The perimeter of the minor segment \(=\) chord \(+\) arc length. Arc length \(= \dfrac{\theta}{360^\circ}\times 2\pi r\):
\[\text{arc} = \frac{79.6}{360}\times 2(3.142)(15.62) = 0.2211 \times 98.16 = 21.7\text{ cm}.\]
\[\text{Perimeter} = 20 + 21.7 = 41.7\text{ cm}.\]
Tambaya 3 Rahoto
(a) Given that \(\sin x = \frac{5}{13}, 0° < x < 90°\), find \(\frac{\cos x - 2\sin x}{2\tan x}\).
(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :
(i) length of ladder LA (ii) angle which LA makes with the ground.
(a) Given \(\sin x = \dfrac{5}{13}\) with \(0^\circ < x < 90^\circ\). Using a 5-12-13 right triangle (\(\sqrt{13^2 - 5^2} = 12\)):
\[\cos x = \frac{12}{13}, \qquad \tan x = \frac{5}{12}.\]
Now evaluate \(\dfrac{\cos x - 2\sin x}{2\tan x}\):
\[\cos x - 2\sin x = \frac{12}{13} - \frac{10}{13} = \frac{2}{13}, \qquad 2\tan x = \frac{10}{12} = \frac{5}{6}.\]
\[\frac{\cos x - 2\sin x}{2\tan x} = \frac{2/13}{5/6} = \frac{2}{13}\times\frac{6}{5} = \frac{12}{65}.\]
(b) The point \(L\) is 9.6 m up the vertical pole, with foot \(F\). Ladder \(LB = 12\) m reaches the ground at \(B\); \(A\) and \(B\) are 10 m apart on opposite sides of the pole.
Horizontal distance \(FB\):
\[FB = \sqrt{12^2 - 9.6^2} = \sqrt{144 - 92.16} = \sqrt{51.84} = 7.2\text{ m}.\]
Since \(A\) and \(B\) are on opposite sides, \(FA + FB = 10\), so \(FA = 10 - 7.2 = 2.8\text{ m}\).
(i) \(LA = \sqrt{FA^2 + 9.6^2} = \sqrt{2.8^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10\text{ m}\) (2 s.f.).
(ii) Let \(\theta\) be the angle \(LA\) makes with the ground:
\[\tan\theta = \frac{9.6}{2.8} = 3.4286 \Rightarrow \theta = 73.7^\circ \approx 74^\circ.\]
Bayanin Amsa
(a) Given \(\sin x = \dfrac{5}{13}\) with \(0^\circ < x < 90^\circ\). Using a 5-12-13 right triangle (\(\sqrt{13^2 - 5^2} = 12\)):
\[\cos x = \frac{12}{13}, \qquad \tan x = \frac{5}{12}.\]
Now evaluate \(\dfrac{\cos x - 2\sin x}{2\tan x}\):
\[\cos x - 2\sin x = \frac{12}{13} - \frac{10}{13} = \frac{2}{13}, \qquad 2\tan x = \frac{10}{12} = \frac{5}{6}.\]
\[\frac{\cos x - 2\sin x}{2\tan x} = \frac{2/13}{5/6} = \frac{2}{13}\times\frac{6}{5} = \frac{12}{65}.\]
(b) The point \(L\) is 9.6 m up the vertical pole, with foot \(F\). Ladder \(LB = 12\) m reaches the ground at \(B\); \(A\) and \(B\) are 10 m apart on opposite sides of the pole.
Horizontal distance \(FB\):
\[FB = \sqrt{12^2 - 9.6^2} = \sqrt{144 - 92.16} = \sqrt{51.84} = 7.2\text{ m}.\]
Since \(A\) and \(B\) are on opposite sides, \(FA + FB = 10\), so \(FA = 10 - 7.2 = 2.8\text{ m}\).
(i) \(LA = \sqrt{FA^2 + 9.6^2} = \sqrt{2.8^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10\text{ m}\) (2 s.f.).
(ii) Let \(\theta\) be the angle \(LA\) makes with the ground:
\[\tan\theta = \frac{9.6}{2.8} = 3.4286 \Rightarrow \theta = 73.7^\circ \approx 74^\circ.\]
Tambaya 4 Rahoto
(a) Copy and complete the table of values for the equation \(y = 2x^{2} - 7x - 9\) for \(-3 \leq x \leq 6\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| y | 13 | -9 | -14 | -12 | 6 |
(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 4 units on the y- axis, draw the graphs of \(y = 2x^{2} - 7x - 9\) for \(-3 \leq x \leq 6\).
(c) Use the graph to estimate the :
(i) roots of the equation \(2x^{2} - 7x = 26\);
(ii) coordinates of the minimum point of y;
(iii) range of values for which \(2x^{2} - 7x < 9\).
(a) Completing the table for \(y = 2x^{2} - 7x - 9\). Evaluate at each x:
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|
| y | 30 | 13 | 0 | -9 | -14 | -15 | -12 | -5 | 6 | 21 |
Sample working: at \(x=-1,\ y=2+7-9=0\); at \(x=2,\ y=8-14-9=-15\).
(b) Graph. Using 2 cm to 1 unit (x-axis) and 2 cm to 4 units (y-axis), plot and join with a smooth parabola whose lowest point is near \(x=1.75\).
(c)(i) Roots of \(2x^{2}-7x=26\). Since \(2x^{2}-7x = y+9\), this becomes \(y+9=26\Rightarrow y=17\). Draw \(y=17\) and read:
\[x \approx -2.26 \quad\text{and}\quad x \approx 5.76.\]
(c)(ii) Minimum point. The turning point is at \(x=\dfrac{7}{4}=1.75\), where \(y=2(1.75)^2-7(1.75)-9=-15.125\). Minimum point \(\approx (1.75,\ -15.1)\).
(c)(iii) Range for \(2x^{2}-7x<9\). This is \(y<0\). The curve is below the x-axis between its roots \(x=-1\) and \(x=4.5\) (since \(2x^2-7x-9=(2x-9)(x+1)\)). Hence
\[-1 < x < 4.5.\]
Bayanin Amsa
(a) Completing the table for \(y = 2x^{2} - 7x - 9\). Evaluate at each x:
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|---|---|
| y | 30 | 13 | 0 | -9 | -14 | -15 | -12 | -5 | 6 | 21 |
Sample working: at \(x=-1,\ y=2+7-9=0\); at \(x=2,\ y=8-14-9=-15\).
(b) Graph. Using 2 cm to 1 unit (x-axis) and 2 cm to 4 units (y-axis), plot and join with a smooth parabola whose lowest point is near \(x=1.75\).
(c)(i) Roots of \(2x^{2}-7x=26\). Since \(2x^{2}-7x = y+9\), this becomes \(y+9=26\Rightarrow y=17\). Draw \(y=17\) and read:
\[x \approx -2.26 \quad\text{and}\quad x \approx 5.76.\]
(c)(ii) Minimum point. The turning point is at \(x=\dfrac{7}{4}=1.75\), where \(y=2(1.75)^2-7(1.75)-9=-15.125\). Minimum point \(\approx (1.75,\ -15.1)\).
(c)(iii) Range for \(2x^{2}-7x<9\). This is \(y<0\). The curve is below the x-axis between its roots \(x=-1\) and \(x=4.5\) (since \(2x^2-7x-9=(2x-9)(x+1)\)). Hence
\[-1 < x < 4.5.\]
Tambaya 5 Rahoto
(a) The operation (*) is defined on the set of real numbers, R, by \(x * y = \frac{x + y}{2}, x, y \in R\).
(i) Evaluate \(3 * \frac{2}{5}\).
(ii) If \(8 * y = 8\frac{1}{4}\), find the value of y.
(b) In \(\Delta ABC, \overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}\). If P is the midpoint of \(\overline{AB}\), express \(\overline{CP}\) as a column vector.
(a) The operation is \(x * y = \dfrac{x + y}{2}\).
(i) \(3 * \dfrac{2}{5} = \dfrac{3 + \tfrac{2}{5}}{2} = \dfrac{\tfrac{17}{5}}{2} = \dfrac{17}{10} = 1\tfrac{7}{10}\).
(ii) \(8 * y = 8\tfrac{1}{4}\) means \(\dfrac{8 + y}{2} = \dfrac{33}{4}\).
\[8 + y = \frac{33}{2} = 16.5 \Rightarrow y = 8.5 = 8\tfrac{1}{2}.\]
(b) Take \(A\) as the reference point. Then \(\overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}\).
\(P\) is the midpoint of \(\overline{AB}\), so
\[\overline{AP} = \frac{1}{2}\overline{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}.\]
Then
\[\overline{CP} = \overline{AP} - \overline{AC} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} - \begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}.\]
Therefore \(\overline{CP} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}\).
Bayanin Amsa
(a) The operation is \(x * y = \dfrac{x + y}{2}\).
(i) \(3 * \dfrac{2}{5} = \dfrac{3 + \tfrac{2}{5}}{2} = \dfrac{\tfrac{17}{5}}{2} = \dfrac{17}{10} = 1\tfrac{7}{10}\).
(ii) \(8 * y = 8\tfrac{1}{4}\) means \(\dfrac{8 + y}{2} = \dfrac{33}{4}\).
\[8 + y = \frac{33}{2} = 16.5 \Rightarrow y = 8.5 = 8\tfrac{1}{2}.\]
(b) Take \(A\) as the reference point. Then \(\overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}\).
\(P\) is the midpoint of \(\overline{AB}\), so
\[\overline{AP} = \frac{1}{2}\overline{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}.\]
Then
\[\overline{CP} = \overline{AP} - \overline{AC} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} - \begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}.\]
Therefore \(\overline{CP} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}\).
Tambaya 6 Rahoto
(a) A manufacturing company requires 3 hours of direct labour to process N87.00 worth of raw materials. If the company uses N30,450.00 worth of raw materials, what amount should it budget at N18.25 per hour?
(b) An investor invested Nx in bank M at the rate of 6% simple interest per annum and Ny in bank N at the rate of 8% simple interest per annum. If a total of N8,000,000.00 was invested in the two banks and the investor received a total of N2,320,000.00 as interest from the two banks after 4 years, calculate the:
(i) values of x and y
(ii) interest paid by the second bank.
(a) Processing \(\text{N}87.00\) worth of raw materials needs 3 hours of labour. For \(\text{N}30{,}450.00\) worth:
\[\text{Hours} = \frac{30{,}450}{87}\times 3 = 350 \times 3 = 1050\text{ hours}.\]
Budget at \(\text{N}18.25\) per hour:
\[1050 \times 18.25 = \text{N}19{,}162.50.\]
(b) Let \(\text{N}x\) be invested at 6% and \(\text{N}y\) at 8%, with \(x + y = 8{,}000{,}000\) ... (1)
Simple interest over 4 years: \(I = \dfrac{P R T}{100}\).
- Bank M: \(\dfrac{x(6)(4)}{100} = 0.24x\)
- Bank N: \(\dfrac{y(8)(4)}{100} = 0.32y\)
Total interest: \(0.24x + 0.32y = 2{,}320{,}000\) ... (2)
From (1), \(x = 8{,}000{,}000 - y\). Substitute into (2):
\[0.24(8{,}000{,}000 - y) + 0.32y = 2{,}320{,}000.\]
\[1{,}920{,}000 + 0.08y = 2{,}320{,}000 \Rightarrow 0.08y = 400{,}000 \Rightarrow y = 5{,}000{,}000.\]
Then \(x = 8{,}000{,}000 - 5{,}000{,}000 = 3{,}000{,}000\).
(i) \(x = \text{N}3{,}000{,}000\) and \(y = \text{N}5{,}000{,}000\).
(ii) Interest paid by the second bank (bank N) \(= 0.32y = 0.32 \times 5{,}000{,}000 = \text{N}1{,}600{,}000\).
Bayanin Amsa
(a) Processing \(\text{N}87.00\) worth of raw materials needs 3 hours of labour. For \(\text{N}30{,}450.00\) worth:
\[\text{Hours} = \frac{30{,}450}{87}\times 3 = 350 \times 3 = 1050\text{ hours}.\]
Budget at \(\text{N}18.25\) per hour:
\[1050 \times 18.25 = \text{N}19{,}162.50.\]
(b) Let \(\text{N}x\) be invested at 6% and \(\text{N}y\) at 8%, with \(x + y = 8{,}000{,}000\) ... (1)
Simple interest over 4 years: \(I = \dfrac{P R T}{100}\).
- Bank M: \(\dfrac{x(6)(4)}{100} = 0.24x\)
- Bank N: \(\dfrac{y(8)(4)}{100} = 0.32y\)
Total interest: \(0.24x + 0.32y = 2{,}320{,}000\) ... (2)
From (1), \(x = 8{,}000{,}000 - y\). Substitute into (2):
\[0.24(8{,}000{,}000 - y) + 0.32y = 2{,}320{,}000.\]
\[1{,}920{,}000 + 0.08y = 2{,}320{,}000 \Rightarrow 0.08y = 400{,}000 \Rightarrow y = 5{,}000{,}000.\]
Then \(x = 8{,}000{,}000 - 5{,}000{,}000 = 3{,}000{,}000\).
(i) \(x = \text{N}3{,}000{,}000\) and \(y = \text{N}5{,}000{,}000\).
(ii) Interest paid by the second bank (bank N) \(= 0.32y = 0.32 \times 5{,}000{,}000 = \text{N}1{,}600{,}000\).
Tambaya 7 Rahoto
(a) The angle of depression of a point P on the ground from the top T of a building is 23.6°. If the distance from P to the foot of the building is 50m, calculate, correct to the nearest metre, the height if the building.
(b) 
In the diagram, \(PT // SU, QS // TR, /SR/ = 6cm\) and \(/RU/ = 10 cm\). If the area of \(\Delta TRU = 45 cm^{2}\), calculate the area of the trapezium QTUS.
(a) Let the building have height \(h\) and foot \(B\), with \(P\) on the ground \(50\ \text{m}\) from \(B\). The angle of depression from the top \(T\) to \(P\) equals the angle of elevation from \(P\) to \(T\) (alternate angles), so this angle at \(P\) is \(23.6^\circ\).
\[\tan 23.6^\circ = \frac{h}{50}\]\[h = 50 \times \tan 23.6^\circ = 50 \times 0.4369 = 21.85\ \text{m}\]Height \(\approx 22\ \text{m}\) (to the nearest metre).
(b) From the diagram, \(PT \parallel SU\) and \(QS \parallel TR\), with \(S\), \(R\), \(U\) on the bottom line where \(|SR| = 6\ \text{cm}\) and \(|RU| = 10\ \text{cm}\).
Find the height between the parallel lines. Triangle \(TRU\) has base \(RU = 10\ \text{cm}\) on the bottom line, and apex \(T\) on the top line, so its height is the perpendicular distance \(H\) between \(PT\) and \(SU\):
\[\text{Area of } \Delta TRU = \frac{1}{2} \times RU \times H = 45\]\[\frac{1}{2} \times 10 \times H = 45 \Rightarrow H = 9\ \text{cm}\]Find \(QT\). Since \(QT \parallel SR\) (both on the parallel top and bottom lines) and \(QS \parallel TR\), the figure \(QTRS\) is a parallelogram, so:
\[QT = SR = 6\ \text{cm}\]Length of \(SU\):
\[SU = SR + RU = 6 + 10 = 16\ \text{cm}\]Area of trapezium \(QTUS\) (parallel sides \(QT = 6\ \text{cm}\) and \(SU = 16\ \text{cm}\), height \(H = 9\ \text{cm}\)):
\[\text{Area} = \frac{1}{2}(QT + SU) \times H = \frac{1}{2}(6 + 16) \times 9\]\[= \frac{1}{2} \times 22 \times 9 = 99\ \text{cm}^2\]Area of trapezium \(QTUS = 99\ \text{cm}^2\)
Bayanin Amsa
(a) Let the building have height \(h\) and foot \(B\), with \(P\) on the ground \(50\ \text{m}\) from \(B\). The angle of depression from the top \(T\) to \(P\) equals the angle of elevation from \(P\) to \(T\) (alternate angles), so this angle at \(P\) is \(23.6^\circ\).
\[\tan 23.6^\circ = \frac{h}{50}\]\[h = 50 \times \tan 23.6^\circ = 50 \times 0.4369 = 21.85\ \text{m}\]Height \(\approx 22\ \text{m}\) (to the nearest metre).
(b) From the diagram, \(PT \parallel SU\) and \(QS \parallel TR\), with \(S\), \(R\), \(U\) on the bottom line where \(|SR| = 6\ \text{cm}\) and \(|RU| = 10\ \text{cm}\).
Find the height between the parallel lines. Triangle \(TRU\) has base \(RU = 10\ \text{cm}\) on the bottom line, and apex \(T\) on the top line, so its height is the perpendicular distance \(H\) between \(PT\) and \(SU\):
\[\text{Area of } \Delta TRU = \frac{1}{2} \times RU \times H = 45\]\[\frac{1}{2} \times 10 \times H = 45 \Rightarrow H = 9\ \text{cm}\]Find \(QT\). Since \(QT \parallel SR\) (both on the parallel top and bottom lines) and \(QS \parallel TR\), the figure \(QTRS\) is a parallelogram, so:
\[QT = SR = 6\ \text{cm}\]Length of \(SU\):
\[SU = SR + RU = 6 + 10 = 16\ \text{cm}\]Area of trapezium \(QTUS\) (parallel sides \(QT = 6\ \text{cm}\) and \(SU = 16\ \text{cm}\), height \(H = 9\ \text{cm}\)):
\[\text{Area} = \frac{1}{2}(QT + SU) \times H = \frac{1}{2}(6 + 16) \times 9\]\[= \frac{1}{2} \times 22 \times 9 = 99\ \text{cm}^2\]Area of trapezium \(QTUS = 99\ \text{cm}^2\)
Tambaya 8 Rahoto
| Marks | 1 | 2 | 3 | 4 | 5 |
| Number of students | \(m + 2\) | \(m - 1\) | \(2m - 3\) | \(m + 5\) | \(3m - 4\) |
The table shows the distribution of marks scored by some students in a test.
(a) If the mean mark is \(3\frac{6}{23}\), find the value of m.
(b) Find the : (i) interquartile range
(ii) probability of selecting a student who scored at least 4 marks in the test.
(a) Finding m. Form the totals in terms of m.
\[\sum f=(m+2)+(m-1)+(2m-3)+(m+5)+(3m-4)=8m-1.\]
\[\sum fx=1(m+2)+2(m-1)+3(2m-3)+4(m+5)+5(3m-4)=28m-9.\]
The mean is \(3\frac{6}{23}=\frac{75}{23}\), so
\[\frac{28m-9}{8m-1}=\frac{75}{23}.\]
\[23(28m-9)=75(8m-1)\Rightarrow 644m-207=600m-75\Rightarrow 44m=132\Rightarrow m=3.\]
Substituting \(m=3\) gives the frequencies:
| Mark | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency | 5 | 2 | 3 | 8 | 5 |
| Cumulative | 5 | 7 | 10 | 18 | 23 |
(b)(i) Interquartile range. \(N=23\). \(Q_1\) is the \(\frac{N+1}{4}=6\text{th}\) value \(=2\); \(Q_3\) is the \(\frac{3(N+1)}{4}=18\text{th}\) value \(=4\).
\[\text{IQR}=Q_3-Q_1=4-2=2.\]
(b)(ii) Probability of at least 4 marks.
\[P(\ge4)=\frac{8+5}{23}=\frac{13}{23}.\]
Bayanin Amsa
(a) Finding m. Form the totals in terms of m.
\[\sum f=(m+2)+(m-1)+(2m-3)+(m+5)+(3m-4)=8m-1.\]
\[\sum fx=1(m+2)+2(m-1)+3(2m-3)+4(m+5)+5(3m-4)=28m-9.\]
The mean is \(3\frac{6}{23}=\frac{75}{23}\), so
\[\frac{28m-9}{8m-1}=\frac{75}{23}.\]
\[23(28m-9)=75(8m-1)\Rightarrow 644m-207=600m-75\Rightarrow 44m=132\Rightarrow m=3.\]
Substituting \(m=3\) gives the frequencies:
| Mark | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency | 5 | 2 | 3 | 8 | 5 |
| Cumulative | 5 | 7 | 10 | 18 | 23 |
(b)(i) Interquartile range. \(N=23\). \(Q_1\) is the \(\frac{N+1}{4}=6\text{th}\) value \(=2\); \(Q_3\) is the \(\frac{3(N+1)}{4}=18\text{th}\) value \(=4\).
\[\text{IQR}=Q_3-Q_1=4-2=2.\]
(b)(ii) Probability of at least 4 marks.
\[P(\ge4)=\frac{8+5}{23}=\frac{13}{23}.\]
Tambaya 9 Rahoto
If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the
(a) first term;
(b) sum of the first fifteen terms.
For an A.P. with first term \(a\) and common difference \(d\):
Sixth term: \(a + 5d = 37\) ... (1)
Sum of first six terms: \(S_6 = \dfrac{6}{2}(2a + 5d) = 3(2a + 5d) = 147\), so \(2a + 5d = 49\) ... (2)
(a) Subtract (1) from (2):
\[(2a + 5d) - (a + 5d) = 49 - 37 \Rightarrow a = 12.\]
The first term is \(a = \mathbf{12}\).
From (1): \(12 + 5d = 37 \Rightarrow 5d = 25 \Rightarrow d = 5\).
(b) Sum of the first fifteen terms:
\[S_{15} = \frac{15}{2}\big(2a + 14d\big) = \frac{15}{2}\big(2(12) + 14(5)\big) = \frac{15}{2}(24 + 70) = \frac{15}{2}(94) = 705.\]
Therefore \(S_{15} = \mathbf{705}\).
Bayanin Amsa
For an A.P. with first term \(a\) and common difference \(d\):
Sixth term: \(a + 5d = 37\) ... (1)
Sum of first six terms: \(S_6 = \dfrac{6}{2}(2a + 5d) = 3(2a + 5d) = 147\), so \(2a + 5d = 49\) ... (2)
(a) Subtract (1) from (2):
\[(2a + 5d) - (a + 5d) = 49 - 37 \Rightarrow a = 12.\]
The first term is \(a = \mathbf{12}\).
From (1): \(12 + 5d = 37 \Rightarrow 5d = 25 \Rightarrow d = 5\).
(b) Sum of the first fifteen terms:
\[S_{15} = \frac{15}{2}\big(2a + 14d\big) = \frac{15}{2}\big(2(12) + 14(5)\big) = \frac{15}{2}(24 + 70) = \frac{15}{2}(94) = 705.\]
Therefore \(S_{15} = \mathbf{705}\).
Tambaya 10 Rahoto
Out of 120 customers in a shop, 45 bought both bags and shoes. If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:
(a) Illustrate the this information in a diagram;
(b) find the number of customers who bought shoes;
(c) calculate the probability that a customer selected at random bought bags.
Let \(b\) be the number who bought bags and \(s\) the number who bought shoes. Every customer bought at least one item, so \(n(\text{bags} \cup \text{shoes}) = 120\), with \(n(\text{both}) = 45\).
Using \(n(B \cup S) = b + s - n(\text{both})\):
\[120 = b + s - 45 \Rightarrow b + s = 165.\]
Also 11 more bought shoes than bags: \(s = b + 11\).
(a) Diagram. A two-circle Venn diagram (Bags and Shoes) with the overlap \(= 45\), bags-only \(= b - 45\), shoes-only \(= s - 45\).
(b) Substitute \(s = b + 11\) into \(b + s = 165\):
\[b + (b + 11) = 165 \Rightarrow 2b = 154 \Rightarrow b = 77, \quad s = 88.\]
The number who bought shoes is \(\mathbf{88}\).
(c) The number who bought bags is 77, so
\[P(\text{bought bags}) = \frac{77}{120}.\]
Checking the regions: bags-only \(= 32\), both \(= 45\), shoes-only \(= 43\); total \(= 120\).
Bayanin Amsa
Let \(b\) be the number who bought bags and \(s\) the number who bought shoes. Every customer bought at least one item, so \(n(\text{bags} \cup \text{shoes}) = 120\), with \(n(\text{both}) = 45\).
Using \(n(B \cup S) = b + s - n(\text{both})\):
\[120 = b + s - 45 \Rightarrow b + s = 165.\]
Also 11 more bought shoes than bags: \(s = b + 11\).
(a) Diagram. A two-circle Venn diagram (Bags and Shoes) with the overlap \(= 45\), bags-only \(= b - 45\), shoes-only \(= s - 45\).
(b) Substitute \(s = b + 11\) into \(b + s = 165\):
\[b + (b + 11) = 165 \Rightarrow 2b = 154 \Rightarrow b = 77, \quad s = 88.\]
The number who bought shoes is \(\mathbf{88}\).
(c) The number who bought bags is 77, so
\[P(\text{bought bags}) = \frac{77}{120}.\]
Checking the regions: bags-only \(= 32\), both \(= 45\), shoes-only \(= 43\); total \(= 120\).
Tambaya 11 Rahoto
(a) Solve the equation : \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\)
(b) 
In the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).
(a) Solve the equation.
\[\frac{2}{3}(3x-5)-\frac{3}{5}(2x-3)=3\]Multiply every term by \(15\) (the LCM of \(3\) and \(5\)):
\[15\cdot\frac{2}{3}(3x-5)-15\cdot\frac{3}{5}(2x-3)=15\cdot3\]\[10(3x-5)-9(2x-3)=45\]\[30x-50-18x+27=45\]\[12x-23=45\]\[12x=68\]\[x=\frac{68}{12}=\frac{17}{3}=\mathbf{5\tfrac{2}{3}}\](b) Find \(m+n\).
From the diagram, \(P,Q,R\) lie on a straight line and \(P,U,T,S\) lie on a straight line, with \(\angle STQ=m\), \(\angle TUQ=80^{\circ}\), \(\angle UPQ=r\), \(\angle PQU=n\) and \(\angle RQT=88^{\circ}\).
At \(U\): \(P,U,T\) are collinear, so \(\angle QUP\) and \(\angle QUT\) are angles on a straight line:
\[\angle QUP=180^{\circ}-80^{\circ}=100^{\circ}\]At \(Q\): \(P,Q,R\) are collinear, so the three angles on that line satisfy
\[\angle PQU+\angle UQT+\angle TQR=180^{\circ}\]\[n+\angle UQT+88^{\circ}=180^{\circ}\Rightarrow\angle UQT=92^{\circ}-n\]Triangle \(UQT\):
\[\angle QUT+\angle UQT+\angle UTQ=180^{\circ}\]\[80^{\circ}+(92^{\circ}-n)+\angle UTQ=180^{\circ}\Rightarrow\angle UTQ=8^{\circ}+n\]At \(T\): \(U,T,S\) are collinear, so \(\angle UTQ\) and \(\angle STQ(=m)\) are angles on a straight line:
\[m+\angle UTQ=180^{\circ}\]\[m+(8^{\circ}+n)=180^{\circ}\]\[m+n=\mathbf{172^{\circ}}\]Bayanin Amsa
(a) Solve the equation.
\[\frac{2}{3}(3x-5)-\frac{3}{5}(2x-3)=3\]Multiply every term by \(15\) (the LCM of \(3\) and \(5\)):
\[15\cdot\frac{2}{3}(3x-5)-15\cdot\frac{3}{5}(2x-3)=15\cdot3\]\[10(3x-5)-9(2x-3)=45\]\[30x-50-18x+27=45\]\[12x-23=45\]\[12x=68\]\[x=\frac{68}{12}=\frac{17}{3}=\mathbf{5\tfrac{2}{3}}\](b) Find \(m+n\).
From the diagram, \(P,Q,R\) lie on a straight line and \(P,U,T,S\) lie on a straight line, with \(\angle STQ=m\), \(\angle TUQ=80^{\circ}\), \(\angle UPQ=r\), \(\angle PQU=n\) and \(\angle RQT=88^{\circ}\).
At \(U\): \(P,U,T\) are collinear, so \(\angle QUP\) and \(\angle QUT\) are angles on a straight line:
\[\angle QUP=180^{\circ}-80^{\circ}=100^{\circ}\]At \(Q\): \(P,Q,R\) are collinear, so the three angles on that line satisfy
\[\angle PQU+\angle UQT+\angle TQR=180^{\circ}\]\[n+\angle UQT+88^{\circ}=180^{\circ}\Rightarrow\angle UQT=92^{\circ}-n\]Triangle \(UQT\):
\[\angle QUT+\angle UQT+\angle UTQ=180^{\circ}\]\[80^{\circ}+(92^{\circ}-n)+\angle UTQ=180^{\circ}\Rightarrow\angle UTQ=8^{\circ}+n\]At \(T\): \(U,T,S\) are collinear, so \(\angle UTQ\) and \(\angle STQ(=m)\) are angles on a straight line:
\[m+\angle UTQ=180^{\circ}\]\[m+(8^{\circ}+n)=180^{\circ}\]\[m+n=\mathbf{172^{\circ}}\]Tambaya 12 Rahoto
(a) If \((y - 1)\log_{10}4 = y\log_{10}16\), without using Mathematics tables or calculator, find the value of y.
(b) When I walk from my house at 4km/h, I will get to my office 30mins later than when I walk at 5km/h. Calculate the distance between my house and office.
(a) Given \((y - 1)\log_{10}4 = y\log_{10}16\).
Since \(16 = 4^2\), we have \(\log_{10}16 = 2\log_{10}4\). Substituting:
\[(y - 1)\log_{10}4 = y(2\log_{10}4).\]
Divide both sides by \(\log_{10}4\) (which is not zero):
\[y - 1 = 2y \Rightarrow -1 = y \Rightarrow y = -1.\]
(b) Let the distance from house to office be \(d\) km. Time at 4 km/h is \(\dfrac{d}{4}\) h; time at 5 km/h is \(\dfrac{d}{5}\) h. Walking at the slower speed takes 30 minutes \(\left(= \tfrac{1}{2}\text{ h}\right)\) longer:
\[\frac{d}{4} - \frac{d}{5} = \frac{1}{2}.\]
\[d\left(\frac{5 - 4}{20}\right) = \frac{1}{2} \Rightarrow \frac{d}{20} = \frac{1}{2} \Rightarrow d = 10.\]
The distance between the house and the office is \(\mathbf{10\text{ km}}\).
Bayanin Amsa
(a) Given \((y - 1)\log_{10}4 = y\log_{10}16\).
Since \(16 = 4^2\), we have \(\log_{10}16 = 2\log_{10}4\). Substituting:
\[(y - 1)\log_{10}4 = y(2\log_{10}4).\]
Divide both sides by \(\log_{10}4\) (which is not zero):
\[y - 1 = 2y \Rightarrow -1 = y \Rightarrow y = -1.\]
(b) Let the distance from house to office be \(d\) km. Time at 4 km/h is \(\dfrac{d}{4}\) h; time at 5 km/h is \(\dfrac{d}{5}\) h. Walking at the slower speed takes 30 minutes \(\left(= \tfrac{1}{2}\text{ h}\right)\) longer:
\[\frac{d}{4} - \frac{d}{5} = \frac{1}{2}.\]
\[d\left(\frac{5 - 4}{20}\right) = \frac{1}{2} \Rightarrow \frac{d}{20} = \frac{1}{2} \Rightarrow d = 10.\]
The distance between the house and the office is \(\mathbf{10\text{ km}}\).
Tambaya 13 Rahoto
(a) Using completing the square method, solve, correct to 2 decimal places, the equation \(3y^{2} - 5y + 2 = 0\).
(b) Given that \(M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}, N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\) and \(MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}\), find the matrix N.
(a) Solve \(3y^2 - 5y + 2 = 0\) by completing the square. Divide through by 3:
\[y^2 - \frac{5}{3}y + \frac{2}{3} = 0 \Rightarrow y^2 - \frac{5}{3}y = -\frac{2}{3}.\]
Half the coefficient of \(y\) is \(\dfrac{5}{6}\); add \(\left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}\) to both sides:
\[\left(y - \frac{5}{6}\right)^2 = -\frac{2}{3} + \frac{25}{36} = \frac{-24 + 25}{36} = \frac{1}{36}.\]
\[y - \frac{5}{6} = \pm\frac{1}{6} \Rightarrow y = \frac{5}{6} + \frac{1}{6} = 1 \quad\text{or}\quad y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}.\]
Hence \(y = 1.00\) or \(y = 0.67\) (2 d.p.).
(b) Let \(N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\). Then
\[MN = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} m & x \\ n & y \end{pmatrix} = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}.\]
First column: \(m + 2n = 2\) and \(4m + 3n = 3\). From the first, \(m = 2 - 2n\); substituting: \(4(2 - 2n) + 3n = 3 \Rightarrow 8 - 5n = 3 \Rightarrow n = 1,\; m = 0\).
Second column: \(x + 2y = 1\) and \(4x + 3y = 4\). From the first, \(x = 1 - 2y\); substituting: \(4(1 - 2y) + 3y = 4 \Rightarrow 4 - 5y = 4 \Rightarrow y = 0,\; x = 1\).
\[N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\]
Bayanin Amsa
(a) Solve \(3y^2 - 5y + 2 = 0\) by completing the square. Divide through by 3:
\[y^2 - \frac{5}{3}y + \frac{2}{3} = 0 \Rightarrow y^2 - \frac{5}{3}y = -\frac{2}{3}.\]
Half the coefficient of \(y\) is \(\dfrac{5}{6}\); add \(\left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}\) to both sides:
\[\left(y - \frac{5}{6}\right)^2 = -\frac{2}{3} + \frac{25}{36} = \frac{-24 + 25}{36} = \frac{1}{36}.\]
\[y - \frac{5}{6} = \pm\frac{1}{6} \Rightarrow y = \frac{5}{6} + \frac{1}{6} = 1 \quad\text{or}\quad y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}.\]
Hence \(y = 1.00\) or \(y = 0.67\) (2 d.p.).
(b) Let \(N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\). Then
\[MN = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} m & x \\ n & y \end{pmatrix} = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}.\]
First column: \(m + 2n = 2\) and \(4m + 3n = 3\). From the first, \(m = 2 - 2n\); substituting: \(4(2 - 2n) + 3n = 3 \Rightarrow 8 - 5n = 3 \Rightarrow n = 1,\; m = 0\).
Second column: \(x + 2y = 1\) and \(4x + 3y = 4\). From the first, \(x = 1 - 2y\); substituting: \(4(1 - 2y) + 3y = 4 \Rightarrow 4 - 5y = 4 \Rightarrow y = 0,\; x = 1\).
\[N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\]
