WAEC SSCE - General Mathematics - 2020 (Essay)
Tambaya 1 Rahoto
The table shows the distribution of marks obtained by students in an examination.
| Marks (%) | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw the cumulative frequency curve for the distribution.
(c) Using the curve, find correct to one decimal place, the:
(i) median mark;
(ii) lowest mark for the distinction if 5% of the students passed with distinction
(a) Cumulative Frequency Table
| Marks (%) | Upper Class Boundary | Frequency | Cumulative Frequency |
|---|---|---|---|
| 0 – 9 | 9.5 | 7 | 7 |
| 10 – 19 | 19.5 | 11 | 18 |
| 20 – 29 | 29.5 | 17 | 35 |
| 30 – 39 | 39.5 | 20 | 55 |
| 40 – 49 | 49.5 | 29 | 84 |
| 50 – 59 | 59.5 | 34 | 118 |
| 60 – 69 | 69.5 | 30 | 148 |
| 70 – 79 | 79.5 | 25 | 173 |
| 80 – 89 | 89.5 | 21 | 194 |
| 90 – 99 | 99.5 | 6 | 200 |
(b) Cumulative Frequency Curve
Using the upper class boundaries, plot the points:
(9.5, 7), (19.5, 18), (29.5, 35), (39.5, 55), (49.5, 84),
(59.5, 118), (69.5, 148), (79.5, 173), (89.5, 194) and (99.5, 200).
Join the points with a smooth cumulative frequency curve (ogive).
(c)
(i) Median mark
Total frequency, N = 200.
Median position = N⁄2 = 200⁄2 = 100th value.
The 100th value lies in the class 50
Bayanin Amsa
(a) Cumulative Frequency Table
| Marks (%) | Upper Class Boundary | Frequency | Cumulative Frequency |
|---|---|---|---|
| 0 – 9 | 9.5 | 7 | 7 |
| 10 – 19 | 19.5 | 11 | 18 |
| 20 – 29 | 29.5 | 17 | 35 |
| 30 – 39 | 39.5 | 20 | 55 |
| 40 – 49 | 49.5 | 29 | 84 |
| 50 – 59 | 59.5 | 34 | 118 |
| 60 – 69 | 69.5 | 30 | 148 |
| 70 – 79 | 79.5 | 25 | 173 |
| 80 – 89 | 89.5 | 21 | 194 |
| 90 – 99 | 99.5 | 6 | 200 |
(b) Cumulative Frequency Curve
Using the upper class boundaries, plot the points:
(9.5, 7), (19.5, 18), (29.5, 35), (39.5, 55), (49.5, 84),
(59.5, 118), (69.5, 148), (79.5, 173), (89.5, 194) and (99.5, 200).
Join the points with a smooth cumulative frequency curve (ogive).
(c)
(i) Median mark
Total frequency, N = 200.
Median position = N⁄2 = 200⁄2 = 100th value.
The 100th value lies in the class 50
Tambaya 2 Rahoto
(a) Given that P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
(i) Make Q the subject of the relation;
(ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that \(\frac{x + 2y}{5}\) = x - 2y, find x : y
(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).
Raise both sides to the power \(\tfrac{3}{2}\):
\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]
\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]
(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):
- \(ms = 15 \times 0.2 = 3\)
- \(rk = 10 \times 4 = 40\)
- \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)
\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]
(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:
\[x + 2y = 5(x - 2y) = 5x - 10y.\]
\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]
Therefore \(x : y = 3 : 1\).
Bayanin Amsa
(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).
Raise both sides to the power \(\tfrac{3}{2}\):
\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]
\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]
(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):
- \(ms = 15 \times 0.2 = 3\)
- \(rk = 10 \times 4 = 40\)
- \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)
\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]
(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:
\[x + 2y = 5(x - 2y) = 5x - 10y.\]
\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]
Therefore \(x : y = 3 : 1\).
Tambaya 3 Rahoto
(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius is 14. Calculate, correct to three significant figures, the surface area of the structure, [Take \(\pi = \frac{22}{7}\)]
(b) Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.
(a) Reading the diagram. The structure is a cone (apex \(L\), base diameter \(MN\), centre \(O\)) sitting on a hemisphere of the same base radius. The cone has vertical height \(h = 48\text{ m}\) and base radius \(r = 14\text{ m}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the flat circular join is inside the solid and is not counted.
Step 1: Slant height of the cone.
\[l = \sqrt{r^{2}+h^{2}} = \sqrt{14^{2}+48^{2}} = \sqrt{196+2304} = \sqrt{2500} = 50\text{ m}.\]
Step 2: Curved surface area of the cone \(= \pi r l\):
\[= \frac{22}{7}\times 14\times 50 = 22\times 2\times 50 = 2200\text{ m}^2.\]
Step 3: Curved surface area of the hemisphere \(= 2\pi r^{2}\):
\[= 2\times\frac{22}{7}\times 14^{2} = 2\times\frac{22}{7}\times 196 = 2\times 22\times 28 = 1232\text{ m}^2.\]
Step 4: Total surface area of the structure.
\[= 2200 + 1232 = 3432\text{ m}^2 \approx 3430\text{ m}^2\ (3\text{ s.f.}).\]
(b) Ages problem. Let Musah's present age be \(M\) and Sesay's present age be \(S\).
Sum of present ages:
\[M + S = 100.\]
Five years ago Musah was twice as old as Sesay:
\[M - 5 = 2(S - 5) \;\Rightarrow\; M - 5 = 2S - 10 \;\Rightarrow\; M = 2S - 5.\]
Substitute into the sum:
\[(2S - 5) + S = 100 \;\Rightarrow\; 3S = 105 \;\Rightarrow\; S = 35.\]
Answers: (a) surface area \(= 3430\text{ m}^2\) (3 s.f.); (b) Sesay's present age is \(\mathbf{35}\) years.
Bayanin Amsa
(a) Reading the diagram. The structure is a cone (apex \(L\), base diameter \(MN\), centre \(O\)) sitting on a hemisphere of the same base radius. The cone has vertical height \(h = 48\text{ m}\) and base radius \(r = 14\text{ m}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the flat circular join is inside the solid and is not counted.
Step 1: Slant height of the cone.
\[l = \sqrt{r^{2}+h^{2}} = \sqrt{14^{2}+48^{2}} = \sqrt{196+2304} = \sqrt{2500} = 50\text{ m}.\]
Step 2: Curved surface area of the cone \(= \pi r l\):
\[= \frac{22}{7}\times 14\times 50 = 22\times 2\times 50 = 2200\text{ m}^2.\]
Step 3: Curved surface area of the hemisphere \(= 2\pi r^{2}\):
\[= 2\times\frac{22}{7}\times 14^{2} = 2\times\frac{22}{7}\times 196 = 2\times 22\times 28 = 1232\text{ m}^2.\]
Step 4: Total surface area of the structure.
\[= 2200 + 1232 = 3432\text{ m}^2 \approx 3430\text{ m}^2\ (3\text{ s.f.}).\]
(b) Ages problem. Let Musah's present age be \(M\) and Sesay's present age be \(S\).
Sum of present ages:
\[M + S = 100.\]
Five years ago Musah was twice as old as Sesay:
\[M - 5 = 2(S - 5) \;\Rightarrow\; M - 5 = 2S - 10 \;\Rightarrow\; M = 2S - 5.\]
Substitute into the sum:
\[(2S - 5) + S = 100 \;\Rightarrow\; 3S = 105 \;\Rightarrow\; S = 35.\]
Answers: (a) surface area \(= 3430\text{ m}^2\) (3 s.f.); (b) Sesay's present age is \(\mathbf{35}\) years.
Tambaya 4 Rahoto
(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate. correct to the nearest whole number, the:
(i) distance between cyclist X and Y;
(ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
Bayanin Amsa
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
Tambaya 5 Rahoto
(a) Copy and complete the table of values for the relation \(y = 3 \sin 2x\).
| x | \(o^o\) | \(15^o\) | \(30^o\) | \(45^o\) | \(60^o\) | \(75^o\) | \(90^o\) | \(105^o\) | \(120^o\) | \(135^o\) | \(^o\) |
| y | 0.0 | 1.5 | -2.6 |
(b) Using a scale of 2 cm to 15° on the x-axis and 2cm to I unit on the y-axis, draw the graph of \(y = 3 \sin 2x\) for \(0° \geq x \geq 150°\).
(c) Use the graph to find the truth set of;
(i) \(3 \sin 2x + 2 = 0\);
(ii ) \(\frac{3}{2} \sin 2x = 0.25\).
(a) Completing the table for \(y=3\sin 2x\) (angles in degrees). For example \(x=30^\circ:\;3\sin 60^\circ=3(0.866)=2.6\).
| x | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| y | 0.0 | 1.5 | 2.6 | 3.0 | 2.6 | 1.5 | 0.0 | -1.5 | -2.6 | -3.0 | -2.6 |
(b) Plot these points and join with a smooth sine curve for \(0^\circ\le x\le 150^\circ\).
(c)(i) \(3\sin 2x+2=0\). This means \(3\sin 2x=-2\), so draw the line \(y=-2\) and read the crossing:
\[ \sin 2x=-\tfrac{2}{3} \Rightarrow 2x=221.8^\circ \Rightarrow x\approx 111^\circ \]Truth set \(=\{x\approx 111^\circ\}\).
(ii) \(\frac{3}{2}\sin 2x=0.25\). Multiply by 2: \(3\sin 2x=0.5\), so draw the line \(y=0.5\):
\[ \sin 2x=\tfrac{1}{6} \Rightarrow 2x=9.6^\circ \text{ or } 170.4^\circ \Rightarrow x\approx 5^\circ \text{ or } 85^\circ \]Truth set \(=\{x\approx 5^\circ,\; 85^\circ\}\).
Bayanin Amsa
(a) Completing the table for \(y=3\sin 2x\) (angles in degrees). For example \(x=30^\circ:\;3\sin 60^\circ=3(0.866)=2.6\).
| x | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| y | 0.0 | 1.5 | 2.6 | 3.0 | 2.6 | 1.5 | 0.0 | -1.5 | -2.6 | -3.0 | -2.6 |
(b) Plot these points and join with a smooth sine curve for \(0^\circ\le x\le 150^\circ\).
(c)(i) \(3\sin 2x+2=0\). This means \(3\sin 2x=-2\), so draw the line \(y=-2\) and read the crossing:
\[ \sin 2x=-\tfrac{2}{3} \Rightarrow 2x=221.8^\circ \Rightarrow x\approx 111^\circ \]Truth set \(=\{x\approx 111^\circ\}\).
(ii) \(\frac{3}{2}\sin 2x=0.25\). Multiply by 2: \(3\sin 2x=0.5\), so draw the line \(y=0.5\):
\[ \sin 2x=\tfrac{1}{6} \Rightarrow 2x=9.6^\circ \text{ or } 170.4^\circ \Rightarrow x\approx 5^\circ \text{ or } 85^\circ \]Truth set \(=\{x\approx 5^\circ,\; 85^\circ\}\).
Tambaya 6 Rahoto
(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE.
(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate
\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)
(a) Finding \(\widehat{CDE}\). From the diagram \(A,B,C,D,E\) lie on the circle with centre \(O\), and the lines \(AD\) and \(EB\) both pass through \(O\), so \(AD\) and \(EB\) are diameters. Also \(|BC|=|CD|\) and \(\widehat{BCD}=108^\circ\).
Equal chords cut equal arcs, so \(\text{arc }BC=\text{arc }CD\); call each \(a\).
\(\widehat{BCD}=108^\circ\) is the angle at the circumference standing on the arc \(BAED\) (the arc from \(B\) to \(D\) not through \(C\)):
\[\text{arc }BAED=2\times108^\circ=216^\circ\;\Rightarrow\;\text{arc }BCD=360^\circ-216^\circ=144^\circ.\]So \(2a=144^\circ\Rightarrow a=72^\circ\); thus \(\text{arc }BC=\text{arc }CD=72^\circ.\)
Since \(AD\) is a diameter, arc \(ABCD\) (semicircle) \(=180^\circ\), so \(\text{arc }AB=180^\circ-(72^\circ+72^\circ)=36^\circ.\) Since \(EB\) is a diameter, arc \(EAB=180^\circ\), so \(\text{arc }AE=180^\circ-36^\circ=144^\circ,\) and \(\text{arc }ED=180^\circ-144^\circ=36^\circ.\)
\(\widehat{CDE}\) stands on the arc \(CBAE\) (from \(C\) to \(E\) not through \(D\)):
\[\text{arc }CBAE=72^\circ+36^\circ+144^\circ=252^\circ,\qquad \widehat{CDE}=\tfrac12(252^\circ)=\boxed{126^\circ.}\](Check: \(\widehat{CDE}=\widehat{CDA}+\widehat{ADE}=\tfrac12(108^\circ)+\tfrac12(144^\circ)=54^\circ+72^\circ=126^\circ.\))
(b) Evaluate \(\dfrac{\cos^2 x-\sin x}{\sin^2 x+\cos x}\) given \(\tan x=\sqrt3,\ 0^\circ\le x\le 90^\circ.\)
\(\tan x=\sqrt3\Rightarrow x=60^\circ,\) so \(\sin x=\dfrac{\sqrt3}{2},\ \cos x=\dfrac12.\)
\[\text{Numerator}=\left(\tfrac12\right)^2-\tfrac{\sqrt3}{2}=\tfrac14-\tfrac{\sqrt3}{2}=\frac{1-2\sqrt3}{4}.\]\[\text{Denominator}=\left(\tfrac{\sqrt3}{2}\right)^2+\tfrac12=\tfrac34+\tfrac12=\tfrac54.\]\[\frac{\frac{1-2\sqrt3}{4}}{\frac54}=\frac{1-2\sqrt3}{5}\approx-0.49.\]Bayanin Amsa
(a) Finding \(\widehat{CDE}\). From the diagram \(A,B,C,D,E\) lie on the circle with centre \(O\), and the lines \(AD\) and \(EB\) both pass through \(O\), so \(AD\) and \(EB\) are diameters. Also \(|BC|=|CD|\) and \(\widehat{BCD}=108^\circ\).
Equal chords cut equal arcs, so \(\text{arc }BC=\text{arc }CD\); call each \(a\).
\(\widehat{BCD}=108^\circ\) is the angle at the circumference standing on the arc \(BAED\) (the arc from \(B\) to \(D\) not through \(C\)):
\[\text{arc }BAED=2\times108^\circ=216^\circ\;\Rightarrow\;\text{arc }BCD=360^\circ-216^\circ=144^\circ.\]So \(2a=144^\circ\Rightarrow a=72^\circ\); thus \(\text{arc }BC=\text{arc }CD=72^\circ.\)
Since \(AD\) is a diameter, arc \(ABCD\) (semicircle) \(=180^\circ\), so \(\text{arc }AB=180^\circ-(72^\circ+72^\circ)=36^\circ.\) Since \(EB\) is a diameter, arc \(EAB=180^\circ\), so \(\text{arc }AE=180^\circ-36^\circ=144^\circ,\) and \(\text{arc }ED=180^\circ-144^\circ=36^\circ.\)
\(\widehat{CDE}\) stands on the arc \(CBAE\) (from \(C\) to \(E\) not through \(D\)):
\[\text{arc }CBAE=72^\circ+36^\circ+144^\circ=252^\circ,\qquad \widehat{CDE}=\tfrac12(252^\circ)=\boxed{126^\circ.}\](Check: \(\widehat{CDE}=\widehat{CDA}+\widehat{ADE}=\tfrac12(108^\circ)+\tfrac12(144^\circ)=54^\circ+72^\circ=126^\circ.\))
(b) Evaluate \(\dfrac{\cos^2 x-\sin x}{\sin^2 x+\cos x}\) given \(\tan x=\sqrt3,\ 0^\circ\le x\le 90^\circ.\)
\(\tan x=\sqrt3\Rightarrow x=60^\circ,\) so \(\sin x=\dfrac{\sqrt3}{2},\ \cos x=\dfrac12.\)
\[\text{Numerator}=\left(\tfrac12\right)^2-\tfrac{\sqrt3}{2}=\tfrac14-\tfrac{\sqrt3}{2}=\frac{1-2\sqrt3}{4}.\]\[\text{Denominator}=\left(\tfrac{\sqrt3}{2}\right)^2+\tfrac12=\tfrac34+\tfrac12=\tfrac54.\]\[\frac{\frac{1-2\sqrt3}{4}}{\frac54}=\frac{1-2\sqrt3}{5}\approx-0.49.\]Tambaya 7 Rahoto
(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40°, find: < ODE.
(b) ABCD is a parallelogram in which |\(\overline{CD}\)| = 7 cm, I\(\overline{AD}\)I = 5 cm and < ADC= 125°.
(i) Illustrate the information in a diagram.
(ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\)). Evaluate (2x\(^2\) - 2x).
(a) Finding \( \angle ODE \) from the diagram
Reading the diagram: \(AB\) is a tangent touching the circle at \(A\); \(C\), \(O\) and \(B\) lie on one straight line (so \(CB\) passes through the centre \(O\)); \(E\) is the point where this line \(CB\) meets the circle on the right, so \(CE\) is a diameter. \(CD \parallel AB\) and \( \angle ABE = 40^\circ \).
Step 1: Use the tangent. A radius is perpendicular to a tangent at the point of contact, so \( \angle OAB = 90^\circ \).
In \( \triangle OAB \), \( \angle ABO = 40^\circ \), hence
\[ \angle AOB = 180^\circ - 90^\circ - 40^\circ = 50^\circ. \]
Step 2: Use the parallel chord. The line \(CB\) is a transversal cutting the parallel lines \(AB\) and \(CD\). By alternate angles,
\[ \angle DCB = \angle ABE = 40^\circ, \] so \( \angle DCO = 40^\circ \).
Step 3: Base angles of an isosceles triangle. In \( \triangle OCD \), \(OC = OD\) (both radii), so it is isosceles with
\[ \angle ODC = \angle OCD = 40^\circ. \]
Step 4: Angle in a semicircle. Since \(CE\) is a diameter and \(D\) lies on the circle, the angle it subtends is a right angle:
\[ \angle CDE = 90^\circ. \]
Step 5: Combine. The radius \(OD\) lies inside \( \angle CDE \), so
\[ \angle ODE = \angle CDE - \angle ODC = 90^\circ - 40^\circ = 50^\circ. \]
\( \angle ODE = 50^\circ \).
(b) Parallelogram \(ABCD\)
(i) Illustration. Draw parallelogram \(ABCD\) with vertices labelled in order. Mark side \(DC = 7\ \text{cm}\) along the base and side \(AD = 5\ \text{cm}\) meeting it at \(D\), with the interior angle \( \angle ADC = 125^\circ \) between them. The opposite sides are equal and parallel: \(AB = DC = 7\ \text{cm}\), \(BC = AD = 5\ \text{cm}\), and \( \angle ABC = 125^\circ \), while \( \angle DAB = \angle BCD = 55^\circ \).
(ii) Area. For a parallelogram, area equals the product of two adjacent sides and the sine of the included angle:
\[ \text{Area} = |DC| \times |AD| \times \sin(\angle ADC). \]
\[ \text{Area} = 7 \times 5 \times \sin 125^\circ = 35 \times 0.8192 = 28.67\ \text{cm}^2. \]
Area \( \approx 28.7\ \text{cm}^2 \) (to one decimal place).
(c) Evaluate \( 2x^2 - 2x \) when \( x = \tfrac{1}{2}(1 - \sqrt{2}) \)
First compute \( x^2 \):
\[ x^2 = \left(\frac{1-\sqrt{2}}{2}\right)^2 = \frac{(1-\sqrt{2})^2}{4} = \frac{1 - 2\sqrt{2} + 2}{4} = \frac{3 - 2\sqrt{2}}{4}. \]
Then
\[ 2x^2 = \frac{3 - 2\sqrt{2}}{2}, \qquad 2x = 1 - \sqrt{2} = \frac{2 - 2\sqrt{2}}{2}. \]
Therefore
\[ 2x^2 - 2x = \frac{3 - 2\sqrt{2}}{2} - \frac{2 - 2\sqrt{2}}{2} = \frac{3 - 2\sqrt{2} - 2 + 2\sqrt{2}}{2} = \frac{1}{2}. \]
\( 2x^2 - 2x = \dfrac{1}{2} \).
Bayanin Amsa
(a) Finding \( \angle ODE \) from the diagram
Reading the diagram: \(AB\) is a tangent touching the circle at \(A\); \(C\), \(O\) and \(B\) lie on one straight line (so \(CB\) passes through the centre \(O\)); \(E\) is the point where this line \(CB\) meets the circle on the right, so \(CE\) is a diameter. \(CD \parallel AB\) and \( \angle ABE = 40^\circ \).
Step 1: Use the tangent. A radius is perpendicular to a tangent at the point of contact, so \( \angle OAB = 90^\circ \).
In \( \triangle OAB \), \( \angle ABO = 40^\circ \), hence
\[ \angle AOB = 180^\circ - 90^\circ - 40^\circ = 50^\circ. \]
Step 2: Use the parallel chord. The line \(CB\) is a transversal cutting the parallel lines \(AB\) and \(CD\). By alternate angles,
\[ \angle DCB = \angle ABE = 40^\circ, \] so \( \angle DCO = 40^\circ \).
Step 3: Base angles of an isosceles triangle. In \( \triangle OCD \), \(OC = OD\) (both radii), so it is isosceles with
\[ \angle ODC = \angle OCD = 40^\circ. \]
Step 4: Angle in a semicircle. Since \(CE\) is a diameter and \(D\) lies on the circle, the angle it subtends is a right angle:
\[ \angle CDE = 90^\circ. \]
Step 5: Combine. The radius \(OD\) lies inside \( \angle CDE \), so
\[ \angle ODE = \angle CDE - \angle ODC = 90^\circ - 40^\circ = 50^\circ. \]
\( \angle ODE = 50^\circ \).
(b) Parallelogram \(ABCD\)
(i) Illustration. Draw parallelogram \(ABCD\) with vertices labelled in order. Mark side \(DC = 7\ \text{cm}\) along the base and side \(AD = 5\ \text{cm}\) meeting it at \(D\), with the interior angle \( \angle ADC = 125^\circ \) between them. The opposite sides are equal and parallel: \(AB = DC = 7\ \text{cm}\), \(BC = AD = 5\ \text{cm}\), and \( \angle ABC = 125^\circ \), while \( \angle DAB = \angle BCD = 55^\circ \).
(ii) Area. For a parallelogram, area equals the product of two adjacent sides and the sine of the included angle:
\[ \text{Area} = |DC| \times |AD| \times \sin(\angle ADC). \]
\[ \text{Area} = 7 \times 5 \times \sin 125^\circ = 35 \times 0.8192 = 28.67\ \text{cm}^2. \]
Area \( \approx 28.7\ \text{cm}^2 \) (to one decimal place).
(c) Evaluate \( 2x^2 - 2x \) when \( x = \tfrac{1}{2}(1 - \sqrt{2}) \)
First compute \( x^2 \):
\[ x^2 = \left(\frac{1-\sqrt{2}}{2}\right)^2 = \frac{(1-\sqrt{2})^2}{4} = \frac{1 - 2\sqrt{2} + 2}{4} = \frac{3 - 2\sqrt{2}}{4}. \]
Then
\[ 2x^2 = \frac{3 - 2\sqrt{2}}{2}, \qquad 2x = 1 - \sqrt{2} = \frac{2 - 2\sqrt{2}}{2}. \]
Therefore
\[ 2x^2 - 2x = \frac{3 - 2\sqrt{2}}{2} - \frac{2 - 2\sqrt{2}}{2} = \frac{3 - 2\sqrt{2} - 2 + 2\sqrt{2}}{2} = \frac{1}{2}. \]
\( 2x^2 - 2x = \dfrac{1}{2} \).
Tambaya 8 Rahoto
(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at an open market and \(\frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had N222,000.00 left, find:
(i) her monthly income.
(ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A. P.) is 4m - 2n. If the ninth term of the progression is 2m - 8n. find the common difference in terms of m and n.
(a)(i) Monthly income. At the mall and market she spent \(\tfrac{1}{4} + \tfrac{1}{3} = \tfrac{7}{12}\) of her income, leaving \(\tfrac{5}{12}\). Of this remainder she spent \(\tfrac{2}{5}\) at the mechanic, so she kept \(\tfrac{3}{5}\) of the remainder:
\[\tfrac{3}{5}\times\tfrac{5}{12} = \tfrac{1}{4}\text{ of the income} = \text{N}222{,}000.\]
\[\text{Income} = 222{,}000\times4 = \text{N}888{,}000.\]
(a)(ii) Amount at the open market.
\[\tfrac{1}{3}\times888{,}000 = \text{N}296{,}000.\]
(b) Common difference. Let the first term be a and common difference d.
\[T_3 = a + 2d = 4m - 2n,\qquad T_9 = a + 8d = 2m - 8n.\]
Subtracting the first from the second:
\[6d = (2m - 8n) - (4m - 2n) = -2m - 6n \Rightarrow d = \frac{-2m - 6n}{6} = -\frac{m + 3n}{3}.\]
Bayanin Amsa
(a)(i) Monthly income. At the mall and market she spent \(\tfrac{1}{4} + \tfrac{1}{3} = \tfrac{7}{12}\) of her income, leaving \(\tfrac{5}{12}\). Of this remainder she spent \(\tfrac{2}{5}\) at the mechanic, so she kept \(\tfrac{3}{5}\) of the remainder:
\[\tfrac{3}{5}\times\tfrac{5}{12} = \tfrac{1}{4}\text{ of the income} = \text{N}222{,}000.\]
\[\text{Income} = 222{,}000\times4 = \text{N}888{,}000.\]
(a)(ii) Amount at the open market.
\[\tfrac{1}{3}\times888{,}000 = \text{N}296{,}000.\]
(b) Common difference. Let the first term be a and common difference d.
\[T_3 = a + 2d = 4m - 2n,\qquad T_9 = a + 8d = 2m - 8n.\]
Subtracting the first from the second:
\[6d = (2m - 8n) - (4m - 2n) = -2m - 6n \Rightarrow d = \frac{-2m - 6n}{6} = -\frac{m + 3n}{3}.\]
Tambaya 9 Rahoto
(a) If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of \(\mu\) = {x: \(1 \leq x \leq 10\)} find A′ \(\cap\) B′ \(\cap\) C′
(b) Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?
(a) The universal set is \(\mu = \{1,2,3,4,5,6,7,8,9,10\}\).
- \(A = \{\text{multiples of }2\} = \{2,4,6,8,10\}\)
- \(B = \{\text{multiples of }3\} = \{3,6,9\}\)
- \(C = \{\text{factors of }6\} = \{1,2,3,6\}\)
By De Morgan's law, \(A' \cap B' \cap C' = (A \cup B \cup C)'\).
\[A \cup B \cup C = \{1,2,3,4,6,8,9,10\}.\]
Therefore the elements of \(\mu\) not in this union are:
\[A' \cap B' \cap C' = \{5, 7\}.\]
(b) Buying singly, one ticket costs \(\$18.50\). The bulk price for 5 tickets is \(\$80.00\), shared equally among the 5 people:
\[\text{Cost per person} = \frac{80.00}{5} = \$16.00.\]
Each person's saving compared with buying a single ticket:
\[18.50 - 16.00 = \$2.50.\]
Each person would save \(\mathbf{\$2.50}\).
Bayanin Amsa
(a) The universal set is \(\mu = \{1,2,3,4,5,6,7,8,9,10\}\).
- \(A = \{\text{multiples of }2\} = \{2,4,6,8,10\}\)
- \(B = \{\text{multiples of }3\} = \{3,6,9\}\)
- \(C = \{\text{factors of }6\} = \{1,2,3,6\}\)
By De Morgan's law, \(A' \cap B' \cap C' = (A \cup B \cup C)'\).
\[A \cup B \cup C = \{1,2,3,4,6,8,9,10\}.\]
Therefore the elements of \(\mu\) not in this union are:
\[A' \cap B' \cap C' = \{5, 7\}.\]
(b) Buying singly, one ticket costs \(\$18.50\). The bulk price for 5 tickets is \(\$80.00\), shared equally among the 5 people:
\[\text{Cost per person} = \frac{80.00}{5} = \$16.00.\]
Each person's saving compared with buying a single ticket:
\[18.50 - 16.00 = \$2.50.\]
Each person would save \(\mathbf{\$2.50}\).
Tambaya 10 Rahoto
(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50°. Find:
(I) < MNP
(ii) < POQ
(b) Find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12)
(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).
From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.
(i) Find \(\angle MNP\).
\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:
\[\angle ONM=\angle OMN=50^{\circ}\]Angle at the centre of triangle \(OMN\):
\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:
\[\angle NOP=\angle MON=80^{\circ}\]The central angle standing on the minor arc \(MNP\) is therefore
\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is
\[360^{\circ}-160^{\circ}=200^{\circ}.\]The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:
\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\](ii) Find \(\angle POQ\).
\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:
\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\](b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).
Make \(y\) the subject to read off the gradient:
\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):
\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).
Bayanin Amsa
(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).
From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.
(i) Find \(\angle MNP\).
\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:
\[\angle ONM=\angle OMN=50^{\circ}\]Angle at the centre of triangle \(OMN\):
\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:
\[\angle NOP=\angle MON=80^{\circ}\]The central angle standing on the minor arc \(MNP\) is therefore
\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is
\[360^{\circ}-160^{\circ}=200^{\circ}.\]The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:
\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\](ii) Find \(\angle POQ\).
\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:
\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\](b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).
Make \(y\) the subject to read off the gradient:
\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):
\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).
Tambaya 11 Rahoto
The total surface area of a cone of slant height 1cm and base radius rcm is 224\(\pi\) cm\(^2\). If r : 1 = 2.5, find:
(a) correct to one decimal place, the value of r
(b) correct to the nearest whole number, the volume of the cone [Take \(\pi\) = \(\frac{22}{7}\)]
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Bayanin Amsa
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Tambaya 12 Rahoto
A die was rolled a number of times. The outcomes are as shown in the table
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
If the probability of obtaining 2 is 0.15, find the:
(a) value of m;
(b) number of times the die was rolled;
(c) probability of obtaining an even number.
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
(a) Value of m. The total number of rolls is \(N=32+m+25+40+28+45=170+m\). Since \(P(2)=0.15\):
\[ \frac{m}{170+m}=0.15 \;\Rightarrow\; m=0.15(170+m)=25.5+0.15m \] \[ 0.85m=25.5 \;\Rightarrow\; m=30 \](b) Number of times the die was rolled.
\[ N=170+30=200 \](c) Probability of an even number. Even outcomes are 2, 4 and 6:
\[ 30+40+45=115 \] \[ P(\text{even})=\frac{115}{200}=\frac{23}{40}=0.575 \]Bayanin Amsa
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
(a) Value of m. The total number of rolls is \(N=32+m+25+40+28+45=170+m\). Since \(P(2)=0.15\):
\[ \frac{m}{170+m}=0.15 \;\Rightarrow\; m=0.15(170+m)=25.5+0.15m \] \[ 0.85m=25.5 \;\Rightarrow\; m=30 \](b) Number of times the die was rolled.
\[ N=170+30=200 \](c) Probability of an even number. Even outcomes are 2, 4 and 6:
\[ 30+40+45=115 \] \[ P(\text{even})=\frac{115}{200}=\frac{23}{40}=0.575 \]
