WAEC SSCE - General Mathematics - 1999 (Essay)
Tambaya 1 Rahoto
On a graph sheet, using a scale of 2cm to 2 units on both axes,
(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);
(b) construct the locus L of points equidistant from P and Q;
(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;
(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\(^{2}\).
The scale gives \(1\) coordinate unit \(=1\text{ cm}\). Therefore \(PQ=7\text{ cm}\), since \(P(-5,3)\) and \(Q(2,3)\) have the same \(y\)-coordinate.
Locus \(L\): Points equidistant from \(P\) and \(Q\) lie on the perpendicular bisector of \(PQ\). The midpoint of \(PQ\) is
\[ \left(\frac{-5+2}{2},\frac{3+3}{2}\right)=(-1.5,3). \]
Since \(PQ\) is horizontal, its perpendicular bisector is the vertical line \(x=-1.5\).
Construction of \(R\) and \(S\): Draw arcs of radius \(5\text{ cm}\), centred at \(P\) and at \(Q\). Their two intersections are \(R\) and \(S\). Each intersection is \(5\text{ cm}\) from both \(P\) and \(Q\), so \(PR=RQ=QS=SP=5\text{ cm}\), giving a rhombus.
Half of \(PQ\) is \(3.5\text{ cm}\). Using Pythagoras in the right triangle from the midpoint of \(PQ\) to \(R\):
\[ \text{vertical distance}=\sqrt{5^2-3.5^2} =\sqrt{25-12.25} =\sqrt{12.75} =\frac{\sqrt{51}}{2}\approx3.57. \]
Therefore, the coordinates are
\[ R=\left(-1.5,\;3+\frac{\sqrt{51}}{2}\right)\approx(-1.5,6.57), \]
\[ S=\left(-1.5,\;3-\frac{\sqrt{51}}{2}\right)\approx(-1.5,-0.57). \]
The diagonal \(PQ=7\text{ cm}\), and the other diagonal is
\[ RS=2\left(\frac{\sqrt{51}}{2}\right)=\sqrt{51}\text{ cm}. \]
Hence the area is
\[ \text{Area}=\frac{1}{2}\times PQ\times RS =\frac{1}{2}\times7\times\sqrt{51} \approx24.74\text{ cm}^2. \]
The supplied reference coordinates \((-1.6,6.5)\) and \((-1.6,0.6)\) are not consistent with the perpendicular bisector \(x=-1.5\), and the positive \(y\)-coordinate for \(S\) is incorrect. A construction read from a graph may reasonably give approximately \(R(-1.5,6.6)\) and \(S(-1.5,-0.6)\).
Bayanin Amsa
The scale gives \(1\) coordinate unit \(=1\text{ cm}\). Therefore \(PQ=7\text{ cm}\), since \(P(-5,3)\) and \(Q(2,3)\) have the same \(y\)-coordinate.
Locus \(L\): Points equidistant from \(P\) and \(Q\) lie on the perpendicular bisector of \(PQ\). The midpoint of \(PQ\) is
\[ \left(\frac{-5+2}{2},\frac{3+3}{2}\right)=(-1.5,3). \]
Since \(PQ\) is horizontal, its perpendicular bisector is the vertical line \(x=-1.5\).
Construction of \(R\) and \(S\): Draw arcs of radius \(5\text{ cm}\), centred at \(P\) and at \(Q\). Their two intersections are \(R\) and \(S\). Each intersection is \(5\text{ cm}\) from both \(P\) and \(Q\), so \(PR=RQ=QS=SP=5\text{ cm}\), giving a rhombus.
Half of \(PQ\) is \(3.5\text{ cm}\). Using Pythagoras in the right triangle from the midpoint of \(PQ\) to \(R\):
\[ \text{vertical distance}=\sqrt{5^2-3.5^2} =\sqrt{25-12.25} =\sqrt{12.75} =\frac{\sqrt{51}}{2}\approx3.57. \]
Therefore, the coordinates are
\[ R=\left(-1.5,\;3+\frac{\sqrt{51}}{2}\right)\approx(-1.5,6.57), \]
\[ S=\left(-1.5,\;3-\frac{\sqrt{51}}{2}\right)\approx(-1.5,-0.57). \]
The diagonal \(PQ=7\text{ cm}\), and the other diagonal is
\[ RS=2\left(\frac{\sqrt{51}}{2}\right)=\sqrt{51}\text{ cm}. \]
Hence the area is
\[ \text{Area}=\frac{1}{2}\times PQ\times RS =\frac{1}{2}\times7\times\sqrt{51} \approx24.74\text{ cm}^2. \]
The supplied reference coordinates \((-1.6,6.5)\) and \((-1.6,0.6)\) are not consistent with the perpendicular bisector \(x=-1.5\), and the positive \(y\)-coordinate for \(S\) is incorrect. A construction read from a graph may reasonably give approximately \(R(-1.5,6.6)\) and \(S(-1.5,-0.6)\).
Tambaya 2 Rahoto
(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.
(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.
(a) \(3y-2x=21\) and \(4y+5x=5\). Multiply the first by \(5\) and the second by \(2\): \[15y-10x=105,\qquad 8y+10x=10.\] Adding: \(23y=115\Rightarrow y=5\). Then \(3(5)-2x=21\Rightarrow-2x=6\Rightarrow x=-3\). So \(x=-3,\;y=5\).
(b) Card \(6\) wins \(\$60\); cards \(1\text{–}5\) lose \(\$10\times(\text{number})\).
- (i) A loss occurs whenever cards \(1,2,3,4,5\) are picked: \[P(\text{losing})=\frac{5}{6}.\]
- (ii) Losing exactly \(\$20\) in two picks: the outcomes per card are \(-10,-20,-30,-40,-50,+60\). The only pair of picks totalling \(-\$20\) is card \(1\) followed by card \(1\) \((-10-10=-20)\). \[P=\frac16\times\frac16=\frac{1}{36}.\]
Bayanin Amsa
(a) \(3y-2x=21\) and \(4y+5x=5\). Multiply the first by \(5\) and the second by \(2\): \[15y-10x=105,\qquad 8y+10x=10.\] Adding: \(23y=115\Rightarrow y=5\). Then \(3(5)-2x=21\Rightarrow-2x=6\Rightarrow x=-3\). So \(x=-3,\;y=5\).
(b) Card \(6\) wins \(\$60\); cards \(1\text{–}5\) lose \(\$10\times(\text{number})\).
- (i) A loss occurs whenever cards \(1,2,3,4,5\) are picked: \[P(\text{losing})=\frac{5}{6}.\]
- (ii) Losing exactly \(\$20\) in two picks: the outcomes per card are \(-10,-20,-30,-40,-50,+60\). The only pair of picks totalling \(-\$20\) is card \(1\) followed by card \(1\) \((-10-10=-20)\). \[P=\frac16\times\frac16=\frac{1}{36}.\]
Tambaya 3 Rahoto
(a) Simplify \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
(b) Evaluate \(\log_{5} (\frac{3}{5}) + 3 \log_{5} (\frac{5}{2}) - \log_{5} (\frac{81}{8})\).
(a) \[\left(\frac{4}{25}\right)^{-\frac12}=\left(\frac{25}{4}\right)^{\frac12}=\frac52,\qquad 2^{4}=16,\qquad \left(\frac{15}{2}\right)^{-2}=\left(\frac{2}{15}\right)^{2}=\frac{4}{225}.\] So \[\frac52\times16\div\frac{4}{225}=40\times\frac{225}{4}=2250.\]
(b) \[\log_{5}\!\frac35+3\log_{5}\!\frac52-\log_{5}\!\frac{81}{8}=\log_{5}\!\left(\frac35\cdot\left(\frac52\right)^{3}\cdot\frac{8}{81}\right)=\log_{5}\!\left(\frac35\cdot\frac{125}{8}\cdot\frac{8}{81}\right)=\log_{5}\!\frac{25}{27}.\] Hence the value is \[\log_{5}\frac{25}{27}=2-3\log_{5}3\approx-0.048.\]
Bayanin Amsa
(a) \[\left(\frac{4}{25}\right)^{-\frac12}=\left(\frac{25}{4}\right)^{\frac12}=\frac52,\qquad 2^{4}=16,\qquad \left(\frac{15}{2}\right)^{-2}=\left(\frac{2}{15}\right)^{2}=\frac{4}{225}.\] So \[\frac52\times16\div\frac{4}{225}=40\times\frac{225}{4}=2250.\]
(b) \[\log_{5}\!\frac35+3\log_{5}\!\frac52-\log_{5}\!\frac{81}{8}=\log_{5}\!\left(\frac35\cdot\left(\frac52\right)^{3}\cdot\frac{8}{81}\right)=\log_{5}\!\left(\frac35\cdot\frac{125}{8}\cdot\frac{8}{81}\right)=\log_{5}\!\frac{25}{27}.\] Hence the value is \[\log_{5}\frac{25}{27}=2-3\log_{5}3\approx-0.048.\]
Tambaya 4 Rahoto
The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(a) Find the cost when there are 44 pupils.
(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?
Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.
\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]
(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]
(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.
Bayanin Amsa
Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.
\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]
(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]
(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.
Tambaya 5 Rahoto
(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?
(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;
(i) the value of x ; (ii) the area of lawn.
(a) Let the cost price be \(C\). Marked to gain \(20\%\): marked \(=1.2C\). Sold at \(10\%\) reduction: \[0.9\times1.2C=1.08C=864\Rightarrow C=\frac{864}{1.08}=\text{N}800.\] The original cost was N800.
(b) Length \((x+5)\), width \((x-2)\), diagonal \((x+6)\). By Pythagoras: \[(x+5)^{2}+(x-2)^{2}=(x+6)^{2}.\] \[x^{2}+10x+25+x^{2}-4x+4=x^{2}+12x+36\Rightarrow x^{2}-6x-7=0\Rightarrow(x-7)(x+1)=0.\] Taking the positive value, \(x=7\).
(ii) Area \(=(x+5)(x-2)=12\times5=60\text{ m}^{2}.\)
Bayanin Amsa
(a) Let the cost price be \(C\). Marked to gain \(20\%\): marked \(=1.2C\). Sold at \(10\%\) reduction: \[0.9\times1.2C=1.08C=864\Rightarrow C=\frac{864}{1.08}=\text{N}800.\] The original cost was N800.
(b) Length \((x+5)\), width \((x-2)\), diagonal \((x+6)\). By Pythagoras: \[(x+5)^{2}+(x-2)^{2}=(x+6)^{2}.\] \[x^{2}+10x+25+x^{2}-4x+4=x^{2}+12x+36\Rightarrow x^{2}-6x-7=0\Rightarrow(x-7)(x+1)=0.\] Taking the positive value, \(x=7\).
(ii) Area \(=(x+5)(x-2)=12\times5=60\text{ m}^{2}.\)
Tambaya 6 Rahoto
(a) Copy and complete the following table of values for \(y = 9 \cos x + 5 \sin x\) to one decimal place.
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 10.3 | -0.2 | -5.3 | -10.3 |
(b) Using a scale of 2cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 9 \cos x + 5 \sin x\) for \(0° \leq x \leq 210°\).
(c) Use your graph to solve the equation: (i) \(9\cos x + 5\sin x = 0\); (ii) \(9\cos x+ 5\sin x = 3.5\), correct to the nearest degree.
(d) Find the maximum value of y correct to one decimal place.
(a) Completing the table of values
For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):
\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 9.0 | 10.3 | 8.8 | 5.0 | -0.2 | -5.3 | -9.0 | -10.3 |
(b) Graph of \(y = 9\cos x + 5\sin x\)
Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).
(c) Solving from the graph
(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:
\[ x \approx 119^\circ \](Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))
(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:
\[ x \approx 99^\circ \](Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))
(d) Maximum value of y
The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):
\[ y_{\max}=\sqrt{81+25}=\sqrt{106}\approx 10.3 \]The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).
Bayanin Amsa
(a) Completing the table of values
For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):
\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 9.0 | 10.3 | 8.8 | 5.0 | -0.2 | -5.3 | -9.0 | -10.3 |
(b) Graph of \(y = 9\cos x + 5\sin x\)
Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).
(c) Solving from the graph
(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:
\[ x \approx 119^\circ \](Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))
(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:
\[ x \approx 99^\circ \](Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))
(d) Maximum value of y
The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):
\[ y_{\max}=\sqrt{81+25}=\sqrt{106}\approx 10.3 \]The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).
Tambaya 7 Rahoto
In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.
(a) How high up the wall, correct to the nearest metre, does the ladder reach?
(b) If the foot (F) of the ladder is pulled from the wall to F\(^{1}\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\(^{1}\).
(ii) Calculate, correct to the nearest degree, \(QF^{1}T^{1}\).
Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.
(a) Height reached up the wall. In right triangle \(TQF\):
\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\](b)(i) Distance the top slides down. First the original foot distance:
\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):
\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]Distance slid down:
\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\](b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):
\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]Bayanin Amsa
Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.
(a) Height reached up the wall. In right triangle \(TQF\):
\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\](b)(i) Distance the top slides down. First the original foot distance:
\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):
\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]Distance slid down:
\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\](b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):
\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]Tambaya 8 Rahoto
The table gives the frequency distribution of marks obtained by a group of students in a test.
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | \(x - 1\) | \(x\) | 9 | 4 | 1 |
If the mean is 5,
(a) Calculate the value of x;
(b) Find the : (i) mode ; (ii) median of the distribution.
(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x-1 | x | 9 | 4 | 1 |
(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):
\[ \Sigma fx = 15+4(x-1)+5x+54+28+8 = 101+9x \]Since mean \(=5\):
\[ \frac{101+9x}{18+2x}=5 \Rightarrow 101+9x=90+10x \Rightarrow x=11 \]So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).
(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.
(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.
(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):
\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]Bayanin Amsa
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x-1 | x | 9 | 4 | 1 |
(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):
\[ \Sigma fx = 15+4(x-1)+5x+54+28+8 = 101+9x \]Since mean \(=5\):
\[ \frac{101+9x}{18+2x}=5 \Rightarrow 101+9x=90+10x \Rightarrow x=11 \]So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).
(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.
(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.
(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):
\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]Tambaya 9 Rahoto
ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37° and |BC| = 5 cm. Calculate :
(a) |AC|, correct to 3 significant figures ;
(b) < PBA.
In \(\triangle PBC\), the right angle is at \(C\), \(|BC|=5\text{ cm}\), \(\angle PBC=37^{\circ}\), and \(P\) is the midpoint of \(AC\).
(a) \[\tan37^{\circ}=\frac{PC}{BC}\Rightarrow PC=5\tan37^{\circ}=5\times0.7536=3.768\text{ cm}.\] Since \(P\) is the midpoint, \(|AC|=2\times PC=7.536\approx\) 7.54 cm (3 s.f.).
(b) In \(\triangle ABC\), \[\tan(\angle ABC)=\frac{AC}{BC}=\frac{7.536}{5}=1.5072\Rightarrow\angle ABC=56.44^{\circ}.\] Therefore \[\angle PBA=\angle ABC-\angle PBC=56.44^{\circ}-37^{\circ}\approx19.4^{\circ}.\]
Bayanin Amsa
In \(\triangle PBC\), the right angle is at \(C\), \(|BC|=5\text{ cm}\), \(\angle PBC=37^{\circ}\), and \(P\) is the midpoint of \(AC\).
(a) \[\tan37^{\circ}=\frac{PC}{BC}\Rightarrow PC=5\tan37^{\circ}=5\times0.7536=3.768\text{ cm}.\] Since \(P\) is the midpoint, \(|AC|=2\times PC=7.536\approx\) 7.54 cm (3 s.f.).
(b) In \(\triangle ABC\), \[\tan(\angle ABC)=\frac{AC}{BC}=\frac{7.536}{5}=1.5072\Rightarrow\angle ABC=56.44^{\circ}.\] Therefore \[\angle PBA=\angle ABC-\angle PBC=56.44^{\circ}-37^{\circ}\approx19.4^{\circ}.\]
Tambaya 10 Rahoto
(a) Given that \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). Without using mathematical tables or calculator, evaluate \(\sin 105°\), leaving your answer in the surd form.
(You may use 105° = 60° + 45°)
(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?
(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \(2\) and \(\frac{2}{9}\) respectively. Find :
(i) the common difference ; (ii) the 5th term.
(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]
(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.
(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]
Bayanin Amsa
(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]
(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.
(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]
Tambaya 11 Rahoto
(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\(^{3}\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \(\pi = \frac{22}{7}\)]
(b) 
In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.
(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).
(i) Volume of metal used.
The metal is the hollow shell between the outer and inner cylinders:
\[V_{\text{metal}}=\pi(R^2-r^2)\times L\]\[R^2-r^2=5^2-3.5^2=25-12.25=12.75\text{ cm}^2\]\[V_{\text{metal}}=\frac{22}{7}\times 12.75\times 2800\]\[=\frac{22}{7}\times 35700=22\times 5100=112200\text{ cm}^3\]The volume of metal is \(112200\text{ cm}^3\).
(ii) Volume of water the pipe holds when full.
This is the inner cylinder's volume:
\[V_{\text{water}}=\pi r^2 L=\frac{22}{7}\times 3.5^2\times 2800\]\[=\frac{22}{7}\times 12.25\times 2800=\frac{22}{7}\times 34300=22\times 4900=107800\text{ cm}^3\]Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):
\[\frac{107800}{1000}=107.8\text{ litres}\]The pipe holds \(107.8\) litres (to 1 d.p.).
(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.
By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):
\[\angle PMN=\angle MLN \quad\text{...(1)}\]Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:
\[\angle PMN=\angle MNL \quad\text{...(2)}\]From (1) and (2):
\[\angle MLN=\angle MNL\]In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).
Therefore triangle \(LMN\) is isosceles. (Q.E.D.)
Bayanin Amsa
(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).
(i) Volume of metal used.
The metal is the hollow shell between the outer and inner cylinders:
\[V_{\text{metal}}=\pi(R^2-r^2)\times L\]\[R^2-r^2=5^2-3.5^2=25-12.25=12.75\text{ cm}^2\]\[V_{\text{metal}}=\frac{22}{7}\times 12.75\times 2800\]\[=\frac{22}{7}\times 35700=22\times 5100=112200\text{ cm}^3\]The volume of metal is \(112200\text{ cm}^3\).
(ii) Volume of water the pipe holds when full.
This is the inner cylinder's volume:
\[V_{\text{water}}=\pi r^2 L=\frac{22}{7}\times 3.5^2\times 2800\]\[=\frac{22}{7}\times 12.25\times 2800=\frac{22}{7}\times 34300=22\times 4900=107800\text{ cm}^3\]Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):
\[\frac{107800}{1000}=107.8\text{ litres}\]The pipe holds \(107.8\) litres (to 1 d.p.).
(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.
By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):
\[\angle PMN=\angle MLN \quad\text{...(1)}\]Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:
\[\angle PMN=\angle MNL \quad\text{...(2)}\]From (1) and (2):
\[\angle MLN=\angle MNL\]In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).
Therefore triangle \(LMN\) is isosceles. (Q.E.D.)
Tambaya 12 Rahoto
In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :
(a) |AB| ;
(b) the area of the triangle BCD ;
(c) |CD| ;
(d) perimeter of the trapezium.
Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).
(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):
\[|AB|^2 = |BD|^2 - |AD|^2 = 17^2 - 15^2 = 289 - 225 = 64,\]\[|AB| = \sqrt{64} = \mathbf{8\text{ cm}}.\](b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):
\[\text{Area} = \tfrac{1}{2}\times |BC| \times |AB| = \tfrac{1}{2}\times 9 \times 8 = \mathbf{36\text{ cm}^2}.\](c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then
\[|CD| = \sqrt{(15-9)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = \mathbf{10\text{ cm}}.\](d) Perimeter of the trapezium.
\[P = |AB| + |BC| + |CD| + |DA| = 8 + 9 + 10 + 15 = \mathbf{42\text{ cm}}.\]Bayanin Amsa
Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).
(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):
\[|AB|^2 = |BD|^2 - |AD|^2 = 17^2 - 15^2 = 289 - 225 = 64,\]\[|AB| = \sqrt{64} = \mathbf{8\text{ cm}}.\](b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):
\[\text{Area} = \tfrac{1}{2}\times |BC| \times |AB| = \tfrac{1}{2}\times 9 \times 8 = \mathbf{36\text{ cm}^2}.\](c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then
\[|CD| = \sqrt{(15-9)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = \mathbf{10\text{ cm}}.\](d) Perimeter of the trapezium.
\[P = |AB| + |BC| + |CD| + |DA| = 8 + 9 + 10 + 15 = \mathbf{42\text{ cm}}.\]Tambaya 13 Rahoto
(a) If \(\varepsilon\) is the set \({1, 2, 3,..., 19, 20}\) and A, B and C are subsets of \(\varepsilon\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;
(b) Find : (i) \(A \cap B\) ; (ii) \(A \cap C\) ; (iii) \(B \cup C\).
(c) Using your results in (b), show that \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\).
(a) With \(\varepsilon=\{1,2,\dots,20\}\):
- (i) \(A=\{\text{multiples of }5\}=\{5,10,15,20\}.\)
- (ii) \(B=\{\text{multiples of }4\}=\{4,8,12,16,20\}.\)
- (iii) \(C=\{\text{multiples of }3\}=\{3,6,9,12,15,18\}.\)
(b)
- (i) \(A\cap B=\{20\}.\)
- (ii) \(A\cap C=\{15\}.\)
- (iii) \(B\cup C=\{3,4,6,8,9,12,15,16,18,20\}.\)
(c) \[(A\cap B)\cup(A\cap C)=\{20\}\cup\{15\}=\{15,20\}.\] Also \(A\cap(B\cup C)\): from \(A=\{5,10,15,20\}\), those in \(B\cup C\) are \(\{15,20\}\). Since both sides equal \(\{15,20\}\), the identity \((A\cap B)\cup(A\cap C)=A\cap(B\cup C)\) is verified.
Bayanin Amsa
(a) With \(\varepsilon=\{1,2,\dots,20\}\):
- (i) \(A=\{\text{multiples of }5\}=\{5,10,15,20\}.\)
- (ii) \(B=\{\text{multiples of }4\}=\{4,8,12,16,20\}.\)
- (iii) \(C=\{\text{multiples of }3\}=\{3,6,9,12,15,18\}.\)
(b)
- (i) \(A\cap B=\{20\}.\)
- (ii) \(A\cap C=\{15\}.\)
- (iii) \(B\cup C=\{3,4,6,8,9,12,15,16,18,20\}.\)
(c) \[(A\cap B)\cup(A\cap C)=\{20\}\cup\{15\}=\{15,20\}.\] Also \(A\cap(B\cup C)\): from \(A=\{5,10,15,20\}\), those in \(B\cup C\) are \(\{15,20\}\). Since both sides equal \(\{15,20\}\), the identity \((A\cap B)\cup(A\cap C)=A\cap(B\cup C)\) is verified.
