WAEC SSCE - General Mathematics - 2021 (Essay)
Tambaya 1 Rahoto
In the diagram, PQRS is a circle. = . ?SPR = 26° and the interior angles of PQS are in the ratio 2:3 :3.
Calculate: (i) PQR; (ii) RPQ; (iii) PRQ
(b) The coordinates of two points P and Q in a plane are (7, 3) and (5, x) respectively, where X is a real number.
If |PQ| = 29 units, find the value of x.
(a) Circle \(PQRS\) with \(|PS|=|PQ|\), \(\angle SPR=26^{\circ}\) and the interior angles of \(\triangle PQS\) in the ratio \(2:3:3\).
Step 1: Angles of triangle \(PQS\).
The three angles add up to \(180^{\circ}\). With \(2+3+3=8\) parts, one part is \(\dfrac{180^{\circ}}{8}=22.5^{\circ}\). Taking the angles at \(P:Q:S\) as \(2:3:3\):
\[\angle SPQ=2\times 22.5^{\circ}=45^{\circ}\]\[\angle PQS=3\times 22.5^{\circ}=67.5^{\circ}\]\[\angle PSQ=3\times 22.5^{\circ}=67.5^{\circ}\](The two equal angles \(67.5^{\circ}\) match the equal chords \(|PS|=|PQ|\) shown by the tick marks.)
Step 2: Arcs of the circle. Using "inscribed angle = half its arc":
- \(\angle SPR=26^{\circ}\) stands on arc \(SR\), so arc \(SR=52^{\circ}\).
- \(\angle PSQ=67.5^{\circ}\) stands on arc \(PQ\), so arc \(PQ=135^{\circ}\).
- \(\angle PQS=67.5^{\circ}\) stands on arc \(PS\), so arc \(PS=135^{\circ}\).
- The four arcs total \(360^{\circ}\): arc \(RQ=360^{\circ}-(135^{\circ}+52^{\circ}+135^{\circ})=38^{\circ}\).
(i) \(\angle PQR\).
\[\angle PQR=\angle PQS+\angle SQR\]\(\angle SQR\) stands on arc \(SR=52^{\circ}\), so \(\angle SQR=26^{\circ}\). Hence:
\[\angle PQR=67.5^{\circ}+26^{\circ}=93.5^{\circ}\](ii) \(\angle RPQ\).
\[\angle RPQ=\angle SPQ-\angle SPR=45^{\circ}-26^{\circ}=19^{\circ}\](iii) \(\angle PRQ\). \(\angle PRQ\) and \(\angle PSQ\) both stand on the same chord \(PQ\) (angles in the same segment), so they are equal:
\[\angle PRQ=\angle PSQ=67.5^{\circ}\]Check: in cyclic quadrilateral \(PQRS\), \(\angle PQR+\angle PSR=93.5^{\circ}+(67.5^{\circ}+19^{\circ})=93.5^{\circ}+86.5^{\circ}=180^{\circ}\). Correct.
(b) Points \(P(7,3)\) and \(Q(5,x)\) with \(|PQ|=\sqrt{29}\) units.
Distance formula:
\[|PQ|^2=(7-5)^2+(3-x)^2\]\[29=2^2+(3-x)^2\]\[29=4+(3-x)^2\]\[(3-x)^2=25\]\[3-x=\pm 5\]So \(3-x=5\Rightarrow x=-2\), or \(3-x=-5\Rightarrow x=8\).
\[x=8 \quad\text{or}\quad x=-2\]Bayanin Amsa
(a) Circle \(PQRS\) with \(|PS|=|PQ|\), \(\angle SPR=26^{\circ}\) and the interior angles of \(\triangle PQS\) in the ratio \(2:3:3\).
Step 1: Angles of triangle \(PQS\).
The three angles add up to \(180^{\circ}\). With \(2+3+3=8\) parts, one part is \(\dfrac{180^{\circ}}{8}=22.5^{\circ}\). Taking the angles at \(P:Q:S\) as \(2:3:3\):
\[\angle SPQ=2\times 22.5^{\circ}=45^{\circ}\]\[\angle PQS=3\times 22.5^{\circ}=67.5^{\circ}\]\[\angle PSQ=3\times 22.5^{\circ}=67.5^{\circ}\](The two equal angles \(67.5^{\circ}\) match the equal chords \(|PS|=|PQ|\) shown by the tick marks.)
Step 2: Arcs of the circle. Using "inscribed angle = half its arc":
- \(\angle SPR=26^{\circ}\) stands on arc \(SR\), so arc \(SR=52^{\circ}\).
- \(\angle PSQ=67.5^{\circ}\) stands on arc \(PQ\), so arc \(PQ=135^{\circ}\).
- \(\angle PQS=67.5^{\circ}\) stands on arc \(PS\), so arc \(PS=135^{\circ}\).
- The four arcs total \(360^{\circ}\): arc \(RQ=360^{\circ}-(135^{\circ}+52^{\circ}+135^{\circ})=38^{\circ}\).
(i) \(\angle PQR\).
\[\angle PQR=\angle PQS+\angle SQR\]\(\angle SQR\) stands on arc \(SR=52^{\circ}\), so \(\angle SQR=26^{\circ}\). Hence:
\[\angle PQR=67.5^{\circ}+26^{\circ}=93.5^{\circ}\](ii) \(\angle RPQ\).
\[\angle RPQ=\angle SPQ-\angle SPR=45^{\circ}-26^{\circ}=19^{\circ}\](iii) \(\angle PRQ\). \(\angle PRQ\) and \(\angle PSQ\) both stand on the same chord \(PQ\) (angles in the same segment), so they are equal:
\[\angle PRQ=\angle PSQ=67.5^{\circ}\]Check: in cyclic quadrilateral \(PQRS\), \(\angle PQR+\angle PSR=93.5^{\circ}+(67.5^{\circ}+19^{\circ})=93.5^{\circ}+86.5^{\circ}=180^{\circ}\). Correct.
(b) Points \(P(7,3)\) and \(Q(5,x)\) with \(|PQ|=\sqrt{29}\) units.
Distance formula:
\[|PQ|^2=(7-5)^2+(3-x)^2\]\[29=2^2+(3-x)^2\]\[29=4+(3-x)^2\]\[(3-x)^2=25\]\[3-x=\pm 5\]So \(3-x=5\Rightarrow x=-2\), or \(3-x=-5\Rightarrow x=8\).
\[x=8 \quad\text{or}\quad x=-2\]Tambaya 2 Rahoto
(a) Mr Sarfo borrowed $25,000 from Afiak financial services at 21% simple interest per annum for 3 years if he was able to pay back the loan in two years at equal yearly installments how much did he pay each year?
(b) Two consecutive numbers are such that the sum of thrice the smaller and twice the larger is 17.
find correct through three significant figures the smaller number as a percentage of the sum of the two numbers
(a) Loan repayment
The loan is cleared over the 2 years in which it is actually held. Simple interest for 2 years:
\[I=\frac{P\,R\,T}{100}=\frac{25000\times 21\times 2}{100}=\$10{,}500.\]
Total amount to repay \(=25000+10500=\$35{,}500\). Paid in two equal yearly installments:
\[\text{Each installment}=\frac{35500}{2}=\$17{,}750.\]
(b) Consecutive numbers
Let the numbers be \(n\) (smaller) and \(n+1\) (larger).
\[3n+2(n+1)=17\Rightarrow 5n+2=17\Rightarrow n=3,\ \text{larger}=4.\]
Sum of the two numbers \(=3+4=7\). Smaller as a percentage of the sum:
\[\frac{3}{7}\times100\%=42.857\ldots\%\approx 42.9\%\ (3\text{ s.f.}).\]
Bayanin Amsa
(a) Loan repayment
The loan is cleared over the 2 years in which it is actually held. Simple interest for 2 years:
\[I=\frac{P\,R\,T}{100}=\frac{25000\times 21\times 2}{100}=\$10{,}500.\]
Total amount to repay \(=25000+10500=\$35{,}500\). Paid in two equal yearly installments:
\[\text{Each installment}=\frac{35500}{2}=\$17{,}750.\]
(b) Consecutive numbers
Let the numbers be \(n\) (smaller) and \(n+1\) (larger).
\[3n+2(n+1)=17\Rightarrow 5n+2=17\Rightarrow n=3,\ \text{larger}=4.\]
Sum of the two numbers \(=3+4=7\). Smaller as a percentage of the sum:
\[\frac{3}{7}\times100\%=42.857\ldots\%\approx 42.9\%\ (3\text{ s.f.}).\]
Tambaya 3 Rahoto
The table shows the distribution of the number of hours per day spent in studying by 50 students.
| Number of hours per day | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Number of students | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Calculate, correct to two decimal places,
the: (a) mean; (b) standard deviation.
Given distribution.
| Hours per day (x) | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| Students (f) | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Total number of students: \(\sum f = 50\).
(a) Mean. Compute \(\sum fx\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| f | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
| fx | 20 | 35 | 30 | 63 | 96 | 36 | 30 | 55 |
\[\sum fx = 20+35+30+63+96+36+30+55 = 365.\]
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{365}{50} = 7.30.\]
(b) Standard deviation. Compute \(\sum fx^{2}\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| fx\(^2\) | 80 | 175 | 180 | 441 | 768 | 324 | 300 | 605 |
\[\sum fx^{2} = 80+175+180+441+768+324+300+605 = 2873.\]
\[\text{Variance} = \frac{\sum fx^{2}}{\sum f} - \bar{x}^{2} = \frac{2873}{50} - (7.3)^{2} = 57.46 - 53.29 = 4.17.\]
\[\text{S.D.} = \sqrt{4.17} = 2.04 \text{ (2 d.p.).}\]
The mean study time is 7.30 hours and the standard deviation is 2.04 hours.
Bayanin Amsa
Given distribution.
| Hours per day (x) | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| Students (f) | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Total number of students: \(\sum f = 50\).
(a) Mean. Compute \(\sum fx\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| f | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
| fx | 20 | 35 | 30 | 63 | 96 | 36 | 30 | 55 |
\[\sum fx = 20+35+30+63+96+36+30+55 = 365.\]
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{365}{50} = 7.30.\]
(b) Standard deviation. Compute \(\sum fx^{2}\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| fx\(^2\) | 80 | 175 | 180 | 441 | 768 | 324 | 300 | 605 |
\[\sum fx^{2} = 80+175+180+441+768+324+300+605 = 2873.\]
\[\text{Variance} = \frac{\sum fx^{2}}{\sum f} - \bar{x}^{2} = \frac{2873}{50} - (7.3)^{2} = 57.46 - 53.29 = 4.17.\]
\[\text{S.D.} = \sqrt{4.17} = 2.04 \text{ (2 d.p.).}\]
The mean study time is 7.30 hours and the standard deviation is 2.04 hours.
Tambaya 4 Rahoto
In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm,
\(\overline{|QU|}\) = 12 cm and ?PUQ= ?STR =90°. If the area of PQR = 20 cm\(^2\),
calculate, correct to the nearest whole number, the:
(a) perimeter; (b) area; of the trapezium
Reading the diagram. The trapezium has \(QR \parallel PS\). \(QU\) and \(RT\) are drawn perpendicular to \(PS\), so \(QU = RT = 12\text{ cm}\) is the vertical height of the trapezium. From the figure: \(|PU| = 5\text{ cm}\), \(|QU| = 12\text{ cm}\), \(\angle PUQ = \angle STR = 90^\circ\), the angle at \(S\) is \(50^\circ\), and the area of \(\triangle PQR = 20\text{ cm}^2\).
Step 1: The slant side \(PQ\). Triangle \(PUQ\) is right-angled at \(U\):
\[|PQ| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}.\]Step 2: The top side \(QR\). Since \(QU \perp PS\), \(RT \perp PS\) and \(QR \parallel PS\), the figure \(QURT\) is a rectangle, so \(QR = UT\). Vertices \(Q\) and \(R\) both lie \(12\text{ cm}\) above line \(PS\), while \(P\) lies on \(PS\); hence the height of \(\triangle PQR\) on base \(QR\) is \(12\text{ cm}\):
\[\text{Area }\triangle PQR = \tfrac{1}{2}\times QR \times 12 = 6\,QR.\]\[6\,QR = 20 \;\Rightarrow\; QR = \tfrac{20}{6} = 3.33\text{ cm}.\]Step 3: The right-hand side using \(\angle S = 50^\circ\). Triangle \(STR\) is right-angled at \(T\) with \(RT = 12\):
\[|TS| = \frac{RT}{\tan 50^\circ} = \frac{12}{1.1918} = 10.07\text{ cm},\qquad |RS| = \frac{RT}{\sin 50^\circ} = \frac{12}{0.7660} = 15.67\text{ cm}.\]Step 4: The base \(PS\).
\[|PS| = |PU| + |UT| + |TS| = 5 + 3.33 + 10.07 = 18.40\text{ cm}.\](a) Perimeter of the trapezium.
\[P = |PQ| + |QR| + |RS| + |SP| = 13 + 3.33 + 15.67 + 18.40 = 50.40\text{ cm}.\]Perimeter \(\approx \mathbf{50\text{ cm}}\) (nearest whole number).
(b) Area of the trapezium. Parallel sides \(QR = 3.33\) and \(PS = 18.40\), height \(12\):
\[A = \tfrac{1}{2}(QR + PS)\times h = \tfrac{1}{2}(3.33 + 18.40)\times 12 = \tfrac{1}{2}(21.73)(12) = 130.4\text{ cm}^2.\]Area \(\approx \mathbf{130\text{ cm}^2}\) (nearest whole number).
Bayanin Amsa
Reading the diagram. The trapezium has \(QR \parallel PS\). \(QU\) and \(RT\) are drawn perpendicular to \(PS\), so \(QU = RT = 12\text{ cm}\) is the vertical height of the trapezium. From the figure: \(|PU| = 5\text{ cm}\), \(|QU| = 12\text{ cm}\), \(\angle PUQ = \angle STR = 90^\circ\), the angle at \(S\) is \(50^\circ\), and the area of \(\triangle PQR = 20\text{ cm}^2\).
Step 1: The slant side \(PQ\). Triangle \(PUQ\) is right-angled at \(U\):
\[|PQ| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}.\]Step 2: The top side \(QR\). Since \(QU \perp PS\), \(RT \perp PS\) and \(QR \parallel PS\), the figure \(QURT\) is a rectangle, so \(QR = UT\). Vertices \(Q\) and \(R\) both lie \(12\text{ cm}\) above line \(PS\), while \(P\) lies on \(PS\); hence the height of \(\triangle PQR\) on base \(QR\) is \(12\text{ cm}\):
\[\text{Area }\triangle PQR = \tfrac{1}{2}\times QR \times 12 = 6\,QR.\]\[6\,QR = 20 \;\Rightarrow\; QR = \tfrac{20}{6} = 3.33\text{ cm}.\]Step 3: The right-hand side using \(\angle S = 50^\circ\). Triangle \(STR\) is right-angled at \(T\) with \(RT = 12\):
\[|TS| = \frac{RT}{\tan 50^\circ} = \frac{12}{1.1918} = 10.07\text{ cm},\qquad |RS| = \frac{RT}{\sin 50^\circ} = \frac{12}{0.7660} = 15.67\text{ cm}.\]Step 4: The base \(PS\).
\[|PS| = |PU| + |UT| + |TS| = 5 + 3.33 + 10.07 = 18.40\text{ cm}.\](a) Perimeter of the trapezium.
\[P = |PQ| + |QR| + |RS| + |SP| = 13 + 3.33 + 15.67 + 18.40 = 50.40\text{ cm}.\]Perimeter \(\approx \mathbf{50\text{ cm}}\) (nearest whole number).
(b) Area of the trapezium. Parallel sides \(QR = 3.33\) and \(PS = 18.40\), height \(12\):
\[A = \tfrac{1}{2}(QR + PS)\times h = \tfrac{1}{2}(3.33 + 18.40)\times 12 = \tfrac{1}{2}(21.73)(12) = 130.4\text{ cm}^2.\]Area \(\approx \mathbf{130\text{ cm}^2}\) (nearest whole number).
Tambaya 5 Rahoto
The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270".
Calculate, correct: (a) two significant figures, |YZ|
(b) The nearest degree, the bearing of Y from Z
Take \(X\) as origin. \(Y\) is 15 km due south of \(X\); \(Z\) is 20 km on bearing \(270^{\circ}\), i.e. due west of \(X\). So \(\angle YXZ=90^{\circ}\) (south and west are perpendicular).
(a) \(|YZ|\) by Pythagoras:
\[|YZ|=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25\text{ km (2 s.f.)}.\]
(b) Bearing of \(Y\) from \(Z\)
Place \(Z=(-20,0)\), \(Y=(0,-15)\). The vector \(Z\!\to\!Y=(20,-15)\): from \(Z\), \(Y\) lies east and south. The angle east of south:
\[\tan\theta=\frac{20}{15}=1.333\Rightarrow\theta=53.13^{\circ}\ \text{(east of due south)}.\]
Bearing \(=180^{\circ}-53.13^{\circ}=126.87^{\circ}\approx \mathbf{127^{\circ}}\) (nearest degree).
Bayanin Amsa
Take \(X\) as origin. \(Y\) is 15 km due south of \(X\); \(Z\) is 20 km on bearing \(270^{\circ}\), i.e. due west of \(X\). So \(\angle YXZ=90^{\circ}\) (south and west are perpendicular).
(a) \(|YZ|\) by Pythagoras:
\[|YZ|=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25\text{ km (2 s.f.)}.\]
(b) Bearing of \(Y\) from \(Z\)
Place \(Z=(-20,0)\), \(Y=(0,-15)\). The vector \(Z\!\to\!Y=(20,-15)\): from \(Z\), \(Y\) lies east and south. The angle east of south:
\[\tan\theta=\frac{20}{15}=1.333\Rightarrow\theta=53.13^{\circ}\ \text{(east of due south)}.\]
Bearing \(=180^{\circ}-53.13^{\circ}=126.87^{\circ}\approx \mathbf{127^{\circ}}\) (nearest degree).
Tambaya 6 Rahoto
(a) Copy and complete the table of values for the relation \(y=2x^2-x-2\) for \(4 \le x \le 4\).
| x | -4 | -3 | -2 | -2 | 0 | 1 | 2 | 3 | 4 |
| y | 19 | -2 | 26 |
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of \(y=2x^2-x-2\) for \(4 \le x \le 4\).
(c) On the same axes, draw the graph of \(y=2x+3\).
(d) Use the graph to find the: (i) roots of the equation \(2x-3r-5\ 0\); (i) range of values of \(x\) for which \(2x^2-x-2<0\).
(a) Completed table for \(y = 2x^{2} - x - 2\), \(-4 \leq x \leq 4\). Substituting each value of \(x\):
\(x=-4:\ 2(16)+4-2=34\). \(x=-3:\ 2(9)+3-2=19\). \(x=-2:\ 2(4)+2-2=8\). \(x=-1:\ 2(1)+1-2=1\). \(x=0:\ -2\). \(x=1:\ 2-1-2=-1\). \(x=2:\ 8-2-2=4\). \(x=3:\ 18-3-2=13\). \(x=4:\ 32-4-2=26\).
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) Graph. Plot the nine points (2 cm to 1 unit on the x-axis, 2 cm to 5 units on the y-axis) and join with a smooth upward (minimum) parabola; its lowest point is near \(x=0.25,\ y=-2.1\).
(c) Line \(y = 2x + 3\). Plot two points, e.g. \((0,3)\) and \((2,7)\), and draw the straight line on the same axes.
(d)(i) Roots of \(2x^{2} - 3x - 5 = 0\). Where the curve meets the line, \(2x^{2}-x-2 = 2x+3\), which rearranges to \(2x^{2}-3x-5 = 0\). So the x-coordinates of the two intersection points are the required roots. From the graph \(x = -1\) and \(x = 2.5\). (Check: \(2x^2-3x-5=(2x-5)(x+1)=0\Rightarrow x=2.5,\ -1\).)
(d)(ii) Range of \(x\) for which \(2x^{2} - x - 2 < 0\). This is where the curve lies below the x-axis (\(y<0\)). The curve crosses \(y=0\) at \(x=\frac{1\pm\sqrt{17}}{4}\), i.e. \(x \approx -0.8\) and \(x \approx 1.3\). Hence \(y<0\) between them:
\[-0.8 < x < 1.3.\]
Bayanin Amsa
(a) Completed table for \(y = 2x^{2} - x - 2\), \(-4 \leq x \leq 4\). Substituting each value of \(x\):
\(x=-4:\ 2(16)+4-2=34\). \(x=-3:\ 2(9)+3-2=19\). \(x=-2:\ 2(4)+2-2=8\). \(x=-1:\ 2(1)+1-2=1\). \(x=0:\ -2\). \(x=1:\ 2-1-2=-1\). \(x=2:\ 8-2-2=4\). \(x=3:\ 18-3-2=13\). \(x=4:\ 32-4-2=26\).
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) Graph. Plot the nine points (2 cm to 1 unit on the x-axis, 2 cm to 5 units on the y-axis) and join with a smooth upward (minimum) parabola; its lowest point is near \(x=0.25,\ y=-2.1\).
(c) Line \(y = 2x + 3\). Plot two points, e.g. \((0,3)\) and \((2,7)\), and draw the straight line on the same axes.
(d)(i) Roots of \(2x^{2} - 3x - 5 = 0\). Where the curve meets the line, \(2x^{2}-x-2 = 2x+3\), which rearranges to \(2x^{2}-3x-5 = 0\). So the x-coordinates of the two intersection points are the required roots. From the graph \(x = -1\) and \(x = 2.5\). (Check: \(2x^2-3x-5=(2x-5)(x+1)=0\Rightarrow x=2.5,\ -1\).)
(d)(ii) Range of \(x\) for which \(2x^{2} - x - 2 < 0\). This is where the curve lies below the x-axis (\(y<0\)). The curve crosses \(y=0\) at \(x=\frac{1\pm\sqrt{17}}{4}\), i.e. \(x \approx -0.8\) and \(x \approx 1.3\). Hence \(y<0\) between them:
\[-0.8 < x < 1.3.\]
Tambaya 7 Rahoto
.(a) In APQR, ∠PQR= 90°. If its area is 216cm\(^2\) and |PQ|:|QR| is 3:4, find |PR|.
(b) The present ages of a man and his son are 47 years and 17 years respectively. In how many years would the man's age be twice that of his son?
(a) In \(\triangle PQR\), \(\angle PQR=90^{\circ}\), so \(PQ\) and \(QR\) are the perpendicular sides. Let \(PQ=3k,\ QR=4k\).
\[\text{Area}=\tfrac12\,PQ\cdot QR=216\Rightarrow\tfrac12(3k)(4k)=216\Rightarrow 6k^{2}=216\Rightarrow k^{2}=36\Rightarrow k=6.\]
So \(PQ=18\text{ cm},\ QR=24\text{ cm}\). By Pythagoras:
\[|PR|=\sqrt{18^{2}+24^{2}}=\sqrt{324+576}=\sqrt{900}=30\text{ cm}.\]
(b) Let the man's age be twice his son's after \(x\) years:
\[47+x=2(17+x)\Rightarrow 47+x=34+2x\Rightarrow x=13.\]
In 13 years' time the man will be twice as old as his son (60 and 30).
Bayanin Amsa
(a) In \(\triangle PQR\), \(\angle PQR=90^{\circ}\), so \(PQ\) and \(QR\) are the perpendicular sides. Let \(PQ=3k,\ QR=4k\).
\[\text{Area}=\tfrac12\,PQ\cdot QR=216\Rightarrow\tfrac12(3k)(4k)=216\Rightarrow 6k^{2}=216\Rightarrow k^{2}=36\Rightarrow k=6.\]
So \(PQ=18\text{ cm},\ QR=24\text{ cm}\). By Pythagoras:
\[|PR|=\sqrt{18^{2}+24^{2}}=\sqrt{324+576}=\sqrt{900}=30\text{ cm}.\]
(b) Let the man's age be twice his son's after \(x\) years:
\[47+x=2(17+x)\Rightarrow 47+x=34+2x\Rightarrow x=13.\]
In 13 years' time the man will be twice as old as his son (60 and 30).
Tambaya 8 Rahoto
In the diagram, \(\overline{AD}\) is a diameter of a circle with Centre O. If ABD is a triangle in a semi-circle ∠OAB=34",
find: (a) ∠OAB (b) ∠OCB
In the diagram, \(\overline{AD}\) is a diameter of the circle with centre \(O\), \(B\) lies on the circle, and \(BC\) is a tangent to the circle at \(B\). It is given that \(\angle OAB = 34^\circ\).
(a) \(\angle OAB\)
\(OA\) and \(OB\) are radii, so triangle \(OAB\) is isosceles and \(\angle OBA = \angle OAB\):
\[ \angle OAB = 34^\circ \](b) \(\angle OCB\)
Since \(AD\) is a diameter, the angle in the semicircle is a right angle:
\[ \angle ABD = 90^\circ \]The angle subtended at the centre is twice the angle at the circumference on the same arc \(BD\), so
\[ \angle BOC = \angle BOD = 2 \times \angle OAB = 2 \times 34^\circ = 68^\circ \]A tangent is perpendicular to the radius at the point of contact, so at \(B\):
\[ \angle OBC = 90^\circ \]In triangle \(OBC\), the angles sum to \(180^\circ\):
\[ \angle BOC + \angle OBC + \angle OCB = 180^\circ \] \[ 68^\circ + 90^\circ + \angle OCB = 180^\circ \] \[ \angle OCB = 180^\circ - 158^\circ = 22^\circ \]Bayanin Amsa
In the diagram, \(\overline{AD}\) is a diameter of the circle with centre \(O\), \(B\) lies on the circle, and \(BC\) is a tangent to the circle at \(B\). It is given that \(\angle OAB = 34^\circ\).
(a) \(\angle OAB\)
\(OA\) and \(OB\) are radii, so triangle \(OAB\) is isosceles and \(\angle OBA = \angle OAB\):
\[ \angle OAB = 34^\circ \](b) \(\angle OCB\)
Since \(AD\) is a diameter, the angle in the semicircle is a right angle:
\[ \angle ABD = 90^\circ \]The angle subtended at the centre is twice the angle at the circumference on the same arc \(BD\), so
\[ \angle BOC = \angle BOD = 2 \times \angle OAB = 2 \times 34^\circ = 68^\circ \]A tangent is perpendicular to the radius at the point of contact, so at \(B\):
\[ \angle OBC = 90^\circ \]In triangle \(OBC\), the angles sum to \(180^\circ\):
\[ \angle BOC + \angle OBC + \angle OCB = 180^\circ \] \[ 68^\circ + 90^\circ + \angle OCB = 180^\circ \] \[ \angle OCB = 180^\circ - 158^\circ = 22^\circ \]Tambaya 9 Rahoto
.(a) In a class of 80 students,\(\frac{3}{4}\) study Biology and \(\frac{3}{5}\) study Physics.
If each student studies at least one of the subjects: (i) draw a Venn diagram to represent this information
(ii) how many students study both subjects
(iii) find the fraction of the class that study Biology but not Physics.
(b) Johnson and Jocatol Ltd. owned a business office with floor measuring 15m by 8 m which was to be carpeted.
The cost of carpeting was Gh¢ 890.00 per square metre. If a total of GH 216,120.00 was spent on painting and carpeting, how much was the cost of painting?
(a) In a class of \(n=80\) students, let \(B\) be the set who study Biology and \(P\) the set who study Physics.
Number studying Biology \(=\tfrac{3}{4}\times80=60\).
Number studying Physics \(=\tfrac{3}{5}\times80=48\).
Since every student studies at least one subject, no one lies outside the two sets, i.e. \(n(B\cup P)=80\).
(i) Venn diagram. Let \(x\) be the number who study both subjects. Then Biology-only \(=60-x\) and Physics-only \(=48-x\):
(ii) How many study both subjects. Using the total,
\[ (60-x)+x+(48-x)=80 \] \[ 108-x=80\;\Rightarrow\;x=108-80=28. \]So 28 students study both Biology and Physics. Hence Biology-only \(=60-28=32\) and Physics-only \(=48-28=20\). Check: \(32+28+20=80\). \(\checkmark\)
(iii) Fraction that study Biology but not Physics. These are the Biology-only students:
\[ \frac{60-28}{80}=\frac{32}{80}=\frac{2}{5}. \]The fraction of the class that study Biology but not Physics is \(\dfrac{2}{5}\).
(b) Area of the office floor \(=15\text{ m}\times8\text{ m}=120\text{ m}^{2}\).
Cost of carpeting \(=120\text{ m}^{2}\times\text{GH}\cent890.00=\text{GH}\cent106{,}800.00\).
Total spent on painting and carpeting \(=\text{GH}\cent216{,}120.00\). Therefore
\[ \text{Cost of painting}=216{,}120.00-106{,}800.00=\text{GH}\cent109{,}320.00. \]The cost of painting was GH\(\cent\)109,320.00.
Bayanin Amsa
(a) In a class of \(n=80\) students, let \(B\) be the set who study Biology and \(P\) the set who study Physics.
Number studying Biology \(=\tfrac{3}{4}\times80=60\).
Number studying Physics \(=\tfrac{3}{5}\times80=48\).
Since every student studies at least one subject, no one lies outside the two sets, i.e. \(n(B\cup P)=80\).
(i) Venn diagram. Let \(x\) be the number who study both subjects. Then Biology-only \(=60-x\) and Physics-only \(=48-x\):
(ii) How many study both subjects. Using the total,
\[ (60-x)+x+(48-x)=80 \] \[ 108-x=80\;\Rightarrow\;x=108-80=28. \]So 28 students study both Biology and Physics. Hence Biology-only \(=60-28=32\) and Physics-only \(=48-28=20\). Check: \(32+28+20=80\). \(\checkmark\)
(iii) Fraction that study Biology but not Physics. These are the Biology-only students:
\[ \frac{60-28}{80}=\frac{32}{80}=\frac{2}{5}. \]The fraction of the class that study Biology but not Physics is \(\dfrac{2}{5}\).
(b) Area of the office floor \(=15\text{ m}\times8\text{ m}=120\text{ m}^{2}\).
Cost of carpeting \(=120\text{ m}^{2}\times\text{GH}\cent890.00=\text{GH}\cent106{,}800.00\).
Total spent on painting and carpeting \(=\text{GH}\cent216{,}120.00\). Therefore
\[ \text{Cost of painting}=216{,}120.00-106{,}800.00=\text{GH}\cent109{,}320.00. \]The cost of painting was GH\(\cent\)109,320.00.
Tambaya 10 Rahoto
(a) A man shared his property among his children as follows:
| Child's name | Ann | Afia | Kojo | Nuno | Akom |
| Percentage share | 5 | 15 | 10 | 45 | 25 |
Represent the information on a pie chart
(b) A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability th a girl takes away two red beads, one after the other, from the box.
(a) Pie chart of the property share
The shares sum to \(5+15+10+45+25 = 100\). A full circle is \(360^\circ\), so each child's sector angle is their percentage of \(360^\circ\), i.e. percentage \(\times \frac{360}{100} = \text{percentage} \times 3.6^\circ\):
| Child | Percentage share (%) | Sector angle |
|---|---|---|
| Ann | 5 | \(\frac{5}{100}\times 360^\circ = 18^\circ\) |
| Afia | 15 | \(\frac{15}{100}\times 360^\circ = 54^\circ\) |
| Kojo | 10 | \(\frac{10}{100}\times 360^\circ = 36^\circ\) |
| Nuno | 45 | \(\frac{45}{100}\times 360^\circ = 162^\circ\) |
| Akom | 25 | \(\frac{25}{100}\times 360^\circ = 90^\circ\) |
| Total | 100 | \(360^\circ\) |
Check: \(18^\circ+54^\circ+36^\circ+162^\circ+90^\circ = 360^\circ\). Drawing a circle and marking each sector with a protractor gives:
(b) Probability of taking two red beads, one after the other
The box holds \(5 + 3 + 4 = 12\) identical beads, of which 5 are red. The two beads are taken one after the other without replacement.
For the first draw, 5 of the 12 beads are red:
\[P(\text{1st red}) = \frac{5}{12}.\]After one red bead is removed, 4 red beads remain out of 11 beads in total:
\[P(\text{2nd red}\mid\text{1st red}) = \frac{4}{11}.\]Since both events must happen, multiply the probabilities:
\[P(\text{two red}) = \frac{5}{12}\times\frac{4}{11} = \frac{20}{132} = \frac{5}{33}\approx 0.1515.\]Therefore the probability of taking two red beads one after the other is \(\dfrac{5}{33}\) (about \(0.152\)).
Bayanin Amsa
(a) Pie chart of the property share
The shares sum to \(5+15+10+45+25 = 100\). A full circle is \(360^\circ\), so each child's sector angle is their percentage of \(360^\circ\), i.e. percentage \(\times \frac{360}{100} = \text{percentage} \times 3.6^\circ\):
| Child | Percentage share (%) | Sector angle |
|---|---|---|
| Ann | 5 | \(\frac{5}{100}\times 360^\circ = 18^\circ\) |
| Afia | 15 | \(\frac{15}{100}\times 360^\circ = 54^\circ\) |
| Kojo | 10 | \(\frac{10}{100}\times 360^\circ = 36^\circ\) |
| Nuno | 45 | \(\frac{45}{100}\times 360^\circ = 162^\circ\) |
| Akom | 25 | \(\frac{25}{100}\times 360^\circ = 90^\circ\) |
| Total | 100 | \(360^\circ\) |
Check: \(18^\circ+54^\circ+36^\circ+162^\circ+90^\circ = 360^\circ\). Drawing a circle and marking each sector with a protractor gives:
(b) Probability of taking two red beads, one after the other
The box holds \(5 + 3 + 4 = 12\) identical beads, of which 5 are red. The two beads are taken one after the other without replacement.
For the first draw, 5 of the 12 beads are red:
\[P(\text{1st red}) = \frac{5}{12}.\]After one red bead is removed, 4 red beads remain out of 11 beads in total:
\[P(\text{2nd red}\mid\text{1st red}) = \frac{4}{11}.\]Since both events must happen, multiply the probabilities:
\[P(\text{two red}) = \frac{5}{12}\times\frac{4}{11} = \frac{20}{132} = \frac{5}{33}\approx 0.1515.\]Therefore the probability of taking two red beads one after the other is \(\dfrac{5}{33}\) (about \(0.152\)).
Tambaya 11 Rahoto
(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms, find, correct to: (i) two significant figures, the distance between the two farms, (ii) the nearest degree, the bearing of Manu's farm from Dogbe's.
(b) A ladder 10 m long leaned against a vertical wall xm high. The distance between the wall and the foot of the ladder is 2 m longer than the height of the wall.
Calculate the value of x
(a) The cottage \(C\) is at bearing \(200^{\circ}\) from Dogbe's farm \(D\) and \(110^{\circ}\) from Manu's farm \(M\). Reversing bearings, from the cottage:
- Bearing of \(D\) from \(C=200^{\circ}-180^{\circ}=020^{\circ}\), with \(|CD|=5\) km.
- Bearing of \(M\) from \(C=110^{\circ}+180^{\circ}=290^{\circ}\), with \(|CM|=3\) km.
Angle at the cottage \(=290^{\circ}-020^{\circ}=270^{\circ}\Rightarrow\) reflex; the angle between the two directions is \(360^{\circ}-270^{\circ}=90^{\circ}\).
(i) With \(\angle DCM=90^{\circ}\):
\[|DM|=\sqrt{5^{2}+3^{2}}=\sqrt{34}=5.83\approx\mathbf{5.8\text{ km}}\ (2\text{ s.f.}).\]
(ii) Using coordinates \(D=(1.71,4.70),\ M=(-2.82,1.03)\): \(D\!\to\!M=(-4.53,-3.67)\) points south-west.
\[\tan\alpha=\frac{4.53}{3.67}=1.234\Rightarrow\alpha=51^{\circ};\quad\text{bearing}=180^{\circ}+51^{\circ}=\mathbf{231^{\circ}}.\]
(b) Ladder 10 m, wall \(x\) m, foot distance \((x+2)\) m:
\[x^{2}+(x+2)^{2}=10^{2}\Rightarrow 2x^{2}+4x+4=100\Rightarrow x^{2}+2x-48=0\Rightarrow (x+8)(x-6)=0.\]
Since \(x>0,\ \mathbf{x=6\text{ m}}.\)
Bayanin Amsa
(a) The cottage \(C\) is at bearing \(200^{\circ}\) from Dogbe's farm \(D\) and \(110^{\circ}\) from Manu's farm \(M\). Reversing bearings, from the cottage:
- Bearing of \(D\) from \(C=200^{\circ}-180^{\circ}=020^{\circ}\), with \(|CD|=5\) km.
- Bearing of \(M\) from \(C=110^{\circ}+180^{\circ}=290^{\circ}\), with \(|CM|=3\) km.
Angle at the cottage \(=290^{\circ}-020^{\circ}=270^{\circ}\Rightarrow\) reflex; the angle between the two directions is \(360^{\circ}-270^{\circ}=90^{\circ}\).
(i) With \(\angle DCM=90^{\circ}\):
\[|DM|=\sqrt{5^{2}+3^{2}}=\sqrt{34}=5.83\approx\mathbf{5.8\text{ km}}\ (2\text{ s.f.}).\]
(ii) Using coordinates \(D=(1.71,4.70),\ M=(-2.82,1.03)\): \(D\!\to\!M=(-4.53,-3.67)\) points south-west.
\[\tan\alpha=\frac{4.53}{3.67}=1.234\Rightarrow\alpha=51^{\circ};\quad\text{bearing}=180^{\circ}+51^{\circ}=\mathbf{231^{\circ}}.\]
(b) Ladder 10 m, wall \(x\) m, foot distance \((x+2)\) m:
\[x^{2}+(x+2)^{2}=10^{2}\Rightarrow 2x^{2}+4x+4=100\Rightarrow x^{2}+2x-48=0\Rightarrow (x+8)(x-6)=0.\]
Since \(x>0,\ \mathbf{x=6\text{ m}}.\)
Tambaya 12 Rahoto
(a) On Sam's first birthday celebration, his grandfather deposited an amount of $1,000.00 in a bank at compound interest of 4% per annum.
Find how much is in the account if Sam is 4 years old.
(b)
In the diagram, ABCD are points on the circle with centre O. If \(|AB| = |BC|\) and \(\angle ADC = 50^\circ\), find \(\angle BAD\).
(a) Compound interest. \(\$1000\) at \(4\%\) per annum, from the 1st birthday until age 4 is 3 years of growth (multiplier \(1.04\) each year):
End of yr 2: \(\frac{104}{100}\times1000=\$1040.00\); yr 3: \(\frac{104}{100}\times1040=\$1081.60\); yr 4: \(\frac{104}{100}\times1081.60=\$1124.86\).
\[A=1000\Big(1+\tfrac{4}{100}\Big)^{3}=1000(1.04)^{3}=\mathbf{\$1124.86}.\]
(b) Circle geometry. \(A,B,C,D\) lie on a circle centre \(O\), \(|AB|=|BC|\), \(\angle ADC=50^{\circ}\), and \(AC\) subtends \(\angle ADC\) with \(\angle ACD=90^{\circ}\) (angle in a semicircle, \(AD\) a diameter).
In \(\triangle ACD\): \(\angle CAD=180^{\circ}-50^{\circ}-90^{\circ}=40^{\circ}\).
\(ABCD\) is cyclic, so \(\angle ABC=180^{\circ}-\angle ADC=130^{\circ}\). Since \(|AB|=|BC|\), \(\triangle ABC\) is isosceles with \(\angle BAC=\angle BCA\):
\[2\angle BAC+130^{\circ}=180^{\circ}\Rightarrow\angle BAC=25^{\circ}.\]
\[\angle BAD=\angle CAD+\angle BAC=40^{\circ}+25^{\circ}=\mathbf{65^{\circ}}.\]
Bayanin Amsa
(a) Compound interest. \(\$1000\) at \(4\%\) per annum, from the 1st birthday until age 4 is 3 years of growth (multiplier \(1.04\) each year):
End of yr 2: \(\frac{104}{100}\times1000=\$1040.00\); yr 3: \(\frac{104}{100}\times1040=\$1081.60\); yr 4: \(\frac{104}{100}\times1081.60=\$1124.86\).
\[A=1000\Big(1+\tfrac{4}{100}\Big)^{3}=1000(1.04)^{3}=\mathbf{\$1124.86}.\]
(b) Circle geometry. \(A,B,C,D\) lie on a circle centre \(O\), \(|AB|=|BC|\), \(\angle ADC=50^{\circ}\), and \(AC\) subtends \(\angle ADC\) with \(\angle ACD=90^{\circ}\) (angle in a semicircle, \(AD\) a diameter).
In \(\triangle ACD\): \(\angle CAD=180^{\circ}-50^{\circ}-90^{\circ}=40^{\circ}\).
\(ABCD\) is cyclic, so \(\angle ABC=180^{\circ}-\angle ADC=130^{\circ}\). Since \(|AB|=|BC|\), \(\triangle ABC\) is isosceles with \(\angle BAC=\angle BCA\):
\[2\angle BAC+130^{\circ}=180^{\circ}\Rightarrow\angle BAC=25^{\circ}.\]
\[\angle BAD=\angle CAD+\angle BAC=40^{\circ}+25^{\circ}=\mathbf{65^{\circ}}.\]
Tambaya 13 Rahoto
A man left town am at 10:00 AM and traveled by car to town N at an average speed of 72 km/h.
He spent 2hours for a meeting and returned through town M by bus at an average speed of 40KM/H.
If the distance covered by the bus was 2km longer than that of the car and he arrived at town M at 1 :55PM.
calculate distance from M to N.
Setting up the time equation
Total time from 10:00 a.m. to 1:55 p.m. \(=3\text{ h }55\text{ min}=3\tfrac{55}{60}\text{ h}=3\tfrac{11}{12}\text{ h}\).
Removing the 2-hour meeting, the driving+bus time is
\[3\tfrac{11}{12}-2=1\tfrac{11}{12}\text{ h}=\tfrac{23}{12}\text{ h}.\]
Let the car distance \(M\!\to\!N\) be \(d\) km (at 72 km/h). The bus distance is \((d+2)\) km (at 40 km/h). Then
\[\frac{d}{72}+\frac{d+2}{40}=\frac{23}{12}.\]
Multiply through by 360:
\[5d+9(d+2)=690\Rightarrow 14d+18=690\Rightarrow 14d=672\Rightarrow d=48.\]
Check: car \(=48/72=40\) min, bus \(=50/40=1\text{ h }15\text{ min}\); driving \(=1\text{ h }55\text{ min}\); with the 2 h meeting the total is 3 h 55 min, arriving 1:55 p.m. \(\checkmark\)
Distance from \(M\) to \(N\) \(= 48\) km.
Bayanin Amsa
Setting up the time equation
Total time from 10:00 a.m. to 1:55 p.m. \(=3\text{ h }55\text{ min}=3\tfrac{55}{60}\text{ h}=3\tfrac{11}{12}\text{ h}\).
Removing the 2-hour meeting, the driving+bus time is
\[3\tfrac{11}{12}-2=1\tfrac{11}{12}\text{ h}=\tfrac{23}{12}\text{ h}.\]
Let the car distance \(M\!\to\!N\) be \(d\) km (at 72 km/h). The bus distance is \((d+2)\) km (at 40 km/h). Then
\[\frac{d}{72}+\frac{d+2}{40}=\frac{23}{12}.\]
Multiply through by 360:
\[5d+9(d+2)=690\Rightarrow 14d+18=690\Rightarrow 14d=672\Rightarrow d=48.\]
Check: car \(=48/72=40\) min, bus \(=50/40=1\text{ h }15\text{ min}\); driving \(=1\text{ h }55\text{ min}\); with the 2 h meeting the total is 3 h 55 min, arriving 1:55 p.m. \(\checkmark\)
Distance from \(M\) to \(N\) \(= 48\) km.
