WAEC SSCE - General Mathematics - 2003 (Essay)
Question 1 Report
In the diagram, PQRS is a circle with centre O and radius 7cm. SQ and PR intersect at K and < SKR = 90°. If the length of the arc SR is four times that of arc PQ, find the length of the arc SR. [Take \(\pi = \frac{22}{7}\)].
Reading the diagram. S is at the top and Q at the bottom, with centre O lying on SQ, so SQ is a vertical chord passing through the centre. P is on the left and R on the right, with PR a horizontal chord. The two chords SQ and PR cross at the interior point K, where \(\angle SKR = 90^\circ\). The radius is \(7\text{ cm}\).
Step 1: Apply the intersecting-chords angle rule. When two chords cross inside a circle, the angle between them equals half the sum of the two arcs it intercepts. The angle \(\angle SKR\) intercepts arc \(SR\), while its vertically opposite angle intercepts arc \(PQ\). Hence
\[\angle SKR = \tfrac{1}{2}\left(\text{arc } SR + \text{arc } PQ\right).\]
Step 2: Form an equation in the arc measures. Let arc \(PQ = x^\circ\). Since arc \(SR\) is four times arc \(PQ\), arc \(SR = 4x^\circ\). With \(\angle SKR = 90^\circ\):
\[90 = \tfrac{1}{2}(4x + x) \;\Rightarrow\; 180 = 5x \;\Rightarrow\; x = 36.\]
So the arc \(SR\) subtends an angle of \(4x = 4(36) = 144^\circ\) at the centre.
Step 3: Convert the arc angle to a length. Using \(\text{arc length} = \dfrac{\theta}{360}\times 2\pi r\) with \(\theta = 144^\circ\), \(r = 7\), \(\pi = \tfrac{22}{7}\):
\[\text{arc } SR = \frac{144}{360}\times 2\times\frac{22}{7}\times 7 = \frac{144}{360}\times 44 = 0.4\times 44 = 17.6\text{ cm}.\]
Answer: The length of arc \(SR\) is \(\mathbf{17.6\text{ cm}}\).
Answer Details
Reading the diagram. S is at the top and Q at the bottom, with centre O lying on SQ, so SQ is a vertical chord passing through the centre. P is on the left and R on the right, with PR a horizontal chord. The two chords SQ and PR cross at the interior point K, where \(\angle SKR = 90^\circ\). The radius is \(7\text{ cm}\).
Step 1: Apply the intersecting-chords angle rule. When two chords cross inside a circle, the angle between them equals half the sum of the two arcs it intercepts. The angle \(\angle SKR\) intercepts arc \(SR\), while its vertically opposite angle intercepts arc \(PQ\). Hence
\[\angle SKR = \tfrac{1}{2}\left(\text{arc } SR + \text{arc } PQ\right).\]
Step 2: Form an equation in the arc measures. Let arc \(PQ = x^\circ\). Since arc \(SR\) is four times arc \(PQ\), arc \(SR = 4x^\circ\). With \(\angle SKR = 90^\circ\):
\[90 = \tfrac{1}{2}(4x + x) \;\Rightarrow\; 180 = 5x \;\Rightarrow\; x = 36.\]
So the arc \(SR\) subtends an angle of \(4x = 4(36) = 144^\circ\) at the centre.
Step 3: Convert the arc angle to a length. Using \(\text{arc length} = \dfrac{\theta}{360}\times 2\pi r\) with \(\theta = 144^\circ\), \(r = 7\), \(\pi = \tfrac{22}{7}\):
\[\text{arc } SR = \frac{144}{360}\times 2\times\frac{22}{7}\times 7 = \frac{144}{360}\times 44 = 0.4\times 44 = 17.6\text{ cm}.\]
Answer: The length of arc \(SR\) is \(\mathbf{17.6\text{ cm}}\).
Question 2 Report
(a) A = {1, 2, 5, 7} and B = {1, 3, 6, 7} are subsets of the universal set U = {1, 2, 3,...., 10}. Find (i) \(A'\) ; (ii) \((A \cap B)'\) ; (iii) \((A \cup B)'\) ; (iv) the subsets of B each of which has three elements.
(b) Write down the 15th term of the sequence, \(\frac{2}{1 \times 3}, \frac{2}{2 \times 4}, \frac{4}{3 \times 5}, \frac{5}{4 \times 6},...\).
(c) An Arithmetic Progression (A.P) has 3 as its first term and 4 as the common difference, (i) write an expression in its simplest form for the nth term ; (ii) find the least term of the A.P that is greater than 100.
(a) \(U=\{1,2,3,4,5,6,7,8,9,10\}\), \(A=\{1,2,5,7\}\), \(B=\{1,3,6,7\}\).
(i) \(A'=U\setminus A=\{3,4,6,8,9,10\}\).
(ii) \(A\cap B=\{1,7\}\), so \((A\cap B)'=\{2,3,4,5,6,8,9,10\}\).
(iii) \(A\cup B=\{1,2,3,5,6,7\}\), so \((A\cup B)'=\{4,8,9,10\}\).
(iv) The three-element subsets of \(B=\{1,3,6,7\}\) are \(\{1,3,6\},\ \{1,3,7\},\ \{1,6,7\},\ \{3,6,7\}\).
(b) The sequence \(\dfrac{2}{1\times3},\dfrac{3}{2\times4},\dfrac{4}{3\times5},\dfrac{5}{4\times6},\ldots\) has \(n\)th term \(\dfrac{n+1}{n(n+2)}\).
\[\text{15th term}=\frac{15+1}{15(15+2)}=\frac{16}{15\times17}=\frac{16}{255}\](c) A.P. with \(a=3,\ d=4\).
(i) \(n\)th term \(=a+(n-1)d=3+(n-1)4=4n-1\).
(ii) Least term greater than 100: \(4n-1>100\Rightarrow 4n>101\Rightarrow n>25.25\), so \(n=26\).
\[T_{26}=4(26)-1=103\]Least term above 100 is \(103\).
Answer Details
(a) \(U=\{1,2,3,4,5,6,7,8,9,10\}\), \(A=\{1,2,5,7\}\), \(B=\{1,3,6,7\}\).
(i) \(A'=U\setminus A=\{3,4,6,8,9,10\}\).
(ii) \(A\cap B=\{1,7\}\), so \((A\cap B)'=\{2,3,4,5,6,8,9,10\}\).
(iii) \(A\cup B=\{1,2,3,5,6,7\}\), so \((A\cup B)'=\{4,8,9,10\}\).
(iv) The three-element subsets of \(B=\{1,3,6,7\}\) are \(\{1,3,6\},\ \{1,3,7\},\ \{1,6,7\},\ \{3,6,7\}\).
(b) The sequence \(\dfrac{2}{1\times3},\dfrac{3}{2\times4},\dfrac{4}{3\times5},\dfrac{5}{4\times6},\ldots\) has \(n\)th term \(\dfrac{n+1}{n(n+2)}\).
\[\text{15th term}=\frac{15+1}{15(15+2)}=\frac{16}{15\times17}=\frac{16}{255}\](c) A.P. with \(a=3,\ d=4\).
(i) \(n\)th term \(=a+(n-1)d=3+(n-1)4=4n-1\).
(ii) Least term greater than 100: \(4n-1>100\Rightarrow 4n>101\Rightarrow n>25.25\), so \(n=26\).
\[T_{26}=4(26)-1=103\]Least term above 100 is \(103\).
Question 3 Report
(a) Simplify : \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\)
(b) Solve the following equations correct to one decimal place.
(i) \(\tan (\theta + 25)° = 5.145\)
(ii) \(5\cos \theta - 1 = 0\), where \(0° \leq \theta \leq 90°\).
(a) Write \(625=5^{4}\) and \(25=5^{2}\).
\[625^{\frac{3}{8}}\times5^{\frac{1}{2}}\div25=(5^{4})^{\frac{3}{8}}\times5^{\frac{1}{2}}\div5^{2}=5^{\frac{12}{8}}\times5^{\frac{1}{2}}\times5^{-2}\]Add the indices: \(\dfrac{12}{8}+\dfrac{1}{2}-2=\dfrac{3}{2}+\dfrac{1}{2}-2=2-2=0\).
\[=5^{0}=1\]\(=1\)
(b)(i) \(\tan(\theta+25)^{\circ}=5.145\).
\((\theta+25)^{\circ}=\tan^{-1}5.145=79.0^{\circ}\), so \(\theta=79.0-25=54.0^{\circ}\).
\(\theta\approx 54.0^{\circ}\)
(ii) \(5\cos\theta-1=0\Rightarrow \cos\theta=\dfrac{1}{5}=0.2\).
\(\theta=\cos^{-1}0.2=78.5^{\circ}\) (within \(0^{\circ}\le\theta\le90^{\circ}\)).
\(\theta\approx 78.5^{\circ}\)
Answer Details
(a) Write \(625=5^{4}\) and \(25=5^{2}\).
\[625^{\frac{3}{8}}\times5^{\frac{1}{2}}\div25=(5^{4})^{\frac{3}{8}}\times5^{\frac{1}{2}}\div5^{2}=5^{\frac{12}{8}}\times5^{\frac{1}{2}}\times5^{-2}\]Add the indices: \(\dfrac{12}{8}+\dfrac{1}{2}-2=\dfrac{3}{2}+\dfrac{1}{2}-2=2-2=0\).
\[=5^{0}=1\]\(=1\)
(b)(i) \(\tan(\theta+25)^{\circ}=5.145\).
\((\theta+25)^{\circ}=\tan^{-1}5.145=79.0^{\circ}\), so \(\theta=79.0-25=54.0^{\circ}\).
\(\theta\approx 54.0^{\circ}\)
(ii) \(5\cos\theta-1=0\Rightarrow \cos\theta=\dfrac{1}{5}=0.2\).
\(\theta=\cos^{-1}0.2=78.5^{\circ}\) (within \(0^{\circ}\le\theta\le90^{\circ}\)).
\(\theta\approx 78.5^{\circ}\)
Question 4 Report
Without using Mathematical tables or a calculator, simplify :
(a) \(\sqrt{50} - 3\sqrt{2}(2\sqrt{2} - 5) - 5\sqrt{32}\)
(b) \(\frac{1}{2} \log_{10} \frac{25}{4} - 2 \log_{10} \frac{4}{5} + \log_{10} \frac{320}{125}\).
(a) Simplify \(\sqrt{50}-3\sqrt{2}(2\sqrt{2}-5)-5\sqrt{32}\).
\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), so \(5\sqrt{32}=20\sqrt{2}\).
Expand the bracket: \(3\sqrt{2}(2\sqrt{2}-5)=6(\sqrt{2}\cdot\sqrt{2})-15\sqrt{2}=6(2)-15\sqrt{2}=12-15\sqrt{2}\).
Combine:
\[5\sqrt{2}-(12-15\sqrt{2})-20\sqrt{2}=5\sqrt{2}-12+15\sqrt{2}-20\sqrt{2}=(5+15-20)\sqrt{2}-12=-12\]\(=-12\)
(b) \(\tfrac{1}{2}\log_{10}\tfrac{25}{4}-2\log_{10}\tfrac{4}{5}+\log_{10}\tfrac{320}{125}\).
Bring the coefficients inside as powers: \(\tfrac{1}{2}\log\tfrac{25}{4}=\log\left(\tfrac{25}{4}\right)^{1/2}=\log\tfrac{5}{2}\); \(\;2\log\tfrac{4}{5}=\log\tfrac{16}{25}\).
\[\log\frac{5}{2}-\log\frac{16}{25}+\log\frac{320}{125}=\log\!\left(\frac{5}{2}\times\frac{25}{16}\times\frac{320}{125}\right)\]\(\dfrac{5}{2}\times\dfrac{25}{16}=\dfrac{125}{32}\); then \(\dfrac{125}{32}\times\dfrac{320}{125}=\dfrac{320}{32}=10\).
\[=\log_{10}10=1\]\(=1\)
Answer Details
(a) Simplify \(\sqrt{50}-3\sqrt{2}(2\sqrt{2}-5)-5\sqrt{32}\).
\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), so \(5\sqrt{32}=20\sqrt{2}\).
Expand the bracket: \(3\sqrt{2}(2\sqrt{2}-5)=6(\sqrt{2}\cdot\sqrt{2})-15\sqrt{2}=6(2)-15\sqrt{2}=12-15\sqrt{2}\).
Combine:
\[5\sqrt{2}-(12-15\sqrt{2})-20\sqrt{2}=5\sqrt{2}-12+15\sqrt{2}-20\sqrt{2}=(5+15-20)\sqrt{2}-12=-12\]\(=-12\)
(b) \(\tfrac{1}{2}\log_{10}\tfrac{25}{4}-2\log_{10}\tfrac{4}{5}+\log_{10}\tfrac{320}{125}\).
Bring the coefficients inside as powers: \(\tfrac{1}{2}\log\tfrac{25}{4}=\log\left(\tfrac{25}{4}\right)^{1/2}=\log\tfrac{5}{2}\); \(\;2\log\tfrac{4}{5}=\log\tfrac{16}{25}\).
\[\log\frac{5}{2}-\log\frac{16}{25}+\log\frac{320}{125}=\log\!\left(\frac{5}{2}\times\frac{25}{16}\times\frac{320}{125}\right)\]\(\dfrac{5}{2}\times\dfrac{25}{16}=\dfrac{125}{32}\); then \(\dfrac{125}{32}\times\dfrac{320}{125}=\dfrac{320}{32}=10\).
\[=\log_{10}10=1\]\(=1\)
Question 5 Report
An aeroplane flies due west for 3 hours from P (lat. 50°N, long. 60°W) to a point Q at an average speed of 600km/h. The aeroplane then flies due south from Q to a point Y 500km away. Calculate, correct to 3 significant figures,
(a) the longitude of Q ;
(b) the latitude of Y . [Take the radius of the earth = 6400km and \(\pi = \frac{22}{7}\)].
Distance flown due west \(=\text{speed}\times\text{time}=600\times3=1800\) km, along the parallel of latitude \(50^{\circ}N\).
(a) Longitude of Q. Distance along a parallel \(=\dfrac{\theta}{360}\times2\pi R\cos(\text{lat})\), where \(\theta\) is the change in longitude.
\[1800=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\cos50^{\circ}\]\(2\times\tfrac{22}{7}\times6400=40228.6\); \(\cos50^{\circ}=0.6428\); product \(=25859\).
\[\theta=\frac{1800\times360}{25859}\approx 25.1^{\circ}\]Flying due west from long. \(60^{\circ}W\) increases the western longitude: \(60+25.1=85.1^{\circ}W\).
Longitude of Q \(\approx 85.1^{\circ}W\) (3 s.f.).
(b) Latitude of Y. Flying due south is along a meridian: distance \(=\dfrac{\phi}{360}\times2\pi R\).
\[500=\frac{\phi}{360}\times40228.6\Rightarrow \phi=\frac{500\times360}{40228.6}\approx 4.47^{\circ}\]Moving south from \(50^{\circ}N\): latitude of \(Y=50-4.47\approx 45.5^{\circ}N\).
Latitude of Y \(\approx 45.5^{\circ}N\) (3 s.f.).
Answer Details
Distance flown due west \(=\text{speed}\times\text{time}=600\times3=1800\) km, along the parallel of latitude \(50^{\circ}N\).
(a) Longitude of Q. Distance along a parallel \(=\dfrac{\theta}{360}\times2\pi R\cos(\text{lat})\), where \(\theta\) is the change in longitude.
\[1800=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\cos50^{\circ}\]\(2\times\tfrac{22}{7}\times6400=40228.6\); \(\cos50^{\circ}=0.6428\); product \(=25859\).
\[\theta=\frac{1800\times360}{25859}\approx 25.1^{\circ}\]Flying due west from long. \(60^{\circ}W\) increases the western longitude: \(60+25.1=85.1^{\circ}W\).
Longitude of Q \(\approx 85.1^{\circ}W\) (3 s.f.).
(b) Latitude of Y. Flying due south is along a meridian: distance \(=\dfrac{\phi}{360}\times2\pi R\).
\[500=\frac{\phi}{360}\times40228.6\Rightarrow \phi=\frac{500\times360}{40228.6}\approx 4.47^{\circ}\]Moving south from \(50^{\circ}N\): latitude of \(Y=50-4.47\approx 45.5^{\circ}N\).
Latitude of Y \(\approx 45.5^{\circ}N\) (3 s.f.).
Question 6 Report
(a) Simplify : \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\)
(b)
In the diagram, YPF is a straight line. < XPY = 44°, < MPF = 46°, < XYP = < MFP = 90°, /XY/ = 7cm and /MP/ = 9 cm.
(i) Calculate, correct to 3 significant figures, /XM/ and /YF/ ; (ii) Find < XMP.
(a) Simplify the compound fraction
\[\frac{\tfrac{1}{3}c^{2} - \tfrac{2}{3}cd}{\tfrac{1}{2}d^{2} - \tfrac{1}{4}cd}\]Factorise numerator and denominator.
\[\text{Numerator} = \tfrac{1}{3}c^{2} - \tfrac{2}{3}cd = \tfrac{1}{3}c\,(c - 2d)\]\[\text{Denominator} = \tfrac{1}{2}d^{2} - \tfrac{1}{4}cd = \tfrac{1}{4}d\,(2d - c)\]Write \((c - 2d) = -(2d - c)\):
\[\frac{\tfrac{1}{3}c\,(c-2d)}{\tfrac{1}{4}d\,(2d-c)} = \frac{\tfrac{1}{3}c\,[-(2d-c)]}{\tfrac{1}{4}d\,(2d-c)} = -\frac{\tfrac{1}{3}c}{\tfrac{1}{4}d}\]\[= -\frac{1}{3}c \times \frac{4}{d} = \boxed{-\frac{4c}{3d}}\](b) Two right-angled triangles on the straight line \(YPF\)
From the diagram, \(YPF\) is a straight line, \(\angle XYP = \angle MFP = 90^\circ\), \(\angle XPY = 44^\circ\), \(\angle MPF = 46^\circ\), \(|XY| = 7\text{ cm}\) and \(|MP| = 9\text{ cm}\).
(i) Calculate \(|XM|\) and \(|YF|\).
Triangle \(XYP\) (right-angled at \(Y\)):
\[\tan 44^\circ = \frac{|XY|}{|YP|} \Rightarrow |YP| = \frac{7}{\tan 44^\circ} = \frac{7}{0.9657} = 7.249\text{ cm}\]\[\sin 44^\circ = \frac{|XY|}{|XP|} \Rightarrow |XP| = \frac{7}{\sin 44^\circ} = \frac{7}{0.6947} = 10.08\text{ cm}\]Triangle \(MPF\) (right-angled at \(F\), hypotenuse \(MP = 9\)):
\[|PF| = 9\cos 46^\circ = 9(0.6947) = 6.252\text{ cm}\]Length \(YF\): since \(Y\), \(P\), \(F\) are collinear,
\[|YF| = |YP| + |PF| = 7.249 + 6.252 = 13.5\text{ cm (3 s.f.)}\]Length \(XM\): at \(P\) the angles on the straight line give
\[\angle XPM = 180^\circ - 44^\circ - 46^\circ = 90^\circ\]so triangle \(XPM\) is right-angled at \(P\) with \(|XP| = 10.08\) and \(|PM| = 9\):
\[|XM| = \sqrt{|XP|^{2} + |PM|^{2}} = \sqrt{10.08^{2} + 9^{2}} = \sqrt{101.5 + 81} = \sqrt{182.5} = 13.5\text{ cm (3 s.f.)}\](ii) Find \(\angle XMP\). In the right-angled triangle \(XPM\) (right angle at \(P\)),
\[\tan(\angle XMP) = \frac{|XP|}{|PM|} = \frac{10.08}{9} = 1.1196\]\[\angle XMP = \tan^{-1}(1.1196) = 48.2^\circ\]Hence \(|XM| \approx 13.5\text{ cm}\), \(|YF| \approx 13.5\text{ cm}\) and \(\angle XMP \approx 48.2^\circ\).
Answer Details
(a) Simplify the compound fraction
\[\frac{\tfrac{1}{3}c^{2} - \tfrac{2}{3}cd}{\tfrac{1}{2}d^{2} - \tfrac{1}{4}cd}\]Factorise numerator and denominator.
\[\text{Numerator} = \tfrac{1}{3}c^{2} - \tfrac{2}{3}cd = \tfrac{1}{3}c\,(c - 2d)\]\[\text{Denominator} = \tfrac{1}{2}d^{2} - \tfrac{1}{4}cd = \tfrac{1}{4}d\,(2d - c)\]Write \((c - 2d) = -(2d - c)\):
\[\frac{\tfrac{1}{3}c\,(c-2d)}{\tfrac{1}{4}d\,(2d-c)} = \frac{\tfrac{1}{3}c\,[-(2d-c)]}{\tfrac{1}{4}d\,(2d-c)} = -\frac{\tfrac{1}{3}c}{\tfrac{1}{4}d}\]\[= -\frac{1}{3}c \times \frac{4}{d} = \boxed{-\frac{4c}{3d}}\](b) Two right-angled triangles on the straight line \(YPF\)
From the diagram, \(YPF\) is a straight line, \(\angle XYP = \angle MFP = 90^\circ\), \(\angle XPY = 44^\circ\), \(\angle MPF = 46^\circ\), \(|XY| = 7\text{ cm}\) and \(|MP| = 9\text{ cm}\).
(i) Calculate \(|XM|\) and \(|YF|\).
Triangle \(XYP\) (right-angled at \(Y\)):
\[\tan 44^\circ = \frac{|XY|}{|YP|} \Rightarrow |YP| = \frac{7}{\tan 44^\circ} = \frac{7}{0.9657} = 7.249\text{ cm}\]\[\sin 44^\circ = \frac{|XY|}{|XP|} \Rightarrow |XP| = \frac{7}{\sin 44^\circ} = \frac{7}{0.6947} = 10.08\text{ cm}\]Triangle \(MPF\) (right-angled at \(F\), hypotenuse \(MP = 9\)):
\[|PF| = 9\cos 46^\circ = 9(0.6947) = 6.252\text{ cm}\]Length \(YF\): since \(Y\), \(P\), \(F\) are collinear,
\[|YF| = |YP| + |PF| = 7.249 + 6.252 = 13.5\text{ cm (3 s.f.)}\]Length \(XM\): at \(P\) the angles on the straight line give
\[\angle XPM = 180^\circ - 44^\circ - 46^\circ = 90^\circ\]so triangle \(XPM\) is right-angled at \(P\) with \(|XP| = 10.08\) and \(|PM| = 9\):
\[|XM| = \sqrt{|XP|^{2} + |PM|^{2}} = \sqrt{10.08^{2} + 9^{2}} = \sqrt{101.5 + 81} = \sqrt{182.5} = 13.5\text{ cm (3 s.f.)}\](ii) Find \(\angle XMP\). In the right-angled triangle \(XPM\) (right angle at \(P\)),
\[\tan(\angle XMP) = \frac{|XP|}{|PM|} = \frac{10.08}{9} = 1.1196\]\[\angle XMP = \tan^{-1}(1.1196) = 48.2^\circ\]Hence \(|XM| \approx 13.5\text{ cm}\), \(|YF| \approx 13.5\text{ cm}\) and \(\angle XMP \approx 48.2^\circ\).
Question 7 Report
In the diagram, PQT is a straight line and SQ // RT.
(a) Join QR and show that : (i) < RPS = < QRT ; (ii) < PRS = < QTR.
(b) ABC is a triangle. The sides AB and AC are produced to D and E respectively such that < DBC = 132° and < ECD = 96°. Show that \(\Delta\) ABC is isosceles.
(a) P, Q, R, S lie on the circle, PQT is a straight line (T on PQ produced beyond Q), and \(SQ\parallel RT\). Join QR.
(i) Show that \(\angle RPS=\angle QRT\).
- \(\angle RPS\) and \(\angle RQS\) both stand on the same chord \(RS\) and P, Q are in the same segment, so \(\angle RPS=\angle RQS\) (angles in the same segment).
- \(SQ\parallel RT\) with QR as transversal gives \(\angle RQS=\angle QRT\) (alternate angles).
Therefore \(\angle RPS=\angle QRT\), as required.
(ii) Show that \(\angle PRS=\angle QTR\).
- \(\angle PRS\) and \(\angle PQS\) both stand on the same chord \(PS\) with R, Q in the same segment, so \(\angle PRS=\angle PQS\) (angles in the same segment).
- The straight line PQT is a transversal cutting the parallel lines \(SQ\) and \(RT\) at Q and T, so \(\angle PQS=\angle QTR\) (corresponding angles).
Therefore \(\angle PRS=\angle QTR\), as required.
(b) In triangle ABC, AB is produced to D and AC is produced to E, with \(\angle DBC=132^\circ\) and the exterior angle at C equal to \(96^\circ\).
\(\angle DBC\) is the exterior angle at B, and it is the supplement of the interior angle \(\angle ABC\) (angles on the straight line ABD):
\[\angle ABC=180^\circ-132^\circ=48^\circ\]The exterior angle at C is the supplement of the interior angle \(\angle ACB\) (angles on the straight line ACE):
\[\angle ACB=180^\circ-96^\circ=84^\circ\]The three interior angles of triangle ABC sum to \(180^\circ\):
\[\angle BAC=180^\circ-48^\circ-84^\circ=48^\circ\]Hence \(\angle BAC=\angle ABC=48^\circ\). Since two angles are equal, the sides opposite them are equal, i.e. \(BC=AC\). Therefore triangle ABC is isosceles.
Answer Details
(a) P, Q, R, S lie on the circle, PQT is a straight line (T on PQ produced beyond Q), and \(SQ\parallel RT\). Join QR.
(i) Show that \(\angle RPS=\angle QRT\).
- \(\angle RPS\) and \(\angle RQS\) both stand on the same chord \(RS\) and P, Q are in the same segment, so \(\angle RPS=\angle RQS\) (angles in the same segment).
- \(SQ\parallel RT\) with QR as transversal gives \(\angle RQS=\angle QRT\) (alternate angles).
Therefore \(\angle RPS=\angle QRT\), as required.
(ii) Show that \(\angle PRS=\angle QTR\).
- \(\angle PRS\) and \(\angle PQS\) both stand on the same chord \(PS\) with R, Q in the same segment, so \(\angle PRS=\angle PQS\) (angles in the same segment).
- The straight line PQT is a transversal cutting the parallel lines \(SQ\) and \(RT\) at Q and T, so \(\angle PQS=\angle QTR\) (corresponding angles).
Therefore \(\angle PRS=\angle QTR\), as required.
(b) In triangle ABC, AB is produced to D and AC is produced to E, with \(\angle DBC=132^\circ\) and the exterior angle at C equal to \(96^\circ\).
\(\angle DBC\) is the exterior angle at B, and it is the supplement of the interior angle \(\angle ABC\) (angles on the straight line ABD):
\[\angle ABC=180^\circ-132^\circ=48^\circ\]The exterior angle at C is the supplement of the interior angle \(\angle ACB\) (angles on the straight line ACE):
\[\angle ACB=180^\circ-96^\circ=84^\circ\]The three interior angles of triangle ABC sum to \(180^\circ\):
\[\angle BAC=180^\circ-48^\circ-84^\circ=48^\circ\]Hence \(\angle BAC=\angle ABC=48^\circ\). Since two angles are equal, the sides opposite them are equal, i.e. \(BC=AC\). Therefore triangle ABC is isosceles.
Question 8 Report
The marks obtained by 40 students in an examination are as follows :
85 77 87 74 77 78 79 89 95 90 78 73 86 83 91 74 84 81 83 75 77 70 81 69 75 63 76 87 61 78 69 96 65 80 84 80 77 74 88 72.
(a) Copy and complete the table for the distribution using the above data.
| Class Boundaries | Tally | Frequency |
| 59.5 - 64.5 | ||
| 64.5 - 69.5 | ||
| 69.5 - 74.5 | ||
| 74.5 - 79.5 | ||
| 79.5 - 84.5 | ||
| 84.5 - 89.5 | ||
| 89.5 - 94.5 | ||
| 94.5 - 99.5 |
(b) Draw a histogram to represent the distribution.
(c) Using your histogram, estimate the modal mark.
(d) If a student is chosen at random, find the probability that the student obtains a mark greater than 79.
(a) Completed frequency distribution
Sorting the 40 marks into the given class boundaries and tallying each mark:
| Class Boundaries | Tally | Frequency |
|---|---|---|
| 59.5 - 64.5 | || | 2 |
| 64.5 - 69.5 | ||| | 3 |
| 69.5 - 74.5 | |||| | | 6 |
| 74.5 - 79.5 | |||| |||| | | 11 |
| 79.5 - 84.5 | |||| ||| | 8 |
| 84.5 - 89.5 | |||| | | 6 |
| 89.5 - 94.5 | || | 2 |
| 94.5 - 99.5 | || | 2 |
| Total | 40 |
(Check: \(2+3+6+11+8+6+2+2 = 40\) students.)
(b) Histogram
Since all class widths are equal (5 marks), the bar heights equal the frequencies. The class boundaries are plotted on the horizontal axis and frequency on the vertical axis, with the bars drawn adjoining (no gaps):
(c) Modal mark from the histogram
The modal class is 74.5 - 79.5, the tallest bar (frequency 11). To read the mode, straight lines are drawn inside the modal bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. The vertical through their point of intersection meets the mark axis at the mode (shown dashed above), giving mode \(\approx 77.6\).
This agrees with the standard formula \[\text{Mode} = L + \left(\frac{f_1 - f_0}{(f_1-f_0)+(f_1-f_2)}\right)c,\] where \(L = 74.5,\ f_1 = 11,\ f_0 = 6,\ f_2 = 8,\ c = 5\): \[\text{Mode} = 74.5 + \left(\frac{11-6}{(11-6)+(11-8)}\right)\times 5 = 74.5 + \frac{5}{8}\times 5 = 74.5 + 3.125 \approx 77.6.\] So the modal mark is approximately 77.6 (about 78 marks).
(d) Probability of a mark greater than 79
Marks greater than 79 fall in the four highest classes (79.5 and above), giving \[8 + 6 + 2 + 2 = 18 \text{ students.}\] With 40 students in all, \[P(\text{mark} > 79) = \frac{18}{40} = \frac{9}{20} = 0.45.\]
Answer Details
(a) Completed frequency distribution
Sorting the 40 marks into the given class boundaries and tallying each mark:
| Class Boundaries | Tally | Frequency |
|---|---|---|
| 59.5 - 64.5 | || | 2 |
| 64.5 - 69.5 | ||| | 3 |
| 69.5 - 74.5 | |||| | | 6 |
| 74.5 - 79.5 | |||| |||| | | 11 |
| 79.5 - 84.5 | |||| ||| | 8 |
| 84.5 - 89.5 | |||| | | 6 |
| 89.5 - 94.5 | || | 2 |
| 94.5 - 99.5 | || | 2 |
| Total | 40 |
(Check: \(2+3+6+11+8+6+2+2 = 40\) students.)
(b) Histogram
Since all class widths are equal (5 marks), the bar heights equal the frequencies. The class boundaries are plotted on the horizontal axis and frequency on the vertical axis, with the bars drawn adjoining (no gaps):
(c) Modal mark from the histogram
The modal class is 74.5 - 79.5, the tallest bar (frequency 11). To read the mode, straight lines are drawn inside the modal bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. The vertical through their point of intersection meets the mark axis at the mode (shown dashed above), giving mode \(\approx 77.6\).
This agrees with the standard formula \[\text{Mode} = L + \left(\frac{f_1 - f_0}{(f_1-f_0)+(f_1-f_2)}\right)c,\] where \(L = 74.5,\ f_1 = 11,\ f_0 = 6,\ f_2 = 8,\ c = 5\): \[\text{Mode} = 74.5 + \left(\frac{11-6}{(11-6)+(11-8)}\right)\times 5 = 74.5 + \frac{5}{8}\times 5 = 74.5 + 3.125 \approx 77.6.\] So the modal mark is approximately 77.6 (about 78 marks).
(d) Probability of a mark greater than 79
Marks greater than 79 fall in the four highest classes (79.5 and above), giving \[8 + 6 + 2 + 2 = 18 \text{ students.}\] With 40 students in all, \[P(\text{mark} > 79) = \frac{18}{40} = \frac{9}{20} = 0.45.\]
Question 9 Report
Using ruler and a pair of compasses only,
(a) construct a quadrilateral PXYQ such that /PX/ = 9.9 cm, /QX/ = 10.2 cm, < QPZ = 75°, /QY/ = 10.4 cm and PQ // XY.
(b) Construct the (i) locus \(l_{1}\) of points equidistant from X and Y ; (ii) locus \(l_{2}\) of points equidistant from QY and YX.
(c) Locate M, the point of intersection of \(l_{1}\) and \(l_{2}\).
(d) Measure /PM/.
The accurate scale drawing of the whole construction is shown below. All construction arcs and the two loci are kept visible, and the required lengths and angle are transferred from the data:
(a) Constructing quadrilateral \(PXYQ\)
- Draw a base line and mark \(PX = 9.9\) cm.
- At \(P\), construct \(\angle QPX = 75^{\circ}\): first construct \(60^{\circ}\), then \(90^{\circ}\), and bisect the \(60^{\circ}\)-\(90^{\circ}\) interval to add \(15^{\circ}\), giving \(60^{\circ}+15^{\circ}=75^{\circ}\).
- With centre \(X\) and radius \(10.2\) cm, draw an arc to cut the \(75^{\circ}\) arm at \(Q\), so that \(|QX| = 10.2\) cm.
- Through \(X\), construct a line parallel to \(PQ\) (copy \(\angle QPX\) at \(X\) on the same side), giving the direction of \(XY\) with \(PQ \parallel XY\).
- With centre \(Q\) and radius \(10.4\) cm, draw an arc to cut the parallel line at \(Y\), so that \(|QY| = 10.4\) cm.
- Join \(XY\), \(YQ\) and \(QP\) to complete the quadrilateral \(PXYQ\).
(b) The two loci
(i) Locus \(l_{1}\) - points equidistant from \(X\) and \(Y\) form the perpendicular bisector of \(XY\). With centre \(X\) and then centre \(Y\), using the same radius (greater than \(\tfrac{1}{2}|XY|\)), draw arcs above and below \(XY\); the line through their intersections is \(l_{1}\).
(ii) Locus \(l_{2}\) - points equidistant from the lines \(QY\) and \(YX\) lie on the bisector of \(\angle QYX\). With centre \(Y\), draw an arc cutting \(YQ\) and \(YX\); from those two cuts, using equal radii, draw arcs to meet, and join \(Y\) to that meeting point to obtain \(l_{2}\).
(c) Locating \(M\)
\(M\) is the point where \(l_{1}\) and \(l_{2}\) cross, as marked on the diagram.
(d) Measuring \(|PM|\)
Joining \(P\) to \(M\) and measuring with the ruler:
\(|PM| = 9.7\text{ cm}\)
Answer Details
The accurate scale drawing of the whole construction is shown below. All construction arcs and the two loci are kept visible, and the required lengths and angle are transferred from the data:
(a) Constructing quadrilateral \(PXYQ\)
- Draw a base line and mark \(PX = 9.9\) cm.
- At \(P\), construct \(\angle QPX = 75^{\circ}\): first construct \(60^{\circ}\), then \(90^{\circ}\), and bisect the \(60^{\circ}\)-\(90^{\circ}\) interval to add \(15^{\circ}\), giving \(60^{\circ}+15^{\circ}=75^{\circ}\).
- With centre \(X\) and radius \(10.2\) cm, draw an arc to cut the \(75^{\circ}\) arm at \(Q\), so that \(|QX| = 10.2\) cm.
- Through \(X\), construct a line parallel to \(PQ\) (copy \(\angle QPX\) at \(X\) on the same side), giving the direction of \(XY\) with \(PQ \parallel XY\).
- With centre \(Q\) and radius \(10.4\) cm, draw an arc to cut the parallel line at \(Y\), so that \(|QY| = 10.4\) cm.
- Join \(XY\), \(YQ\) and \(QP\) to complete the quadrilateral \(PXYQ\).
(b) The two loci
(i) Locus \(l_{1}\) - points equidistant from \(X\) and \(Y\) form the perpendicular bisector of \(XY\). With centre \(X\) and then centre \(Y\), using the same radius (greater than \(\tfrac{1}{2}|XY|\)), draw arcs above and below \(XY\); the line through their intersections is \(l_{1}\).
(ii) Locus \(l_{2}\) - points equidistant from the lines \(QY\) and \(YX\) lie on the bisector of \(\angle QYX\). With centre \(Y\), draw an arc cutting \(YQ\) and \(YX\); from those two cuts, using equal radii, draw arcs to meet, and join \(Y\) to that meeting point to obtain \(l_{2}\).
(c) Locating \(M\)
\(M\) is the point where \(l_{1}\) and \(l_{2}\) cross, as marked on the diagram.
(d) Measuring \(|PM|\)
Joining \(P\) to \(M\) and measuring with the ruler:
\(|PM| = 9.7\text{ cm}\)
Question 10 Report
The table shows the number of suitcases possessed by a group of travellers.
| No. of suitcases | 0 | 1 | 2 | 3 | 4 | 5 |
| Travellers | 2 | 7 | 7 | 2 | 3 | 9 |
(a) Calculate the (i) median (ii) mean, correct to the nearest whole number.
(b) Draw a bar chart to represent the information.
Given data.
| No. of suitcases (x) | 0 | 1 | 2 | 3 | 4 | 5 |
| Travellers (f) | 2 | 7 | 7 | 2 | 3 | 9 |
Total number of travellers: \(\sum f = 2+7+7+2+3+9 = 30\).
(a)(i) Median
With \(N = 30\) values (even), the median is the mean of the 15th and 16th values. Building the cumulative frequencies:
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| Cumulative f | 2 | 9 | 16 | 18 | 21 | 30 |
Positions 10 to 16 correspond to \(x = 2\); hence both the 15th and 16th values equal 2.
\[\text{Median} = \frac{2+2}{2} = 2 \text{ suitcases.}\](a)(ii) Mean
\[\sum fx = (0)(2)+(1)(7)+(2)(7)+(3)(2)+(4)(3)+(5)(9) = 0+7+14+6+12+45 = 84.\]\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{84}{30} = 2.8 \approx 3 \text{ (nearest whole number).}\](b) Bar chart
The number of suitcases (0 to 5) is placed on the horizontal axis and the number of travellers on the vertical axis. Each bar has equal width with equal gaps, and the heights are 2, 7, 7, 2, 3 and 9 respectively.
The tallest bar occurs at 5 suitcases, with a height of 9 travellers.
Answer Details
Given data.
| No. of suitcases (x) | 0 | 1 | 2 | 3 | 4 | 5 |
| Travellers (f) | 2 | 7 | 7 | 2 | 3 | 9 |
Total number of travellers: \(\sum f = 2+7+7+2+3+9 = 30\).
(a)(i) Median
With \(N = 30\) values (even), the median is the mean of the 15th and 16th values. Building the cumulative frequencies:
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| Cumulative f | 2 | 9 | 16 | 18 | 21 | 30 |
Positions 10 to 16 correspond to \(x = 2\); hence both the 15th and 16th values equal 2.
\[\text{Median} = \frac{2+2}{2} = 2 \text{ suitcases.}\](a)(ii) Mean
\[\sum fx = (0)(2)+(1)(7)+(2)(7)+(3)(2)+(4)(3)+(5)(9) = 0+7+14+6+12+45 = 84.\]\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{84}{30} = 2.8 \approx 3 \text{ (nearest whole number).}\](b) Bar chart
The number of suitcases (0 to 5) is placed on the horizontal axis and the number of travellers on the vertical axis. Each bar has equal width with equal gaps, and the heights are 2, 7, 7, 2, 3 and 9 respectively.
The tallest bar occurs at 5 suitcases, with a height of 9 travellers.
Question 11 Report
(a) If p varies directly as \(r^{2}\) and p = 3.2 when r = 4, find the value of p when r = 6.5.
(b) Solve the simultaneous equations :
\(\frac{x}{2} + \frac{y}{4} = 1 ; \frac{x}{3} - \frac{y}{4} = \frac{-1}{6}\)
(a) \(p\propto r^{2}\Rightarrow p=kr^{2}\).
Using \(p=3.2,\ r=4\): \(3.2=k(16)\Rightarrow k=0.2\).
When \(r=6.5\): \(p=0.2(6.5)^{2}=0.2(42.25)=8.45\).
\(p=8.45\)
(b) Clear the fractions.
\(\dfrac{x}{2}+\dfrac{y}{4}=1\) \(\times4\Rightarrow 2x+y=4\) ... (1)
\(\dfrac{x}{3}-\dfrac{y}{4}=-\dfrac{1}{6}\) \(\times12\Rightarrow 4x-3y=-2\) ... (2)
From (1): \(y=4-2x\). Substitute into (2):
\[4x-3(4-2x)=-2\Rightarrow 4x-12+6x=-2\Rightarrow 10x=10\Rightarrow x=1\]Then \(y=4-2(1)=2\).
\(x=1,\ y=2\)
Answer Details
(a) \(p\propto r^{2}\Rightarrow p=kr^{2}\).
Using \(p=3.2,\ r=4\): \(3.2=k(16)\Rightarrow k=0.2\).
When \(r=6.5\): \(p=0.2(6.5)^{2}=0.2(42.25)=8.45\).
\(p=8.45\)
(b) Clear the fractions.
\(\dfrac{x}{2}+\dfrac{y}{4}=1\) \(\times4\Rightarrow 2x+y=4\) ... (1)
\(\dfrac{x}{3}-\dfrac{y}{4}=-\dfrac{1}{6}\) \(\times12\Rightarrow 4x-3y=-2\) ... (2)
From (1): \(y=4-2x\). Substitute into (2):
\[4x-3(4-2x)=-2\Rightarrow 4x-12+6x=-2\Rightarrow 10x=10\Rightarrow x=1\]Then \(y=4-2(1)=2\).
\(x=1,\ y=2\)
Question 12 Report
The table below shows the values of the relation \(y = 11 - 2x - 2x^{2}\) for \(-4 \leq x \leq 3\).
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | -13 | 11 |
(a) Copy and complete the table.
(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 \(y = 11 - 2x - 2x^{2}\).
(c) Use your graph to find : (i) the roots of the equation \(11 - 2x - 2x^{2} = 0\) ; (ii) the values of x for which \(3 - 2x - 2x^{2} = 0\) ; (iii) the gradient of the curve at x = 1.
(a) Copy and complete the table
Using \(y = 11 - 2x - 2x^{2}\), build the value from its parts \(x^{2}\), \(11\), \(-2x\) and \(-2x^{2}\) for each \(x\):
\(x=-3:\ 11-2(-3)-2(9)=11+6-18=-1\)
\(x=-2:\ 11+4-8=7\)
\(x=-1:\ 11+2-2=11\)
\(x=1:\ 11-2-2=7\)
\(x=2:\ 11-4-8=-1\)
\(x=3:\ 11-6-18=-13\)
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| \(x^{2}\) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
| \(11\) | 11 | 11 | 11 | 11 | 11 | 11 | 11 | 11 |
| \(-2x\) | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 |
| \(-2x^{2}\) | -32 | -18 | -8 | -2 | 0 | -2 | -8 | -18 |
| \(y\) | -13 | -1 | 7 | 11 | 11 | 7 | -1 | -13 |
(b) The graph of \(y = 11 - 2x - 2x^{2}\)
Plotting the eight points \((-4,-13),(-3,-1),(-2,7),(-1,11),(0,11),(1,7),(2,-1),(3,-13)\) with a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, and joining them with a smooth curve, gives a downward (maximum) parabola with its peak midway between \(x=-1\) and \(x=0\), at about \((-0.5,\ 11.5)\).
(c) Using the graph
(i) Roots of \(11 - 2x - 2x^{2} = 0\). These are the \(x\)-values where the curve crosses the \(x\)-axis (\(y=0\)). Reading off the graph:
\[x \approx -2.9 \quad \text{or} \quad x \approx 1.9\](ii) Values of \(x\) for which \(3 - 2x - 2x^{2} = 0\). Subtracting this equation from the curve:
\[y - (3 - 2x - 2x^{2}) = (11 - 2x - 2x^{2}) - (3 - 2x - 2x^{2}) = 8\]So where \(3 - 2x - 2x^{2}=0\) the curve has \(y = 8\). Drawing the horizontal line \(y = 8\) (shown dashed) and reading where it meets the curve:
\[x \approx -1.8 \quad \text{or} \quad x \approx 0.8\](iii) Gradient of the curve at \(x = 1\). Draw the tangent to the curve at the point \((1,\ 7)\) (shown as the straight line touching the curve). Taking two points on this tangent, e.g. \((0,\ 13)\) and \((2,\ 1)\):
\[\text{gradient} = \frac{1 - 13}{2 - 0} = \frac{-12}{2} = -6\]Hence the gradient of the curve at \(x = 1\) is \(-6\).
Answer Details
(a) Copy and complete the table
Using \(y = 11 - 2x - 2x^{2}\), build the value from its parts \(x^{2}\), \(11\), \(-2x\) and \(-2x^{2}\) for each \(x\):
\(x=-3:\ 11-2(-3)-2(9)=11+6-18=-1\)
\(x=-2:\ 11+4-8=7\)
\(x=-1:\ 11+2-2=11\)
\(x=1:\ 11-2-2=7\)
\(x=2:\ 11-4-8=-1\)
\(x=3:\ 11-6-18=-13\)
| \(x\) | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| \(x^{2}\) | 16 | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
| \(11\) | 11 | 11 | 11 | 11 | 11 | 11 | 11 | 11 |
| \(-2x\) | 8 | 6 | 4 | 2 | 0 | -2 | -4 | -6 |
| \(-2x^{2}\) | -32 | -18 | -8 | -2 | 0 | -2 | -8 | -18 |
| \(y\) | -13 | -1 | 7 | 11 | 11 | 7 | -1 | -13 |
(b) The graph of \(y = 11 - 2x - 2x^{2}\)
Plotting the eight points \((-4,-13),(-3,-1),(-2,7),(-1,11),(0,11),(1,7),(2,-1),(3,-13)\) with a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, and joining them with a smooth curve, gives a downward (maximum) parabola with its peak midway between \(x=-1\) and \(x=0\), at about \((-0.5,\ 11.5)\).
(c) Using the graph
(i) Roots of \(11 - 2x - 2x^{2} = 0\). These are the \(x\)-values where the curve crosses the \(x\)-axis (\(y=0\)). Reading off the graph:
\[x \approx -2.9 \quad \text{or} \quad x \approx 1.9\](ii) Values of \(x\) for which \(3 - 2x - 2x^{2} = 0\). Subtracting this equation from the curve:
\[y - (3 - 2x - 2x^{2}) = (11 - 2x - 2x^{2}) - (3 - 2x - 2x^{2}) = 8\]So where \(3 - 2x - 2x^{2}=0\) the curve has \(y = 8\). Drawing the horizontal line \(y = 8\) (shown dashed) and reading where it meets the curve:
\[x \approx -1.8 \quad \text{or} \quad x \approx 0.8\](iii) Gradient of the curve at \(x = 1\). Draw the tangent to the curve at the point \((1,\ 7)\) (shown as the straight line touching the curve). Taking two points on this tangent, e.g. \((0,\ 13)\) and \((2,\ 1)\):
\[\text{gradient} = \frac{1 - 13}{2 - 0} = \frac{-12}{2} = -6\]Hence the gradient of the curve at \(x = 1\) is \(-6\).
Question 13 Report
The table shows the monthly contributions and expenditure pattern of an employee in 1999.
| Item | Percentage |
| Pension | 5 |
| Income Tax | 25 |
| Food | 40 |
| Transport | 10 |
| Rent | 12.5 |
| Others | 7.5 |
(a) Draw a pie chart to illustrate the data.
(b) If the employee's gross monthly salary was N10,800.00, calculate (i) the pension contribution of the employee ; (ii) the income tax paid by the employee.
(c) If the pension contribution and income tax were deducted from the gross monthly salary, before payment, calculate the take- home pay of the employee.
The percentages account for the whole salary, and they sum to
\[5+25+40+10+12.5+7.5 = 100.\](a) Pie chart
The whole circle is \(360^\circ\), representing \(100\%\). Each item's sector angle is its percentage multiplied by \(\dfrac{360^\circ}{100} = 3.6^\circ\) per unit:
| Item | Percentage (%) | Sector angle |
|---|---|---|
| Pension | 5 | \(\frac{5}{100}\times360^\circ = 18^\circ\) |
| Income Tax | 25 | \(\frac{25}{100}\times360^\circ = 90^\circ\) |
| Food | 40 | \(\frac{40}{100}\times360^\circ = 144^\circ\) |
| Transport | 10 | \(\frac{10}{100}\times360^\circ = 36^\circ\) |
| Rent | 12.5 | \(\frac{12.5}{100}\times360^\circ = 45^\circ\) |
| Others | 7.5 | \(\frac{7.5}{100}\times360^\circ = 27^\circ\) |
| Total | 100 | \(360^\circ\) |
Check: \(18^\circ+90^\circ+144^\circ+36^\circ+45^\circ+27^\circ = 360^\circ.\) The completed pie chart is:
(b) Salary calculations (gross monthly salary \(=\) N10,800.00)
(i) Pension contribution \(= 5\%\) of N10,800, i.e. the \(18^\circ\) sector:
\[\frac{5}{100}\times 10800 \;=\; \frac{18}{360}\times 10800 \;=\; \text{N}540.00.\](ii) Income tax \(= 25\%\) of N10,800, i.e. the \(90^\circ\) sector:
\[\frac{25}{100}\times 10800 \;=\; \frac{90}{360}\times 10800 \;=\; \text{N}2{,}700.00.\](c) Take-home pay
Deduct the pension and income tax from the gross salary. Together they are \((5+25)\% = 30\%\), so the take-home pay is \(100\% - 30\% = 70\%\):
\[10800 - 540 - 2700 \;=\; \frac{70}{100}\times 10800 \;=\; \text{N}7{,}560.00.\]The employee's take-home pay is N7,560.00.
Answer Details
The percentages account for the whole salary, and they sum to
\[5+25+40+10+12.5+7.5 = 100.\](a) Pie chart
The whole circle is \(360^\circ\), representing \(100\%\). Each item's sector angle is its percentage multiplied by \(\dfrac{360^\circ}{100} = 3.6^\circ\) per unit:
| Item | Percentage (%) | Sector angle |
|---|---|---|
| Pension | 5 | \(\frac{5}{100}\times360^\circ = 18^\circ\) |
| Income Tax | 25 | \(\frac{25}{100}\times360^\circ = 90^\circ\) |
| Food | 40 | \(\frac{40}{100}\times360^\circ = 144^\circ\) |
| Transport | 10 | \(\frac{10}{100}\times360^\circ = 36^\circ\) |
| Rent | 12.5 | \(\frac{12.5}{100}\times360^\circ = 45^\circ\) |
| Others | 7.5 | \(\frac{7.5}{100}\times360^\circ = 27^\circ\) |
| Total | 100 | \(360^\circ\) |
Check: \(18^\circ+90^\circ+144^\circ+36^\circ+45^\circ+27^\circ = 360^\circ.\) The completed pie chart is:
(b) Salary calculations (gross monthly salary \(=\) N10,800.00)
(i) Pension contribution \(= 5\%\) of N10,800, i.e. the \(18^\circ\) sector:
\[\frac{5}{100}\times 10800 \;=\; \frac{18}{360}\times 10800 \;=\; \text{N}540.00.\](ii) Income tax \(= 25\%\) of N10,800, i.e. the \(90^\circ\) sector:
\[\frac{25}{100}\times 10800 \;=\; \frac{90}{360}\times 10800 \;=\; \text{N}2{,}700.00.\](c) Take-home pay
Deduct the pension and income tax from the gross salary. Together they are \((5+25)\% = 30\%\), so the take-home pay is \(100\% - 30\% = 70\%\):
\[10800 - 540 - 2700 \;=\; \frac{70}{100}\times 10800 \;=\; \text{N}7{,}560.00.\]The employee's take-home pay is N7,560.00.
