WAEC SSCE - General Mathematics - 2014 (Essay)
Ajụjụ 1 Ripọtì
(a) Without using tables or calculator, simplify : \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\), leaving the answer in standard form (scientific notation).
(b) 
In the diagram, \(\overline{EF}\) is parallel to \(\overline{GH}\). If \(< AEF = 3x°, < ABC = 120°\) and \(< CHG = 7x°\), find the value of \(< GHB\).
(a) Simplify in standard form.
Write every number in standard form first, then combine powers of 10.
Numerator:
\[0.6\times32\times0.004=(6\times10^{-1})(3.2\times10^{1})(4\times10^{-3})\]\[=(6\times3.2\times4)\times10^{-1+1-3}=76.8\times10^{-3}=7.68\times10^{-2}\]Denominator:
\[1.2\times0.008\times0.16=(1.2\times10^{0})(8\times10^{-3})(1.6\times10^{-1})\]\[=(1.2\times8\times1.6)\times10^{0-3-1}=15.36\times10^{-4}=1.536\times10^{-3}\]Divide:
\[\frac{7.68\times10^{-2}}{1.536\times10^{-3}}=\frac{7.68}{1.536}\times10^{-2-(-3)}=5\times10^{1}\]So the value is \(\mathbf{5.0\times10^{1}}\) (that is, \(50\)).
(b) Find \(\angle GHB\).
From the diagram, \(\overline{EF}\parallel\overline{GH}\). The line \(A\,E\,B\) is straight, it bends at \(B\) where \(\angle ABC=120^{\circ}\), and the line then continues straight through to \(H\). The marked angles are \(\angle AEF=3x^{\circ}\) and \(\angle CHG=7x^{\circ}\).
Draw a line \(BK\) through \(B\) parallel to both \(\overline{EF}\) and \(\overline{GH}\).
Since \(A,E,B\) are collinear, \(\angle FEB=180^{\circ}-3x\). With \(EF\parallel BK\), the co-interior angles give
\[\angle EBK=180^{\circ}-\angle FEB=3x^{\circ}\]With \(BK\parallel GH\) and \(BH\) a transversal, alternate angles give
\[\angle KBH=\angle GHB=7x^{\circ}\]The ray \(BK\) lies inside \(\angle EBH\), so
\[\angle EBH=\angle EBK+\angle KBH\]\[120=3x+7x=10x\]\[x=12\]Therefore
\[\angle GHB=7x=7(12)=\mathbf{84^{\circ}}\]Akọwa Nkọwa
(a) Simplify in standard form.
Write every number in standard form first, then combine powers of 10.
Numerator:
\[0.6\times32\times0.004=(6\times10^{-1})(3.2\times10^{1})(4\times10^{-3})\]\[=(6\times3.2\times4)\times10^{-1+1-3}=76.8\times10^{-3}=7.68\times10^{-2}\]Denominator:
\[1.2\times0.008\times0.16=(1.2\times10^{0})(8\times10^{-3})(1.6\times10^{-1})\]\[=(1.2\times8\times1.6)\times10^{0-3-1}=15.36\times10^{-4}=1.536\times10^{-3}\]Divide:
\[\frac{7.68\times10^{-2}}{1.536\times10^{-3}}=\frac{7.68}{1.536}\times10^{-2-(-3)}=5\times10^{1}\]So the value is \(\mathbf{5.0\times10^{1}}\) (that is, \(50\)).
(b) Find \(\angle GHB\).
From the diagram, \(\overline{EF}\parallel\overline{GH}\). The line \(A\,E\,B\) is straight, it bends at \(B\) where \(\angle ABC=120^{\circ}\), and the line then continues straight through to \(H\). The marked angles are \(\angle AEF=3x^{\circ}\) and \(\angle CHG=7x^{\circ}\).
Draw a line \(BK\) through \(B\) parallel to both \(\overline{EF}\) and \(\overline{GH}\).
Since \(A,E,B\) are collinear, \(\angle FEB=180^{\circ}-3x\). With \(EF\parallel BK\), the co-interior angles give
\[\angle EBK=180^{\circ}-\angle FEB=3x^{\circ}\]With \(BK\parallel GH\) and \(BH\) a transversal, alternate angles give
\[\angle KBH=\angle GHB=7x^{\circ}\]The ray \(BK\) lies inside \(\angle EBH\), so
\[\angle EBH=\angle EBK+\angle KBH\]\[120=3x+7x=10x\]\[x=12\]Therefore
\[\angle GHB=7x=7(12)=\mathbf{84^{\circ}}\]Ajụjụ 2 Ripọtì
(a) Copy and complete the following table for multiplication modulo 11.
| \(\otimes\) | 1 | 5 | 9 | 10 |
| 1 | 1 | 5 | 9 | 10 |
| 5 | 5 | |||
| 9 | 9 | |||
| 10 | 10 |
Use the table to : (i) evaluate \((9 \otimes 5) \otimes (10 \otimes 10)\);
(ii) find the truth set of :(1) \(10 \otimes m = 2\); (2) \(n \otimes n = 4\)
(b) When a fraction is reduced to its lowest term, it is equal to \(\frac{3}{4}\). The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.
(a) Multiplication table modulo 11 (multiply, then take the remainder on division by 11).
For example \(5\otimes 5 = 25 = 2(11)+3 \equiv 3\), \(9\otimes 9 = 81 = 7(11)+4 \equiv 4\), \(10\otimes 10 = 100 = 9(11)+1 \equiv 1\).
| \(\otimes\) | 1 | 5 | 9 | 10 |
|---|---|---|---|---|
| 1 | 1 | 5 | 9 | 10 |
| 5 | 5 | 3 | 1 | 6 |
| 9 | 9 | 1 | 4 | 2 |
| 10 | 10 | 6 | 2 | 1 |
(i) \((9\otimes 5)\otimes(10\otimes 10) = 1 \otimes 1 = \mathbf{1}\).
(ii)(1) \(10\otimes m = 2\): from the row for 10, \(10\otimes 9 = 2\), so the truth set is \(\{9\}\).
(ii)(2) \(n\otimes n = 4\): scanning the leading diagonal, only \(9\otimes 9 = 4\), so the truth set is \(\{9\}\).
(b) The fraction. In lowest terms it equals \(\tfrac{3}{4}\), so let it be \(\dfrac{3k}{4k}\).
Numerator doubled is 34 more than the denominator:
\[2(3k) = 4k + 34 \;\Rightarrow\; 6k - 4k = 34 \;\Rightarrow\; 2k = 34 \;\Rightarrow\; k = 17\] \[\text{Fraction} = \frac{3(17)}{4(17)} = \frac{51}{68}\]Check: \(2(51) = 102 = 68 + 34\). The fraction is \(\mathbf{\tfrac{51}{68}}\).
Akọwa Nkọwa
(a) Multiplication table modulo 11 (multiply, then take the remainder on division by 11).
For example \(5\otimes 5 = 25 = 2(11)+3 \equiv 3\), \(9\otimes 9 = 81 = 7(11)+4 \equiv 4\), \(10\otimes 10 = 100 = 9(11)+1 \equiv 1\).
| \(\otimes\) | 1 | 5 | 9 | 10 |
|---|---|---|---|---|
| 1 | 1 | 5 | 9 | 10 |
| 5 | 5 | 3 | 1 | 6 |
| 9 | 9 | 1 | 4 | 2 |
| 10 | 10 | 6 | 2 | 1 |
(i) \((9\otimes 5)\otimes(10\otimes 10) = 1 \otimes 1 = \mathbf{1}\).
(ii)(1) \(10\otimes m = 2\): from the row for 10, \(10\otimes 9 = 2\), so the truth set is \(\{9\}\).
(ii)(2) \(n\otimes n = 4\): scanning the leading diagonal, only \(9\otimes 9 = 4\), so the truth set is \(\{9\}\).
(b) The fraction. In lowest terms it equals \(\tfrac{3}{4}\), so let it be \(\dfrac{3k}{4k}\).
Numerator doubled is 34 more than the denominator:
\[2(3k) = 4k + 34 \;\Rightarrow\; 6k - 4k = 34 \;\Rightarrow\; 2k = 34 \;\Rightarrow\; k = 17\] \[\text{Fraction} = \frac{3(17)}{4(17)} = \frac{51}{68}\]Check: \(2(51) = 102 = 68 + 34\). The fraction is \(\mathbf{\tfrac{51}{68}}\).
Ajụjụ 3 Ripọtì
(a) Simplify : \(3\sqrt{75} - \sqrt{12} + \sqrt{108}\), leaving the answer in surd form (radicals).
(b) If \(124_{n} = 232_{five}\), find n.
(a) Simplify \(3\sqrt{75} - \sqrt{12} + \sqrt{108}\)
Reduce each surd to lowest terms by taking out perfect squares:
\[\sqrt{75} = \sqrt{25\times 3} = 5\sqrt{3} \;\Rightarrow\; 3\sqrt{75} = 15\sqrt{3}\]
\[\sqrt{12} = \sqrt{4\times 3} = 2\sqrt{3}\]
\[\sqrt{108} = \sqrt{36\times 3} = 6\sqrt{3}\]
Combine the like surds:
\[15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3} = 19\sqrt{3}\]
(b) If \(124_{n} = 232_{five}\), find \(n\).
Convert the right side to base ten:
\[232_{five} = 2(5^{2}) + 3(5) + 2 = 50 + 15 + 2 = 67\]
Express the left side in base ten:
\[124_{n} = 1(n^{2}) + 2(n) + 4 = n^{2} + 2n + 4\]
Set them equal and solve:
\[n^{2} + 2n + 4 = 67 \;\Rightarrow\; n^{2} + 2n - 63 = 0\]
\[(n+9)(n-7) = 0\]
Since a base must be positive, \(n = 7\).
Akọwa Nkọwa
(a) Simplify \(3\sqrt{75} - \sqrt{12} + \sqrt{108}\)
Reduce each surd to lowest terms by taking out perfect squares:
\[\sqrt{75} = \sqrt{25\times 3} = 5\sqrt{3} \;\Rightarrow\; 3\sqrt{75} = 15\sqrt{3}\]
\[\sqrt{12} = \sqrt{4\times 3} = 2\sqrt{3}\]
\[\sqrt{108} = \sqrt{36\times 3} = 6\sqrt{3}\]
Combine the like surds:
\[15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3} = 19\sqrt{3}\]
(b) If \(124_{n} = 232_{five}\), find \(n\).
Convert the right side to base ten:
\[232_{five} = 2(5^{2}) + 3(5) + 2 = 50 + 15 + 2 = 67\]
Express the left side in base ten:
\[124_{n} = 1(n^{2}) + 2(n) + 4 = n^{2} + 2n + 4\]
Set them equal and solve:
\[n^{2} + 2n + 4 = 67 \;\Rightarrow\; n^{2} + 2n - 63 = 0\]
\[(n+9)(n-7) = 0\]
Since a base must be positive, \(n = 7\).
Ajụjụ 4 Ripọtì
(a) Given that \(5 \cos (x + 8.5)° - 1 = 0, 0° \leq x \leq 90°\), calculate, correct to the nearest degree, the value of x.
(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P : (i) represent this information in a diagram;
(ii) calculate the distance between Q and R, correct to two decimal places ; (iii) find the bearing of R from Q, correct to the nearest degree.
(a) Solving \(5\cos(x+8.5)^\circ - 1 = 0\).
\[\cos(x+8.5)^\circ = \frac{1}{5} = 0.2\]
\[x + 8.5 = \cos^{-1}(0.2) = 78.46^\circ\]
\[x = 78.46 - 8.5 = 69.96^\circ \approx 70^\circ\]
(b) Reading the bearings: the bearing of \(Q\) from \(P\) is \(150^\circ\) and the bearing of \(P\) from \(R\) is \(015^\circ\). Since the bearing of \(P\) from \(R\) is \(015^\circ\), the bearing of \(R\) from \(P\) is \(015^\circ + 180^\circ = 195^\circ\). With \(|PQ| = 24\) km and \(|PR| = 32\) km:
(i) Diagram. At \(P\), draw the north line. \(Q\) lies on a bearing of \(150^\circ\) at 24 km; \(R\) lies on a bearing of \(195^\circ\) at 32 km. The angle \(\angle QPR = 195^\circ - 150^\circ = 45^\circ\).
(ii) Distance \(|QR|\). By the cosine rule in \(\triangle PQR\):
\[|QR|^2 = 24^2 + 32^2 - 2(24)(32)\cos 45^\circ\]
\[= 576 + 1024 - 1536(0.7071) = 1600 - 1086.11 = 513.89\]
\[|QR| = \sqrt{513.89} = 22.67\text{ km (2 d.p.)}\]
(iii) Bearing of \(R\) from \(Q\). Taking \(P\) as origin with north as the \(y\)-axis: \(Q = (24\sin150^\circ,\,24\cos150^\circ) = (12,\,-20.78)\) and \(R = (32\sin195^\circ,\,32\cos195^\circ) = (-8.28,\,-30.91)\).
Displacement from \(Q\) to \(R\): \((-20.28,\,-10.13)\), which points south and west (third quadrant).
\[\theta = \tan^{-1}\!\left(\frac{20.28}{10.13}\right) = 63.5^\circ \text{ west of south}\]
\[\text{Bearing of }R\text{ from }Q = 180^\circ + 63.5^\circ = 243.5^\circ \approx 243^\circ\]
Akọwa Nkọwa
(a) Solving \(5\cos(x+8.5)^\circ - 1 = 0\).
\[\cos(x+8.5)^\circ = \frac{1}{5} = 0.2\]
\[x + 8.5 = \cos^{-1}(0.2) = 78.46^\circ\]
\[x = 78.46 - 8.5 = 69.96^\circ \approx 70^\circ\]
(b) Reading the bearings: the bearing of \(Q\) from \(P\) is \(150^\circ\) and the bearing of \(P\) from \(R\) is \(015^\circ\). Since the bearing of \(P\) from \(R\) is \(015^\circ\), the bearing of \(R\) from \(P\) is \(015^\circ + 180^\circ = 195^\circ\). With \(|PQ| = 24\) km and \(|PR| = 32\) km:
(i) Diagram. At \(P\), draw the north line. \(Q\) lies on a bearing of \(150^\circ\) at 24 km; \(R\) lies on a bearing of \(195^\circ\) at 32 km. The angle \(\angle QPR = 195^\circ - 150^\circ = 45^\circ\).
(ii) Distance \(|QR|\). By the cosine rule in \(\triangle PQR\):
\[|QR|^2 = 24^2 + 32^2 - 2(24)(32)\cos 45^\circ\]
\[= 576 + 1024 - 1536(0.7071) = 1600 - 1086.11 = 513.89\]
\[|QR| = \sqrt{513.89} = 22.67\text{ km (2 d.p.)}\]
(iii) Bearing of \(R\) from \(Q\). Taking \(P\) as origin with north as the \(y\)-axis: \(Q = (24\sin150^\circ,\,24\cos150^\circ) = (12,\,-20.78)\) and \(R = (32\sin195^\circ,\,32\cos195^\circ) = (-8.28,\,-30.91)\).
Displacement from \(Q\) to \(R\): \((-20.28,\,-10.13)\), which points south and west (third quadrant).
\[\theta = \tan^{-1}\!\left(\frac{20.28}{10.13}\right) = 63.5^\circ \text{ west of south}\]
\[\text{Bearing of }R\text{ from }Q = 180^\circ + 63.5^\circ = 243.5^\circ \approx 243^\circ\]
Ajụjụ 5 Ripọtì
(a) If \(\frac{3}{2p - \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{4}p + 1}\), find p.
(b) A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.
(a) Solve \(\dfrac{3}{2p-\tfrac{1}{2}} = \dfrac{\tfrac{1}{3}}{\tfrac{1}{4}p + 1}\)
Cross multiply:
\[3\left(\tfrac{1}{4}p + 1\right) = \tfrac{1}{3}\left(2p - \tfrac{1}{2}\right)\]
Expand both sides:
\[\tfrac{3}{4}p + 3 = \tfrac{2}{3}p - \tfrac{1}{6}\]
Collect the \(p\) terms and constants:
\[\tfrac{3}{4}p - \tfrac{2}{3}p = -\tfrac{1}{6} - 3\]
\[\left(\tfrac{9-8}{12}\right)p = -\tfrac{19}{6} \;\Rightarrow\; \tfrac{1}{12}p = -\tfrac{19}{6}\]
\[p = -\tfrac{19}{6}\times 12 = -38\]
\(p = -38\).
(b) Actual percentage profit
The marked price of GH¢760.00 gives a 20% profit, so it is \(120\%\) of the cost price:
\[\text{Cost price} = \frac{760}{1.20} = \text{GH¢}633.33\]
Sold at a 5% discount off the marked price:
\[\text{Selling price} = 760\times 0.95 = \text{GH¢}722.00\]
Actual profit \(= 722.00 - 633.33 = \text{GH¢}88.67\).
\[\text{Percentage profit} = \frac{88.67}{633.33}\times 100 \approx 14\%\]
Actual percentage profit \(\approx 14\%\) (to 2 s.f.).
Akọwa Nkọwa
(a) Solve \(\dfrac{3}{2p-\tfrac{1}{2}} = \dfrac{\tfrac{1}{3}}{\tfrac{1}{4}p + 1}\)
Cross multiply:
\[3\left(\tfrac{1}{4}p + 1\right) = \tfrac{1}{3}\left(2p - \tfrac{1}{2}\right)\]
Expand both sides:
\[\tfrac{3}{4}p + 3 = \tfrac{2}{3}p - \tfrac{1}{6}\]
Collect the \(p\) terms and constants:
\[\tfrac{3}{4}p - \tfrac{2}{3}p = -\tfrac{1}{6} - 3\]
\[\left(\tfrac{9-8}{12}\right)p = -\tfrac{19}{6} \;\Rightarrow\; \tfrac{1}{12}p = -\tfrac{19}{6}\]
\[p = -\tfrac{19}{6}\times 12 = -38\]
\(p = -38\).
(b) Actual percentage profit
The marked price of GH¢760.00 gives a 20% profit, so it is \(120\%\) of the cost price:
\[\text{Cost price} = \frac{760}{1.20} = \text{GH¢}633.33\]
Sold at a 5% discount off the marked price:
\[\text{Selling price} = 760\times 0.95 = \text{GH¢}722.00\]
Actual profit \(= 722.00 - 633.33 = \text{GH¢}88.67\).
\[\text{Percentage profit} = \frac{88.67}{633.33}\times 100 \approx 14\%\]
Actual percentage profit \(\approx 14\%\) (to 2 s.f.).
Ajụjụ 6 Ripọtì
(a) Solve : \((x - 2)(x - 3) = 12\).
(b) 
In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take \(\pi = \frac{22}{7}\)].
(a) Solve \((x-2)(x-3)=12\).
Expand the left side:
\[x^2-5x+6=12\]Bring every term to one side:
\[x^2-5x-6=0\]Factorise. Two numbers whose product is \(-6\) and whose sum is \(-5\) are \(-6\) and \(+1\):
\[(x-6)(x+1)=0\]Hence \(x-6=0\) or \(x+1=0\), giving:
\[x=6 \quad\text{or}\quad x=-1\](b) Area of the shaded (overlap) portion.
From the diagram the two circles have centre \(M\) and \(N\), each of radius \(r=7\text{ cm}\), and \(\angle PMQ=\angle PNQ=60^{\circ}\). The shaded lens is the region common to both circles. It is made up of two equal circular segments cut off by the common chord \(PQ\): one from circle \(M\) and one from circle \(N\).
Area of one segment = area of sector \(-\) area of triangle.
Area of sector \(PMQ\):
\[\frac{60}{360}\times\frac{22}{7}\times 7^2=\frac{1}{6}\times\frac{22}{7}\times 49=\frac{154}{6}=25.667\text{ cm}^2\]Area of triangle \(PMQ\) (two radii with included angle \(60^{\circ}\)):
\[\frac{1}{2}\times 7\times 7\times\sin 60^{\circ}=24.5\times 0.8660=21.217\text{ cm}^2\]Area of one segment:
\[25.667-21.217=4.450\text{ cm}^2\]The shaded lens is two such segments (one per circle):
\[2\times 4.450=8.90\text{ cm}^2\]Correct to the nearest whole number, the area of the shaded portion is \(9\text{ cm}^2\).
Akọwa Nkọwa
(a) Solve \((x-2)(x-3)=12\).
Expand the left side:
\[x^2-5x+6=12\]Bring every term to one side:
\[x^2-5x-6=0\]Factorise. Two numbers whose product is \(-6\) and whose sum is \(-5\) are \(-6\) and \(+1\):
\[(x-6)(x+1)=0\]Hence \(x-6=0\) or \(x+1=0\), giving:
\[x=6 \quad\text{or}\quad x=-1\](b) Area of the shaded (overlap) portion.
From the diagram the two circles have centre \(M\) and \(N\), each of radius \(r=7\text{ cm}\), and \(\angle PMQ=\angle PNQ=60^{\circ}\). The shaded lens is the region common to both circles. It is made up of two equal circular segments cut off by the common chord \(PQ\): one from circle \(M\) and one from circle \(N\).
Area of one segment = area of sector \(-\) area of triangle.
Area of sector \(PMQ\):
\[\frac{60}{360}\times\frac{22}{7}\times 7^2=\frac{1}{6}\times\frac{22}{7}\times 49=\frac{154}{6}=25.667\text{ cm}^2\]Area of triangle \(PMQ\) (two radii with included angle \(60^{\circ}\)):
\[\frac{1}{2}\times 7\times 7\times\sin 60^{\circ}=24.5\times 0.8660=21.217\text{ cm}^2\]Area of one segment:
\[25.667-21.217=4.450\text{ cm}^2\]The shaded lens is two such segments (one per circle):
\[2\times 4.450=8.90\text{ cm}^2\]Correct to the nearest whole number, the area of the shaded portion is \(9\text{ cm}^2\).
Ajụjụ 7 Ripọtì
A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :
(a) both contracts ;
(b) exactly one of the contracts ;
(c) neither of the contracts?
Setup. The contracts are independent. Given:
\[P(\text{win }X) = 0.5,\qquad P(\text{not win }Y) = 0.3 \Rightarrow P(\text{win }Y) = 0.7\]
Also \(P(\text{not win }X) = 0.5\).
(a) Wins both contracts
\[P(X\cap Y) = 0.5\times 0.7 = 0.35\]
(b) Wins exactly one contract
Either win \(X\) and lose \(Y\), or lose \(X\) and win \(Y\):
\[P = (0.5\times 0.3) + (0.5\times 0.7) = 0.15 + 0.35 = 0.50\]
(c) Wins neither contract
\[P(\text{not }X\cap\text{not }Y) = 0.5\times 0.3 = 0.15\]
Check: \(0.35 + 0.50 + 0.15 = 1.00\), confirming all outcomes are accounted for.
Akọwa Nkọwa
Setup. The contracts are independent. Given:
\[P(\text{win }X) = 0.5,\qquad P(\text{not win }Y) = 0.3 \Rightarrow P(\text{win }Y) = 0.7\]
Also \(P(\text{not win }X) = 0.5\).
(a) Wins both contracts
\[P(X\cap Y) = 0.5\times 0.7 = 0.35\]
(b) Wins exactly one contract
Either win \(X\) and lose \(Y\), or lose \(X\) and win \(Y\):
\[P = (0.5\times 0.3) + (0.5\times 0.7) = 0.15 + 0.35 = 0.50\]
(c) Wins neither contract
\[P(\text{not }X\cap\text{not }Y) = 0.5\times 0.3 = 0.15\]
Check: \(0.35 + 0.50 + 0.15 = 1.00\), confirming all outcomes are accounted for.
Ajụjụ 8 Ripọtì
(a) Solve the simultaneous equation : \(\frac{1}{x} + \frac{1}{y} = 5 ; \frac{1}{y} - \frac{1}{x} = 1\).
(b) A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.
(a) Solve \(\dfrac{1}{x}+\dfrac{1}{y}=5\) and \(\dfrac{1}{y}-\dfrac{1}{x}=1\)
Let \(u=\dfrac{1}{x}\) and \(v=\dfrac{1}{y}\). Then
\[u+v = 5 \quad\text{and}\quad v-u = 1\]
Add the two equations:
\[2v = 6 \;\Rightarrow\; v = 3 \;\Rightarrow\; \frac{1}{y}=3 \;\Rightarrow\; y = \frac{1}{3}\]
Then \(u = 5 - v = 2\), so \(\dfrac{1}{x}=2 \Rightarrow x = \dfrac{1}{2}\).
\(x = \tfrac{1}{2},\; y = \tfrac{1}{3}\).
(b) Length of good surface
Total distance \(48\) km in \(45\) minutes \(= 0.75\) h. Let good surface \(= g\) km at \(72\) km/h and bad surface \(= b\) km at \(48\) km/h.
Distance: \(g + b = 48\).
Time: \(\dfrac{g}{72} + \dfrac{b}{48} = 0.75\). Multiply through by \(144\):
\[2g + 3b = 108\]
Substitute \(b = 48 - g\):
\[2g + 3(48 - g) = 108 \;\Rightarrow\; 2g + 144 - 3g = 108 \;\Rightarrow\; -g = -36\]
\[g = 36\]
The good surface is 36 km.
Akọwa Nkọwa
(a) Solve \(\dfrac{1}{x}+\dfrac{1}{y}=5\) and \(\dfrac{1}{y}-\dfrac{1}{x}=1\)
Let \(u=\dfrac{1}{x}\) and \(v=\dfrac{1}{y}\). Then
\[u+v = 5 \quad\text{and}\quad v-u = 1\]
Add the two equations:
\[2v = 6 \;\Rightarrow\; v = 3 \;\Rightarrow\; \frac{1}{y}=3 \;\Rightarrow\; y = \frac{1}{3}\]
Then \(u = 5 - v = 2\), so \(\dfrac{1}{x}=2 \Rightarrow x = \dfrac{1}{2}\).
\(x = \tfrac{1}{2},\; y = \tfrac{1}{3}\).
(b) Length of good surface
Total distance \(48\) km in \(45\) minutes \(= 0.75\) h. Let good surface \(= g\) km at \(72\) km/h and bad surface \(= b\) km at \(48\) km/h.
Distance: \(g + b = 48\).
Time: \(\dfrac{g}{72} + \dfrac{b}{48} = 0.75\). Multiply through by \(144\):
\[2g + 3b = 108\]
Substitute \(b = 48 - g\):
\[2g + 3(48 - g) = 108 \;\Rightarrow\; 2g + 144 - 3g = 108 \;\Rightarrow\; -g = -36\]
\[g = 36\]
The good surface is 36 km.
Ajụjụ 9 Ripọtì
(a) Copy and complete the table of values for the relation \(y = 2 \sin x + 1\)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° |
| y | 1.0 | 2.7 | 0.0 | -0.7 |
(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 2 \sin x + 1, 0° \leq x \leq 270°\).
(c) Use the graph to find the values of x for which \(\sin x = \frac{1}{4}\).
(a) For \(y=2\sin x+1\), the completed table is:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | 1.0 | 2.0 | 2.7 | 3.0 | 2.7 | 2.0 | 1.0 | 0.0 | -0.7 | -1.0 |
For example, \(y(60^\circ)=2\sin60^\circ+1=2(0.866)+1\approx2.7\).
(b) The required graph, plotted using the stated scales, is shown below.
(c) If \(\sin x=\frac14\), then
\[y=2\sin x+1=2\left(\frac14\right)+1=1.5.\]
Draw the horizontal line \(y=1.5\) and read its two points of intersection with the curve. Hence,
\[\boxed{x\approx15^\circ\text{ or }168^\circ}.\]
Akọwa Nkọwa
(a) For \(y=2\sin x+1\), the completed table is:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | 1.0 | 2.0 | 2.7 | 3.0 | 2.7 | 2.0 | 1.0 | 0.0 | -0.7 | -1.0 |
For example, \(y(60^\circ)=2\sin60^\circ+1=2(0.866)+1\approx2.7\).
(b) The required graph, plotted using the stated scales, is shown below.
(c) If \(\sin x=\frac14\), then
\[y=2\sin x+1=2\left(\frac14\right)+1=1.5.\]
Draw the horizontal line \(y=1.5\) and read its two points of intersection with the curve. Hence,
\[\boxed{x\approx15^\circ\text{ or }168^\circ}.\]
Ajụjụ 10 Ripọtì
(a) 
In the Venn diagram, P, Qand R are subsets of the universal set U. If n(U) = 125, find : (i) the value of x ; (ii) n(\(P \cup Q \cap R'\)).
(b)
In the diagram, O is the centre of the circle. If WX is parallel to YZ and < WXY = 50°, find the value of (i) , WYZ
(ii) < YEZ.
(a)(i) Adding all the regions of the Venn diagram and equating to \( n(U) = 125 \):
\[ (16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125 \]
\[ 45 + 20x = 125 \implies 20x = 80 \implies x = 4 \]
(a)(ii) \( n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x) = 22 + 4x \)
\[ = 22 + 4(4) = 22 + 16 = \mathbf{38} \]
(b) XW is a diameter, so \( \angle XYW = 90^\circ \) (angle in a semicircle).
In triangle XYW: \( \angle XWY + 90^\circ + 50^\circ = 180^\circ \), so \( \angle XWY = 40^\circ \).
(i) Since \( WX \parallel YZ \), \( \angle WYZ = \angle XWY = \mathbf{40^\circ} \) (alternate angles).
(ii) \( \angle WOZ = 2 \times \angle WYZ = 2 \times 40^\circ = 80^\circ \) (angle at centre is twice angle at circumference). In triangle WEO, \( \angle WEO = 180^\circ - (40^\circ + 80^\circ) = 60^\circ \), therefore \( \angle YEZ = \mathbf{60^\circ} \) (vertically opposite angles).
Akọwa Nkọwa
(a)(i) Adding all the regions of the Venn diagram and equating to \( n(U) = 125 \):
\[ (16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125 \]
\[ 45 + 20x = 125 \implies 20x = 80 \implies x = 4 \]
(a)(ii) \( n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x) = 22 + 4x \)
\[ = 22 + 4(4) = 22 + 16 = \mathbf{38} \]
(b) XW is a diameter, so \( \angle XYW = 90^\circ \) (angle in a semicircle).
In triangle XYW: \( \angle XWY + 90^\circ + 50^\circ = 180^\circ \), so \( \angle XWY = 40^\circ \).
(i) Since \( WX \parallel YZ \), \( \angle WYZ = \angle XWY = \mathbf{40^\circ} \) (alternate angles).
(ii) \( \angle WOZ = 2 \times \angle WYZ = 2 \times 40^\circ = 80^\circ \) (angle at centre is twice angle at circumference). In triangle WEO, \( \angle WEO = 180^\circ - (40^\circ + 80^\circ) = 60^\circ \), therefore \( \angle YEZ = \mathbf{60^\circ} \) (vertically opposite angles).
Ajụjụ 11 Ripọtì
| Scores | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | 13 | 11 | 9 | 10 |
The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the :
(a) Mean deviation of the distribution ; (b) probability that a score selected at random is at least a 4.
(a) Mean deviation. First find the mean.
| Score (x) | 1 | 2 | 3 | 4 | 5 | 6 | Total |
|---|---|---|---|---|---|---|---|
| Freq (f) | 2 | 5 | 13 | 11 | 9 | 10 | 50 |
| fx | 2 | 10 | 39 | 44 | 45 | 60 | 200 |
| x | \(|x-\bar{x}|\) | \(f|x-\bar{x}|\) |
|---|---|---|
| 1 | 3 | 6 |
| 2 | 2 | 10 |
| 3 | 1 | 13 |
| 4 | 0 | 0 |
| 5 | 1 | 9 |
| 6 | 2 | 20 |
| Total | 58 |
(b) Probability of at least a 4. Scores 4, 5 or 6:
\[P(\ge 4) = \frac{11 + 9 + 10}{50} = \frac{30}{50} = \frac{3}{5} = 0.6\]Akọwa Nkọwa
(a) Mean deviation. First find the mean.
| Score (x) | 1 | 2 | 3 | 4 | 5 | 6 | Total |
|---|---|---|---|---|---|---|---|
| Freq (f) | 2 | 5 | 13 | 11 | 9 | 10 | 50 |
| fx | 2 | 10 | 39 | 44 | 45 | 60 | 200 |
| x | \(|x-\bar{x}|\) | \(f|x-\bar{x}|\) |
|---|---|---|
| 1 | 3 | 6 |
| 2 | 2 | 10 |
| 3 | 1 | 13 |
| 4 | 0 | 0 |
| 5 | 1 | 9 |
| 6 | 2 | 20 |
| Total | 58 |
(b) Probability of at least a 4. Scores 4, 5 or 6:
\[P(\ge 4) = \frac{11 + 9 + 10}{50} = \frac{30}{50} = \frac{3}{5} = 0.6\]Ajụjụ 12 Ripọtì
(a) 
In the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\(cm^{2}\), calculate the value of r. [Take \(\pi = \frac{22}{7}\)].
(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \(< PQR = 66°\) and \(< PSQ = 109°\), calculate the value of \(< RQS\).
(a) Shaded segment of the circle.
From the diagram, \(O\) is the centre, \(OX=OY=r\), \(\angle XOY=90^\circ\), and the shaded part is the segment lying between chord \(XY\) and the arc \(XKY\).
\[\text{Shaded area}=\text{area of sector } XOY-\text{area of }\triangle XOY.\]
Sector \(XOY\) subtends \(90^\circ\), so it is a quarter of the circle:
\[\text{sector}=\frac{90}{360}\pi r^{2}=\frac{1}{4}\pi r^{2}.\]
Triangle \(XOY\) is right-angled at \(O\) with legs \(r\) and \(r\):
\[\triangle XOY=\frac{1}{2}\,r\times r=\frac{1}{2}r^{2}.\]
Hence
\[504=\frac{1}{4}\pi r^{2}-\frac{1}{2}r^{2}=r^{2}\left(\frac{1}{4}\times\frac{22}{7}-\frac{1}{2}\right)=r^{2}\left(\frac{22}{28}-\frac{14}{28}\right)=r^{2}\left(\frac{8}{28}\right)=\frac{2}{7}r^{2}.\]
\[r^{2}=504\times\frac{7}{2}=1764\quad\Rightarrow\quad r=\sqrt{1764}=\boxed{42\text{ cm}}.\]
(b) Two isosceles triangles on a common base PQ.
Triangles \(PQR\) and \(PQS\) stand on opposite sides of the common base \(PQ\).
In \(\triangle PQR\) (isosceles with \(RP=RQ\)), the base angles are equal, and \(\angle PQR=66^\circ\) is a base angle, so
\[\angle RPQ=\angle RQP=66^\circ.\]
In \(\triangle PQS\) (isosceles with \(SP=SQ\)), \(\angle PSQ=109^\circ\) is the apex angle, so the base angles are
\[\angle SPQ=\angle SQP=\frac{180^\circ-109^\circ}{2}=\frac{71^\circ}{2}=35.5^\circ.\]
Since \(R\) and \(S\) are on opposite sides of \(PQ\), the angle \(RQS\) is the sum of the two base angles at \(Q\):
\[\angle RQS=\angle RQP+\angle PQS=66^\circ+35.5^\circ=\boxed{101.5^\circ}.\]
Akọwa Nkọwa
(a) Shaded segment of the circle.
From the diagram, \(O\) is the centre, \(OX=OY=r\), \(\angle XOY=90^\circ\), and the shaded part is the segment lying between chord \(XY\) and the arc \(XKY\).
\[\text{Shaded area}=\text{area of sector } XOY-\text{area of }\triangle XOY.\]
Sector \(XOY\) subtends \(90^\circ\), so it is a quarter of the circle:
\[\text{sector}=\frac{90}{360}\pi r^{2}=\frac{1}{4}\pi r^{2}.\]
Triangle \(XOY\) is right-angled at \(O\) with legs \(r\) and \(r\):
\[\triangle XOY=\frac{1}{2}\,r\times r=\frac{1}{2}r^{2}.\]
Hence
\[504=\frac{1}{4}\pi r^{2}-\frac{1}{2}r^{2}=r^{2}\left(\frac{1}{4}\times\frac{22}{7}-\frac{1}{2}\right)=r^{2}\left(\frac{22}{28}-\frac{14}{28}\right)=r^{2}\left(\frac{8}{28}\right)=\frac{2}{7}r^{2}.\]
\[r^{2}=504\times\frac{7}{2}=1764\quad\Rightarrow\quad r=\sqrt{1764}=\boxed{42\text{ cm}}.\]
(b) Two isosceles triangles on a common base PQ.
Triangles \(PQR\) and \(PQS\) stand on opposite sides of the common base \(PQ\).
In \(\triangle PQR\) (isosceles with \(RP=RQ\)), the base angles are equal, and \(\angle PQR=66^\circ\) is a base angle, so
\[\angle RPQ=\angle RQP=66^\circ.\]
In \(\triangle PQS\) (isosceles with \(SP=SQ\)), \(\angle PSQ=109^\circ\) is the apex angle, so the base angles are
\[\angle SPQ=\angle SQP=\frac{180^\circ-109^\circ}{2}=\frac{71^\circ}{2}=35.5^\circ.\]
Since \(R\) and \(S\) are on opposite sides of \(PQ\), the angle \(RQS\) is the sum of the two base angles at \(Q\):
\[\angle RQS=\angle RQP+\angle PQS=66^\circ+35.5^\circ=\boxed{101.5^\circ}.\]
Ajụjụ 13 Ripọtì
(a) Two functions, f and g, are defined by \(f : x \to 2x^{2} - 1\) and \(g : x \to 3x + 2\) where x is a real number.
(i) If \(f(x - 1) - 7 = 0\), find the values of x.
(ii) Evaluate : \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)}\).
(b) An operation, \((\ast)\) is defined on the set R, of real numbers, by \(m \ast n = \frac{-n}{m^{2} + 1}\), where \(m, n \in R\). If \(-3, -10 \in R\), show whether or not \(\ast\) is commutative.
(a)(i) \(f(x) = 2x^2 - 1\), so \(f(x-1) = 2(x-1)^2 - 1\).
\[2(x-1)^2 - 1 - 7 = 0 \;\Rightarrow\; 2(x-1)^2 = 8 \;\Rightarrow\; (x-1)^2 = 4\]
\[x - 1 = \pm 2 \;\Rightarrow\; x = 3 \text{ or } x = -1\]
(ii) Evaluate each term:
- \(f\!\left(-\tfrac12\right) = 2\left(\tfrac14\right) - 1 = -\tfrac12\)
- \(g(3) = 3(3) + 2 = 11\)
- \(f(4) = 2(16) - 1 = 31\)
- \(g(5) = 3(5) + 2 = 17\)
\[\frac{f\!\left(-\tfrac12\right)\cdot g(3)}{f(4) - g(5)} = \frac{\left(-\tfrac12\right)(11)}{31 - 17} = \frac{-\tfrac{11}{2}}{14} = -\frac{11}{28}\]
(b) \(m \ast n = \dfrac{-n}{m^2 + 1}\). Test with \(m = -3,\, n = -10\):
\[(-3)\ast(-10) = \frac{-(-10)}{(-3)^2 + 1} = \frac{10}{10} = 1\]
\[(-10)\ast(-3) = \frac{-(-3)}{(-10)^2 + 1} = \frac{3}{101}\]
Since \(1 \neq \dfrac{3}{101}\), we have \((-3)\ast(-10) \neq (-10)\ast(-3)\). Therefore the operation \(\ast\) is not commutative.
Akọwa Nkọwa
(a)(i) \(f(x) = 2x^2 - 1\), so \(f(x-1) = 2(x-1)^2 - 1\).
\[2(x-1)^2 - 1 - 7 = 0 \;\Rightarrow\; 2(x-1)^2 = 8 \;\Rightarrow\; (x-1)^2 = 4\]
\[x - 1 = \pm 2 \;\Rightarrow\; x = 3 \text{ or } x = -1\]
(ii) Evaluate each term:
- \(f\!\left(-\tfrac12\right) = 2\left(\tfrac14\right) - 1 = -\tfrac12\)
- \(g(3) = 3(3) + 2 = 11\)
- \(f(4) = 2(16) - 1 = 31\)
- \(g(5) = 3(5) + 2 = 17\)
\[\frac{f\!\left(-\tfrac12\right)\cdot g(3)}{f(4) - g(5)} = \frac{\left(-\tfrac12\right)(11)}{31 - 17} = \frac{-\tfrac{11}{2}}{14} = -\frac{11}{28}\]
(b) \(m \ast n = \dfrac{-n}{m^2 + 1}\). Test with \(m = -3,\, n = -10\):
\[(-3)\ast(-10) = \frac{-(-10)}{(-3)^2 + 1} = \frac{10}{10} = 1\]
\[(-10)\ast(-3) = \frac{-(-3)}{(-10)^2 + 1} = \frac{3}{101}\]
Since \(1 \neq \dfrac{3}{101}\), we have \((-3)\ast(-10) \neq (-10)\ast(-3)\). Therefore the operation \(\ast\) is not commutative.
