WAEC SSCE - General Mathematics - 2000 (Essay)
Question 1 Rapport
(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.
(a) Since \(\cos x = 0.7431\) and \(0^\circ < x < 90^\circ\), from tables \(x = 42^\circ\).
(i) \(\sin42^\circ = 0.6691\), so
\[2\sin x = 2\times0.6691 = 1.338.\]
(ii) \(\dfrac{x}{2} = 21^\circ\), so
\[\tan\frac{x}{2} = \tan21^\circ = 0.3839.\]
(b) The five interior angles of a pentagon sum to \((5-2)\times180^\circ = 540^\circ\). With ratio \(2:3:4:4:5\), the total number of parts is \(2+3+4+4+5 = 18\).
\[\text{One part} = \frac{540^\circ}{18} = 30^\circ.\]
\[\text{Largest angle} = 5\times30^\circ = 150^\circ.\]
Détails de la réponse
(a) Since \(\cos x = 0.7431\) and \(0^\circ < x < 90^\circ\), from tables \(x = 42^\circ\).
(i) \(\sin42^\circ = 0.6691\), so
\[2\sin x = 2\times0.6691 = 1.338.\]
(ii) \(\dfrac{x}{2} = 21^\circ\), so
\[\tan\frac{x}{2} = \tan21^\circ = 0.3839.\]
(b) The five interior angles of a pentagon sum to \((5-2)\times180^\circ = 540^\circ\). With ratio \(2:3:4:4:5\), the total number of parts is \(2+3+4+4+5 = 18\).
\[\text{One part} = \frac{540^\circ}{18} = 30^\circ.\]
\[\text{Largest angle} = 5\times30^\circ = 150^\circ.\]
Question 2 Rapport
(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 \(\pi = \frac{22}{7}\)].
(a) Common difference. With \(a = 8\), the 7th and 9th terms are \(8 + 6d\) and \(8 + 8d\), and their ratio is \(5 : 8\):
\[\frac{8 + 6d}{8 + 8d} = \frac{5}{8}\Rightarrow 8(8 + 6d) = 5(8 + 8d).\]
\[64 + 48d = 40 + 40d \Rightarrow 8d = -24 \Rightarrow d = -3.\]
(b)(i) Volume of the sphere (\(r = 2\) cm):
\[V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\times\tfrac{22}{7}\times8 = \frac{704}{21} = 33.52\text{ cm}^3.\]
(ii) Density:
\[\rho = \frac{\text{mass}}{\text{volume}} = \frac{11.2}{33.52} = 0.334\text{ g/cm}^3.\]
(iii) Mass of a sphere of radius 3 cm (same material). Mass varies as \(r^3\):
\[\text{mass} = 11.2\times\left(\frac{3}{2}\right)^3 = 11.2\times\frac{27}{8} = 37.8\text{ g}.\]
Détails de la réponse
(a) Common difference. With \(a = 8\), the 7th and 9th terms are \(8 + 6d\) and \(8 + 8d\), and their ratio is \(5 : 8\):
\[\frac{8 + 6d}{8 + 8d} = \frac{5}{8}\Rightarrow 8(8 + 6d) = 5(8 + 8d).\]
\[64 + 48d = 40 + 40d \Rightarrow 8d = -24 \Rightarrow d = -3.\]
(b)(i) Volume of the sphere (\(r = 2\) cm):
\[V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\times\tfrac{22}{7}\times8 = \frac{704}{21} = 33.52\text{ cm}^3.\]
(ii) Density:
\[\rho = \frac{\text{mass}}{\text{volume}} = \frac{11.2}{33.52} = 0.334\text{ g/cm}^3.\]
(iii) Mass of a sphere of radius 3 cm (same material). Mass varies as \(r^3\):
\[\text{mass} = 11.2\times\left(\frac{3}{2}\right)^3 = 11.2\times\frac{27}{8} = 37.8\text{ g}.\]
Question 3 Rapport
(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 \(= 350\times75 = \text{N}26{,}250\).
\[\text{Total sales} = 21{,}375 + 26{,}250 = \text{N}47{,}625.\]
\[\text{Profit} = 47{,}625 - 37{,}500 = \text{N}10{,}125.\]
\[\%\text{ profit} = \frac{10{,}125}{37{,}500}\times100 = 27\%.\]
Détails de la réponse
(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 4 Rapport
(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:
\[1\tfrac{19}{81} = \frac{81 + 19}{81} = \frac{100}{81},\qquad \sqrt{\frac{100}{81}} = \frac{10}{9}.\]
Détails de la réponse
(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 5 Rapport
(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 \(\mathbf{52.5\text{ m}}\).
Détails de la réponse
(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 6 Rapport
(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 \(\angle ONR=20^\circ\):
\[\frac{\sin\angle ORN}{|ON|}=\frac{\sin\angle ONR}{|OR|}\]\[\sin\angle ORN=\frac{|ON|\sin 20^\circ}{|OR|}=\frac{3\times0.3420}{5}=0.2052\]\[\angle ORN=\sin^{-1}(0.2052)\approx 11.8^\circ\approx\mathbf{12^\circ}.\]Détails de la réponse
(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 7 Rapport
(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 \(\sqrt{x^2 + 1} = \tfrac{5}{4}\):
\[x^2 + 1 = \frac{25}{16} \Rightarrow x^2 = \frac{25}{16} - 1 = \frac{9}{16}.\]
\[x = \frac{3}{4}\quad(\text{positive value}).\]
Détails de la réponse
(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 8 Rapport
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?
Reading the pie chart. The whole circle (\(360^\circ\)) represents all \(200\) pupils, so each degree stands for \(\frac{200}{360}=\frac{5}{9}\) of a pupil. The sectors read off the diagram are:
| Class (marks) | Angle |
|---|---|
| 31 - 40 | \(54^\circ\) |
| 41 - 50 | \(90^\circ\) |
| 51 - 60 | \(108^\circ\) |
| 61 - 70 | \(54^\circ\) |
| 71 - 80 | \(36^\circ\) |
| 81 - 90 | \(18^\circ\) |
Check: \(54+90+108+54+36+18=360^\circ\). Good.
(a)(i) Pupils scoring between 41 and 50. This is the \(90^\circ\) sector.
\[\frac{90}{360}\times 200 = 50 \text{ pupils.}\](a)(ii) Pupils scoring above 80. This is the \(81\text{-}90\) sector, \(18^\circ\).
\[\frac{18}{360}\times 200 = 10 \text{ pupils.}\](b) Fraction scoring at most 50 marks. “At most 50” means the \(31\text{-}40\) and \(41\text{-}50\) classes together: \(54^\circ+90^\circ=144^\circ\).
\[\text{Fraction}=\frac{144}{360}=\frac{2}{5}.\](c) Modal class. The modal class is the one with the greatest sector angle (largest frequency), which is \(108^\circ\).
Therefore the modal class is 51 - 60.
Détails de la réponse
Reading the pie chart. The whole circle (\(360^\circ\)) represents all \(200\) pupils, so each degree stands for \(\frac{200}{360}=\frac{5}{9}\) of a pupil. The sectors read off the diagram are:
| Class (marks) | Angle |
|---|---|
| 31 - 40 | \(54^\circ\) |
| 41 - 50 | \(90^\circ\) |
| 51 - 60 | \(108^\circ\) |
| 61 - 70 | \(54^\circ\) |
| 71 - 80 | \(36^\circ\) |
| 81 - 90 | \(18^\circ\) |
Check: \(54+90+108+54+36+18=360^\circ\). Good.
(a)(i) Pupils scoring between 41 and 50. This is the \(90^\circ\) sector.
\[\frac{90}{360}\times 200 = 50 \text{ pupils.}\](a)(ii) Pupils scoring above 80. This is the \(81\text{-}90\) sector, \(18^\circ\).
\[\frac{18}{360}\times 200 = 10 \text{ pupils.}\](b) Fraction scoring at most 50 marks. “At most 50” means the \(31\text{-}40\) and \(41\text{-}50\) classes together: \(54^\circ+90^\circ=144^\circ\).
\[\text{Fraction}=\frac{144}{360}=\frac{2}{5}.\](c) Modal class. The modal class is the one with the greatest sector angle (largest frequency), which is \(108^\circ\).
Therefore the modal class is 51 - 60.
Question 9 Rapport
(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.
Détails de la réponse
| 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.
Question 10 Rapport
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 \(60^\circ\).
Détails de la réponse
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 11 Rapport
(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 \(\angle BCD=90^{\circ}\).
Détails de la réponse
(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 12 Rapport
(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:
\[\text{Area }MPQN=\text{Area }\Delta LPQ-\text{Area }\Delta LMN=32-18=14\text{ cm}^{2}\]
Détails de la réponse
(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 13 Rapport
(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) Completing the table for \(y=x^2-2x-5\). For example, when \(x=-3\), \(y=9+6-5=10\), and when \(x=2\), \(y=4-4-5=-5\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|---|
| y | 10 | 3 | -2 | -5 | -6 | -5 | -2 | 3 | 10 |
(b) and (c) The plotted curve is the parabola \(y=x^2-2x-5\). The straight line is \(y=2x+3\), which passes through \((0,3)\) and \((3,9)\).
(d) At a point of intersection, both graphs have the same \(y\)-value:
\[ x^2-2x-5=2x+3 \] \[ x^2-4x-8=0 \](e) The roots are the \(x\)-coordinates of the intersections. From the graph, they are approximately \(x=-1.5\) and \(x=5.5\).
The more accurate solutions are \(x=2\pm2\sqrt{3}\), giving \(x\approx-1.46\) and \(x\approx5.46\). The supplied reference answer uses \(y=2x-3\), but this conflicts with the question stem, which states \(y=2x+3\). Therefore \(x^2-4x-8=0\) is the equation consistent with the stated line.
Détails de la réponse
(a) Completing the table for \(y=x^2-2x-5\). For example, when \(x=-3\), \(y=9+6-5=10\), and when \(x=2\), \(y=4-4-5=-5\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|---|
| y | 10 | 3 | -2 | -5 | -6 | -5 | -2 | 3 | 10 |
(b) and (c) The plotted curve is the parabola \(y=x^2-2x-5\). The straight line is \(y=2x+3\), which passes through \((0,3)\) and \((3,9)\).
(d) At a point of intersection, both graphs have the same \(y\)-value:
\[ x^2-2x-5=2x+3 \] \[ x^2-4x-8=0 \](e) The roots are the \(x\)-coordinates of the intersections. From the graph, they are approximately \(x=-1.5\) and \(x=5.5\).
The more accurate solutions are \(x=2\pm2\sqrt{3}\), giving \(x\approx-1.46\) and \(x\approx5.46\). The supplied reference answer uses \(y=2x-3\), but this conflicts with the question stem, which states \(y=2x+3\). Therefore \(x^2-4x-8=0\) is the equation consistent with the stated line.
