WAEC SSCE - General Mathematics - 2001 (Essay)
Tambaya 1 Rahoto
(a) Simplify : \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years' time, the ratio of their ages will be 7 : 2. Find their present ages.
(a) \(\dfrac{1}{3^{5n}}\times 9^{n-1}\times 27^{n+1}\). Convert everything to base 3:
\(9^{n-1}=3^{2(n-1)}=3^{2n-2}\); \(27^{n+1}=3^{3(n+1)}=3^{3n+3}\); \(\dfrac{1}{3^{5n}}=3^{-5n}\).
Add exponents: \(-5n+(2n-2)+(3n+3)= (-5n+2n+3n)+(-2+3)=0+1=1\).
Result \(=3^{1}=\mathbf{3}\).
(b) Let woman \(=w\), daughter \(=d\). \(w+d=46\).
In 4 years: \(\dfrac{w+4}{d+4}=\dfrac{7}{2}\Rightarrow 2(w+4)=7(d+4)\Rightarrow 2w+8=7d+28\Rightarrow 2w-7d=20\).
Substitute \(w=46-d\): \(2(46-d)-7d=20\Rightarrow 92-9d=20\Rightarrow 9d=72\Rightarrow d=8\), \(w=38\).
Woman is 38 years, daughter is 8 years. Check: in 4 years \(42:12=7:2\). \(\checkmark\)
Bayanin Amsa
(a) \(\dfrac{1}{3^{5n}}\times 9^{n-1}\times 27^{n+1}\). Convert everything to base 3:
\(9^{n-1}=3^{2(n-1)}=3^{2n-2}\); \(27^{n+1}=3^{3(n+1)}=3^{3n+3}\); \(\dfrac{1}{3^{5n}}=3^{-5n}\).
Add exponents: \(-5n+(2n-2)+(3n+3)= (-5n+2n+3n)+(-2+3)=0+1=1\).
Result \(=3^{1}=\mathbf{3}\).
(b) Let woman \(=w\), daughter \(=d\). \(w+d=46\).
In 4 years: \(\dfrac{w+4}{d+4}=\dfrac{7}{2}\Rightarrow 2(w+4)=7(d+4)\Rightarrow 2w+8=7d+28\Rightarrow 2w-7d=20\).
Substitute \(w=46-d\): \(2(46-d)-7d=20\Rightarrow 92-9d=20\Rightarrow 9d=72\Rightarrow d=8\), \(w=38\).
Woman is 38 years, daughter is 8 years. Check: in 4 years \(42:12=7:2\). \(\checkmark\)
Tambaya 2 Rahoto
The sides of a rectangular floor are xm and (x + 7)m. The diagonal is (x + 8)m. Calculate, in metres :
(a) the value of x ;
(b) the area of the floor.
Sides \(x\) and \((x+7)\), diagonal \((x+8)\). By Pythagoras:
\(x^{2}+(x+7)^{2}=(x+8)^{2}\).
\(x^{2}+x^{2}+14x+49=x^{2}+16x+64\).
\(2x^{2}+14x+49-x^{2}-16x-64=0 \Rightarrow x^{2}-2x-15=0\).
\((x-5)(x+3)=0 \Rightarrow x=5\) (reject \(x=-3\), a length must be positive).
(a) \(x=5\ \text{m}\).
(b) Area \(=x(x+7)=5\times 12=\mathbf{60\ m^{2}}\).
Bayanin Amsa
Sides \(x\) and \((x+7)\), diagonal \((x+8)\). By Pythagoras:
\(x^{2}+(x+7)^{2}=(x+8)^{2}\).
\(x^{2}+x^{2}+14x+49=x^{2}+16x+64\).
\(2x^{2}+14x+49-x^{2}-16x-64=0 \Rightarrow x^{2}-2x-15=0\).
\((x-5)(x+3)=0 \Rightarrow x=5\) (reject \(x=-3\), a length must be positive).
(a) \(x=5\ \text{m}\).
(b) Area \(=x(x+7)=5\times 12=\mathbf{60\ m^{2}}\).
Tambaya 3 Rahoto
(a) The roots of the equation \(2x^{2} + (p + 1)x + 9 = 0\), are 1 and 3, where p and q are constants. Find the values of p and q.
(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth's surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800km above the surface of the earth. [Take the radius of the earth as 6,400km].
(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).
Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).
Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).
\(p=-9,\ q=6\).
(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):
\(k=80\times 6400^{2}\).
At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):
\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).
Weight \(\approx \mathbf{63\ kg}\).
Bayanin Amsa
(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).
Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).
Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).
\(p=-9,\ q=6\).
(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):
\(k=80\times 6400^{2}\).
At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):
\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).
Weight \(\approx \mathbf{63\ kg}\).
Tambaya 4 Rahoto
(a)
In the diagram, MN || ST, NP || QT and < STQ = 70°. Find x.
(b)
In the diagram above, AC is a straight line, |BC| = |BD|, \(\stackrel\frown{BCD} = 50°\) and \(\stackrel\frown{BAD} = 55°\). Find \(\stackrel\frown{BDA}\).
(a) Find \(x\).
From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).
Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give
\[\angle MWQ=\angle STQ=70^{\circ}\]Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:
\[x+\angle MWQ=180^{\circ}\]\[x=180^{\circ}-70^{\circ}=\mathbf{110^{\circ}}\](b) Find \(\angle BDA\).
From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).
Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:
\[\angle BDC=\angle BCD=50^{\circ}\]\[\angle DBC=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):
\[\angle DBA=180^{\circ}-80^{\circ}=100^{\circ}\]In triangle \(ABD\):
\[\angle BDA=180^{\circ}-\angle BAD-\angle DBA=180^{\circ}-55^{\circ}-100^{\circ}=\mathbf{25^{\circ}}\]Bayanin Amsa
(a) Find \(x\).
From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).
Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give
\[\angle MWQ=\angle STQ=70^{\circ}\]Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:
\[x+\angle MWQ=180^{\circ}\]\[x=180^{\circ}-70^{\circ}=\mathbf{110^{\circ}}\](b) Find \(\angle BDA\).
From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).
Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:
\[\angle BDC=\angle BCD=50^{\circ}\]\[\angle DBC=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):
\[\angle DBA=180^{\circ}-80^{\circ}=100^{\circ}\]In triangle \(ABD\):
\[\angle BDA=180^{\circ}-\angle BAD-\angle DBA=180^{\circ}-55^{\circ}-100^{\circ}=\mathbf{25^{\circ}}\]Tambaya 5 Rahoto
(a) Draw the table of values for the relation \(y = x^{2}\) for the interval \(-3 \leq x \leq 4\).
(b) Using a scale of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw the graphs of : (i) \(y = x^{2}\) ; (ii) \(y = 2x + 3\) for \(-3 \leq x \leq 4\).
(c) Use your graph to find : (i) the roots of the equation \(x^{2} = 2x + 3\) ; (ii) the gradient of \(y = x^{2}\) at x = -2.
(a) Table of values for \(y=x^2\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=x^2\) | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
(b) Values for the straight line \(y=2x+3\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=2x+3\) | \(-3\) | \(-1\) | 1 | 3 | 5 | 7 | 9 | 11 |
Using the stated scales, plot the points and draw smooth curve \(y=x^2\) and the straight line \(y=2x+3\). A tangent to the curve at \((-2,4)\) is also shown for part (c)(ii).
(c)(i) The graphs intersect at \((-1,1)\) and \((3,9)\). Hence, the roots of \(x^2=2x+3\) are
\[\boxed{x=-1\text{ and }x=3}.\]
(c)(ii) From the tangent at \(x=-2\), take the points \((-3,8)\) and \((-1,0)\):
\[\text{gradient}=\frac{0-8}{-1-(-3)}=\frac{-8}{2}=\boxed{-4}.\]
Bayanin Amsa
(a) Table of values for \(y=x^2\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=x^2\) | 9 | 4 | 1 | 0 | 1 | 4 | 9 | 16 |
(b) Values for the straight line \(y=2x+3\)
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=2x+3\) | \(-3\) | \(-1\) | 1 | 3 | 5 | 7 | 9 | 11 |
Using the stated scales, plot the points and draw smooth curve \(y=x^2\) and the straight line \(y=2x+3\). A tangent to the curve at \((-2,4)\) is also shown for part (c)(ii).
(c)(i) The graphs intersect at \((-1,1)\) and \((3,9)\). Hence, the roots of \(x^2=2x+3\) are
\[\boxed{x=-1\text{ and }x=3}.\]
(c)(ii) From the tangent at \(x=-2\), take the points \((-3,8)\) and \((-1,0)\):
\[\text{gradient}=\frac{0-8}{-1-(-3)}=\frac{-8}{2}=\boxed{-4}.\]
Tambaya 6 Rahoto
Given that \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\), calculate without using mathematical tables or calculator, the value of :
(a) \(\log_{10} 54\) ;
(b) \(\log_{10} 0.24\).
Given \(\log_{10}2=0.3010,\ \log_{10}3=0.4771\).
(a) \(54=2\times 27=2\times 3^{3}\).
\(\log_{10}54=\log_{10}2+3\log_{10}3=0.3010+3(0.4771)=0.3010+1.4313=\mathbf{1.7323}\).
(b) \(0.24=\dfrac{24}{100}=\dfrac{2^{3}\times 3}{100}\).
\(\log_{10}24=3\log_{10}2+\log_{10}3=0.9030+0.4771=1.3801\).
\(\log_{10}0.24=\log_{10}24-\log_{10}100=1.3801-2=\mathbf{-0.6199}\) (i.e. \(\bar{1}.3801\)).
Bayanin Amsa
Given \(\log_{10}2=0.3010,\ \log_{10}3=0.4771\).
(a) \(54=2\times 27=2\times 3^{3}\).
\(\log_{10}54=\log_{10}2+3\log_{10}3=0.3010+3(0.4771)=0.3010+1.4313=\mathbf{1.7323}\).
(b) \(0.24=\dfrac{24}{100}=\dfrac{2^{3}\times 3}{100}\).
\(\log_{10}24=3\log_{10}2+\log_{10}3=0.9030+0.4771=1.3801\).
\(\log_{10}0.24=\log_{10}24-\log_{10}100=1.3801-2=\mathbf{-0.6199}\) (i.e. \(\bar{1}.3801\)).
Tambaya 7 Rahoto
(a) In a market survey, 100 traders sell fruits, 40 sell apples, 46 oranges, 50 mangoes, 14 apples and oranges, 15 apples and mangoes and 10 sell the three types of fruits. Each of the 100 traders sells at least one of the three fruits.
(i) Represent the information in a Venn diagram ; (ii) Find the number that sell oranges and mangoes only.
(b) Find the value of x for which \(312_{four} + 52_{x} = 96_{ten}\)
(a) Apples \(A=40\), Oranges \(O=46\), Mangoes \(M=50\). \(A\cap O=14\), \(A\cap M=15\), all three \(=10\); every trader sells at least one, total \(=100\).
(i) Venn diagram. Three intersecting circles; centre (all three) \(=10\).
(ii) Let \(O\cap M\) (total) \(=y\). By the inclusion-exclusion principle:
\(|A|+|O|+|M|-|A\cap O|-|A\cap M|-|O\cap M|+|A\cap O\cap M|=100\).
\(40+46+50-14-15-y+10=100 \Rightarrow 117-y=100 \Rightarrow y=17\).
Oranges and mangoes only \(=y-10=17-10=\mathbf{7}\).
(b) \(312_{four}+52_x=96_{ten}\).
\(312_{four}=3(16)+1(4)+2=54\). So \(52_x=96-54=42\).
\(52_x=5x+2=42 \Rightarrow 5x=40 \Rightarrow x=\mathbf{8}\).
Bayanin Amsa
(a) Apples \(A=40\), Oranges \(O=46\), Mangoes \(M=50\). \(A\cap O=14\), \(A\cap M=15\), all three \(=10\); every trader sells at least one, total \(=100\).
(i) Venn diagram. Three intersecting circles; centre (all three) \(=10\).
(ii) Let \(O\cap M\) (total) \(=y\). By the inclusion-exclusion principle:
\(|A|+|O|+|M|-|A\cap O|-|A\cap M|-|O\cap M|+|A\cap O\cap M|=100\).
\(40+46+50-14-15-y+10=100 \Rightarrow 117-y=100 \Rightarrow y=17\).
Oranges and mangoes only \(=y-10=17-10=\mathbf{7}\).
(b) \(312_{four}+52_x=96_{ten}\).
\(312_{four}=3(16)+1(4)+2=54\). So \(52_x=96-54=42\).
\(52_x=5x+2=42 \Rightarrow 5x=40 \Rightarrow x=\mathbf{8}\).
Tambaya 8 Rahoto
The diagram above shows the bar charts representing the number of vehicles manufactured by a company in January, February and March, 1992.
(a) How many vehicles were produced in February?
(b) What fraction of the vehicles manufactured in February were cars?
(c) How many buses were produced altogether from January to March, 1992?
(d) What is the ratio in the lowest term of the number of lorries produced in February to that in March?
Read the height of every bar against the vertical scale, which is marked in steps of \(5\) vehicles. Match each shading to the key: fine dots are cars, solid black is lorries, diagonal cross-hatching is buses. Reading each of the three bars in each month group gives the table below.
| Month | Cars | Lorries | Buses | Monthly total |
|---|---|---|---|---|
| January | 10 | 15 | 20 | 45 |
| February | 40 | 25 | 30 | 95 |
| March | 5 | 35 | 50 | 90 |
The same data drawn to scale:
(a) Vehicles produced in February. Add the three February bars:
\[ 40 + 25 + 30 = 95 \text{ vehicles.} \]
(b) Fraction of February vehicles that were cars. Cars in February over the February total:
\[ \frac{40}{95} = \frac{8}{19}. \]
(c) Buses produced altogether, January to March. Add the three bus readings:
\[ 20 + 30 + 50 = 100 \text{ buses.} \]
(d) Ratio of lorries in February to lorries in March. The lorry bars read \(25\) and \(35\). Divide both by their highest common factor, \(5\):
\[ 25 : 35 = 5 : 7. \]
The key point when reading a grouped bar chart is to fix the scale first (each square here is \(5\) vehicles) and to match every bar to the correct shading in the key before adding. Miscounting the scale or reading the wrong shading is what leads to a wrong total such as treating February as \(75\) instead of \(95\).
Bayanin Amsa
Read the height of every bar against the vertical scale, which is marked in steps of \(5\) vehicles. Match each shading to the key: fine dots are cars, solid black is lorries, diagonal cross-hatching is buses. Reading each of the three bars in each month group gives the table below.
| Month | Cars | Lorries | Buses | Monthly total |
|---|---|---|---|---|
| January | 10 | 15 | 20 | 45 |
| February | 40 | 25 | 30 | 95 |
| March | 5 | 35 | 50 | 90 |
The same data drawn to scale:
(a) Vehicles produced in February. Add the three February bars:
\[ 40 + 25 + 30 = 95 \text{ vehicles.} \]
(b) Fraction of February vehicles that were cars. Cars in February over the February total:
\[ \frac{40}{95} = \frac{8}{19}. \]
(c) Buses produced altogether, January to March. Add the three bus readings:
\[ 20 + 30 + 50 = 100 \text{ buses.} \]
(d) Ratio of lorries in February to lorries in March. The lorry bars read \(25\) and \(35\). Divide both by their highest common factor, \(5\):
\[ 25 : 35 = 5 : 7. \]
The key point when reading a grouped bar chart is to fix the scale first (each square here is \(5\) vehicles) and to match every bar to the correct shading in the key before adding. Miscounting the scale or reading the wrong shading is what leads to a wrong total such as treating February as \(75\) instead of \(95\).
Tambaya 9 Rahoto
The number of child births recorded in 50 maternity centres of a local government in August 1993 are as follows :
50 99 81 86 69 85 93 63 92 65 77 74 76 71 90 74 81 94 67 75 95 81 68 105 99 68 75 75 76 73 79 74 80 69 74 62 74 80 79 68 79 75 75 71 83 75 80 85 81 82
(a) Construct a frequency distribution table, using class intervals 45 - 54, 55 - 64, etc.
(b) Draw the histogram for the distribution
(c) Use your histogram to estimate the mode.
(d) Calculate the mean number of births.
(a) Frequency distribution
| Number of births | Class midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 45–54 | 49.5 | 1 | 49.5 |
| 55–64 | 59.5 | 2 | 119.0 |
| 65–74 | 69.5 | 15 | 1042.5 |
| 75–84 | 79.5 | 21 | 1669.5 |
| 85–94 | 89.5 | 7 | 626.5 |
| 95–104 | 99.5 | 3 | 298.5 |
| 105–114 | 109.5 | 1 | 109.5 |
| Total | 50 | 3915 |
(b) Histogram
The continuous class boundaries are \(44.5\text{–}54.5,\ 54.5\text{–}64.5,\ldots,\ 104.5\text{–}114.5\). Since all class widths are 10, the heights of the adjacent bars are their frequencies.
(c) Estimated mode
The modal class is \(75\text{–}84\). From the histogram, or by interpolation within this class,
\[\text{Mode}=L+\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\times c\]
\[=74.5+\frac{21-15}{(21-15)+(21-7)}\times10\]
\[=74.5+\frac{6}{20}\times10=\boxed{77.5\text{ births (approximately)}}.\]
(d) Mean number of births
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3915}{50}=\boxed{78.3\text{ births}}.\]
Bayanin Amsa
(a) Frequency distribution
| Number of births | Class midpoint, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|
| 45–54 | 49.5 | 1 | 49.5 |
| 55–64 | 59.5 | 2 | 119.0 |
| 65–74 | 69.5 | 15 | 1042.5 |
| 75–84 | 79.5 | 21 | 1669.5 |
| 85–94 | 89.5 | 7 | 626.5 |
| 95–104 | 99.5 | 3 | 298.5 |
| 105–114 | 109.5 | 1 | 109.5 |
| Total | 50 | 3915 |
(b) Histogram
The continuous class boundaries are \(44.5\text{–}54.5,\ 54.5\text{–}64.5,\ldots,\ 104.5\text{–}114.5\). Since all class widths are 10, the heights of the adjacent bars are their frequencies.
(c) Estimated mode
The modal class is \(75\text{–}84\). From the histogram, or by interpolation within this class,
\[\text{Mode}=L+\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\times c\]
\[=74.5+\frac{21-15}{(21-15)+(21-7)}\times10\]
\[=74.5+\frac{6}{20}\times10=\boxed{77.5\text{ births (approximately)}}.\]
(d) Mean number of births
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{3915}{50}=\boxed{78.3\text{ births}}.\]
Tambaya 10 Rahoto
(a) A boy blew his rubber balloon to a spherical shape. The balloon burst when its diameter was 15 cm. Calculate, correct to the nearest whole number, the volume of air in the balloon at the point of bursting. [Take \(\pi = \frac{22}{7}\)]
(b) A point X is on latitude 28°N and longitude 105°W. Y is another point on the same latitude as X but on longitude 35°E. (i) Calculate, correct to three significant figures, the distance between X and Y along latitude 28°N ; (ii) How far is X from the equator? [Take \(\pi = \frac{22}{7}\) and radius of the earth = 6,400km].
(a) Volume of the balloon. Diameter \(=15\text{ cm}\Rightarrow r=7.5\text{ cm}\).
\(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times\dfrac{22}{7}\times 7.5^{3}=\dfrac{88}{21}\times 421.875=\dfrac{37125}{21}=1767.86\).
Volume \(\approx \mathbf{1768\ cm^{3}}\).
(b) X: \(28^{\circ}\)N, \(105^{\circ}\)W; Y: \(28^{\circ}\)N, \(35^{\circ}\)E. Difference in longitude \(=105+35=140^{\circ}\).
(i) Distance X to Y along latitude \(28^{\circ}\)N:
\(d=\dfrac{140}{360}\times 2\pi R\cos 28^{\circ}=\dfrac{140}{360}\times 2\times\dfrac{22}{7}\times 6400\times\cos 28^{\circ}\).
\(=\dfrac{140}{360}\times 40228.57\times 0.8829\approx 13{,}800\text{ km}\ (\mathbf{1.38\times 10^{4}\ km})\).
(ii) Distance of X from the equator (along its meridian):
\(d=\dfrac{28}{360}\times 2\pi R=\dfrac{28}{360}\times 40228.57\approx \mathbf{3130\ km}\).
Bayanin Amsa
(a) Volume of the balloon. Diameter \(=15\text{ cm}\Rightarrow r=7.5\text{ cm}\).
\(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times\dfrac{22}{7}\times 7.5^{3}=\dfrac{88}{21}\times 421.875=\dfrac{37125}{21}=1767.86\).
Volume \(\approx \mathbf{1768\ cm^{3}}\).
(b) X: \(28^{\circ}\)N, \(105^{\circ}\)W; Y: \(28^{\circ}\)N, \(35^{\circ}\)E. Difference in longitude \(=105+35=140^{\circ}\).
(i) Distance X to Y along latitude \(28^{\circ}\)N:
\(d=\dfrac{140}{360}\times 2\pi R\cos 28^{\circ}=\dfrac{140}{360}\times 2\times\dfrac{22}{7}\times 6400\times\cos 28^{\circ}\).
\(=\dfrac{140}{360}\times 40228.57\times 0.8829\approx 13{,}800\text{ km}\ (\mathbf{1.38\times 10^{4}\ km})\).
(ii) Distance of X from the equator (along its meridian):
\(d=\dfrac{28}{360}\times 2\pi R=\dfrac{28}{360}\times 40228.57\approx \mathbf{3130\ km}\).
Tambaya 11 Rahoto
(a) The probabilities that three boys pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that :
(i) all three boys pass ; (ii) none of the boys pass ; (iii) only two of the boys pass.
(b) A shop-keeper marks a television set for sale at N36,000 so as to make a profit of 20% on the cost price. When he sells it, he allows a discount of 5% of the marked price. Calculate the actual percentage profit.
(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).
(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).
(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).
(iii) Exactly two pass:
- 1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
- 1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
- 2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)
Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).
(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).
Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).
Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).
Bayanin Amsa
(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).
(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).
(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).
(iii) Exactly two pass:
- 1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
- 1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
- 2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)
Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).
(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).
Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).
Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).
Tambaya 12 Rahoto
In the diagram, three points A, B and C are on the same horizontal ground. B is 15m from A, on a bearing of 053°, C is 18m from B on a bearing of 161°. A vertical pole with top T is erected at B such that < ATB = 58°. Calculate, correct to three significant figures,
(a) the length of AC.
(b) the bearing of C from A ;
(c) the height of the pole BT.
From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).
(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):
\[053^\circ+180^\circ=233^\circ\]The bearing of C from B is \(161^\circ\), so
\[\angle ABC=233^\circ-161^\circ=72^\circ\]Apply the cosine rule to triangle ABC:
\[AC^2=AB^2+BC^2-2\,(AB)(BC)\cos(\angle ABC)\]\[AC^2=15^2+18^2-2(15)(18)\cos72^\circ\]\[AC^2=225+324-540(0.30902)=382.13\]\[AC=19.5\text{ m (3 s.f.)}\](b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):
\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\]\[\sin\angle BAC=\frac{18\sin72^\circ}{19.548}=\frac{17.119}{19.548}=0.87575\]\[\angle BAC=61.1^\circ\]C lies clockwise of B as seen from A, so the bearing of C from A is
\[053^\circ+61.1^\circ=114^\circ\ (\text{3 s.f.})\](c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:
\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]Bayanin Amsa
From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).
(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):
\[053^\circ+180^\circ=233^\circ\]The bearing of C from B is \(161^\circ\), so
\[\angle ABC=233^\circ-161^\circ=72^\circ\]Apply the cosine rule to triangle ABC:
\[AC^2=AB^2+BC^2-2\,(AB)(BC)\cos(\angle ABC)\]\[AC^2=15^2+18^2-2(15)(18)\cos72^\circ\]\[AC^2=225+324-540(0.30902)=382.13\]\[AC=19.5\text{ m (3 s.f.)}\](b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):
\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\]\[\sin\angle BAC=\frac{18\sin72^\circ}{19.548}=\frac{17.119}{19.548}=0.87575\]\[\angle BAC=61.1^\circ\]C lies clockwise of B as seen from A, so the bearing of C from A is
\[053^\circ+61.1^\circ=114^\circ\ (\text{3 s.f.})\](c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:
\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]
