WAEC SSCE - General Mathematics - 2008 (Essay)
Question 1 Report
(a) A rectangular field is l metres long and b metres wide. Its perimeter is 280 metres. If the length is two and a half times the breadth, find the values of l and b.
(b) The base of a pyramid is a 4.5 metres rectangle. The height of the pyramid is 4 metres. Calculate its volume.
(a) Let the length be \(l\) and the breadth \(b\). The perimeter is \(280\) m:
\[2(l + b) = 280 \;\Rightarrow\; l + b = 140\]
The length is two and a half times the breadth: \(l = 2.5b\). Substitute:
\[2.5b + b = 140 \;\Rightarrow\; 3.5b = 140 \;\Rightarrow\; b = 40\text{ m}, \quad l = 2.5(40) = 100\text{ m}\]
(b) Taking the base as a square of side \(4.5\) m and height \(4\) m, the volume of a pyramid is one-third of base area times height:
\[V = \frac13 \times (\text{base area}) \times h = \frac13 \times (4.5 \times 4.5) \times 4\]
\[= \frac13 \times 20.25 \times 4 = \frac{81}{3} = 27\text{ m}^3\]
Answer Details
(a) Let the length be \(l\) and the breadth \(b\). The perimeter is \(280\) m:
\[2(l + b) = 280 \;\Rightarrow\; l + b = 140\]
The length is two and a half times the breadth: \(l = 2.5b\). Substitute:
\[2.5b + b = 140 \;\Rightarrow\; 3.5b = 140 \;\Rightarrow\; b = 40\text{ m}, \quad l = 2.5(40) = 100\text{ m}\]
(b) Taking the base as a square of side \(4.5\) m and height \(4\) m, the volume of a pyramid is one-third of base area times height:
\[V = \frac13 \times (\text{base area}) \times h = \frac13 \times (4.5 \times 4.5) \times 4\]
\[= \frac13 \times 20.25 \times 4 = \frac{81}{3} = 27\text{ m}^3\]
Question 2 Report
(a) The triangle ABC has sides AB = 17m, BC = 12m and AC = 10m. Calculate the :
(i) largest angle of the triangle ; (ii) area of the triangle.
(b) From a point T on a horizontal ground, the angle of elevation of the top R of a tower RS, 38m high is 63°. Calculate, correct to the nearest metre, the distance between T and S.
(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:
\[\cos C=\frac{CA^{2}+CB^{2}-AB^{2}}{2\,(CA)(CB)}=\frac{10^{2}+12^{2}-17^{2}}{2(10)(12)}=\frac{100+144-289}{240}=\frac{-45}{240}=-0.1875\]\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).
Largest angle \(\approx 100.8^{\circ}\).
(ii) Area.
\[\text{Area}=\frac{1}{2}(CA)(CB)\sin C=\frac{1}{2}(10)(12)\sin 100.8^{\circ}=60\times0.9823\approx 58.9\text{ m}^{2}\](b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.
\[\tan 63^{\circ}=\frac{RS}{TS}=\frac{38}{TS}\Rightarrow TS=\frac{38}{\tan 63^{\circ}}=\frac{38}{1.9626}\approx 19.4\text{ m}\]Distance \(TS\approx 19\) m (nearest metre).
Answer Details
(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:
\[\cos C=\frac{CA^{2}+CB^{2}-AB^{2}}{2\,(CA)(CB)}=\frac{10^{2}+12^{2}-17^{2}}{2(10)(12)}=\frac{100+144-289}{240}=\frac{-45}{240}=-0.1875\]\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).
Largest angle \(\approx 100.8^{\circ}\).
(ii) Area.
\[\text{Area}=\frac{1}{2}(CA)(CB)\sin C=\frac{1}{2}(10)(12)\sin 100.8^{\circ}=60\times0.9823\approx 58.9\text{ m}^{2}\](b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.
\[\tan 63^{\circ}=\frac{RS}{TS}=\frac{38}{TS}\Rightarrow TS=\frac{38}{\tan 63^{\circ}}=\frac{38}{1.9626}\approx 19.4\text{ m}\]Distance \(TS\approx 19\) m (nearest metre).
Question 3 Report
(a) Using ruler and a pair of compasses only, construct :
(i) a quadrilateral PQRS such that /PQ/ = 7 cm, < QPS = 60°, /PS/ = 6.5 cm, < PQR = 135° and /QS/ = /QR/ ;
(ii) locus, \(l_{1}\) of points equidistant from P and Q ;
(iii) locus, \(l_{2}\) of points equidistant from P and S.
(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect. (ii) With center T and radius /TP/, construct a circle \(l_{3}\).
(a) Construction
- Draw the line segment \(PQ=7\text{ cm}\).
- At \(P\), construct an angle of \(60^\circ\). On this ray, mark \(S\) such that \(PS=6.5\text{ cm}\).
- Join \(Q\) to \(S\). With centre \(Q\) and radius \(QS\), draw an arc. At \(Q\), construct \(\angle PQR=135^\circ\); let this ray meet the arc at \(R\). Join \(R\) to \(S\). Thus \(QR=QS\).
- Construct the perpendicular bisector of \(PQ\) and label it \(l_1\). This is the locus of points equidistant from \(P\) and \(Q\).
- Construct the perpendicular bisector of \(PS\) and label it \(l_2\). This is the locus of points equidistant from \(P\) and \(S\).
(b)
- The point of intersection of \(l_1\) and \(l_2\) is \(T\).
- With centre \(T\) and radius \(TP\), draw the circle \(l_3\), as shown above. The circle passes through \(P\), \(Q\) and \(S\), since \(TP=TQ=TS\).
Answer Details
(a) Construction
- Draw the line segment \(PQ=7\text{ cm}\).
- At \(P\), construct an angle of \(60^\circ\). On this ray, mark \(S\) such that \(PS=6.5\text{ cm}\).
- Join \(Q\) to \(S\). With centre \(Q\) and radius \(QS\), draw an arc. At \(Q\), construct \(\angle PQR=135^\circ\); let this ray meet the arc at \(R\). Join \(R\) to \(S\). Thus \(QR=QS\).
- Construct the perpendicular bisector of \(PQ\) and label it \(l_1\). This is the locus of points equidistant from \(P\) and \(Q\).
- Construct the perpendicular bisector of \(PS\) and label it \(l_2\). This is the locus of points equidistant from \(P\) and \(S\).
(b)
- The point of intersection of \(l_1\) and \(l_2\) is \(T\).
- With centre \(T\) and radius \(TP\), draw the circle \(l_3\), as shown above. The circle passes through \(P\), \(Q\) and \(S\), since \(TP=TQ=TS\).
Question 4 Report
(a) Solve, correct to two decimal places, the equation \(4x^{2} = 11x + 21\).
(b) A man invests £1500 for two years at compound interest. After one year, his money amounts to £1560. Find the :
(i) rate of interest ; (ii) interest for the second year.
(c) A car costs N300,000.00. It depreciates by 25% in the first year and 20% in the second year. Find its value after 2 years.
(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).
Using the formula with \(a=4,\ b=-11,\ c=-21\):
\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).
\(x\approx 4.05\) or \(x\approx -1.30\)
(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).
(i) First-year interest \(=1560-1500=\pounds60\).
\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\](ii) Interest for the second year is charged on \(\pounds1560\):
\[\frac{4}{100}\times1560=\pounds62.40\](c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).
After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).
After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).
Value after 2 years \(=N180{,}000.00\)
Answer Details
(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).
Using the formula with \(a=4,\ b=-11,\ c=-21\):
\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).
\(x\approx 4.05\) or \(x\approx -1.30\)
(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).
(i) First-year interest \(=1560-1500=\pounds60\).
\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\](ii) Interest for the second year is charged on \(\pounds1560\):
\[\frac{4}{100}\times1560=\pounds62.40\](c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).
After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).
After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).
Value after 2 years \(=N180{,}000.00\)
Question 5 Report
(a) Solve the inequality : \(\frac{2}{5}(x - 2) - \frac{1}{6}(x + 5) \leq 0\).
(b) Given that P = \(\frac{x^{2} - y^{2}}{x^{2} + xy}\),
(i) express P in its simplest form ; (ii) find the value of P if x = -4 and y = -6.
(a) \(\dfrac25(x - 2) - \dfrac16(x + 5) \le 0\). Multiply through by the LCM \(30\):
\[12(x - 2) - 5(x + 5) \le 0\]
\[12x - 24 - 5x - 25 \le 0 \;\Rightarrow\; 7x - 49 \le 0 \;\Rightarrow\; 7x \le 49 \;\Rightarrow\; x \le 7\]
(b)(i) Factorise numerator and denominator of \(P = \dfrac{x^2 - y^2}{x^2 + xy}\):
\[P = \frac{(x - y)(x + y)}{x(x + y)} = \frac{x - y}{x}\]
(ii) With \(x = -4,\, y = -6\):
\[P = \frac{-4 - (-6)}{-4} = \frac{2}{-4} = -\frac12\]
Answer Details
(a) \(\dfrac25(x - 2) - \dfrac16(x + 5) \le 0\). Multiply through by the LCM \(30\):
\[12(x - 2) - 5(x + 5) \le 0\]
\[12x - 24 - 5x - 25 \le 0 \;\Rightarrow\; 7x - 49 \le 0 \;\Rightarrow\; 7x \le 49 \;\Rightarrow\; x \le 7\]
(b)(i) Factorise numerator and denominator of \(P = \dfrac{x^2 - y^2}{x^2 + xy}\):
\[P = \frac{(x - y)(x + y)}{x(x + y)} = \frac{x - y}{x}\]
(ii) With \(x = -4,\, y = -6\):
\[P = \frac{-4 - (-6)}{-4} = \frac{2}{-4} = -\frac12\]
Question 6 Report
The table below shows how a man spends his income in a month.
| Items | Amount Spent |
| Food | N4500 |
| House Rent | N3000 |
| Provisions | N2500 |
| Electricity | N2000 |
| Transportation | N5000 |
| Others | N3000 |
(a) Represent the information on a pie chart.
(b) What percentage of his income is spent on transportation?
Total monthly income
\[4500+3000+2500+2000+5000+3000=N20\,000\]
(a) Pie chart
The angle for each sector is:
\[\text{Sector angle}=\frac{\text{amount spent}}{20\,000}\times360^\circ\]
| Item | Amount (₦) | Sector angle |
|---|---|---|
| Food | 4,500 | \(\frac{4500}{20000}\times360^\circ=81^\circ\) |
| House Rent | 3,000 | \(54^\circ\) |
| Provisions | 2,500 | \(45^\circ\) |
| Electricity | 2,000 | \(36^\circ\) |
| Transportation | 5,000 | \(90^\circ\) |
| Others | 3,000 | \(54^\circ\) |
| Total | 20,000 | \(360^\circ\) |
The required pie chart is:
(b)
\[\text{Percentage spent on transportation}=\frac{5000}{20000}\times100\%=25\%\]
Therefore, 25% of his income is spent on transportation.
Answer Details
Total monthly income
\[4500+3000+2500+2000+5000+3000=N20\,000\]
(a) Pie chart
The angle for each sector is:
\[\text{Sector angle}=\frac{\text{amount spent}}{20\,000}\times360^\circ\]
| Item | Amount (₦) | Sector angle |
|---|---|---|
| Food | 4,500 | \(\frac{4500}{20000}\times360^\circ=81^\circ\) |
| House Rent | 3,000 | \(54^\circ\) |
| Provisions | 2,500 | \(45^\circ\) |
| Electricity | 2,000 | \(36^\circ\) |
| Transportation | 5,000 | \(90^\circ\) |
| Others | 3,000 | \(54^\circ\) |
| Total | 20,000 | \(360^\circ\) |
The required pie chart is:
(b)
\[\text{Percentage spent on transportation}=\frac{5000}{20000}\times100\%=25\%\]
Therefore, 25% of his income is spent on transportation.
Question 7 Report
(a) Without using calculator or tables, find the value of \(\log 3.6\) given that \(\log 2 = 0.3010, \log 3 = 0.4771\) and \(\log 5 = 0.6990\).
(b) If all numbers in the equation \(\frac{y}{y + 101} = \frac{11}{10010}\) are in base two, solve for y.
(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):
\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]
\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]
(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes
\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]
\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]
Converting back to base two, \(y = 1_2\).
Answer Details
(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):
\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]
\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]
(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes
\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]
\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]
Converting back to base two, \(y = 1_2\).
Question 8 Report
(a) A pentagon is such that one of its exterior sides is 60°. Two others are (90 - m)° each while the remaining angles are (30 + 2m)° each. Find the value of m.
(b) 
In the diagram, PQR is a straight line, \(\overline{QR} = \sqrt{3} cm\) and \(\overline{SQ} = 2 cm\). Calculate, correct to one decimal place, < PQS.
(a) The exterior angles of any polygon add up to \(360^\circ\). The pentagon has five exterior angles: one is \(60^\circ\), two are \((90-m)^\circ\) each, and the remaining two are \((30+2m)^\circ\) each. Therefore:
\[60+2(90-m)+2(30+2m)=360\]\[60+180-2m+60+4m=360\]\[300+2m=360\]\[2m=60\Rightarrow m=30\](b) From the diagram PQR is a straight line, \(\overline{QR}=\sqrt{3}\text{ cm}\), \(\overline{SQ}=2\text{ cm}\), and the right-angle mark at R shows \(SR\perp QR\). So triangle SQR is right-angled at R, with SQ the hypotenuse.
In right-angled triangle SQR, using \(\angle SQR\):
\[\cos(\angle SQR)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\overline{QR}}{\overline{SQ}}=\frac{\sqrt{3}}{2}\]\[\angle SQR=\cos^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)=30^\circ\]Since PQR is a straight line, \(\angle PQS\) and \(\angle SQR\) are angles on a straight line at Q:
\[\angle PQS=180^\circ-\angle SQR=180^\circ-30^\circ=150.0^\circ\]Answer Details
(a) The exterior angles of any polygon add up to \(360^\circ\). The pentagon has five exterior angles: one is \(60^\circ\), two are \((90-m)^\circ\) each, and the remaining two are \((30+2m)^\circ\) each. Therefore:
\[60+2(90-m)+2(30+2m)=360\]\[60+180-2m+60+4m=360\]\[300+2m=360\]\[2m=60\Rightarrow m=30\](b) From the diagram PQR is a straight line, \(\overline{QR}=\sqrt{3}\text{ cm}\), \(\overline{SQ}=2\text{ cm}\), and the right-angle mark at R shows \(SR\perp QR\). So triangle SQR is right-angled at R, with SQ the hypotenuse.
In right-angled triangle SQR, using \(\angle SQR\):
\[\cos(\angle SQR)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\overline{QR}}{\overline{SQ}}=\frac{\sqrt{3}}{2}\]\[\angle SQR=\cos^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)=30^\circ\]Since PQR is a straight line, \(\angle PQS\) and \(\angle SQR\) are angles on a straight line at Q:
\[\angle PQS=180^\circ-\angle SQR=180^\circ-30^\circ=150.0^\circ\]Question 9 Report
(a) 
In the diagram, AB // CD and BC // FE. \(\stackrel\frown{CDE} = 75°\) and \(\stackrel\frown{DEF} = 26°\). Find the angles marked x and y.
(b) 
The diagram shows a circle ABCD with centre O and radius 7 cm. The reflex angle AOC = 190° and < DAO = 35°. Find :
(i) < ABC ; (ii) < ADC.
(c) Using the diagram in (b) above, calculate, correct to 3 significant figures, the length of : (i) arc ABC ; (ii) the chord AD. [Take \(\pi = 3.142\)].
(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).
Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):
\[\text{angle between } DE \text{ and the } BC\text{-family}=\widehat{DEF}=26^\circ.\]Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is
\[75^\circ-26^\circ=49^\circ.\]Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:
\[\boxed{x=49^\circ.}\]By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):
\[y=360^\circ-49^\circ=\boxed{311^\circ.}\](b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).
(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:
\[\widehat{ABC}=\tfrac12(190^\circ)=95^\circ.\](ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):
\[\widehat{ADC}=\tfrac12(170^\circ)=85^\circ.\](Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)
(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):
\[\text{arc}=\frac{170}{360}\times 2\times 3.142\times 7=\frac{170}{360}\times 43.988=20.77\approx 20.8\text{ cm}.\](c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).
\[AD=2r\sin\!\left(\tfrac{110^\circ}{2}\right)=2(7)\sin 55^\circ=14\times 0.8192=11.47\approx 11.5\text{ cm}.\]Answer Details
(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).
Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):
\[\text{angle between } DE \text{ and the } BC\text{-family}=\widehat{DEF}=26^\circ.\]Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is
\[75^\circ-26^\circ=49^\circ.\]Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:
\[\boxed{x=49^\circ.}\]By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):
\[y=360^\circ-49^\circ=\boxed{311^\circ.}\](b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).
(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:
\[\widehat{ABC}=\tfrac12(190^\circ)=95^\circ.\](ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):
\[\widehat{ADC}=\tfrac12(170^\circ)=85^\circ.\](Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)
(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):
\[\text{arc}=\frac{170}{360}\times 2\times 3.142\times 7=\frac{170}{360}\times 43.988=20.77\approx 20.8\text{ cm}.\](c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).
\[AD=2r\sin\!\left(\tfrac{110^\circ}{2}\right)=2(7)\sin 55^\circ=14\times 0.8192=11.47\approx 11.5\text{ cm}.\]Question 10 Report
(a) Copy and complete the table of values for \(y = 3\sin x + 2\cos x\) for \(0° \leq x \leq 360°\).
| x | 0° | 60° | 120° | 180° | 240° | 300° | 360° |
| y | 2.00 | 2.00 |
(b) Using a scale of 2 cm to 60° on x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 3 \sin x + 2 \cos x\) for \(0° \leq x \leq 360°\).
(c) Use your graph to solve the equation : \(3 \sin x + 2 \cos x = 1.5\).
(d) Find the range of values of x for which \(3\sin x + 2\cos x < -1\).
(a) For \(y=3\sin x+2\cos x\), the completed table is:
| \(x\) | \(0^\circ\) | \(60^\circ\) | \(120^\circ\) | \(180^\circ\) | \(240^\circ\) | \(300^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 3.60 | 1.60 | -2.00 | -3.60 | -1.60 | 2.00 |
For example, at \(x=60^\circ\), \(y=3(0.8660)+2(0.5)=3.598\approx3.60\).
(b) The graph, drawn using the stated scales, is shown below.
(c) Draw the horizontal line \(y=1.5\). Its intersections with the curve give
\[x\approx122^\circ\quad\text{or}\quad x\approx351^\circ.\]
(d) Draw the horizontal line \(y=-1\). The curve is below this line between its two points of intersection. Hence,
\[\boxed{162^\circ<x<310^\circ}.\]
Answer Details
(a) For \(y=3\sin x+2\cos x\), the completed table is:
| \(x\) | \(0^\circ\) | \(60^\circ\) | \(120^\circ\) | \(180^\circ\) | \(240^\circ\) | \(300^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 3.60 | 1.60 | -2.00 | -3.60 | -1.60 | 2.00 |
For example, at \(x=60^\circ\), \(y=3(0.8660)+2(0.5)=3.598\approx3.60\).
(b) The graph, drawn using the stated scales, is shown below.
(c) Draw the horizontal line \(y=1.5\). Its intersections with the curve give
\[x\approx122^\circ\quad\text{or}\quad x\approx351^\circ.\]
(d) Draw the horizontal line \(y=-1\). The curve is below this line between its two points of intersection. Hence,
\[\boxed{162^\circ<x<310^\circ}.\]
Question 11 Report
The ages, in years, of 50 teachers in a school are given below :
21 37 49 27 49 42 26 33 46 40 50 29 23 24 29 31 36 22 27 38 30 26 42 39 34 23 21 32 41 46 46 31 33 29 28 43 47 40 34 44 26 38 34 49 45 27 25 33 39 40
(a) Form a frequency distribution table of the data using the intervals : 21 - 25, 26 - 30, 31 - 35 etc.
(b) Draw the histogram of the distribution
(c) Use your histogram to estimate the mode
(d) Calculate the mean age.
(a) Frequency distribution
| Age (years) | Tally | Class mark, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|---|
| 21 - 25 | ||||| || | 23 | 7 | 161 |
| 26 - 30 | ||||| ||||| | | 28 | 11 | 308 |
| 31 - 35 | ||||| |||| | 33 | 9 | 297 |
| 36 - 40 | ||||| |||| | 38 | 9 | 342 |
| 41 - 45 | ||||| | | 43 | 6 | 258 |
| 46 - 50 | ||||| ||| | 48 | 8 | 384 |
| Total | 50 | 1750 | ||
(b) Histogram
The continuous class boundaries are \(20.5\), \(25.5\), \(30.5\), \(35.5\), \(40.5\), \(45.5\) and \(50.5\). Since all class widths are 5 years, the bar heights are the frequencies.
(c) Modal age
The modal class is \(26\text{ - }30\), for which \(L=25.5\), \(f_1=11\), \(f_0=7\), \(f_2=9\), and class width \(h=5\).
\[\text{Mode}=L+\left(\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\right)h\]
\[=25.5+\left(\frac{11-7}{(11-7)+(11-9)}\right)\times5=25.5+\frac{4}{6}\times5=28.8\]
Hence, the estimated mode is \(28.8\) years.
(d) Mean age
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1750}{50}=35\]
Therefore, the mean age is 35 years.
Answer Details
(a) Frequency distribution
| Age (years) | Tally | Class mark, \(x\) | Frequency, \(f\) | \(fx\) |
|---|---|---|---|---|
| 21 - 25 | ||||| || | 23 | 7 | 161 |
| 26 - 30 | ||||| ||||| | | 28 | 11 | 308 |
| 31 - 35 | ||||| |||| | 33 | 9 | 297 |
| 36 - 40 | ||||| |||| | 38 | 9 | 342 |
| 41 - 45 | ||||| | | 43 | 6 | 258 |
| 46 - 50 | ||||| ||| | 48 | 8 | 384 |
| Total | 50 | 1750 | ||
(b) Histogram
The continuous class boundaries are \(20.5\), \(25.5\), \(30.5\), \(35.5\), \(40.5\), \(45.5\) and \(50.5\). Since all class widths are 5 years, the bar heights are the frequencies.
(c) Modal age
The modal class is \(26\text{ - }30\), for which \(L=25.5\), \(f_1=11\), \(f_0=7\), \(f_2=9\), and class width \(h=5\).
\[\text{Mode}=L+\left(\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\right)h\]
\[=25.5+\left(\frac{11-7}{(11-7)+(11-9)}\right)\times5=25.5+\frac{4}{6}\times5=28.8\]
Hence, the estimated mode is \(28.8\) years.
(d) Mean age
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1750}{50}=35\]
Therefore, the mean age is 35 years.
Question 12 Report
(a) If \(2^{x + y} = 16\) and \(4^{x - y} = \frac{1}{32}\), find the value of x and y.
(b) P, Q and R are related in such a way that \(P \propto \frac{Q^{2}}{R}\). When P = 36, Q = 3 and R = 4. Calculate Q when P = 200 and R = 2.
(a) Express both equations to base 2.
\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)
Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)
Solve the pair:
\[x+y=4,\qquad x-y=-2.5\]Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).
\(x=0.75,\ y=3.25\)
(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).
Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).
When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).
\(Q=5\)
Answer Details
(a) Express both equations to base 2.
\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)
Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)
Solve the pair:
\[x+y=4,\qquad x-y=-2.5\]Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).
\(x=0.75,\ y=3.25\)
(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).
Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).
When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).
\(Q=5\)
Question 13 Report
(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find the values of x and y.
(b)(i) The sum of the second and third terms of a grometric progression is six times the fourth term. Find the two possible values of the common ratio.
(ii) If the second term is 8 and the common ratio is positive, find the first six terms.
(a) \(3,\ x,\ y,\ 18\) are consecutive terms of an A.P. The first term is \(3\) and the fourth term is \(18\):
\[18=3+(4-1)d=3+3d\Rightarrow d=5\]So \(x=3+5=8\) and \(y=8+5=13\).
\(x=8,\ y=13\)
(b)(i) For a G.P. with first term \(a\) and ratio \(r\), terms are \(a,\ ar,\ ar^{2},\ ar^{3},\ldots\)
"Sum of second and third terms \(=6\times\) fourth term":
\[ar+ar^{2}=6ar^{3}\]Divide through by \(ar\ (a,r\neq0)\): \(1+r=6r^{2}\Rightarrow 6r^{2}-r-1=0\).
\[r=\frac{1\pm\sqrt{1+24}}{12}=\frac{1\pm5}{12}\]\(r=\tfrac{1}{2}\) or \(r=-\tfrac{1}{3}\).
(ii) The ratio is positive, so \(r=\tfrac{1}{2}\). The second term is \(ar=8\):
\[a\times\tfrac{1}{2}=8\Rightarrow a=16\]First six terms: \(16,\ 8,\ 4,\ 2,\ 1,\ \tfrac{1}{2}\).
Answer Details
(a) \(3,\ x,\ y,\ 18\) are consecutive terms of an A.P. The first term is \(3\) and the fourth term is \(18\):
\[18=3+(4-1)d=3+3d\Rightarrow d=5\]So \(x=3+5=8\) and \(y=8+5=13\).
\(x=8,\ y=13\)
(b)(i) For a G.P. with first term \(a\) and ratio \(r\), terms are \(a,\ ar,\ ar^{2},\ ar^{3},\ldots\)
"Sum of second and third terms \(=6\times\) fourth term":
\[ar+ar^{2}=6ar^{3}\]Divide through by \(ar\ (a,r\neq0)\): \(1+r=6r^{2}\Rightarrow 6r^{2}-r-1=0\).
\[r=\frac{1\pm\sqrt{1+24}}{12}=\frac{1\pm5}{12}\]\(r=\tfrac{1}{2}\) or \(r=-\tfrac{1}{3}\).
(ii) The ratio is positive, so \(r=\tfrac{1}{2}\). The second term is \(ar=8\):
\[a\times\tfrac{1}{2}=8\Rightarrow a=16\]First six terms: \(16,\ 8,\ 4,\ 2,\ 1,\ \tfrac{1}{2}\).
