WAEC SSCE - General Mathematics - 2000 (Essay)
Question 1 Report
(a) Given the expression \(y = ax^{2} - bx - 12\) , find the values of x when a = 1, b = 2 and y = 3.
(b) If \(\sqrt{x^{2} + 1} = \frac{5}{4}\), find the positive value of x.
(a) Substitute \(a=1,\ b=2,\ y=3\) into \(y = ax^2 - bx - 12\):
\[3 = x^2 - 2x - 12 \Rightarrow x^2 - 2x - 15 = 0.\]
Factorise: \((x - 5)(x + 3) = 0\), giving
\[x = 5\quad\text{or}\quad x = -3.\]
(b) Square both sides of \(\sqrt{x^2 + 1} = \tfrac{5}{4}\):
\[x^2 + 1 = \frac{25}{16} \Rightarrow x^2 = \frac{25}{16} - 1 = \frac{9}{16}.\]
\[x = \frac{3}{4}\quad(\text{positive value}).\]
Answer Details
(a) Substitute \(a=1,\ b=2,\ y=3\) into \(y = ax^2 - bx - 12\):
\[3 = x^2 - 2x - 12 \Rightarrow x^2 - 2x - 15 = 0.\]
Factorise: \((x - 5)(x + 3) = 0\), giving
\[x = 5\quad\text{or}\quad x = -3.\]
(b) Square both sides of \(\sqrt{x^2 + 1} = \tfrac{5}{4}\):
\[x^2 + 1 = \frac{25}{16} \Rightarrow x^2 = \frac{25}{16} - 1 = \frac{9}{16}.\]
\[x = \frac{3}{4}\quad(\text{positive value}).\]
Question 2 Report
(a) A man earns N150,000 per annum. He is allowed a tax free pay on N40,000. If he pays 25 kobo in the naira as tax on his taxable income, how much has he left?
(b) A bookshop has 650 copies of a book for sale. The books were marked at N75 per copy in order to make a profit of 30%. A bookseller bought 300 copies at 5% discount. If the remaining copies are sold at N75 each, calculate the percentage profit the bookshop would make on the whole.
(a) Tax payable. Taxable income \(= 150{,}000 - 40{,}000 = \text{N}110{,}000\). Tax at 25 kobo (\(\text{N}0.25\)) per naira:
\[\text{Tax} = 0.25\times110{,}000 = \text{N}27{,}500.\]
\[\text{Amount left} = 150{,}000 - 27{,}500 = \text{N}122{,}500.\]
(b) Percentage profit on the whole. The 650 copies are marked at N75 to yield 30% profit, so the total cost price is
\[\text{CP} = \frac{650\times75}{1.30} = \frac{48{,}750}{1.30} = \text{N}37{,}500.\]
Sales received: 300 copies at 5% discount \(= 300\times(75\times0.95) = 300\times71.25 = \text{N}21{,}375\); the other 350 copies at N75 \(= 350\times75 = \text{N}26{,}250\).
\[\text{Total sales} = 21{,}375 + 26{,}250 = \text{N}47{,}625.\]
\[\text{Profit} = 47{,}625 - 37{,}500 = \text{N}10{,}125.\]
\[\%\text{ profit} = \frac{10{,}125}{37{,}500}\times100 = 27\%.\]
Answer Details
(a) Tax payable. Taxable income \(= 150{,}000 - 40{,}000 = \text{N}110{,}000\). Tax at 25 kobo (\(\text{N}0.25\)) per naira:
\[\text{Tax} = 0.25\times110{,}000 = \text{N}27{,}500.\]
\[\text{Amount left} = 150{,}000 - 27{,}500 = \text{N}122{,}500.\]
(b) Percentage profit on the whole. The 650 copies are marked at N75 to yield 30% profit, so the total cost price is
\[\text{CP} = \frac{650\times75}{1.30} = \frac{48{,}750}{1.30} = \text{N}37{,}500.\]
Sales received: 300 copies at 5% discount \(= 300\times(75\times0.95) = 300\times71.25 = \text{N}21{,}375\); the other 350 copies at N75 \(= 350\times75 = \text{N}26{,}250\).
\[\text{Total sales} = 21{,}375 + 26{,}250 = \text{N}47{,}625.\]
\[\text{Profit} = 47{,}625 - 37{,}500 = \text{N}10{,}125.\]
\[\%\text{ profit} = \frac{10{,}125}{37{,}500}\times100 = 27\%.\]
Question 3 Report
(a) Given that \(\cos x = 0.7431, 0° < x < 90°\), use tables to find the values of : (i) \(2 \sin x\) ; (ii) \(\tan \frac{x}{2}\).
(b) The interior angles of a pentagon are in ratio 2 : 3 : 4 : 4 : 5. Find the value of the largest angle.
Answer Details
None
Question 4 Report
(a) A cylindrical well of radius 1 metre is dug out to a depth of 8 metres. (i) calculate, in m\(^{3}\), the volume of soil dug out ; (ii) if the soil is used to raise the level of rectangular floor of a room 4m by 12m, calculate, correct to the nearest cm, the thickness of the new layer of soil. [Take \(\pi = \frac{22}{7}\)].
(b) 
The diagram shows a quadrilateral ABCD in which < DAB is a right- angle. |AB| = 3.3 cm, |BC| = 3.9 cm, |CD| = 5.6 cm. (i) find the length of BD. (ii) show that < BCD = 90°.
(a)(i) Volume of soil dug from the well.
The well is a cylinder of radius \(r=1\text{ m}\) and depth \(h=8\text{ m}\).
\[V=\pi r^2 h=\frac{22}{7}\times 1^2\times 8=\frac{176}{7}=25.14\text{ m}^3\]The volume of soil dug out is \(25.14\text{ m}^3\) (to 2 d.p.).
(a)(ii) Thickness of the new layer of soil.
The soil is spread over a rectangular floor \(4\text{ m}\times 12\text{ m}\).
\[\text{Floor area}=4\times 12=48\text{ m}^2\]Let the thickness be \(t\). Volume of the layer = floor area \(\times\) thickness, and this equals the volume of soil:
\[48\times t=\frac{176}{7}\]\[t=\frac{176}{7\times 48}=\frac{176}{336}=0.5238\text{ m}\]Converting to centimetres: \(0.5238\times 100=52.38\text{ cm}\).
Correct to the nearest cm, the thickness is \(52\text{ cm}\).
(b) Quadrilateral \(ABCD\).
From the diagram, \(\angle DAB=90^{\circ}\), \(|AB|=3.3\text{ cm}\), \(|AD|=5.6\text{ cm}\), \(|BC|=3.9\text{ cm}\) and \(|DC|=5.2\text{ cm}\).
(i) Length of \(BD\).
Triangle \(ABD\) is right-angled at \(A\), so by Pythagoras:
\[BD^2=AB^2+AD^2=3.3^2+5.6^2=10.89+31.36=42.25\]\[BD=\sqrt{42.25}=6.5\text{ cm}\](ii) Show that \(\angle BCD=90^{\circ}\).
In triangle \(BCD\), \(|BC|=3.9\), \(|DC|=5.2\) and \(|BD|=6.5\). Test the converse of Pythagoras:
\[BC^2+DC^2=3.9^2+5.2^2=15.21+27.04=42.25\]\[BD^2=6.5^2=42.25\]Since \(BC^2+DC^2=BD^2\), the converse of Pythagoras theorem holds, so the angle opposite \(BD\) is a right angle. Therefore \(\angle BCD=90^{\circ}\).
Answer Details
(a)(i) Volume of soil dug from the well.
The well is a cylinder of radius \(r=1\text{ m}\) and depth \(h=8\text{ m}\).
\[V=\pi r^2 h=\frac{22}{7}\times 1^2\times 8=\frac{176}{7}=25.14\text{ m}^3\]The volume of soil dug out is \(25.14\text{ m}^3\) (to 2 d.p.).
(a)(ii) Thickness of the new layer of soil.
The soil is spread over a rectangular floor \(4\text{ m}\times 12\text{ m}\).
\[\text{Floor area}=4\times 12=48\text{ m}^2\]Let the thickness be \(t\). Volume of the layer = floor area \(\times\) thickness, and this equals the volume of soil:
\[48\times t=\frac{176}{7}\]\[t=\frac{176}{7\times 48}=\frac{176}{336}=0.5238\text{ m}\]Converting to centimetres: \(0.5238\times 100=52.38\text{ cm}\).
Correct to the nearest cm, the thickness is \(52\text{ cm}\).
(b) Quadrilateral \(ABCD\).
From the diagram, \(\angle DAB=90^{\circ}\), \(|AB|=3.3\text{ cm}\), \(|AD|=5.6\text{ cm}\), \(|BC|=3.9\text{ cm}\) and \(|DC|=5.2\text{ cm}\).
(i) Length of \(BD\).
Triangle \(ABD\) is right-angled at \(A\), so by Pythagoras:
\[BD^2=AB^2+AD^2=3.3^2+5.6^2=10.89+31.36=42.25\]\[BD=\sqrt{42.25}=6.5\text{ cm}\](ii) Show that \(\angle BCD=90^{\circ}\).
In triangle \(BCD\), \(|BC|=3.9\), \(|DC|=5.2\) and \(|BD|=6.5\). Test the converse of Pythagoras:
\[BC^2+DC^2=3.9^2+5.2^2=15.21+27.04=42.25\]\[BD^2=6.5^2=42.25\]Since \(BC^2+DC^2=BD^2\), the converse of Pythagoras theorem holds, so the angle opposite \(BD\) is a right angle. Therefore \(\angle BCD=90^{\circ}\).
Question 5 Report
(a) The mean of 1, 2, x, 11, y, 14, arranged in ascending order, is 8 and the median is 9. Find the values of x and y.
(b) 
In the diagram, MN || PQ, |LM| = 3cm and |LP| = 4cm. If the area of \(\Delta\) LMN is 18\(cm^{2}\), find the area of the quadrilateral MPQN.
(a) Find x and y.
The six numbers in ascending order are \(1,\;2,\;x,\;11,\;y,\;14\).
Use the mean. The mean of the six values is 8, so their sum is \(6\times 8=48\):
\[1+2+x+11+y+14=48\]
\[28+x+y=48\Rightarrow x+y=20\quad(1)\]
Use the median. For six values the median is the average of the 3rd and 4th values, which are \(x\) and \(11\):
\[\frac{x+11}{2}=9\Rightarrow x+11=18\Rightarrow x=7\]
Substitute into (1):
\[7+y=20\Rightarrow y=13\]
Check ascending order: \(1,2,7,11,13,14\) is valid. So \(x=7,\;y=13\).
(b) Area of quadrilateral MPQN.
Since \(MN\parallel PQ\), triangles \(LMN\) and \(LPQ\) are similar (equal angles at \(L\), and corresponding angles equal).
From the diagram, \(|LM|=3\text{ cm}\) and \(|MP|=1\text{ cm}\), so:
\[|LP|=|LM|+|MP|=3+1=4\text{ cm}\]
The ratio of corresponding sides is:
\[\frac{|LM|}{|LP|}=\frac{3}{4}\]
The ratio of areas of similar triangles is the square of the ratio of sides:
\[\frac{\text{Area }\Delta LMN}{\text{Area }\Delta LPQ}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\]
Given \(\text{Area }\Delta LMN=18\text{ cm}^{2}\):
\[18=\frac{9}{16}\times\text{Area }\Delta LPQ\]
\[\text{Area }\Delta LPQ=18\times\frac{16}{9}=32\text{ cm}^{2}\]
The quadrilateral \(MPQN\) is the region between the two parallel lines:
\[\text{Area }MPQN=\text{Area }\Delta LPQ-\text{Area }\Delta LMN=32-18=14\text{ cm}^{2}\]
Answer Details
(a) Find x and y.
The six numbers in ascending order are \(1,\;2,\;x,\;11,\;y,\;14\).
Use the mean. The mean of the six values is 8, so their sum is \(6\times 8=48\):
\[1+2+x+11+y+14=48\]
\[28+x+y=48\Rightarrow x+y=20\quad(1)\]
Use the median. For six values the median is the average of the 3rd and 4th values, which are \(x\) and \(11\):
\[\frac{x+11}{2}=9\Rightarrow x+11=18\Rightarrow x=7\]
Substitute into (1):
\[7+y=20\Rightarrow y=13\]
Check ascending order: \(1,2,7,11,13,14\) is valid. So \(x=7,\;y=13\).
(b) Area of quadrilateral MPQN.
Since \(MN\parallel PQ\), triangles \(LMN\) and \(LPQ\) are similar (equal angles at \(L\), and corresponding angles equal).
From the diagram, \(|LM|=3\text{ cm}\) and \(|MP|=1\text{ cm}\), so:
\[|LP|=|LM|+|MP|=3+1=4\text{ cm}\]
The ratio of corresponding sides is:
\[\frac{|LM|}{|LP|}=\frac{3}{4}\]
The ratio of areas of similar triangles is the square of the ratio of sides:
\[\frac{\text{Area }\Delta LMN}{\text{Area }\Delta LPQ}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\]
Given \(\text{Area }\Delta LMN=18\text{ cm}^{2}\):
\[18=\frac{9}{16}\times\text{Area }\Delta LPQ\]
\[\text{Area }\Delta LPQ=18\times\frac{16}{9}=32\text{ cm}^{2}\]
The quadrilateral \(MPQN\) is the region between the two parallel lines:
\[\text{Area }MPQN=\text{Area }\Delta LPQ-\text{Area }\Delta LMN=32-18=14\text{ cm}^{2}\]
Question 6 Report
(a) Evaluate and express your answer in standard form : \(\frac{4.56 \times 3.6}{0.12}\)
(b) Without using mathematical tables or calculator, evaluate \((73.8)^{2} - (26.2)^{2}\).
(c) Simplify \(\sqrt{1\frac{19}{81}}\), expressing your answer in the form \(\frac{a}{b}\) where a and b are positive integers.
(a)
\[\frac{4.56\times3.6}{0.12} = \frac{16.416}{0.12} = 136.8 = 1.368\times10^{2}.\]
(b) Use the difference of two squares \(a^2 - b^2 = (a-b)(a+b)\):
\[(73.8)^2 - (26.2)^2 = (73.8 - 26.2)(73.8 + 26.2) = 47.6\times100 = 4760.\]
(c) Convert the mixed number to an improper fraction:
\[1\tfrac{19}{81} = \frac{81 + 19}{81} = \frac{100}{81},\qquad \sqrt{\frac{100}{81}} = \frac{10}{9}.\]
Answer Details
(a)
\[\frac{4.56\times3.6}{0.12} = \frac{16.416}{0.12} = 136.8 = 1.368\times10^{2}.\]
(b) Use the difference of two squares \(a^2 - b^2 = (a-b)(a+b)\):
\[(73.8)^2 - (26.2)^2 = (73.8 - 26.2)(73.8 + 26.2) = 47.6\times100 = 4760.\]
(c) Convert the mixed number to an improper fraction:
\[1\tfrac{19}{81} = \frac{81 + 19}{81} = \frac{100}{81},\qquad \sqrt{\frac{100}{81}} = \frac{10}{9}.\]
Question 7 Report
(a) The first term of an Arithmetic Progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5 : 8. Calculate the common difference of the progression.
(b) A sphere of radius 2 cm is of mass 11.2g. Find (i) the volume of the sphere ; (ii) the density of the sphere ; (iii) the mass of a sphere of the same material but with radius 3cm. [Take \(\pi = \frac{22}{7}\)].
Question 8 Report
(a) A surveyor walks 100m up a hill which slopes at an angle of 24° to the horizontal. Calculate, correct to the nearest metre, the height through which he rises.
(b) 
In the diagram, ABC is an isosceles triangle. |AB| = |AC| = 5 cm, and |BC| = 8 cm. Calculate, correct to the nearest degree, < BAC.
(c) Two boats, 70 metres apart and on opposite sides of a light-house, are in a straight line with the light-house. The angles of elevation of the top of the light-house from the two boats are 71.6° and 45°. Find the height of the light-house. [Take \(\tan 71.6° = 3\)].
(a) Height risen up the slope
The 100 m is the distance along the slope (the hypotenuse), and the height risen is the vertical (opposite) side of a right triangle whose angle to the horizontal is \(24^\circ\).
Let \(h\) be the height.
\[\sin 24^\circ = \frac{h}{100}\]
\[h = 100 \times \sin 24^\circ = 100 \times 0.4067 = 40.67\text{ m}\]
Correct to the nearest metre, \(h \approx \mathbf{41\text{ m}}\).
(b) Angle BAC of the isosceles triangle
From the diagram, \(|AB| = |AC| = 5\text{ cm}\) and \(|BC| = 8\text{ cm}\). Drop a perpendicular from \(A\) to the midpoint \(M\) of \(BC\). Then \(|BM| = \tfrac{1}{2}\times 8 = 4\text{ cm}\), and \(AM\) bisects \(\angle BAC\).
In right triangle \(ABM\):
\[\sin(\angle BAM) = \frac{BM}{AB} = \frac{4}{5} = 0.8\]
\[\angle BAM = \sin^{-1}(0.8) = 53.13^\circ\]
Therefore
\[\angle BAC = 2 \times 53.13^\circ = 106.26^\circ \approx \mathbf{106^\circ}\]
(c) Height of the light-house
The two boats are on opposite sides of the light-house and in a straight line with its foot, 70 m apart. Let the foot of the light-house be \(F\), the height be \(h\), the horizontal distance to the boat with elevation \(71.6^\circ\) be \(d_1\), and to the boat with elevation \(45^\circ\) be \(d_2\).
\[d_1 + d_2 = 70\]
From the \(71.6^\circ\) boat: \(\tan 71.6^\circ = \dfrac{h}{d_1} = 3\), so \(d_1 = \dfrac{h}{3}\).
From the \(45^\circ\) boat: \(\tan 45^\circ = \dfrac{h}{d_2} = 1\), so \(d_2 = h\).
Substitute:
\[\frac{h}{3} + h = 70\]
\[\frac{h + 3h}{3} = 70 \quad\Rightarrow\quad \frac{4h}{3} = 70\]
\[h = \frac{70 \times 3}{4} = \frac{210}{4} = 52.5\text{ m}\]
The height of the light-house is \(\mathbf{52.5\text{ m}}\).
Answer Details
(a) Height risen up the slope
The 100 m is the distance along the slope (the hypotenuse), and the height risen is the vertical (opposite) side of a right triangle whose angle to the horizontal is \(24^\circ\).
Let \(h\) be the height.
\[\sin 24^\circ = \frac{h}{100}\]
\[h = 100 \times \sin 24^\circ = 100 \times 0.4067 = 40.67\text{ m}\]
Correct to the nearest metre, \(h \approx \mathbf{41\text{ m}}\).
(b) Angle BAC of the isosceles triangle
From the diagram, \(|AB| = |AC| = 5\text{ cm}\) and \(|BC| = 8\text{ cm}\). Drop a perpendicular from \(A\) to the midpoint \(M\) of \(BC\). Then \(|BM| = \tfrac{1}{2}\times 8 = 4\text{ cm}\), and \(AM\) bisects \(\angle BAC\).
In right triangle \(ABM\):
\[\sin(\angle BAM) = \frac{BM}{AB} = \frac{4}{5} = 0.8\]
\[\angle BAM = \sin^{-1}(0.8) = 53.13^\circ\]
Therefore
\[\angle BAC = 2 \times 53.13^\circ = 106.26^\circ \approx \mathbf{106^\circ}\]
(c) Height of the light-house
The two boats are on opposite sides of the light-house and in a straight line with its foot, 70 m apart. Let the foot of the light-house be \(F\), the height be \(h\), the horizontal distance to the boat with elevation \(71.6^\circ\) be \(d_1\), and to the boat with elevation \(45^\circ\) be \(d_2\).
\[d_1 + d_2 = 70\]
From the \(71.6^\circ\) boat: \(\tan 71.6^\circ = \dfrac{h}{d_1} = 3\), so \(d_1 = \dfrac{h}{3}\).
From the \(45^\circ\) boat: \(\tan 45^\circ = \dfrac{h}{d_2} = 1\), so \(d_2 = h\).
Substitute:
\[\frac{h}{3} + h = 70\]
\[\frac{h + 3h}{3} = 70 \quad\Rightarrow\quad \frac{4h}{3} = 70\]
\[h = \frac{70 \times 3}{4} = \frac{210}{4} = 52.5\text{ m}\]
The height of the light-house is \(\mathbf{52.5\text{ m}}\).
Question 9 Report
(a) Copy and complete the following table of values for the relation \(y = x^{2} - 2x - 5\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | -2 | -6 | -2 | 3 | 10 |
(b) Draw the graph of the relation \(y = x^{2} - 2x - 5\); using a scale of 2 cm to 1 unit on the x- axis, and 2 cm to 2 units on the y- axis.
(c) Using the same axes, draw the graph of \(y = 2x + 3\).
(d) Obtain in the form \(ax^{2} + bx + c = 0\) where a, b and c are integers, the equation which is satisfied by the x- coordinate of the points of intersection of the two graphs.
(e) From your graphs, determine the roots of the equation obtained in (d) above.
(a) For \(y=x^2-2x-5\):
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|
| \(y=x^2-2x-5\) | \(10\) | \(3\) | \(-2\) | \(-5\) | \(-6\) | \(-5\) | \(-2\) | \(3\) | \(10\) |
(b) and (c) The required plots of \(y=x^2-2x-5\) and \(y=2x+3\) on the same axes are shown below. The parabola is drawn through the plotted table values, while the straight line is drawn through points such as \((-3,-3)\), \((0,3)\), \((3,9)\) and \((6,15)\).
(d) At each point of intersection, the two \(y\)-values are equal:
\[x^2-4x-8=0.\]
Thus, the required equation is \(\boxed{x^2-4x-8=0}\).
(e) Reading the \(x\)-coordinates of the points of intersection from the graph gives
(The exact values are \(x=2\pm2\sqrt3\), i.e. \(-1.46\) and \(5.46\), respectively.)
Answer Details
(a) For \(y=x^2-2x-5\):
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|
| \(y=x^2-2x-5\) | \(10\) | \(3\) | \(-2\) | \(-5\) | \(-6\) | \(-5\) | \(-2\) | \(3\) | \(10\) |
(b) and (c) The required plots of \(y=x^2-2x-5\) and \(y=2x+3\) on the same axes are shown below. The parabola is drawn through the plotted table values, while the straight line is drawn through points such as \((-3,-3)\), \((0,3)\), \((3,9)\) and \((6,15)\).
(d) At each point of intersection, the two \(y\)-values are equal:
\[x^2-4x-8=0.\]
Thus, the required equation is \(\boxed{x^2-4x-8=0}\).
(e) Reading the \(x\)-coordinates of the points of intersection from the graph gives
(The exact values are \(x=2\pm2\sqrt3\), i.e. \(-1.46\) and \(5.46\), respectively.)
Question 10 Report
(a) Two places X and Y on the equator are on longitudes 67°E and 123°E respectively. (i) What is the distance between them along the equator? (ii) How far from the North pole is X? [Take \(\pi = \frac{22}{7}\) and radius of earth = 6400km].
(b)
I In the diagram, PQR is a circle centre O. N is the mid-point of chord PQ. |PQ| = 8cm, |ON| = 3cm and < ONR = 20°. Calculate the size of < ORN to the nearest degree.
(a) \(X\) is on longitude \(67^\circ E\) and \(Y\) on \(123^\circ E\); both lie on the equator. Take \(\pi=\frac{22}{7}\) and \(R=6400\text{ km}\).
(i) Distance between \(X\) and \(Y\) along the equator. The difference in longitude is
\[\theta=123^\circ-67^\circ=56^\circ.\]Distance along the equator (a great circle):
\[d=\frac{\theta}{360^\circ}\times 2\pi R=\frac{56}{360}\times 2\times\frac{22}{7}\times 6400.\]\[d=\frac{56}{360}\times\frac{2\times22\times6400}{7}=\frac{56}{360}\times 40228.57\]\[d\approx 6257.78\text{ km}\approx\mathbf{6258\text{ km}}.\](ii) Distance of \(X\) from the North pole. The North pole is \(90^\circ\) of latitude from the equator, measured along a meridian (a great circle):
\[d=\frac{90}{360}\times 2\pi R=\frac{1}{4}\times 2\times\frac{22}{7}\times 6400\]\[d=\frac{1}{4}\times 40228.57\approx\mathbf{10057\text{ km}}.\](b) In circle centre \(O\), \(N\) is the mid-point of chord \(PQ\) with \(|PQ|=8\text{ cm}\), \(|ON|=3\text{ cm}\) and \(\angle ONR=20^\circ\). Find \(\angle ORN\).
Since \(N\) is the mid-point of the chord, \(ON\perp PQ\) and
\[|PN|=\tfrac{1}{2}|PQ|=4\text{ cm}.\]The radius is \(|OP|\); by Pythagoras in right triangle \(ONP\):
\[|OP|=\sqrt{|ON|^2+|PN|^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ cm}.\]\(R\) lies on the circle, so \(|OR|=5\text{ cm}\) too. Now apply the sine rule in triangle \(ONR\), where \(|ON|=3\), \(|OR|=5\) and \(\angle ONR=20^\circ\):
\[\frac{\sin\angle ORN}{|ON|}=\frac{\sin\angle ONR}{|OR|}\]\[\sin\angle ORN=\frac{|ON|\sin 20^\circ}{|OR|}=\frac{3\times0.3420}{5}=0.2052\]\[\angle ORN=\sin^{-1}(0.2052)\approx 11.8^\circ\approx\mathbf{12^\circ}.\]Answer Details
(a) \(X\) is on longitude \(67^\circ E\) and \(Y\) on \(123^\circ E\); both lie on the equator. Take \(\pi=\frac{22}{7}\) and \(R=6400\text{ km}\).
(i) Distance between \(X\) and \(Y\) along the equator. The difference in longitude is
\[\theta=123^\circ-67^\circ=56^\circ.\]Distance along the equator (a great circle):
\[d=\frac{\theta}{360^\circ}\times 2\pi R=\frac{56}{360}\times 2\times\frac{22}{7}\times 6400.\]\[d=\frac{56}{360}\times\frac{2\times22\times6400}{7}=\frac{56}{360}\times 40228.57\]\[d\approx 6257.78\text{ km}\approx\mathbf{6258\text{ km}}.\](ii) Distance of \(X\) from the North pole. The North pole is \(90^\circ\) of latitude from the equator, measured along a meridian (a great circle):
\[d=\frac{90}{360}\times 2\pi R=\frac{1}{4}\times 2\times\frac{22}{7}\times 6400\]\[d=\frac{1}{4}\times 40228.57\approx\mathbf{10057\text{ km}}.\](b) In circle centre \(O\), \(N\) is the mid-point of chord \(PQ\) with \(|PQ|=8\text{ cm}\), \(|ON|=3\text{ cm}\) and \(\angle ONR=20^\circ\). Find \(\angle ORN\).
Since \(N\) is the mid-point of the chord, \(ON\perp PQ\) and
\[|PN|=\tfrac{1}{2}|PQ|=4\text{ cm}.\]The radius is \(|OP|\); by Pythagoras in right triangle \(ONP\):
\[|OP|=\sqrt{|ON|^2+|PN|^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ cm}.\]\(R\) lies on the circle, so \(|OR|=5\text{ cm}\) too. Now apply the sine rule in triangle \(ONR\), where \(|ON|=3\), \(|OR|=5\) and \(\angle ONR=20^\circ\):
\[\frac{\sin\angle ORN}{|ON|}=\frac{\sin\angle ONR}{|OR|}\]\[\sin\angle ORN=\frac{|ON|\sin 20^\circ}{|OR|}=\frac{3\times0.3420}{5}=0.2052\]\[\angle ORN=\sin^{-1}(0.2052)\approx 11.8^\circ\approx\mathbf{12^\circ}.\]Question 11 Report
The pie chart shows the distribution of marks scored by 200 pupils in a test.
(a) How many pupils scored : (i) between 41 and 50 marks? ; (ii) above 80 marks ?
(b) What fraction of the pupils scored at most 50 marks?
(c) What is the modal class?
(a)
(i) Pupils who scored between 41 and 50 marks
The sector for 41 to 50 marks has angle \(90^\circ\).
\[\frac{90}{360}\times 200=50\]
Therefore, 50 pupils scored between 41 and 50 marks.
(ii) Pupils who scored above 80 marks
The sector for marks above 80 has angle \(18^\circ\).
\[\frac{18}{360}\times 200=10\]
Therefore, 10 pupils scored above 80 marks.
(b) The angles representing pupils who scored at most 50 marks are \(54^\circ\) and \(90^\circ\).
\[\text{Required fraction}=\frac{54+90}{360}=\frac{144}{360}=\frac{2}{5}\]
Thus, the fraction is \(\boxed{\frac{2}{5}}\).
(c) The modal class is \(\boxed{51\text{–}60\text{ marks}}\), since it has the largest sector in the pie chart.
Answer Details
(a)
(i) Pupils who scored between 41 and 50 marks
The sector for 41 to 50 marks has angle \(90^\circ\).
\[\frac{90}{360}\times 200=50\]
Therefore, 50 pupils scored between 41 and 50 marks.
(ii) Pupils who scored above 80 marks
The sector for marks above 80 has angle \(18^\circ\).
\[\frac{18}{360}\times 200=10\]
Therefore, 10 pupils scored above 80 marks.
(b) The angles representing pupils who scored at most 50 marks are \(54^\circ\) and \(90^\circ\).
\[\text{Required fraction}=\frac{54+90}{360}=\frac{144}{360}=\frac{2}{5}\]
Thus, the fraction is \(\boxed{\frac{2}{5}}\).
(c) The modal class is \(\boxed{51\text{–}60\text{ marks}}\), since it has the largest sector in the pie chart.
Question 12 Report
Using ruler and a pair of compasses only,
(a) construct \(\Delta PQR\) such that |PQ| = 7 cm, |PR| = 6 cm and < PQR = 60°.
(b) locate point M, the mid-point of PQ.
(c) Measure < RMQ.
This question tests accurate ruler-and-compasses construction and the ability to read a required measurement from the finished figure. The three given facts, \(|PQ| = 7\) cm, \(\angle PQR = 60^\circ\) and \(|PR| = 6\) cm, fix the triangle, and the interesting result is the size of \(\angle RMQ\).
Construction steps (ruler and compasses only):
- Draw a base line and mark off \(|PQ| = 7\) cm.
- At \(Q\), construct an angle of \(60^\circ\) using the equilateral-triangle arc method (an arc from \(Q\) cutting \(QP\), then the same radius stepped once round it), and draw the ray \(QR\).
- With centre \(P\) and radius \(6\) cm, draw an arc to cut the \(60^\circ\) ray at \(R\). Join \(PR\) and \(QR\) to complete \(\Delta PQR\).
- (b) Bisect \(PQ\) with equal arcs from \(P\) and \(Q\) above and below the line; the bisector meets \(PQ\) at its mid-point \(M\), so \(|PM| = |MQ| = 3.5\) cm. Join \(RM\).
- (c) Place the protractor at \(M\) and read \(\angle RMQ\).
The accurately drawn figure:
(c) Measurement: \(\angle RMQ = 60^\circ\).
Why the answer is exactly \(60^\circ\). When the \(60^\circ\) ray at \(Q\) is drawn and an arc of radius \(6\) cm is swung from \(P\), the point \(R\) lands where \(PR\) is perpendicular to \(QR\), so \(\angle PRQ = 90^\circ\). In a right-angled triangle the mid-point of the hypotenuse is equidistant from all three vertices, so
\[|MP| = |MQ| = |MR| = 3.5\ \text{cm}.\]Since \(|MQ| = |MR|\), triangle \(MQR\) is isosceles and \(\angle MRQ = \angle MQR = \angle PQR = 60^\circ\). The three angles of triangle \(MQR\) must sum to \(180^\circ\), so
\[\angle RMQ = 180^\circ - 60^\circ - 60^\circ = 60^\circ.\](In fact \(|QR| = 7\cos 60^\circ = 3.5\) cm as well, so \(MQR\) is equilateral, which confirms \(\angle RMQ = 60^\circ\).)
Examination reminder: the marks are awarded for the visible construction arcs (the \(60^\circ\) at \(Q\) and the perpendicular bisector of \(PQ\)) and for a measured value within about \(\pm 1^\circ\) of the true figure. A reading of \(90^\circ\) is a sign that \(RM\) was mistaken for the perpendicular bisector rather than the line to \(R\); measure the angle between \(MR\) and \(MQ\) at \(M\), and the correct value is \(60^\circ\).
Answer Details
This question tests accurate ruler-and-compasses construction and the ability to read a required measurement from the finished figure. The three given facts, \(|PQ| = 7\) cm, \(\angle PQR = 60^\circ\) and \(|PR| = 6\) cm, fix the triangle, and the interesting result is the size of \(\angle RMQ\).
Construction steps (ruler and compasses only):
- Draw a base line and mark off \(|PQ| = 7\) cm.
- At \(Q\), construct an angle of \(60^\circ\) using the equilateral-triangle arc method (an arc from \(Q\) cutting \(QP\), then the same radius stepped once round it), and draw the ray \(QR\).
- With centre \(P\) and radius \(6\) cm, draw an arc to cut the \(60^\circ\) ray at \(R\). Join \(PR\) and \(QR\) to complete \(\Delta PQR\).
- (b) Bisect \(PQ\) with equal arcs from \(P\) and \(Q\) above and below the line; the bisector meets \(PQ\) at its mid-point \(M\), so \(|PM| = |MQ| = 3.5\) cm. Join \(RM\).
- (c) Place the protractor at \(M\) and read \(\angle RMQ\).
The accurately drawn figure:
(c) Measurement: \(\angle RMQ = 60^\circ\).
Why the answer is exactly \(60^\circ\). When the \(60^\circ\) ray at \(Q\) is drawn and an arc of radius \(6\) cm is swung from \(P\), the point \(R\) lands where \(PR\) is perpendicular to \(QR\), so \(\angle PRQ = 90^\circ\). In a right-angled triangle the mid-point of the hypotenuse is equidistant from all three vertices, so
\[|MP| = |MQ| = |MR| = 3.5\ \text{cm}.\]Since \(|MQ| = |MR|\), triangle \(MQR\) is isosceles and \(\angle MRQ = \angle MQR = \angle PQR = 60^\circ\). The three angles of triangle \(MQR\) must sum to \(180^\circ\), so
\[\angle RMQ = 180^\circ - 60^\circ - 60^\circ = 60^\circ.\](In fact \(|QR| = 7\cos 60^\circ = 3.5\) cm as well, so \(MQR\) is equilateral, which confirms \(\angle RMQ = 60^\circ\).)
Examination reminder: the marks are awarded for the visible construction arcs (the \(60^\circ\) at \(Q\) and the perpendicular bisector of \(PQ\)) and for a measured value within about \(\pm 1^\circ\) of the true figure. A reading of \(90^\circ\) is a sign that \(RM\) was mistaken for the perpendicular bisector rather than the line to \(R\); measure the angle between \(MR\) and \(MQ\) at \(M\), and the correct value is \(60^\circ\).
Question 13 Report
(a)
| Limes | Apples | |
| Good | 10 | 8 |
| Bad | 6 | 6 |
The table shows the number of limes and apples of the same size in a bag. If two of the fruits are picked at random, one at a time, without replacement, find the probability that : (i) both are good limes ; (ii) both are bad fruits ; (iii) one is a good apple and the other a bad lime.
(b) Solve the equation \(\log_{3} (4x + 1) - \log_{3} (3x - 5) = 2\).
| Limes | Apples | Total | |
|---|---|---|---|
| Good | 10 | 8 | 18 |
| Bad | 6 | 6 | 12 |
| Total | 16 | 14 | 30 |
There are 30 fruits. Picking is without replacement, so the second denominator is 29.
(i) Both good limes (10 good limes):
\[ \frac{10}{30}\times\frac{9}{29}=\frac{90}{870}=\frac{3}{29}\approx 0.103 \](ii) Both bad fruits (12 bad fruits):
\[ \frac{12}{30}\times\frac{11}{29}=\frac{132}{870}=\frac{22}{145}\approx 0.152 \](iii) One good apple and one bad lime (either order; 8 good apples, 6 bad limes):
\[ \frac{8}{30}\times\frac{6}{29}+\frac{6}{30}\times\frac{8}{29}=\frac{48+48}{870}=\frac{96}{870}=\frac{16}{145}\approx 0.110 \](b) Solve \(\log_3(4x+1)-\log_3(3x-5)=2\). Combine the logs:
\[ \log_3\!\left(\frac{4x+1}{3x-5}\right)=2 \;\Rightarrow\; \frac{4x+1}{3x-5}=3^2=9 \] \[ 4x+1=9(3x-5)=27x-45 \;\Rightarrow\; 46=23x \;\Rightarrow\; x=2 \]Check: \(3x-5=1>0\) and \(4x+1=9>0\), so \(x=2\) is valid.
Answer Details
| Limes | Apples | Total | |
|---|---|---|---|
| Good | 10 | 8 | 18 |
| Bad | 6 | 6 | 12 |
| Total | 16 | 14 | 30 |
There are 30 fruits. Picking is without replacement, so the second denominator is 29.
(i) Both good limes (10 good limes):
\[ \frac{10}{30}\times\frac{9}{29}=\frac{90}{870}=\frac{3}{29}\approx 0.103 \](ii) Both bad fruits (12 bad fruits):
\[ \frac{12}{30}\times\frac{11}{29}=\frac{132}{870}=\frac{22}{145}\approx 0.152 \](iii) One good apple and one bad lime (either order; 8 good apples, 6 bad limes):
\[ \frac{8}{30}\times\frac{6}{29}+\frac{6}{30}\times\frac{8}{29}=\frac{48+48}{870}=\frac{96}{870}=\frac{16}{145}\approx 0.110 \](b) Solve \(\log_3(4x+1)-\log_3(3x-5)=2\). Combine the logs:
\[ \log_3\!\left(\frac{4x+1}{3x-5}\right)=2 \;\Rightarrow\; \frac{4x+1}{3x-5}=3^2=9 \] \[ 4x+1=9(3x-5)=27x-45 \;\Rightarrow\; 46=23x \;\Rightarrow\; x=2 \]Check: \(3x-5=1>0\) and \(4x+1=9>0\), so \(x=2\) is valid.
