WAEC SSCE - General Mathematics - 1998 (Essay)
Tambaya 1 Rahoto
(a) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. What is the probability that the sum of the two numbers will be less than 7 but greater than 3?
(b) 
In the diagram, ABCD is a circle. DAE, CBE, ABF and DCF are straight lines. If y + m = 90°, find the value of x.
(a) One number is chosen from \(\{2,3,4\}\) and one from \(\{1,3,5\}\). There are \(3\times 3 = 9\) equally likely pairs. We need the sum \(S\) to satisfy \(3 < S < 7\), i.e. \(S \in \{4,5,6\}\).
| + | 1 | 3 | 5 |
|---|---|---|---|
| 2 | 3 | 5 | 7 |
| 3 | 4 | 6 | 8 |
| 4 | 5 | 7 | 9 |
The sums strictly between \(3\) and \(7\) are: \((2,3)=5,\;(3,1)=4,\;(3,3)=6,\;(4,1)=5\) - that is \(4\) favourable outcomes.
\[P(3 < S < 7) = \frac{4}{9}.\](b) \(ABCD\) is a circle. \(DAE\) and \(CBE\) are straight lines meeting at the outer point \(E\) (with \(\angle E = m\)); \(ABF\) and \(DCF\) are straight lines meeting at the outer point \(F\) (with \(\angle F = y\)). Let the four arcs be \(AB = p,\; BC = q,\; CD = r,\; DA = s\), so \(p+q+r+s = 360^\circ\).
Angle at each external point = half the difference of the intercepted arcs.
\[m = \angle E = \tfrac12(\text{arc }DC - \text{arc }AB) = \tfrac12(r-p),\qquad y = \angle F = \tfrac12(\text{arc }AD - \text{arc }BC) = \tfrac12(s-q).\]Given \(y + m = 90^\circ\):
\[\tfrac12(r-p) + \tfrac12(s-q) = 90^\circ \;\Rightarrow\; (r+s) - (p+q) = 180^\circ.\]Combine with \(p+q+r+s = 360^\circ\). Adding the two equations gives \(2(r+s) = 540^\circ\), so
\[r+s = 270^\circ,\qquad p+q = 90^\circ.\]At vertex \(B\) the straight lines \(CBE\) and \(ABF\) cross, and \(x\) is the angle \(\angle ABE = \angle FBC\) (vertically opposite), which is the supplement of the interior angle \(\angle ABC\). Now \(\angle ABC\) is inscribed on arc \(ADC\):
\[\angle ABC = \tfrac12(\text{arc }AD + \text{arc }DC) = \tfrac12(s+r) = \tfrac12(270^\circ) = 135^\circ.\]Therefore
\[x = 180^\circ - \angle ABC = 180^\circ - 135^\circ = \boxed{45^\circ}.\]Bayanin Amsa
(a) One number is chosen from \(\{2,3,4\}\) and one from \(\{1,3,5\}\). There are \(3\times 3 = 9\) equally likely pairs. We need the sum \(S\) to satisfy \(3 < S < 7\), i.e. \(S \in \{4,5,6\}\).
| + | 1 | 3 | 5 |
|---|---|---|---|
| 2 | 3 | 5 | 7 |
| 3 | 4 | 6 | 8 |
| 4 | 5 | 7 | 9 |
The sums strictly between \(3\) and \(7\) are: \((2,3)=5,\;(3,1)=4,\;(3,3)=6,\;(4,1)=5\) - that is \(4\) favourable outcomes.
\[P(3 < S < 7) = \frac{4}{9}.\](b) \(ABCD\) is a circle. \(DAE\) and \(CBE\) are straight lines meeting at the outer point \(E\) (with \(\angle E = m\)); \(ABF\) and \(DCF\) are straight lines meeting at the outer point \(F\) (with \(\angle F = y\)). Let the four arcs be \(AB = p,\; BC = q,\; CD = r,\; DA = s\), so \(p+q+r+s = 360^\circ\).
Angle at each external point = half the difference of the intercepted arcs.
\[m = \angle E = \tfrac12(\text{arc }DC - \text{arc }AB) = \tfrac12(r-p),\qquad y = \angle F = \tfrac12(\text{arc }AD - \text{arc }BC) = \tfrac12(s-q).\]Given \(y + m = 90^\circ\):
\[\tfrac12(r-p) + \tfrac12(s-q) = 90^\circ \;\Rightarrow\; (r+s) - (p+q) = 180^\circ.\]Combine with \(p+q+r+s = 360^\circ\). Adding the two equations gives \(2(r+s) = 540^\circ\), so
\[r+s = 270^\circ,\qquad p+q = 90^\circ.\]At vertex \(B\) the straight lines \(CBE\) and \(ABF\) cross, and \(x\) is the angle \(\angle ABE = \angle FBC\) (vertically opposite), which is the supplement of the interior angle \(\angle ABC\). Now \(\angle ABC\) is inscribed on arc \(ADC\):
\[\angle ABC = \tfrac12(\text{arc }AD + \text{arc }DC) = \tfrac12(s+r) = \tfrac12(270^\circ) = 135^\circ.\]Therefore
\[x = 180^\circ - \angle ABC = 180^\circ - 135^\circ = \boxed{45^\circ}.\]Tambaya 2 Rahoto
(a) Given that \(\log_{10} 2 = 0.3010, \log_{10} 7 = 0.8451\) and \(\log_{10} 5 = 0.6990\), evaluate without using logarithm tables:
(i) \(\log_{10} 35\); (ii) \(\log_{10} 2.8\).
(b) Given that \(N^{0.8942} = 2.8\), use your result in (a)(ii) to find the value of N.
(a)(i) \(35 = 5\times7\), so by the addition law of logarithms:
\[\log_{10}35 = \log_{10}5 + \log_{10}7 = 0.6990 + 0.8451 = 1.5441.\]
(a)(ii) \(2.8 = \dfrac{28}{10} = \dfrac{4\times7}{10} = \dfrac{2^2\times7}{10}\):
\[\log_{10}2.8 = 2\log_{10}2 + \log_{10}7 - \log_{10}10 = 2(0.3010) + 0.8451 - 1 = 0.4471.\]
(b) Take logarithms of \(N^{0.8942} = 2.8\):
\[0.8942\log_{10}N = \log_{10}2.8 = 0.4471 \Rightarrow \log_{10}N = \frac{0.4471}{0.8942} = 0.5.\]
\[N = 10^{0.5} = \sqrt{10} = 3.16.\]
Bayanin Amsa
(a)(i) \(35 = 5\times7\), so by the addition law of logarithms:
\[\log_{10}35 = \log_{10}5 + \log_{10}7 = 0.6990 + 0.8451 = 1.5441.\]
(a)(ii) \(2.8 = \dfrac{28}{10} = \dfrac{4\times7}{10} = \dfrac{2^2\times7}{10}\):
\[\log_{10}2.8 = 2\log_{10}2 + \log_{10}7 - \log_{10}10 = 2(0.3010) + 0.8451 - 1 = 0.4471.\]
(b) Take logarithms of \(N^{0.8942} = 2.8\):
\[0.8942\log_{10}N = \log_{10}2.8 = 0.4471 \Rightarrow \log_{10}N = \frac{0.4471}{0.8942} = 0.5.\]
\[N = 10^{0.5} = \sqrt{10} = 3.16.\]
Tambaya 3 Rahoto
The table below shows the number of eggs laid by the chickens in a man's farm in a year.
| No of eggs per year | No of chickens |
| 45 - 49 | 10 |
| 50 - 54 | 36 |
| 55 - 59 | 64 |
| 60 - 64 | 52 |
| 65 - 69 | 28 |
| 70 - 74 | 10 |
(a) Draw a cumulative frequency curve for the distribution.
(b) Use your graph to find the interquartile range.
(c) If a woman buys a chicken from the farm, what is the probability that the chicken lays at least 60 eggs in a year?
(a) Cumulative frequency table
| Number of eggs per year | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 45 - 49 | 44.5 - 49.5 | 10 | 10 |
| 50 - 54 | 49.5 - 54.5 | 36 | 46 |
| 55 - 59 | 54.5 - 59.5 | 64 | 110 |
| 60 - 64 | 59.5 - 64.5 | 52 | 162 |
| 65 - 69 | 64.5 - 69.5 | 28 | 190 |
| 70 - 74 | 69.5 - 74.5 | 10 | 200 |
The total number of chickens is \(N=200\). Plot the upper class boundaries against the cumulative frequencies, beginning with \((44.5,0)\), and join the points with a smooth curve.
(b) Interquartile range
\[Q_1=\frac{N}{4}=\frac{200}{4}=50,\qquad Q_3=\frac{3N}{4}=\frac{3(200)}{4}=150.\]
From the cumulative frequency curve, the corresponding egg numbers are approximately \(Q_1=54\) and \(Q_3=63\).
\[\text{Interquartile range}=Q_3-Q_1=63-54=9\text{ eggs}.\]
(c) Chickens laying at least \(60\) eggs per year:
\[52+28+10=90.\]
\[P(\text{at least }60\text{ eggs})=\frac{90}{200}=\frac{9}{20}=0.45.\]
Bayanin Amsa
(a) Cumulative frequency table
| Number of eggs per year | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 45 - 49 | 44.5 - 49.5 | 10 | 10 |
| 50 - 54 | 49.5 - 54.5 | 36 | 46 |
| 55 - 59 | 54.5 - 59.5 | 64 | 110 |
| 60 - 64 | 59.5 - 64.5 | 52 | 162 |
| 65 - 69 | 64.5 - 69.5 | 28 | 190 |
| 70 - 74 | 69.5 - 74.5 | 10 | 200 |
The total number of chickens is \(N=200\). Plot the upper class boundaries against the cumulative frequencies, beginning with \((44.5,0)\), and join the points with a smooth curve.
(b) Interquartile range
\[Q_1=\frac{N}{4}=\frac{200}{4}=50,\qquad Q_3=\frac{3N}{4}=\frac{3(200)}{4}=150.\]
From the cumulative frequency curve, the corresponding egg numbers are approximately \(Q_1=54\) and \(Q_3=63\).
\[\text{Interquartile range}=Q_3-Q_1=63-54=9\text{ eggs}.\]
(c) Chickens laying at least \(60\) eggs per year:
\[52+28+10=90.\]
\[P(\text{at least }60\text{ eggs})=\frac{90}{200}=\frac{9}{20}=0.45.\]
Tambaya 4 Rahoto
(a) The value of the expression \(2Ax - Kx^{2}\) is 7 when x = 1 and 4 when x = 2. Find the values of the constants A and K.
(b) Solve the equation \(x^{2} - 3x - 1 = 0\), giving your answers correct to 1 decimal place.
(a) Substitute the two conditions into \(2Ax - Kx^2\).
At \(x = 1\): \(2A - K = 7\). \quad(1)
At \(x = 2\): \(4A - 4K = 4\Rightarrow A - K = 1\). \quad(2)
Subtract (2) from (1): \((2A - K) - (A - K) = 7 - 1\Rightarrow A = 6\). Then from (2), \(K = A - 1 = 5\).
\[A = 6,\qquad K = 5.\]
(b) Solve \(x^2 - 3x - 1 = 0\) using the quadratic formula:
\[x = \frac{-(-3)\pm\sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3\pm\sqrt{13}}{2} = \frac{3\pm3.606}{2}.\]
\[x = 3.3\quad\text{or}\quad x = -0.3\ (\text{to 1 d.p.}).\]
Bayanin Amsa
(a) Substitute the two conditions into \(2Ax - Kx^2\).
At \(x = 1\): \(2A - K = 7\). \quad(1)
At \(x = 2\): \(4A - 4K = 4\Rightarrow A - K = 1\). \quad(2)
Subtract (2) from (1): \((2A - K) - (A - K) = 7 - 1\Rightarrow A = 6\). Then from (2), \(K = A - 1 = 5\).
\[A = 6,\qquad K = 5.\]
(b) Solve \(x^2 - 3x - 1 = 0\) using the quadratic formula:
\[x = \frac{-(-3)\pm\sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3\pm\sqrt{13}}{2} = \frac{3\pm3.606}{2}.\]
\[x = 3.3\quad\text{or}\quad x = -0.3\ (\text{to 1 d.p.}).\]
Tambaya 5 Rahoto
Using a ruler and a pair of compasses only, construct (a) triangle QRT with |QR| = 8cm, |RT| = 6cm and |QT| = 4.5cm.
(b) a quadrilateral QRSP which has a common base QR with \(\Delta\)QRT such that QTP is a straight line, PQ || SR, |QP| = 9cm and |RS| = 4.5cm.
(i) Measure |PS| ; (ii) Find the perpendicular distance between RS and PQ ; (iii) What is QRSP?
(a) Construction of \(\triangle QRT\)
- Draw \(QR=8\text{ cm}\).
- With centre \(Q\) and radius \(4.5\text{ cm}\), draw an arc.
- With centre \(R\) and radius \(6\text{ cm}\), draw an arc cutting the first arc at \(T\).
- Join \(QT\) and \(RT\).
(b) Construction of \(QRSP\)
- Produce \(QT\) through \(T\), and mark \(P\) on it such that \(QP=9\text{ cm}\).
- Through \(R\), construct a line parallel to \(QP\). Mark \(S\) on this line so that \(RS=4.5\text{ cm}\).
- Join \(P\) to \(S\). Drop \(RK\perp PQ\), where \(K\) lies on \(PQ\).
(i) From the construction, \[|PS|=6.0\text{ cm}.\]
(ii) The perpendicular distance between \(RS\) and \(PQ\) is \(RK\): \[RK=5.9\text{ cm}.\]
(iii) \(PQ\parallel RS\), while \(PQ=9\text{ cm}\) and \(RS=4.5\text{ cm}\). Thus, \(QRSP\) is a trapezium.
Bayanin Amsa
(a) Construction of \(\triangle QRT\)
- Draw \(QR=8\text{ cm}\).
- With centre \(Q\) and radius \(4.5\text{ cm}\), draw an arc.
- With centre \(R\) and radius \(6\text{ cm}\), draw an arc cutting the first arc at \(T\).
- Join \(QT\) and \(RT\).
(b) Construction of \(QRSP\)
- Produce \(QT\) through \(T\), and mark \(P\) on it such that \(QP=9\text{ cm}\).
- Through \(R\), construct a line parallel to \(QP\). Mark \(S\) on this line so that \(RS=4.5\text{ cm}\).
- Join \(P\) to \(S\). Drop \(RK\perp PQ\), where \(K\) lies on \(PQ\).
(i) From the construction, \[|PS|=6.0\text{ cm}.\]
(ii) The perpendicular distance between \(RS\) and \(PQ\) is \(RK\): \[RK=5.9\text{ cm}.\]
(iii) \(PQ\parallel RS\), while \(PQ=9\text{ cm}\) and \(RS=4.5\text{ cm}\). Thus, \(QRSP\) is a trapezium.
Tambaya 6 Rahoto
A surveyor standing at a point X sights a pole Y due east of him and a tower Z of a building on a bearing of 046°. After walking to a point W, a distance of 180m in the South- East direction, he observes the bearing of Z and Y to be 337° and 050° respectively.
(a) Calculate, correct to the nearest metre : (i) |XY| ; (ii) |ZW|
(b) If N is on XY such that XZ = ZN, find the bearing of Z from N.
Place \(X\) at the origin. \(Y\) is due east, and \(W\) is \(180\,\text{m}\) from \(X\) on bearing \(135^\circ\) (South-East), giving \(W = (180\sin135^\circ,\ 180\cos135^\circ) = (127.28,\ -127.28)\).
(a)(i) Finding \(|XY|\): \(Y = (y, 0)\). From \(W\) the bearing of \(Y\) is \(050^\circ\), so \[\tan50^\circ = \frac{y - 127.28}{0 - (-127.28)} = \frac{y - 127.28}{127.28}.\] \[y - 127.28 = 127.28\tan50^\circ = 151.68 \Rightarrow y = 278.96.\] Therefore \(|XY| \approx \mathbf{279\,\text{m}}\).
(a)(ii) Finding \(|ZW|\): \(Z\) lies on bearing \(046^\circ\) from \(X\): \(Z = t(\sin46^\circ,\cos46^\circ)\). From \(W\), \(Z\) is on bearing \(337^\circ\): \(Z = W + s(\sin337^\circ,\cos337^\circ)\). Solving \[0.7193t + 0.3907s = 127.28,\qquad 0.6947t - 0.9205s = -127.28,\] gives \(t = |XZ| = 72.2\) and \(s = |ZW| \approx \mathbf{193\,\text{m}}\), with \(Z = (51.95,\ 50.17)\).
(b) Bearing of \(Z\) from \(N\): \(N\) is on \(XY\) (the east line) with \(|XZ| = |ZN|\). Since \(Z = (51.95, 50.17)\) and \(|XZ| = 72.2\), \[(n - 51.95)^2 + 50.17^2 = 72.2^2 \Rightarrow n = 103.9\ (\text{taking } n \neq 0).\] So \(N = (103.9, 0)\) and \(\vec{NZ} = (-51.95,\ 50.17)\), which points North-West. \[\text{angle west of north} = \tan^{-1}\!\frac{51.95}{50.17} = 46^\circ \Rightarrow \text{bearing} = 360^\circ - 46^\circ = \mathbf{314^\circ}.\]
Bayanin Amsa
Place \(X\) at the origin. \(Y\) is due east, and \(W\) is \(180\,\text{m}\) from \(X\) on bearing \(135^\circ\) (South-East), giving \(W = (180\sin135^\circ,\ 180\cos135^\circ) = (127.28,\ -127.28)\).
(a)(i) Finding \(|XY|\): \(Y = (y, 0)\). From \(W\) the bearing of \(Y\) is \(050^\circ\), so \[\tan50^\circ = \frac{y - 127.28}{0 - (-127.28)} = \frac{y - 127.28}{127.28}.\] \[y - 127.28 = 127.28\tan50^\circ = 151.68 \Rightarrow y = 278.96.\] Therefore \(|XY| \approx \mathbf{279\,\text{m}}\).
(a)(ii) Finding \(|ZW|\): \(Z\) lies on bearing \(046^\circ\) from \(X\): \(Z = t(\sin46^\circ,\cos46^\circ)\). From \(W\), \(Z\) is on bearing \(337^\circ\): \(Z = W + s(\sin337^\circ,\cos337^\circ)\). Solving \[0.7193t + 0.3907s = 127.28,\qquad 0.6947t - 0.9205s = -127.28,\] gives \(t = |XZ| = 72.2\) and \(s = |ZW| \approx \mathbf{193\,\text{m}}\), with \(Z = (51.95,\ 50.17)\).
(b) Bearing of \(Z\) from \(N\): \(N\) is on \(XY\) (the east line) with \(|XZ| = |ZN|\). Since \(Z = (51.95, 50.17)\) and \(|XZ| = 72.2\), \[(n - 51.95)^2 + 50.17^2 = 72.2^2 \Rightarrow n = 103.9\ (\text{taking } n \neq 0).\] So \(N = (103.9, 0)\) and \(\vec{NZ} = (-51.95,\ 50.17)\), which points North-West. \[\text{angle west of north} = \tan^{-1}\!\frac{51.95}{50.17} = 46^\circ \Rightarrow \text{bearing} = 360^\circ - 46^\circ = \mathbf{314^\circ}.\]
Tambaya 7 Rahoto
The area of a rectangular floor is 13.5m\(^{2}\). One side is 1.5m longer than the other.
(a) Calculate the dimensions of the floor ;
(b) If it costs N250.00 per square metre to carpet the floor and only N2,000.00 is available, what area of the floor can be covered with carpet?
(a) Dimensions. Let the shorter side be \(x\) m; the longer side is \(x + 1.5\) m. The area is 13.5 m²:
\[x(x + 1.5) = 13.5 \Rightarrow x^2 + 1.5x - 13.5 = 0.\]
Multiply through by 2: \(2x^2 + 3x - 27 = 0\), which factorises as \((2x + 9)(x - 3) = 0\).
Taking the positive root, \(x = 3\). So the dimensions are
\[3\text{ m by }4.5\text{ m}.\]
(b) Area that can be carpeted. At N250.00 per square metre with N2,000.00 available:
\[\text{Area} = \frac{2000}{250} = 8\text{ m}^2.\]
So 8 m² of the floor can be covered (the floor is 13.5 m², so it cannot all be carpeted).
Bayanin Amsa
(a) Dimensions. Let the shorter side be \(x\) m; the longer side is \(x + 1.5\) m. The area is 13.5 m²:
\[x(x + 1.5) = 13.5 \Rightarrow x^2 + 1.5x - 13.5 = 0.\]
Multiply through by 2: \(2x^2 + 3x - 27 = 0\), which factorises as \((2x + 9)(x - 3) = 0\).
Taking the positive root, \(x = 3\). So the dimensions are
\[3\text{ m by }4.5\text{ m}.\]
(b) Area that can be carpeted. At N250.00 per square metre with N2,000.00 available:
\[\text{Area} = \frac{2000}{250} = 8\text{ m}^2.\]
So 8 m² of the floor can be covered (the floor is 13.5 m², so it cannot all be carpeted).
Tambaya 8 Rahoto
(a) A radio which a dealer bought for N6,000.00 and marked to give a profit of 30% was reduced in a sales by 10%. Find : (i) the final sales price ; (ii) the percentage profit.
(b) Solve the equation : \(2^{(2x + 1)} - 9(2^{x}) + 4 = 0\).
(a)(i) Final sales price. Marked price for a 30% profit:
\[\text{MP} = 6000\times1.30 = \text{N}7800.\]
Reduced by 10% in the sales:
\[\text{Sales price} = 7800\times0.90 = \text{N}7020.\]
(a)(ii) Percentage profit.
\[\%\text{ profit} = \frac{7020 - 6000}{6000}\times100 = \frac{1020}{6000}\times100 = 17\%.\]
(b) Let \(u = 2^x\). Then \(2^{(2x+1)} = 2\cdot2^{2x} = 2u^2\), so the equation becomes
\[2u^2 - 9u + 4 = 0 \Rightarrow (2u - 1)(u - 4) = 0.\]
\[u = 4\ \Rightarrow\ 2^x = 4\ \Rightarrow\ x = 2;\qquad u = \tfrac{1}{2}\ \Rightarrow\ 2^x = 2^{-1}\ \Rightarrow\ x = -1.\]
Answer: \(x = 2\) or \(x = -1\).
Bayanin Amsa
(a)(i) Final sales price. Marked price for a 30% profit:
\[\text{MP} = 6000\times1.30 = \text{N}7800.\]
Reduced by 10% in the sales:
\[\text{Sales price} = 7800\times0.90 = \text{N}7020.\]
(a)(ii) Percentage profit.
\[\%\text{ profit} = \frac{7020 - 6000}{6000}\times100 = \frac{1020}{6000}\times100 = 17\%.\]
(b) Let \(u = 2^x\). Then \(2^{(2x+1)} = 2\cdot2^{2x} = 2u^2\), so the equation becomes
\[2u^2 - 9u + 4 = 0 \Rightarrow (2u - 1)(u - 4) = 0.\]
\[u = 4\ \Rightarrow\ 2^x = 4\ \Rightarrow\ x = 2;\qquad u = \tfrac{1}{2}\ \Rightarrow\ 2^x = 2^{-1}\ \Rightarrow\ x = -1.\]
Answer: \(x = 2\) or \(x = -1\).
Tambaya 9 Rahoto
A, B and C are subsets of the universal set U such that : \(U = {0, 1, 2, 3,..., 12}; A = {x : 0 \leq x \leq 7}; B = {4, 6, 8, 10, 12}; C = {1 < y < 8}\), where y is a prime number.
(a) Draw a venn diagram to illustrate the information given above;
(b) Find: (i) \((B \cup C)'\); (ii) \(A' \cap B \cap C\).
(a) Venn diagram
First, write the sets in roster form:
\[A=\{0,1,2,3,4,5,6,7\},\quad B=\{4,6,8,10,12\},\quad C=\{2,3,5,7\}.\]
(b)(i)
\[B\cup C=\{2,3,4,5,6,7,8,10,12\}.\]
Therefore, relative to \(U=\{0,1,2,\ldots,12\}\),
\[(B\cup C)'=\{0,1,9,11\}.\]
(b)(ii)
\[A'=\{8,9,10,11,12\}.\]
Since \(B\cap C=\varnothing\),
\[A'\cap B\cap C=\varnothing.\]
Bayanin Amsa
(a) Venn diagram
First, write the sets in roster form:
\[A=\{0,1,2,3,4,5,6,7\},\quad B=\{4,6,8,10,12\},\quad C=\{2,3,5,7\}.\]
(b)(i)
\[B\cup C=\{2,3,4,5,6,7,8,10,12\}.\]
Therefore, relative to \(U=\{0,1,2,\ldots,12\}\),
\[(B\cup C)'=\{0,1,9,11\}.\]
(b)(ii)
\[A'=\{8,9,10,11,12\}.\]
Since \(B\cap C=\varnothing\),
\[A'\cap B\cap C=\varnothing.\]
Tambaya 10 Rahoto
In the diagram, ASRTB represents a piece of string passing over a pulley of radius 10cm in a vertical plane. O is the centre of the pulley and AMB is a horizontal straight line touching the pulley at M. Angle SAB = 90° and angle TBA = 60°.
(a) Calculate (i) the obtuse angle SOT ; (ii) arc SRT ; (iii) |BT|
(b) Find, correct to the nearest cm, the length of the string. (Take \(\pi = \frac{22}{7}\)).
From the diagram, the pulley has centre O and radius \(r = 10\text{ cm}\). The string runs \(A \to S\) (tangent on the left), over the arc \(S \to R \to T\) at the top, then \(T \to B\) (tangent on the right). \(AMB\) is horizontal, tangent to the pulley at M, with \(\angle SAB = 90^\circ\) and \(\angle TBA = 60^\circ\).
(a)(i) Obtuse angle SOT
From A, the tangents \(AS\) and \(AM\) meet the radii at right angles, so in quadrilateral \(ASOM\):
\[\angle SOM = 360^\circ - 90^\circ - 90^\circ - \angle SAM = 360^\circ - 90^\circ - 90^\circ - 90^\circ = 90^\circ\]From B, the tangents \(BT\) and \(BM\) give, in quadrilateral \(BTOM\):
\[\angle TOM = 360^\circ - 90^\circ - 90^\circ - \angle TBM = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ\]Going from S round the bottom through M to T gives \(90^\circ + 120^\circ = 210^\circ\). The angle on the top (through R), the obtuse \(\angle SOT\), is
\[\angle SOT = 360^\circ - 210^\circ = 150^\circ\](a)(ii) Arc SRT
\[\text{arc }SRT = \frac{150}{360}\times 2\pi r = \frac{150}{360}\times 2\times\frac{22}{7}\times 10 = \frac{5}{12}\times\frac{440}{7}\]\[= \frac{2200}{84} = 26.19\text{ cm} \approx 26\text{ cm}\](a)(iii) |BT|
In right triangle \(OTB\), \(OT = 10\text{ cm}\), \(\angle OTB = 90^\circ\), and \(OB\) bisects \(\angle TBM\) so \(\angle TBO = 30^\circ\):
\[\tan 30^\circ = \frac{OT}{BT} \;\Rightarrow\; BT = \frac{10}{\tan 30^\circ} = 10\sqrt{3} = 17.32\text{ cm}\](b) Length of the string
First find \(AS\). In right triangle \(OSA\), \(OS = 10\text{ cm}\), \(\angle OSA = 90^\circ\), and \(OA\) bisects \(\angle SAM\) so \(\angle SAO = 45^\circ\):
\[\tan 45^\circ = \frac{OS}{AS} \;\Rightarrow\; AS = \frac{10}{\tan 45^\circ} = 10\text{ cm}\]The string length is
\[AS + \text{arc }SRT + BT = 10 + 26.19 + 17.32 = 53.51\text{ cm}\]Length of the string \(\approx 54\text{ cm}\).
Summary: \(\angle SOT = 150^\circ\), arc \(SRT \approx 26\text{ cm}\), \(|BT| = 10\sqrt{3}\approx 17.32\text{ cm}\), string \(\approx 54\text{ cm}\).
Bayanin Amsa
From the diagram, the pulley has centre O and radius \(r = 10\text{ cm}\). The string runs \(A \to S\) (tangent on the left), over the arc \(S \to R \to T\) at the top, then \(T \to B\) (tangent on the right). \(AMB\) is horizontal, tangent to the pulley at M, with \(\angle SAB = 90^\circ\) and \(\angle TBA = 60^\circ\).
(a)(i) Obtuse angle SOT
From A, the tangents \(AS\) and \(AM\) meet the radii at right angles, so in quadrilateral \(ASOM\):
\[\angle SOM = 360^\circ - 90^\circ - 90^\circ - \angle SAM = 360^\circ - 90^\circ - 90^\circ - 90^\circ = 90^\circ\]From B, the tangents \(BT\) and \(BM\) give, in quadrilateral \(BTOM\):
\[\angle TOM = 360^\circ - 90^\circ - 90^\circ - \angle TBM = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ\]Going from S round the bottom through M to T gives \(90^\circ + 120^\circ = 210^\circ\). The angle on the top (through R), the obtuse \(\angle SOT\), is
\[\angle SOT = 360^\circ - 210^\circ = 150^\circ\](a)(ii) Arc SRT
\[\text{arc }SRT = \frac{150}{360}\times 2\pi r = \frac{150}{360}\times 2\times\frac{22}{7}\times 10 = \frac{5}{12}\times\frac{440}{7}\]\[= \frac{2200}{84} = 26.19\text{ cm} \approx 26\text{ cm}\](a)(iii) |BT|
In right triangle \(OTB\), \(OT = 10\text{ cm}\), \(\angle OTB = 90^\circ\), and \(OB\) bisects \(\angle TBM\) so \(\angle TBO = 30^\circ\):
\[\tan 30^\circ = \frac{OT}{BT} \;\Rightarrow\; BT = \frac{10}{\tan 30^\circ} = 10\sqrt{3} = 17.32\text{ cm}\](b) Length of the string
First find \(AS\). In right triangle \(OSA\), \(OS = 10\text{ cm}\), \(\angle OSA = 90^\circ\), and \(OA\) bisects \(\angle SAM\) so \(\angle SAO = 45^\circ\):
\[\tan 45^\circ = \frac{OS}{AS} \;\Rightarrow\; AS = \frac{10}{\tan 45^\circ} = 10\text{ cm}\]The string length is
\[AS + \text{arc }SRT + BT = 10 + 26.19 + 17.32 = 53.51\text{ cm}\]Length of the string \(\approx 54\text{ cm}\).
Summary: \(\angle SOT = 150^\circ\), arc \(SRT \approx 26\text{ cm}\), \(|BT| = 10\sqrt{3}\approx 17.32\text{ cm}\), string \(\approx 54\text{ cm}\).
Tambaya 11 Rahoto
(a) Copy and complete the table of values for the relation \(y = 5 - 7x - 6x^{2}\) for \(-3 \leq x \leq 2\).
| x | -3 | -2 | -1 | -0.5 | 0 | 1 | 2 |
| y | -28 | 6 | 5 |
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the :
(i) graph of \(y = 5 - 7x - 6x^{2}\) ; (ii) line \(y = 3\) on the same axis.
(c) Use your graph to find the : (i) roots of the equation \(2 - 7x - 6x^{2} = 0\) ; (ii) maximum value of \(y = 5 - 7x - 6x^{2}\).
(a) For \(y=5-7x-6x^2\), the completed table is:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(-0.5\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
| \(y=5-7x-6x^2\) | \(-28\) | \(-5\) | \(6\) | \(7\) | \(5\) | \(-8\) | \(-33\) |
(b) The required graph of \(y=5-7x-6x^2\) and the line \(y=3\) are shown below. The plotted points from the table lie on the smooth curve.
(c)(i) The roots of \(2-7x-6x^2=0\) are obtained by writing
\[2-7x-6x^2=0\iff 5-7x-6x^2=3.\]
Thus, they are the \(x\)-coordinates where the curve meets \(y=3\). Reading from the graph gives
\[\boxed{x\approx-1.4\text{ or }x\approx0.2}.\]
(c)(ii) The highest point of the curve is approximately \(7\) units. Hence,
\[\boxed{\text{Maximum value of }y=7\text{ (approximately)}}.\]
Bayanin Amsa
(a) For \(y=5-7x-6x^2\), the completed table is:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(-0.5\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|---|---|
| \(y=5-7x-6x^2\) | \(-28\) | \(-5\) | \(6\) | \(7\) | \(5\) | \(-8\) | \(-33\) |
(b) The required graph of \(y=5-7x-6x^2\) and the line \(y=3\) are shown below. The plotted points from the table lie on the smooth curve.
(c)(i) The roots of \(2-7x-6x^2=0\) are obtained by writing
\[2-7x-6x^2=0\iff 5-7x-6x^2=3.\]
Thus, they are the \(x\)-coordinates where the curve meets \(y=3\). Reading from the graph gives
\[\boxed{x\approx-1.4\text{ or }x\approx0.2}.\]
(c)(ii) The highest point of the curve is approximately \(7\) units. Hence,
\[\boxed{\text{Maximum value of }y=7\text{ (approximately)}}.\]
Tambaya 12 Rahoto
The third term of a Geometric Progression (G.P) is 360 and the sixth term is 1215. Find the
(a) common ratio;
(b) first term ;
(c) sum of the first four terms.
(a) Common ratio. For a GP, \(T_3 = ar^2 = 360\) and \(T_6 = ar^5 = 1215\). Dividing:
\[\frac{ar^5}{ar^2} = r^3 = \frac{1215}{360} = \frac{27}{8}\Rightarrow r = \sqrt[3]{\tfrac{27}{8}} = \tfrac{3}{2}.\]
(b) First term.
\[ar^2 = 360 \Rightarrow a\left(\tfrac{3}{2}\right)^2 = 360 \Rightarrow a\times\tfrac{9}{4} = 360 \Rightarrow a = 160.\]
(c) Sum of the first four terms. The terms are \(160, 240, 360, 540\):
\[S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{160\left(\left(\tfrac{3}{2}\right)^4 - 1\right)}{\tfrac{3}{2} - 1} = \frac{160(5.0625 - 1)}{0.5} = 1300.\]
Bayanin Amsa
(a) Common ratio. For a GP, \(T_3 = ar^2 = 360\) and \(T_6 = ar^5 = 1215\). Dividing:
\[\frac{ar^5}{ar^2} = r^3 = \frac{1215}{360} = \frac{27}{8}\Rightarrow r = \sqrt[3]{\tfrac{27}{8}} = \tfrac{3}{2}.\]
(b) First term.
\[ar^2 = 360 \Rightarrow a\left(\tfrac{3}{2}\right)^2 = 360 \Rightarrow a\times\tfrac{9}{4} = 360 \Rightarrow a = 160.\]
(c) Sum of the first four terms. The terms are \(160, 240, 360, 540\):
\[S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{160\left(\left(\tfrac{3}{2}\right)^4 - 1\right)}{\tfrac{3}{2} - 1} = \frac{160(5.0625 - 1)}{0.5} = 1300.\]
