WAEC SSCE - General Mathematics - 2019 (Essay)
Question 1 Rapport
NOT DRAWN TO SCALE
In the diagram, is a chord of a circle with centre 0. 22.42 cm and the perimeter of triangle MON is 55.6 cm. Calculate, correct to the nearest degree. < MON.
(b) T is equidistant from P and Q. The bearing of P from T is 60\(^o\) and the bearing of Q from T is 130\(^o\).
(i) Illustrate the information on a diagram.
(ii) Find the bearing of Q from P.
(a) Angle MON
From the diagram, MN is a chord of the circle with centre O, and \(|MN| = 22.42\ \text{cm}\). OM and ON are radii, so \(|OM| = |ON| = r\).
The perimeter of triangle MON is 55.6 cm:
\[|OM| + |ON| + |MN| = 55.6\]
\[r + r + 22.42 = 55.6\]
\[2r = 33.18 \quad\Rightarrow\quad r = 16.59\ \text{cm}\]
Triangle MON is isosceles with \(OM = ON = 16.59\). Drop the perpendicular from O to the midpoint of MN; it bisects angle MON. Then
\[\sin\!\left(\frac{\angle MON}{2}\right) = \frac{\tfrac{1}{2}|MN|}{r} = \frac{11.21}{16.59} = 0.6757\]
\[\frac{\angle MON}{2} = \sin^{-1}(0.6757) = 42.5^\circ\]
\[\angle MON = 85.0^\circ \approx 85^\circ\]
(b) Bearings
(i) Illustration. T is equidistant from P and Q, so \(|TP| = |TQ|\). From T, the bearing of P is \(060^\circ\) and the bearing of Q is \(130^\circ\). Measuring both angles clockwise from the north line at T, P lies to the north-east and Q further round to the south-east.
The angle between the two bearings at T is
\[\angle PTQ = 130^\circ - 60^\circ = 70^\circ\]
(ii) Bearing of Q from P. Since \(|TP| = |TQ|\), triangle PTQ is isosceles, so the base angles are equal:
\[\angle TPQ = \angle TQP = \frac{180^\circ - 70^\circ}{2} = 55^\circ\]
The bearing of T from P is the reverse of the bearing of P from T:
\[060^\circ + 180^\circ = 240^\circ\]
Q lies \(55^\circ\) from PT, measured back towards the south. Hence the bearing of Q from P is
\[240^\circ - 55^\circ = 185^\circ\]
The bearing of Q from P is \(\mathbf{185^\circ}\).
Détails de la réponse
(a) Angle MON
From the diagram, MN is a chord of the circle with centre O, and \(|MN| = 22.42\ \text{cm}\). OM and ON are radii, so \(|OM| = |ON| = r\).
The perimeter of triangle MON is 55.6 cm:
\[|OM| + |ON| + |MN| = 55.6\]
\[r + r + 22.42 = 55.6\]
\[2r = 33.18 \quad\Rightarrow\quad r = 16.59\ \text{cm}\]
Triangle MON is isosceles with \(OM = ON = 16.59\). Drop the perpendicular from O to the midpoint of MN; it bisects angle MON. Then
\[\sin\!\left(\frac{\angle MON}{2}\right) = \frac{\tfrac{1}{2}|MN|}{r} = \frac{11.21}{16.59} = 0.6757\]
\[\frac{\angle MON}{2} = \sin^{-1}(0.6757) = 42.5^\circ\]
\[\angle MON = 85.0^\circ \approx 85^\circ\]
(b) Bearings
(i) Illustration. T is equidistant from P and Q, so \(|TP| = |TQ|\). From T, the bearing of P is \(060^\circ\) and the bearing of Q is \(130^\circ\). Measuring both angles clockwise from the north line at T, P lies to the north-east and Q further round to the south-east.
The angle between the two bearings at T is
\[\angle PTQ = 130^\circ - 60^\circ = 70^\circ\]
(ii) Bearing of Q from P. Since \(|TP| = |TQ|\), triangle PTQ is isosceles, so the base angles are equal:
\[\angle TPQ = \angle TQP = \frac{180^\circ - 70^\circ}{2} = 55^\circ\]
The bearing of T from P is the reverse of the bearing of P from T:
\[060^\circ + 180^\circ = 240^\circ\]
Q lies \(55^\circ\) from PT, measured back towards the south. Hence the bearing of Q from P is
\[240^\circ - 55^\circ = 185^\circ\]
The bearing of Q from P is \(\mathbf{185^\circ}\).
Question 2 Rapport
The second, fourth and sixth terms of an Arithmetic Progression (AP.) are x - 1, x + 1 and 7 respectively. Find the:
(a) common difference;
(b) first term;
(c) value of x.
Let the A.P. have first term \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).
Second term: \(a+d=x-1\)
Fourth term: \(a+3d=x+1\)
Sixth term: \(a+5d=7\)
(a) Common difference. Subtract the second-term equation from the fourth-term equation:
\[(a+3d)-(a+d)=(x+1)-(x-1)\Rightarrow 2d=2\Rightarrow d=1\]Common difference \(d=1\).
(c) Value of x. Subtract the fourth-term equation from the sixth-term equation:
\[(a+5d)-(a+3d)=7-(x+1)\Rightarrow 2d=6-x\]With \(d=1\): \(2=6-x\Rightarrow x=4\).
(b) First term. From \(a+d=x-1\): \(a+1=4-1=3\Rightarrow a=2\).
First term \(=2\), common difference \(=1\), \(x=4\). (Check: terms are \(2,3,4,5,6,7,\ldots\); 2nd\(=3=x-1\), 4th\(=5=x+1\), 6th\(=7\).)
Détails de la réponse
Let the A.P. have first term \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).
Second term: \(a+d=x-1\)
Fourth term: \(a+3d=x+1\)
Sixth term: \(a+5d=7\)
(a) Common difference. Subtract the second-term equation from the fourth-term equation:
\[(a+3d)-(a+d)=(x+1)-(x-1)\Rightarrow 2d=2\Rightarrow d=1\]Common difference \(d=1\).
(c) Value of x. Subtract the fourth-term equation from the sixth-term equation:
\[(a+5d)-(a+3d)=7-(x+1)\Rightarrow 2d=6-x\]With \(d=1\): \(2=6-x\Rightarrow x=4\).
(b) First term. From \(a+d=x-1\): \(a+1=4-1=3\Rightarrow a=2\).
First term \(=2\), common difference \(=1\), \(x=4\). (Check: terms are \(2,3,4,5,6,7,\ldots\); 2nd\(=3=x-1\), 4th\(=5=x+1\), 6th\(=7\).)
Question 3 Rapport
(a) The curved surface areas of two cones are equal. The base radius of one is 5 cm and its slant height is 12cm. calculate the height of the second cone if its base radius is 6 cm.
(b) Given the matrices A = \(\begin{pmatrix} 2 & 5 \\ -1 & -3 \end{pmatrix}\) and B = \(\begin{pmatrix} 3 & -2 \\ 4 & 1 \end{pmatrix}\), find:
(i) BA;
(ii) the determinant of BA.
(a) Height of the second cone. Curved surface area \(=\pi r l\).
Cone 1: \(\pi(5)(12)=60\pi\). Cone 2 has \(r=6\) and equal C.S.A.:
\(\pi(6)l_2 = 60\pi \Rightarrow l_2 = 10\text{ cm}\).
Height \(h=\sqrt{l_2^{2}-r^{2}}=\sqrt{10^{2}-6^{2}}=\sqrt{100-36}=\sqrt{64}=\mathbf{8\ cm}\).
(b) \(A=\begin{pmatrix}2&5\\-1&-3\end{pmatrix},\ B=\begin{pmatrix}3&-2\\4&1\end{pmatrix}\).
(i) \(BA\):
\(BA=\begin{pmatrix}(3)(2)+(-2)(-1)&(3)(5)+(-2)(-3)\\(4)(2)+(1)(-1)&(4)(5)+(1)(-3)\end{pmatrix}=\begin{pmatrix}8&21\\7&17\end{pmatrix}\).
(ii) Determinant of \(BA\): \((8)(17)-(21)(7)=136-147=\mathbf{-11}\).
Détails de la réponse
(a) Height of the second cone. Curved surface area \(=\pi r l\).
Cone 1: \(\pi(5)(12)=60\pi\). Cone 2 has \(r=6\) and equal C.S.A.:
\(\pi(6)l_2 = 60\pi \Rightarrow l_2 = 10\text{ cm}\).
Height \(h=\sqrt{l_2^{2}-r^{2}}=\sqrt{10^{2}-6^{2}}=\sqrt{100-36}=\sqrt{64}=\mathbf{8\ cm}\).
(b) \(A=\begin{pmatrix}2&5\\-1&-3\end{pmatrix},\ B=\begin{pmatrix}3&-2\\4&1\end{pmatrix}\).
(i) \(BA\):
\(BA=\begin{pmatrix}(3)(2)+(-2)(-1)&(3)(5)+(-2)(-3)\\(4)(2)+(1)(-1)&(4)(5)+(1)(-3)\end{pmatrix}=\begin{pmatrix}8&21\\7&17\end{pmatrix}\).
(ii) Determinant of \(BA\): \((8)(17)-(21)(7)=136-147=\mathbf{-11}\).
Question 4 Rapport
(a) If logo a = 1.3010 and log\(_{10}\)b - 1.4771. find the value of ab
(b)
In the diagram. O is the centre of the circle,< ACB = 39\(^o\) and < CBE = 62\(^o\). Find: (i) the interior angle AOC;
(ii) angle BAC.
(a) Finding \(ab\)
Given \(\log_{10} a = 1.3010\) and \(\log_{10} b = 1.4771\). Using the law \(\log(ab) = \log a + \log b\):
\[\log_{10}(ab) = 1.3010 + 1.4771 = 2.7781\]Taking the antilogarithm,
\[ab = 10^{2.7781} = 600\](As a check, \(a = 10^{1.3010} = 20\) and \(b = 10^{1.4771} = 30\), so \(ab = 600\).) Hence \(ab = 600\).
(b) Circle with centre \(O\)
From the diagram, \(A\), \(B\), \(C\) lie on the circle, \(O\) is the centre, \(ABE\) is a straight line, \(\angle ACB = 39^\circ\) and \(\angle CBE = 62^\circ\).
(i) Interior angle \(AOC\)
Since \(ABE\) is a straight line, \(\angle ABC\) is the supplement of \(\angle CBE\):
\[\angle ABC = 180^\circ - 62^\circ = 118^\circ\]The inscribed angle \(\angle ABC\) stands on chord \(AC\). The angle subtended at the centre by the same chord (the reflex angle at \(O\), on the opposite side of \(AC\) from \(B\)) is twice the inscribed angle:
\[\text{reflex } \angle AOC = 2 \times 118^\circ = 236^\circ\]Therefore the required interior (non-reflex) angle \(AOC\) is
\[\angle AOC = 360^\circ - 236^\circ = \boxed{124^\circ}\](ii) Angle \(BAC\)
In triangle \(ABC\), the three angles sum to \(180^\circ\), with \(\angle ACB = 39^\circ\) and \(\angle ABC = 118^\circ\):
\[\angle BAC = 180^\circ - 118^\circ - 39^\circ = \boxed{23^\circ}\]Hence \(\angle AOC = 124^\circ\) and \(\angle BAC = 23^\circ\).
Détails de la réponse
(a) Finding \(ab\)
Given \(\log_{10} a = 1.3010\) and \(\log_{10} b = 1.4771\). Using the law \(\log(ab) = \log a + \log b\):
\[\log_{10}(ab) = 1.3010 + 1.4771 = 2.7781\]Taking the antilogarithm,
\[ab = 10^{2.7781} = 600\](As a check, \(a = 10^{1.3010} = 20\) and \(b = 10^{1.4771} = 30\), so \(ab = 600\).) Hence \(ab = 600\).
(b) Circle with centre \(O\)
From the diagram, \(A\), \(B\), \(C\) lie on the circle, \(O\) is the centre, \(ABE\) is a straight line, \(\angle ACB = 39^\circ\) and \(\angle CBE = 62^\circ\).
(i) Interior angle \(AOC\)
Since \(ABE\) is a straight line, \(\angle ABC\) is the supplement of \(\angle CBE\):
\[\angle ABC = 180^\circ - 62^\circ = 118^\circ\]The inscribed angle \(\angle ABC\) stands on chord \(AC\). The angle subtended at the centre by the same chord (the reflex angle at \(O\), on the opposite side of \(AC\) from \(B\)) is twice the inscribed angle:
\[\text{reflex } \angle AOC = 2 \times 118^\circ = 236^\circ\]Therefore the required interior (non-reflex) angle \(AOC\) is
\[\angle AOC = 360^\circ - 236^\circ = \boxed{124^\circ}\](ii) Angle \(BAC\)
In triangle \(ABC\), the three angles sum to \(180^\circ\), with \(\angle ACB = 39^\circ\) and \(\angle ABC = 118^\circ\):
\[\angle BAC = 180^\circ - 118^\circ - 39^\circ = \boxed{23^\circ}\]Hence \(\angle AOC = 124^\circ\) and \(\angle BAC = 23^\circ\).
Question 5 Rapport
(a) Solve the inequality \( \frac{1}{3}x - \frac{1}{4}(x + 2) \ge 3x - 1\frac{1}{3} \)
(b) 
In the diagram, ABC is right-angled triangle on a horizontal ground. |AD| is a vertical tower. < BAC = \(90^\circ\), < ACB = \(35^\circ\), < ABD = \(52^\circ\) and |BC| = 66cm.
(a) Solve \(\;\dfrac{1}{3}x-\dfrac{1}{4}(x+2)\ge 3x-1\tfrac{1}{3}.\)
Write \(1\tfrac{1}{3}=\dfrac{4}{3}\) and multiply every term by the LCM of the denominators, \(12\):
\[12\cdot\tfrac{1}{3}x-12\cdot\tfrac{1}{4}(x+2)\ge 12\cdot 3x-12\cdot\tfrac{4}{3}.\]
\[4x-3(x+2)\ge 36x-16.\]
\[4x-3x-6\ge 36x-16.\]
\[x-6\ge 36x-16.\]
Collect like terms:
\[-6+16\ge 36x-x\;\Rightarrow\;10\ge 35x.\]
\[x\le\frac{10}{35}=\frac{2}{7}.\]
\(\therefore x\le\dfrac{2}{7}.\)
(b) In right-angled triangle \(ABC\) on horizontal ground, \(\angle BAC=90^{\circ}\), \(\angle ACB=35^{\circ}\) and \(|BC|=66\text{ m}\). \(AD\) is a vertical tower and \(\angle ABD=52^{\circ}\) is the angle of elevation of the top \(D\) from \(B\).
(i) Height of the tower \(|AD|\).
In triangle \(ABC\), \(BC\) is the hypotenuse and \(AB\) is opposite the \(35^{\circ}\) angle at \(C\):
\[\sin 35^{\circ}=\frac{|AB|}{|BC|}\;\Rightarrow\;|AB|=66\sin 35^{\circ}=66\times0.5736=37.86\text{ m}.\]
Also \(|AC|=66\cos 35^{\circ}=66\times0.8192=54.06\text{ m}.\)
The tower \(AD\) is vertical and \(AB\) horizontal, so triangle \(ABD\) is right-angled at \(A\):
\[\tan 52^{\circ}=\frac{|AD|}{|AB|}\;\Rightarrow\;|AD|=37.86\times\tan 52^{\circ}=37.86\times1.2799=48.45\text{ m}.\]
\(\therefore\) the height of the tower \(|AD|\approx 48.45\text{ m}\) (2 d.p.).
(ii) Angle of elevation of the tower from \(C\).
In right-angled triangle \(ACD\):
\[\tan(\angle ACD)=\frac{|AD|}{|AC|}=\frac{48.45}{54.06}=0.8962,\]
\[\angle ACD=\tan^{-1}(0.8962)=41.87^{\circ}.\]
Adding the horizontal angle \(\angle ACB=35^{\circ}\), the angle of elevation of the tower measured from \(C\) is
\[35^{\circ}+41.87^{\circ}=76.87^{\circ}\text{ (2 d.p.).}\]
Détails de la réponse
(a) Solve \(\;\dfrac{1}{3}x-\dfrac{1}{4}(x+2)\ge 3x-1\tfrac{1}{3}.\)
Write \(1\tfrac{1}{3}=\dfrac{4}{3}\) and multiply every term by the LCM of the denominators, \(12\):
\[12\cdot\tfrac{1}{3}x-12\cdot\tfrac{1}{4}(x+2)\ge 12\cdot 3x-12\cdot\tfrac{4}{3}.\]
\[4x-3(x+2)\ge 36x-16.\]
\[4x-3x-6\ge 36x-16.\]
\[x-6\ge 36x-16.\]
Collect like terms:
\[-6+16\ge 36x-x\;\Rightarrow\;10\ge 35x.\]
\[x\le\frac{10}{35}=\frac{2}{7}.\]
\(\therefore x\le\dfrac{2}{7}.\)
(b) In right-angled triangle \(ABC\) on horizontal ground, \(\angle BAC=90^{\circ}\), \(\angle ACB=35^{\circ}\) and \(|BC|=66\text{ m}\). \(AD\) is a vertical tower and \(\angle ABD=52^{\circ}\) is the angle of elevation of the top \(D\) from \(B\).
(i) Height of the tower \(|AD|\).
In triangle \(ABC\), \(BC\) is the hypotenuse and \(AB\) is opposite the \(35^{\circ}\) angle at \(C\):
\[\sin 35^{\circ}=\frac{|AB|}{|BC|}\;\Rightarrow\;|AB|=66\sin 35^{\circ}=66\times0.5736=37.86\text{ m}.\]
Also \(|AC|=66\cos 35^{\circ}=66\times0.8192=54.06\text{ m}.\)
The tower \(AD\) is vertical and \(AB\) horizontal, so triangle \(ABD\) is right-angled at \(A\):
\[\tan 52^{\circ}=\frac{|AD|}{|AB|}\;\Rightarrow\;|AD|=37.86\times\tan 52^{\circ}=37.86\times1.2799=48.45\text{ m}.\]
\(\therefore\) the height of the tower \(|AD|\approx 48.45\text{ m}\) (2 d.p.).
(ii) Angle of elevation of the tower from \(C\).
In right-angled triangle \(ACD\):
\[\tan(\angle ACD)=\frac{|AD|}{|AC|}=\frac{48.45}{54.06}=0.8962,\]
\[\angle ACD=\tan^{-1}(0.8962)=41.87^{\circ}.\]
Adding the horizontal angle \(\angle ACB=35^{\circ}\), the angle of elevation of the tower measured from \(C\) is
\[35^{\circ}+41.87^{\circ}=76.87^{\circ}\text{ (2 d.p.).}\]
Question 6 Rapport
(a) Ali and Yusif shared N420.000.00 in the ratio 3.,; 5 : 8 respectively. Find the sum of Ali and Yusuf's shares
(b) Solve: 2(\(\frac{1}{8}\))\(^x\) = 32\(^{x - 1}\).
(a) \(\text{N}420{,}000.00\) is shared in the ratio \(3:5:8\) (total \(16\) parts).
One part \(=\dfrac{420000}{16}=\text{N}26{,}250.00\).
Ali (3 parts) \(=\text{N}78{,}750\); Yusuf (5 parts) \(=\text{N}131{,}250\).
Sum of Ali and Yusuf's shares \(=(3+5)\times 26250=8\times 26250=\mathbf{\text{N}210{,}000.00}\).
(b) \(2\left(\dfrac{1}{8}\right)^{x}=32^{x-1}\).
Write in base 2: \(\dfrac{1}{8}=2^{-3},\ 32=2^{5}\).
\(2^{1}\cdot 2^{-3x}=2^{5(x-1)} \Rightarrow 2^{1-3x}=2^{5x-5}\).
\(1-3x=5x-5 \Rightarrow 6=8x \Rightarrow x=\dfrac{3}{4}\).
Détails de la réponse
(a) \(\text{N}420{,}000.00\) is shared in the ratio \(3:5:8\) (total \(16\) parts).
One part \(=\dfrac{420000}{16}=\text{N}26{,}250.00\).
Ali (3 parts) \(=\text{N}78{,}750\); Yusuf (5 parts) \(=\text{N}131{,}250\).
Sum of Ali and Yusuf's shares \(=(3+5)\times 26250=8\times 26250=\mathbf{\text{N}210{,}000.00}\).
(b) \(2\left(\dfrac{1}{8}\right)^{x}=32^{x-1}\).
Write in base 2: \(\dfrac{1}{8}=2^{-3},\ 32=2^{5}\).
\(2^{1}\cdot 2^{-3x}=2^{5(x-1)} \Rightarrow 2^{1-3x}=2^{5x-5}\).
\(1-3x=5x-5 \Rightarrow 6=8x \Rightarrow x=\dfrac{3}{4}\).
Question 7 Rapport
(a) Given that sin y = \(\frac{8}{17}\) find the value of \(\frac{tan y}{1 + 2 tan y}\)
(b) An amount of N300,000.00 was shared among Otobo, Ada and Adeola. Otobo received N60,000.00, Ada received \(\frac{5}{10}\) of the remainder, while the rest went to Adeola. In what ratio was the money shared?
(a) \(\sin y=\dfrac{8}{17}\). Using a \(8\)-\(15\)-\(17\) right triangle, \(\cos y=\dfrac{15}{17}\) and \(\tan y=\dfrac{8}{15}\).
\(\dfrac{\tan y}{1+2\tan y}=\dfrac{\frac{8}{15}}{1+\frac{16}{15}}=\dfrac{\frac{8}{15}}{\frac{31}{15}}=\dfrac{8}{31}\).
(b) \(\text{N}300{,}000\) shared. Otobo \(=\text{N}60{,}000\); remainder \(=\text{N}240{,}000\).
Ada \(=\dfrac{5}{10}\times 240000=\text{N}120{,}000\); Adeola \(=240000-120000=\text{N}120{,}000\).
Ratio Otobo : Ada : Adeola \(=60000:120000:120000=\mathbf{1:2:2}\).
Détails de la réponse
(a) \(\sin y=\dfrac{8}{17}\). Using a \(8\)-\(15\)-\(17\) right triangle, \(\cos y=\dfrac{15}{17}\) and \(\tan y=\dfrac{8}{15}\).
\(\dfrac{\tan y}{1+2\tan y}=\dfrac{\frac{8}{15}}{1+\frac{16}{15}}=\dfrac{\frac{8}{15}}{\frac{31}{15}}=\dfrac{8}{31}\).
(b) \(\text{N}300{,}000\) shared. Otobo \(=\text{N}60{,}000\); remainder \(=\text{N}240{,}000\).
Ada \(=\dfrac{5}{10}\times 240000=\text{N}120{,}000\); Adeola \(=240000-120000=\text{N}120{,}000\).
Ratio Otobo : Ada : Adeola \(=60000:120000:120000=\mathbf{1:2:2}\).
Question 8 Rapport
(a) Copy and complete the table of values for \(y = 2\cos x + 3\sin x\) for \(0^\circ \geq x \geq 360^\circ\)
| x | \(0^\circ\) | \(60^\circ\) | \(120^\circ\) | \(180^\circ\) | \(240^\circ\) | \(300^\circ\) | \(360^\circ\) |
| y | 2.0 | - 3.6 |
(b) Using a scale of 2cm to \(60^\circ\) on the x-axis and 2cm to 1 unit in the y-axis, draw the graph of \(y = 2\cos x + 3\sin x\) for \(0^\circ \geq 360^\circ\)
(c) Using the graph,
(i) Solve \(2\cos x + 3\sin x = -1\)
(ii) Find, correct to one decimal place, the value of y when \(x = 342^\circ\)
(a) Completing the table for \(y = 2\cos x + 3\sin x\). Using \(\sin60^\circ=\cos30^\circ=0.866\):
| x | 0° | 60° | 120° | 180° | 240° | 300° | 360° |
|---|---|---|---|---|---|---|---|
| y | 2.0 | 3.6 | 1.6 | -2.0 | -3.6 | -1.6 | 2.0 |
Sample working: at \(x=60^\circ,\ y=2(0.5)+3(0.866)=1.0+2.6=3.6\); at \(x=300^\circ,\ y=2(0.5)+3(-0.866)=1.0-2.6=-1.6\).
(b) Graph. Using 2 cm to 60° on the x-axis and 2 cm to 1 unit on the y-axis, plot the points and join with a smooth curve. It peaks near \(x=34^\circ\) (about 3.6) and dips near \(x=214^\circ\) (about -3.6).
(c)(i) Solve \(2\cos x + 3\sin x = -1\). Draw the line \(y=-1\). Since \(2\cos x+3\sin x=\sqrt{13}\,\sin(x+33.7^\circ)\), we need \(\sin(x+33.7^\circ)=-0.277\), giving
\[x \approx 162^\circ \quad\text{and}\quad x \approx 310^\circ.\]
(c)(ii) Value of y when \(x=342^\circ\). Read up from \(x=342^\circ\):
\[y = 2\cos342^\circ + 3\sin342^\circ = 2(0.951)+3(-0.309)=1.90-0.93 \approx 1.0.\]
Détails de la réponse
(a) Completing the table for \(y = 2\cos x + 3\sin x\). Using \(\sin60^\circ=\cos30^\circ=0.866\):
| x | 0° | 60° | 120° | 180° | 240° | 300° | 360° |
|---|---|---|---|---|---|---|---|
| y | 2.0 | 3.6 | 1.6 | -2.0 | -3.6 | -1.6 | 2.0 |
Sample working: at \(x=60^\circ,\ y=2(0.5)+3(0.866)=1.0+2.6=3.6\); at \(x=300^\circ,\ y=2(0.5)+3(-0.866)=1.0-2.6=-1.6\).
(b) Graph. Using 2 cm to 60° on the x-axis and 2 cm to 1 unit on the y-axis, plot the points and join with a smooth curve. It peaks near \(x=34^\circ\) (about 3.6) and dips near \(x=214^\circ\) (about -3.6).
(c)(i) Solve \(2\cos x + 3\sin x = -1\). Draw the line \(y=-1\). Since \(2\cos x+3\sin x=\sqrt{13}\,\sin(x+33.7^\circ)\), we need \(\sin(x+33.7^\circ)=-0.277\), giving
\[x \approx 162^\circ \quad\text{and}\quad x \approx 310^\circ.\]
(c)(ii) Value of y when \(x=342^\circ\). Read up from \(x=342^\circ\):
\[y = 2\cos342^\circ + 3\sin342^\circ = 2(0.951)+3(-0.309)=1.90-0.93 \approx 1.0.\]
Question 9 Rapport
A woman bought 130 kg of tomatoes for 4452,000.00. She sold half of the tomatoes at a profit of 30%. The rest of the tomatoes began to go bad, she then reduced the selling price per kg by 12%. Calculate:
(a) the new selling price per kg;
(ii) the percentage profit on the entire sales if she threw away 5 kg of bad tomatoes.
(Reading the garbled cost as \(\text{N}52{,}000.00\) for \(130\text{ kg}\); the figures divide exactly, supporting this.)
Cost price per kg \(=\dfrac{52000}{130}=\text{N}400\).
First half (65 kg) at 30% profit. Selling price \(=400\times 1.30=\text{N}520\) per kg.
Revenue \(=65\times 520=\text{N}33{,}800\).
(a) New selling price per kg. Reduced by 12%: \(520\times(1-0.12)=520\times 0.88=\mathbf{\text{N}457.60}\) per kg.
(ii) Percentage profit on entire sales. Of the remaining 65 kg she throws away 5 kg, selling \(60\text{ kg}\) at \(\text{N}457.60\):
Revenue \(=60\times 457.60=\text{N}27{,}456\).
Total revenue \(=33800+27456=\text{N}61{,}256\); total cost \(=\text{N}52{,}000\).
Profit \(=61256-52000=\text{N}9{,}256\).
Percentage profit \(=\dfrac{9256}{52000}\times 100\approx \mathbf{17.8\%}\).
Détails de la réponse
(Reading the garbled cost as \(\text{N}52{,}000.00\) for \(130\text{ kg}\); the figures divide exactly, supporting this.)
Cost price per kg \(=\dfrac{52000}{130}=\text{N}400\).
First half (65 kg) at 30% profit. Selling price \(=400\times 1.30=\text{N}520\) per kg.
Revenue \(=65\times 520=\text{N}33{,}800\).
(a) New selling price per kg. Reduced by 12%: \(520\times(1-0.12)=520\times 0.88=\mathbf{\text{N}457.60}\) per kg.
(ii) Percentage profit on entire sales. Of the remaining 65 kg she throws away 5 kg, selling \(60\text{ kg}\) at \(\text{N}457.60\):
Revenue \(=60\times 457.60=\text{N}27{,}456\).
Total revenue \(=33800+27456=\text{N}61{,}256\); total cost \(=\text{N}52{,}000\).
Profit \(=61256-52000=\text{N}9{,}256\).
Percentage profit \(=\dfrac{9256}{52000}\times 100\approx \mathbf{17.8\%}\).
Question 10 Rapport
(a) Copy and complete the table of values for the relation \(y = 4x^2 - 8x - 21\), for \(-2.0 \leq x \leq 4.0\)
| x | -2.0 | -1.5 | -1.0 | 0.5 | 0.0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |
| y | 11 | -9 | -21 | -24 | -21 | -9 | 0 |
(b) Using a scale of 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis, draw the graph for the relation \(y = 4x^2 - 8x - 21\)
(ii) Use the graph to find the solution set of
(\(\alpha\)) \(4x^2 - 8x = 3\);
(\(\beta\)) \(4x^2 - 7x - 21 = 0\)
(a) For the missing entry between rm x=-1.0 rm and rm x=0.0 rm , the intended value is rm x=-0.5 rm . The completed table is:
| x | -2.0 | -1.5 | -1.0 | -0.5 | 0.0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| y | 11 | 0 | -9 | -16 | -21 | -24 | -25 | -24 | -21 | -16 | -9 | 0 | 11 |
For example, when rm x=-1.5 rm ,
\[y=4(-1.5)^2-8(-1.5)-21=9+12-21=0.\]
(b)(i) The graph of rm y=4x^2-8x-21 rm is shown below. The plotted curve uses the stated scales: 2 cm represents 1 unit on the rm x rm -axis and 2 cm represents 5 units on the rm y rm -axis.
(b)(ii)( rm rm )
\[4x^2-8x=3\]
Since rm y=4x^2-8x-21 rm , this gives
\[y=3-21=-18.\]
From the intersections of the curve with the line rm y=-18 rm ,
\[\boxed{x\approx -0.35\quad\text{or}\quad x\approx2.3}.\]
(b)(ii)( rm rm )
\[4x^2-7x-21=0.\]
Writing this in terms of the plotted curve,
\[4x^2-8x-21=-x,\]
so draw the line rm y=-x rm . From its intersections with the curve,
\[\boxed{x\approx-1.6\quad\text{or}\quad x\approx3.3}.\]
Détails de la réponse
(a) For the missing entry between rm x=-1.0 rm and rm x=0.0 rm , the intended value is rm x=-0.5 rm . The completed table is:
| x | -2.0 | -1.5 | -1.0 | -0.5 | 0.0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| y | 11 | 0 | -9 | -16 | -21 | -24 | -25 | -24 | -21 | -16 | -9 | 0 | 11 |
For example, when rm x=-1.5 rm ,
\[y=4(-1.5)^2-8(-1.5)-21=9+12-21=0.\]
(b)(i) The graph of rm y=4x^2-8x-21 rm is shown below. The plotted curve uses the stated scales: 2 cm represents 1 unit on the rm x rm -axis and 2 cm represents 5 units on the rm y rm -axis.
(b)(ii)( rm rm )
\[4x^2-8x=3\]
Since rm y=4x^2-8x-21 rm , this gives
\[y=3-21=-18.\]
From the intersections of the curve with the line rm y=-18 rm ,
\[\boxed{x\approx -0.35\quad\text{or}\quad x\approx2.3}.\]
(b)(ii)( rm rm )
\[4x^2-7x-21=0.\]
Writing this in terms of the plotted curve,
\[4x^2-8x-21=-x,\]
so draw the line rm y=-x rm . From its intersections with the curve,
\[\boxed{x\approx-1.6\quad\text{or}\quad x\approx3.3}.\]
Question 11 Rapport
A ladder 11m long leans against a vertical wall at an angle of 75\(^o\) to the ground. The ladder is the pushed 0.2 m up the wall.
(a) Illustrate the information in a diagram.
(b) Find correct to the nearest whole number, the:
(i) new angle which the ladder makes with the ground:
(ii) distance the foot of the ladder has moved from its original position.
(a) A right-angled triangle: the wall is vertical, the ground horizontal, and the \(11\text{ m}\) ladder is the hypotenuse making \(75^{\circ}\) with the ground. The top rests on the wall; pushing it up raises the top by \(0.2\text{ m}\) and moves the foot nearer the wall.
Original position. Height on wall \(=11\sin 75^{\circ}=11(0.9659)=10.625\text{ m}\).
Foot from wall \(=11\cos 75^{\circ}=11(0.2588)=2.847\text{ m}\).
(b)(i) New angle. New height \(=10.625+0.2=10.825\text{ m}\). The ladder length is unchanged:
\(\sin\theta=\dfrac{10.825}{11}=0.9841 \Rightarrow \theta=79.8^{\circ}\approx \mathbf{80^{\circ}}\).
(b)(ii) Distance the foot moved. New foot from wall \(=11\cos 79.8^{\circ}=11(0.1776)=1.953\text{ m}\).
Distance moved \(=2.847-1.953=0.894\text{ m}\approx \mathbf{1\ m}\).
Détails de la réponse
(a) A right-angled triangle: the wall is vertical, the ground horizontal, and the \(11\text{ m}\) ladder is the hypotenuse making \(75^{\circ}\) with the ground. The top rests on the wall; pushing it up raises the top by \(0.2\text{ m}\) and moves the foot nearer the wall.
Original position. Height on wall \(=11\sin 75^{\circ}=11(0.9659)=10.625\text{ m}\).
Foot from wall \(=11\cos 75^{\circ}=11(0.2588)=2.847\text{ m}\).
(b)(i) New angle. New height \(=10.625+0.2=10.825\text{ m}\). The ladder length is unchanged:
\(\sin\theta=\dfrac{10.825}{11}=0.9841 \Rightarrow \theta=79.8^{\circ}\approx \mathbf{80^{\circ}}\).
(b)(ii) Distance the foot moved. New foot from wall \(=11\cos 79.8^{\circ}=11(0.1776)=1.953\text{ m}\).
Distance moved \(=2.847-1.953=0.894\text{ m}\approx \mathbf{1\ m}\).
Question 12 Rapport
(a) Given that 110\(_x\) - 40\(_{five}\). find the value of x
(b) Simplify \(\frac{15}{\sqrt{75}} + \(\sqrt{108}\) + \(\sqrt{432}\), leaving the answer in the form a\(\sqrt{b}\), where a and b are positive integers.
(a) Value of x. \(110_x = 40_{five}\).
Right side: \(40_{five}=4\times 5 + 0 = 20\).
Left side: \(110_x = 1\cdot x^{2} + 1\cdot x + 0 = x^{2}+x\).
\(x^{2}+x=20 \Rightarrow x^{2}+x-20=0 \Rightarrow (x+5)(x-4)=0\).
Since a base is positive, \(x=\mathbf{4}\).
(b) Simplify \(\dfrac{15}{\sqrt{75}}+\sqrt{108}+\sqrt{432}\).
\(\dfrac{15}{\sqrt{75}}=\dfrac{15}{5\sqrt{3}}=\dfrac{3}{\sqrt{3}}=\sqrt{3}\).
\(\sqrt{108}=\sqrt{36\times 3}=6\sqrt{3}\); \(\sqrt{432}=\sqrt{144\times 3}=12\sqrt{3}\).
Sum \(=\sqrt{3}+6\sqrt{3}+12\sqrt{3}=19\sqrt{3}\).
\(a=19,\ b=3\).
Détails de la réponse
(a) Value of x. \(110_x = 40_{five}\).
Right side: \(40_{five}=4\times 5 + 0 = 20\).
Left side: \(110_x = 1\cdot x^{2} + 1\cdot x + 0 = x^{2}+x\).
\(x^{2}+x=20 \Rightarrow x^{2}+x-20=0 \Rightarrow (x+5)(x-4)=0\).
Since a base is positive, \(x=\mathbf{4}\).
(b) Simplify \(\dfrac{15}{\sqrt{75}}+\sqrt{108}+\sqrt{432}\).
\(\dfrac{15}{\sqrt{75}}=\dfrac{15}{5\sqrt{3}}=\dfrac{3}{\sqrt{3}}=\sqrt{3}\).
\(\sqrt{108}=\sqrt{36\times 3}=6\sqrt{3}\); \(\sqrt{432}=\sqrt{144\times 3}=12\sqrt{3}\).
Sum \(=\sqrt{3}+6\sqrt{3}+12\sqrt{3}=19\sqrt{3}\).
\(a=19,\ b=3\).
Question 13 Rapport
Three red balls, five green balls, and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is \(\frac{1}{6}\) find;
(a) The number of blue balls in the sack
(b) the probability of picking a green ball
3 red, 5 green, \(b\) blue. Total \(=8+b\). \(P(\text{red})=\dfrac{3}{8+b}=\dfrac{1}{6}\).
(a) \(3\times 6 = 8+b \Rightarrow 18=8+b \Rightarrow b=\mathbf{10}\) blue balls.
Total balls \(=3+5+10=18\).
(b) \(P(\text{green})=\dfrac{5}{18}\).
Détails de la réponse
3 red, 5 green, \(b\) blue. Total \(=8+b\). \(P(\text{red})=\dfrac{3}{8+b}=\dfrac{1}{6}\).
(a) \(3\times 6 = 8+b \Rightarrow 18=8+b \Rightarrow b=\mathbf{10}\) blue balls.
Total balls \(=3+5+10=18\).
(b) \(P(\text{green})=\dfrac{5}{18}\).
Question 14 Rapport
(a) Without using mathematical tables or calculator, simplify: \(\frac{log_28 + \log_216 - 4 \log_22}{\log_416}\)
(b) If 1342\(_{five}\) - 241\(_{five}\) = x\(_{ten}\), find the value of x.
(a) Evaluate each logarithm to base 2.
\(\log_{2}8=\log_{2}2^{3}=3\), \(\log_{2}16=\log_{2}2^{4}=4\), \(4\log_{2}2=4(1)=4\).
Numerator \(=3+4-4=3\).
Denominator: \(\log_{4}16=\log_{4}4^{2}=2\).
\[\frac{\log_{2}8+\log_{2}16-4\log_{2}2}{\log_{4}16}=\frac{3}{2}\]\(=\dfrac{3}{2}\)
(b) Convert each base-five number to base ten.
\[1342_{five}=1(5^{3})+3(5^{2})+4(5)+2=125+75+20+2=222\]\[241_{five}=2(5^{2})+4(5)+1=50+20+1=71\]Therefore \(x=222-71=151\).
\(x=151\)
Détails de la réponse
(a) Evaluate each logarithm to base 2.
\(\log_{2}8=\log_{2}2^{3}=3\), \(\log_{2}16=\log_{2}2^{4}=4\), \(4\log_{2}2=4(1)=4\).
Numerator \(=3+4-4=3\).
Denominator: \(\log_{4}16=\log_{4}4^{2}=2\).
\[\frac{\log_{2}8+\log_{2}16-4\log_{2}2}{\log_{4}16}=\frac{3}{2}\]\(=\dfrac{3}{2}\)
(b) Convert each base-five number to base ten.
\[1342_{five}=1(5^{3})+3(5^{2})+4(5)+2=125+75+20+2=222\]\[241_{five}=2(5^{2})+4(5)+1=50+20+1=71\]Therefore \(x=222-71=151\).
\(x=151\)
Question 15 Rapport
The ages of a group of athletes are as follows: 18, 16. 18,20, 17, 16, 19, 17, 18, 17 and 13. (a) Find the range of the distribution.
(b) Draw a frequency distribution table for the data.
(c) What is the median age?
(d) Calculate, correct to two decimal places, the;
(i) mean age:
(ii) standard deviation.
Data (11 athletes): 18, 16, 18, 20, 17, 16, 19, 17, 18, 17, 13.
(a) Range \(=\) highest \(-\) lowest \(=20-13=7\).
(b) Frequency distribution table
| Age (x) | 13 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| Frequency (f) | 1 | 2 | 3 | 3 | 1 | 1 |
(c) Median. Arranged: 13, 16, 16, 17, 17, 17, 18, 18, 18, 19, 20. With \(n=11\), the median is the \(6^{th}\) value \(= \mathbf{17}\).
(d)(i) Mean. \(\sum fx = 13+32+51+54+19+20 = 189\), \(\sum f = 11\).
\(\bar{x}=\dfrac{189}{11}=17.1818\ldots \approx \mathbf{17.18}\).
(d)(ii) Standard deviation. \(\sum fx^{2}=169+512+867+972+361+400=3281\).
\(\sigma=\sqrt{\dfrac{\sum fx^{2}}{n}-\bar{x}^{2}}=\sqrt{\dfrac{3281}{11}-\left(\dfrac{189}{11}\right)^{2}}=\sqrt{298.2727-295.2149}=\sqrt{3.0579}\).
\(\sigma \approx \mathbf{1.75}\) (2 d.p.).
Détails de la réponse
Data (11 athletes): 18, 16, 18, 20, 17, 16, 19, 17, 18, 17, 13.
(a) Range \(=\) highest \(-\) lowest \(=20-13=7\).
(b) Frequency distribution table
| Age (x) | 13 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| Frequency (f) | 1 | 2 | 3 | 3 | 1 | 1 |
(c) Median. Arranged: 13, 16, 16, 17, 17, 17, 18, 18, 18, 19, 20. With \(n=11\), the median is the \(6^{th}\) value \(= \mathbf{17}\).
(d)(i) Mean. \(\sum fx = 13+32+51+54+19+20 = 189\), \(\sum f = 11\).
\(\bar{x}=\dfrac{189}{11}=17.1818\ldots \approx \mathbf{17.18}\).
(d)(ii) Standard deviation. \(\sum fx^{2}=169+512+867+972+361+400=3281\).
\(\sigma=\sqrt{\dfrac{\sum fx^{2}}{n}-\bar{x}^{2}}=\sqrt{\dfrac{3281}{11}-\left(\dfrac{189}{11}\right)^{2}}=\sqrt{298.2727-295.2149}=\sqrt{3.0579}\).
\(\sigma \approx \mathbf{1.75}\) (2 d.p.).
Question 16 Rapport
(a) Fred bought a car for $5,600.00 and later sold it at 90% of the cost price. He spent $1,310.00 out of the amount received and invested the rest at 6% per annum simple interest. Calculate the interest earned in 3 years.
(b) Solve the equations 2\(^x\)(4\(^{-7}\)) = 2 and 3\(^{-x}\)(9\(^{2y}\)) = 3 simultaneously.
(a) Simple interest earned
Selling price \(= 90\%\) of \(\$5{,}600.00 = \dfrac{90}{100}\times 5600 = \$5{,}040.00\).
Amount left to invest \(= 5040 - 1310 = \$3{,}730.00\).
Simple interest \(I=\dfrac{PRT}{100}=\dfrac{3730\times 6\times 3}{100}=\dfrac{67140}{100}=\$671.40\).
The interest earned in 3 years is \(\$671.40\).
(b) Simultaneous indices
(The equations are read as \(2^{x}\!\cdot 4^{-y}=2\) and \(3^{-x}\!\cdot 9^{2y}=3\); a single-variable first equation would not give a genuine simultaneous system.)
Equation 1: \(2^{x}\cdot (2^{2})^{-y}=2^{1}\Rightarrow 2^{x-2y}=2^{1}\Rightarrow x-2y=1\).
Equation 2: \(3^{-x}\cdot (3^{2})^{2y}=3^{1}\Rightarrow 3^{-x+4y}=3^{1}\Rightarrow -x+4y=1\).
Adding the two: \((x-2y)+(-x+4y)=1+1\Rightarrow 2y=2\Rightarrow y=1\).
Then \(x-2(1)=1\Rightarrow x=3\).
\(x=3,\ y=1\). Check: \(2^{3}\cdot 4^{-1}=8/4=2\) and \(3^{-3}\cdot 9^{2}=81/27=3\). \(\checkmark\)
Détails de la réponse
(a) Simple interest earned
Selling price \(= 90\%\) of \(\$5{,}600.00 = \dfrac{90}{100}\times 5600 = \$5{,}040.00\).
Amount left to invest \(= 5040 - 1310 = \$3{,}730.00\).
Simple interest \(I=\dfrac{PRT}{100}=\dfrac{3730\times 6\times 3}{100}=\dfrac{67140}{100}=\$671.40\).
The interest earned in 3 years is \(\$671.40\).
(b) Simultaneous indices
(The equations are read as \(2^{x}\!\cdot 4^{-y}=2\) and \(3^{-x}\!\cdot 9^{2y}=3\); a single-variable first equation would not give a genuine simultaneous system.)
Equation 1: \(2^{x}\cdot (2^{2})^{-y}=2^{1}\Rightarrow 2^{x-2y}=2^{1}\Rightarrow x-2y=1\).
Equation 2: \(3^{-x}\cdot (3^{2})^{2y}=3^{1}\Rightarrow 3^{-x+4y}=3^{1}\Rightarrow -x+4y=1\).
Adding the two: \((x-2y)+(-x+4y)=1+1\Rightarrow 2y=2\Rightarrow y=1\).
Then \(x-2(1)=1\Rightarrow x=3\).
\(x=3,\ y=1\). Check: \(2^{3}\cdot 4^{-1}=8/4=2\) and \(3^{-3}\cdot 9^{2}=81/27=3\). \(\checkmark\)
Question 17 Rapport
The cost of dinner for a group of tourist is partly constant and partly varies as the number of tourists present. It costs $740.00 when 20 tourists were present and $960.00 when the number of tourists increased by 10. Find the cost of the dinner when only 15 tourists were present.
The cost \(C\) is partly constant and partly varies as the number of tourists \(n\), so
\[C=a+bn\]where \(a\) (the constant part) and \(b\) (the rate per tourist) are constants.
When \(n=20\), \(C=740\): \(\;740=a+20b\).
When the number increases by 10, \(n=30\), \(C=960\): \(\;960=a+30b\).
Subtract the first equation from the second:
\[960-740=(a+30b)-(a+20b)\Rightarrow 220=10b\Rightarrow b=22\]Then \(a=740-20(22)=740-440=300\).
So \(C=300+22n\). When \(n=15\):
\[C=300+22(15)=300+330=\$630\]The dinner costs \(\$630.00\) when 15 tourists are present.
Détails de la réponse
The cost \(C\) is partly constant and partly varies as the number of tourists \(n\), so
\[C=a+bn\]where \(a\) (the constant part) and \(b\) (the rate per tourist) are constants.
When \(n=20\), \(C=740\): \(\;740=a+20b\).
When the number increases by 10, \(n=30\), \(C=960\): \(\;960=a+30b\).
Subtract the first equation from the second:
\[960-740=(a+30b)-(a+20b)\Rightarrow 220=10b\Rightarrow b=22\]Then \(a=740-20(22)=740-440=300\).
So \(C=300+22n\). When \(n=15\):
\[C=300+22(15)=300+330=\$630\]The dinner costs \(\$630.00\) when 15 tourists are present.
Question 18 Rapport
The data shows the marks obtained by students in a biology test
| 52 | 56 | 25 | 56 | 68 | 73 | 66 | 64 | 56 | 48 |
| 20 | 39 | 9 | 50 | 46 | 54 | 54 | 40 | 50 | 96 |
| 36 | 44 | 18 | 97 | 65 | 21 | 60 | 44 | 54 | 32 |
| 92 | 49 | 37 | 94 | 72 | 88 | 89 | 35 | 59 | 34 |
| 15 | 88 | 53 | 16 | 84 | 52 | 72 | 46 | 60 | 42 |
(a) Construct a frequency distribution table using the class interval 0 - 9, 10 - 19, 19, 20, 29...
(b) Draw a cumulative frequency curve for the distribution
(c) Use the graph to estimate the;
(i) median
(ii) Percentage of students who scored at least 66 marks, correct to the nearest whole number.
(a) Frequency distribution
| Class interval | Class boundaries | Frequency, \(f\) | Cumulative frequency, \(cf\) |
|---|---|---|---|
| 0 - 9 | -0.5 - 9.5 | 1 | 1 |
| 10 - 19 | 9.5 - 19.5 | 3 | 4 |
| 20 - 29 | 19.5 - 29.5 | 3 | 7 |
| 30 - 39 | 29.5 - 39.5 | 6 | 13 |
| 40 - 49 | 39.5 - 49.5 | 8 | 21 |
| 50 - 59 | 49.5 - 59.5 | 12 | 33 |
| 60 - 69 | 59.5 - 69.5 | 6 | 39 |
| 70 - 79 | 69.5 - 79.5 | 3 | 42 |
| 80 - 89 | 79.5 - 89.5 | 4 | 46 |
| 90 - 99 | 89.5 - 99.5 | 4 | 50 |
| Total | 50 | ||
(b) Cumulative frequency curve (less-than ogive)
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\), and joined by a smooth curve.
(c)(i) Median
There are \(50\) students, so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading from \(cf=25\) on the ogive gives a mark of approximately \(53\).
Therefore, the estimated median mark is \(53\) marks.
(c)(ii) Percentage scoring at least 66 marks
From the ogive, the cumulative frequency at \(66\) marks is approximately \(37\). Hence, the estimated number scoring at least \(66\) marks is
\[50-37=13.\]
\[\text{Percentage}=\frac{13}{50}\times100\%=26\%.\]
Therefore, the percentage is \(26\%\), correct to the nearest whole number.
Détails de la réponse
(a) Frequency distribution
| Class interval | Class boundaries | Frequency, \(f\) | Cumulative frequency, \(cf\) |
|---|---|---|---|
| 0 - 9 | -0.5 - 9.5 | 1 | 1 |
| 10 - 19 | 9.5 - 19.5 | 3 | 4 |
| 20 - 29 | 19.5 - 29.5 | 3 | 7 |
| 30 - 39 | 29.5 - 39.5 | 6 | 13 |
| 40 - 49 | 39.5 - 49.5 | 8 | 21 |
| 50 - 59 | 49.5 - 59.5 | 12 | 33 |
| 60 - 69 | 59.5 - 69.5 | 6 | 39 |
| 70 - 79 | 69.5 - 79.5 | 3 | 42 |
| 80 - 89 | 79.5 - 89.5 | 4 | 46 |
| 90 - 99 | 89.5 - 99.5 | 4 | 50 |
| Total | 50 | ||
(b) Cumulative frequency curve (less-than ogive)
The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\), and joined by a smooth curve.
(c)(i) Median
There are \(50\) students, so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading from \(cf=25\) on the ogive gives a mark of approximately \(53\).
Therefore, the estimated median mark is \(53\) marks.
(c)(ii) Percentage scoring at least 66 marks
From the ogive, the cumulative frequency at \(66\) marks is approximately \(37\). Hence, the estimated number scoring at least \(66\) marks is
\[50-37=13.\]
\[\text{Percentage}=\frac{13}{50}\times100\%=26\%.\]
Therefore, the percentage is \(26\%\), correct to the nearest whole number.
Question 19 Rapport
(a) Solve the inequality: \(\frac{1 + 4x}{2}\) -\(\frac{5 + 2x}{7}\) < x -2
(b) If x: y = 3: 5, find the value of \(\frac{2x^2 - y^2}{y^2 - x^2}\)
(a) Solve \(\dfrac{1+4x}{2}-\dfrac{5+2x}{7} \(x<-2.5\) (b) \(x:y=3:5\), so let \(x=3k,\ y=5k\). \(\dfrac{2x^{2}-y^{2}}{y^{2}-x^{2}}=-\dfrac{7}{16}\)
Détails de la réponse
(a) Solve \(\dfrac{1+4x}{2}-\dfrac{5+2x}{7} \(x<-2.5\) (b) \(x:y=3:5\), so let \(x=3k,\ y=5k\). \(\dfrac{2x^{2}-y^{2}}{y^{2}-x^{2}}=-\dfrac{7}{16}\)
Question 20 Rapport
(a)
In the diagram. \(\over{Rs}\) and \(\over{RT}\) are tangent to the circle with centre O, < TUS = 68\(^o\), < SRT = x and < UTO = y. Find the value of x.
(b) Two tanks A and B are filled to capacity with diesel. Tank A holds 600 litres diesel more than tank B. If 100 litres of diesel was pumped cut of each tank, tank A would then contain 3 times as much as tank B. Find the capacity of each tank.
(a) Finding \(x = \angle SRT\).
From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point \(R\), touching the circle (centre \(O\)) at \(T\) and \(S\). \(\angle TUS = 68^\circ\) is the angle at the circumference standing on chord \(TS\).
The angle subtended at the centre by the same chord \(TS\) is twice the angle at the circumference:
\[ \angle TOS = 2 \times \angle TUS = 2 \times 68^\circ = 136^\circ \]
Since a tangent is perpendicular to the radius at the point of contact:
\[ \angle OTR = \angle OSR = 90^\circ \]
Now consider quadrilateral \(RTOS\). Its interior angles sum to \(360^\circ\):
\[ \angle SRT + \angle OTR + \angle TOS + \angle OSR = 360^\circ \]
\[ x + 90^\circ + 136^\circ + 90^\circ = 360^\circ \]
\[ x = 360^\circ - 316^\circ = 44^\circ \]
(b) Capacity of each tank. Let the capacity of tank \(B\) be \(b\) litres. Then tank \(A\) holds \((b + 600)\) litres.
After pumping out \(100\) litres from each, tank \(A\) has \(3\) times as much as tank \(B\):
\[ (b + 600) - 100 = 3\,(b - 100) \]
\[ b + 500 = 3b - 300 \]
\[ 500 + 300 = 3b - b \]
\[ 800 = 2b \;\Rightarrow\; b = 400 \]
So tank \(B = 400\) litres and tank \(A = 400 + 600 = 1000\) litres.
Check: \(1000 - 100 = 900\) and \(3(400 - 100) = 3 \times 300 = 900.\) Correct.
Answers: (a) \(x = 44^\circ\); (b) Tank A \(= 1000\) litres, Tank B \(= 400\) litres.
Détails de la réponse
(a) Finding \(x = \angle SRT\).
From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point \(R\), touching the circle (centre \(O\)) at \(T\) and \(S\). \(\angle TUS = 68^\circ\) is the angle at the circumference standing on chord \(TS\).
The angle subtended at the centre by the same chord \(TS\) is twice the angle at the circumference:
\[ \angle TOS = 2 \times \angle TUS = 2 \times 68^\circ = 136^\circ \]
Since a tangent is perpendicular to the radius at the point of contact:
\[ \angle OTR = \angle OSR = 90^\circ \]
Now consider quadrilateral \(RTOS\). Its interior angles sum to \(360^\circ\):
\[ \angle SRT + \angle OTR + \angle TOS + \angle OSR = 360^\circ \]
\[ x + 90^\circ + 136^\circ + 90^\circ = 360^\circ \]
\[ x = 360^\circ - 316^\circ = 44^\circ \]
(b) Capacity of each tank. Let the capacity of tank \(B\) be \(b\) litres. Then tank \(A\) holds \((b + 600)\) litres.
After pumping out \(100\) litres from each, tank \(A\) has \(3\) times as much as tank \(B\):
\[ (b + 600) - 100 = 3\,(b - 100) \]
\[ b + 500 = 3b - 300 \]
\[ 500 + 300 = 3b - b \]
\[ 800 = 2b \;\Rightarrow\; b = 400 \]
So tank \(B = 400\) litres and tank \(A = 400 + 600 = 1000\) litres.
Check: \(1000 - 100 = 900\) and \(3(400 - 100) = 3 \times 300 = 900.\) Correct.
Answers: (a) \(x = 44^\circ\); (b) Tank A \(= 1000\) litres, Tank B \(= 400\) litres.
Question 21 Rapport
In the diagram, PQRS is a quadrilateral, < PQR = < PRS = 90\(^o\), |PQ| =3cm, |QR| = 4cm and |PS| = 13 cm. Find the area of the quadrilateral.
From the diagram, \(PQRS\) is split by the diagonal \(PR\) into two right-angled triangles: triangle \(PQR\) (right-angled at \(Q\)) and triangle \(PRS\) (right-angled at \(R\)). Given \(|PQ| = 3\) cm, \(|QR| = 4\) cm and \(|PS| = 13\) cm.
Step 1 - find the diagonal \(|PR|\). In right-angled triangle \(PQR\) (\(\angle PQR = 90^{\circ}\)), \(PR\) is the hypotenuse:
\[ |PR| = \sqrt{|PQ|^2 + |QR|^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}. \]
Step 2 - find \(|RS|\). In right-angled triangle \(PRS\) (\(\angle PRS = 90^{\circ}\)), \(PS\) is the hypotenuse:
\[ |RS| = \sqrt{|PS|^2 - |PR|^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}. \]
Step 3 - add the two triangle areas.
\[ \text{Area of } PQR = \tfrac{1}{2}\times |PQ| \times |QR| = \tfrac{1}{2}\times 3 \times 4 = 6\text{ cm}^2. \]
\[ \text{Area of } PRS = \tfrac{1}{2}\times |PR| \times |RS| = \tfrac{1}{2}\times 5 \times 12 = 30\text{ cm}^2. \]
\[ \text{Area of } PQRS = 6 + 30 = 36\text{ cm}^2. \]
Area of the quadrilateral \(= 36\) cm\(^2\).
Détails de la réponse
From the diagram, \(PQRS\) is split by the diagonal \(PR\) into two right-angled triangles: triangle \(PQR\) (right-angled at \(Q\)) and triangle \(PRS\) (right-angled at \(R\)). Given \(|PQ| = 3\) cm, \(|QR| = 4\) cm and \(|PS| = 13\) cm.
Step 1 - find the diagonal \(|PR|\). In right-angled triangle \(PQR\) (\(\angle PQR = 90^{\circ}\)), \(PR\) is the hypotenuse:
\[ |PR| = \sqrt{|PQ|^2 + |QR|^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}. \]
Step 2 - find \(|RS|\). In right-angled triangle \(PRS\) (\(\angle PRS = 90^{\circ}\)), \(PS\) is the hypotenuse:
\[ |RS| = \sqrt{|PS|^2 - |PR|^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}. \]
Step 3 - add the two triangle areas.
\[ \text{Area of } PQR = \tfrac{1}{2}\times |PQ| \times |QR| = \tfrac{1}{2}\times 3 \times 4 = 6\text{ cm}^2. \]
\[ \text{Area of } PRS = \tfrac{1}{2}\times |PR| \times |RS| = \tfrac{1}{2}\times 5 \times 12 = 30\text{ cm}^2. \]
\[ \text{Area of } PQRS = 6 + 30 = 36\text{ cm}^2. \]
Area of the quadrilateral \(= 36\) cm\(^2\).
Question 22 Rapport
A survey of 40 students showed that 23 students study Mathematics, 5 study Mathematics and Physics, 8 study Chemistry and Mathematics, 5 study Physics and Chemistry and 3 study all the three subjects. The number of students who study Physics only is twice the number who study Chemistry only.
(a) Find the number of students who study:
(i) only Physics.
(ii) only one subject
b) What is the probability that a student selected at random studies exactly 2 subjects?
Set notation and the Venn diagram
The numbers studying two subjects include the 3 students who study all three subjects. Therefore, subtract 3 from each two-subject total to find the regions for exactly two subjects.
Finding the regions
- Mathematics and Physics only: \(5-3=2\).
- Mathematics and Chemistry only: \(8-3=5\).
- Physics and Chemistry only: \(5-3=2\).
There are 23 students studying Mathematics in total, so:
\[ \text{Mathematics only}=23-(2+5+3)=13. \]
Let the number studying Chemistry only be \(x\). The question states that Physics only is twice this number, so Physics only is \(2x\).
Using the intended interpretation that all 40 surveyed students are in the three-subject Venn diagram:
\[13+2+5+2+3+x+2x=40\]
\[25+3x=40\]
\[3x=15,\qquad x=5.\]
Therefore:
- Only Physics \(=2x=2(5)=\mathbf{10}\) students.
- Only one subject \(=13+10+5=\mathbf{28}\) students.
Students who study exactly two subjects are in the three pair-only regions:
\[2+5+2=9.\]
Hence the required probability is:
\[P(\text{exactly two subjects})=\frac{9}{40}=0.225.\]
The denominator must be \(40\), the total number of surveyed students. Although the supplied reference includes \(\frac{9}{48}\), this is inconsistent with both the survey total and the decimal \(0.225\); \(\frac{9}{40}\) is the correct probability.
Examination reminder: When a total for two subjects includes students studying all three, subtract the all-three region before counting students who study exactly two subjects.
Détails de la réponse
Set notation and the Venn diagram
The numbers studying two subjects include the 3 students who study all three subjects. Therefore, subtract 3 from each two-subject total to find the regions for exactly two subjects.
Finding the regions
- Mathematics and Physics only: \(5-3=2\).
- Mathematics and Chemistry only: \(8-3=5\).
- Physics and Chemistry only: \(5-3=2\).
There are 23 students studying Mathematics in total, so:
\[ \text{Mathematics only}=23-(2+5+3)=13. \]
Let the number studying Chemistry only be \(x\). The question states that Physics only is twice this number, so Physics only is \(2x\).
Using the intended interpretation that all 40 surveyed students are in the three-subject Venn diagram:
\[13+2+5+2+3+x+2x=40\]
\[25+3x=40\]
\[3x=15,\qquad x=5.\]
Therefore:
- Only Physics \(=2x=2(5)=\mathbf{10}\) students.
- Only one subject \(=13+10+5=\mathbf{28}\) students.
Students who study exactly two subjects are in the three pair-only regions:
\[2+5+2=9.\]
Hence the required probability is:
\[P(\text{exactly two subjects})=\frac{9}{40}=0.225.\]
The denominator must be \(40\), the total number of surveyed students. Although the supplied reference includes \(\frac{9}{48}\), this is inconsistent with both the survey total and the decimal \(0.225\); \(\frac{9}{40}\) is the correct probability.
Examination reminder: When a total for two subjects includes students studying all three, subtract the all-three region before counting students who study exactly two subjects.
Question 23 Rapport
(a) The third and sixth terms of a Geometric Progression (G.P) are and \(\frac{1}{4}\) and \(\frac{1}{32}\) respectively.
Find:
(i) the first term and the common ratio;
(ii) the seventh term.
(b) Given that 2 and -3 are the roots of the equation ax\(^2\) ± bx + c = 0, find the values of a, b and c.
(a) Geometric Progression. \(T_3=ar^{2}=\dfrac{1}{4}\) and \(T_6=ar^{5}=\dfrac{1}{32}\).
Divide: \(\dfrac{ar^{5}}{ar^{2}}=r^{3}=\dfrac{1/32}{1/4}=\dfrac{1}{8}\Rightarrow r=\dfrac{1}{2}\).
Then \(a\left(\tfrac{1}{2}\right)^{2}=\tfrac{1}{4}\Rightarrow \tfrac{a}{4}=\tfrac{1}{4}\Rightarrow a=1\).
(i) First term \(a=1\), common ratio \(r=\dfrac{1}{2}\).
(ii) \(T_7=ar^{6}=1\cdot\left(\dfrac{1}{2}\right)^{6}=\dfrac{1}{64}\).
(b) If \(2\) and \(-3\) are roots of \(ax^{2}+bx+c=0\):
Sum of roots \(=-\dfrac{b}{a}=2+(-3)=-1\); product \(=\dfrac{c}{a}=2\times(-3)=-6\).
Taking \(a=1\): \(b=1,\ c=-6\). So \(\mathbf{a=1,\ b=1,\ c=-6}\) (equation \(x^{2}+x-6=0\)).
Détails de la réponse
(a) Geometric Progression. \(T_3=ar^{2}=\dfrac{1}{4}\) and \(T_6=ar^{5}=\dfrac{1}{32}\).
Divide: \(\dfrac{ar^{5}}{ar^{2}}=r^{3}=\dfrac{1/32}{1/4}=\dfrac{1}{8}\Rightarrow r=\dfrac{1}{2}\).
Then \(a\left(\tfrac{1}{2}\right)^{2}=\tfrac{1}{4}\Rightarrow \tfrac{a}{4}=\tfrac{1}{4}\Rightarrow a=1\).
(i) First term \(a=1\), common ratio \(r=\dfrac{1}{2}\).
(ii) \(T_7=ar^{6}=1\cdot\left(\dfrac{1}{2}\right)^{6}=\dfrac{1}{64}\).
(b) If \(2\) and \(-3\) are roots of \(ax^{2}+bx+c=0\):
Sum of roots \(=-\dfrac{b}{a}=2+(-3)=-1\); product \(=\dfrac{c}{a}=2\times(-3)=-6\).
Taking \(a=1\): \(b=1,\ c=-6\). So \(\mathbf{a=1,\ b=1,\ c=-6}\) (equation \(x^{2}+x-6=0\)).
Question 24 Rapport
The force of attraction F, between two bodies, varies directly as the product of their masses, \(m_1\) and m\(_2\) and inversely as the square of the distance, d, between them. Given that F = 20N, when m\(_1\) = 25kg, m\(_2\) = 10kg and d = 5m, find:
(I) an expression for F in terms of m\(_1\), m\(_2\) and d;
(ii) the distance, d for F = 30N, m\(_1\) = 7.5kg and m\(_2\) = 4kg
(b) Find the value of x in the diagram
(a) Setting up the joint variation
F varies directly as \(m_1 m_2\) and inversely as \(d^2\):
\[F = \frac{k\, m_1 m_2}{d^2}\]
(i) Finding k and the expression
Substitute \(F = 20,\ m_1 = 25,\ m_2 = 10,\ d = 5\):
\[20 = \frac{k(25)(10)}{5^2} = \frac{250k}{25} = 10k\]
\[k = 2\]
Therefore the required expression is
\[F = \frac{2\,m_1 m_2}{d^2}\]
(ii) Finding d
Substitute \(F = 30,\ m_1 = 7.5,\ m_2 = 4\):
\[30 = \frac{2(7.5)(4)}{d^2} = \frac{60}{d^2}\]
\[d^2 = \frac{60}{30} = 2\]
\[d = \sqrt{2} \approx 1.41\ \text{m}\]
(b) Finding x in the diagram
The figure is a five-sided polygon (pentagon) whose interior angles, read from the diagram, are:
\[x,\quad (x+20^\circ),\quad (x+40^\circ),\quad (x+80^\circ),\quad (x+60^\circ)\]
The sum of the interior angles of a pentagon is
\[(5-2)\times 180^\circ = 540^\circ\]
So:
\[x + (x+20^\circ) + (x+40^\circ) + (x+80^\circ) + (x+60^\circ) = 540^\circ\]
\[5x + 200^\circ = 540^\circ\]
\[5x = 340^\circ\]
\[x = 68^\circ\]
Détails de la réponse
(a) Setting up the joint variation
F varies directly as \(m_1 m_2\) and inversely as \(d^2\):
\[F = \frac{k\, m_1 m_2}{d^2}\]
(i) Finding k and the expression
Substitute \(F = 20,\ m_1 = 25,\ m_2 = 10,\ d = 5\):
\[20 = \frac{k(25)(10)}{5^2} = \frac{250k}{25} = 10k\]
\[k = 2\]
Therefore the required expression is
\[F = \frac{2\,m_1 m_2}{d^2}\]
(ii) Finding d
Substitute \(F = 30,\ m_1 = 7.5,\ m_2 = 4\):
\[30 = \frac{2(7.5)(4)}{d^2} = \frac{60}{d^2}\]
\[d^2 = \frac{60}{30} = 2\]
\[d = \sqrt{2} \approx 1.41\ \text{m}\]
(b) Finding x in the diagram
The figure is a five-sided polygon (pentagon) whose interior angles, read from the diagram, are:
\[x,\quad (x+20^\circ),\quad (x+40^\circ),\quad (x+80^\circ),\quad (x+60^\circ)\]
The sum of the interior angles of a pentagon is
\[(5-2)\times 180^\circ = 540^\circ\]
So:
\[x + (x+20^\circ) + (x+40^\circ) + (x+80^\circ) + (x+60^\circ) = 540^\circ\]
\[5x + 200^\circ = 540^\circ\]
\[5x = 340^\circ\]
\[x = 68^\circ\]
Question 25 Rapport
In the diagram, \(\overline{RT}\) and \(\overline{RT}\) are tangent to the circle with centre O. < TUS = 68 °, < SRT = x, and < UTO = y. Find the value of x.
(b) Two tanks A and B am filled to capacity with diesel. Tank A holds 600 litres of diesel more than tank B. If 100 litres of diesel was pumped out of each tank, tank A would then contain 3 times as much diesel as tank B. Find the capacity of each tank.
(a) Finding x
From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point R, touching the circle (centre O) at T and S. U lies on the circle, with the inscribed angle \(\angle TUS = 68^\circ\), and \(\angle SRT = x\).
The inscribed angle \(\angle TUS\) stands on the arc \(TS\) that faces R (the arc not containing U). By the inscribed-angle theorem, the central angle on that same arc is
\[\angle TOS = 2\times \angle TUS = 2\times 68^\circ = 136^\circ\]In quadrilateral \(OTRS\), the radii meet the tangents at right angles, so \(\angle OTR = \angle OSR = 90^\circ\). The angles of the quadrilateral sum to \(360^\circ\):
\[\angle TOS + \angle OTR + \angle SRT + \angle OSR = 360^\circ\]\[136^\circ + 90^\circ + x + 90^\circ = 360^\circ\]\[x = 360^\circ - 316^\circ = 44^\circ\]Therefore \(x = 44^\circ\).
(b) Capacity of the two diesel tanks
Let the capacity of tank B be \(b\) litres. Then tank A holds \(b + 600\) litres.
After pumping out 100 litres from each, tank A contains 3 times as much as tank B:
\[(b + 600) - 100 = 3\big(b - 100\big)\]\[b + 500 = 3b - 300\]\[500 + 300 = 3b - b\]\[800 = 2b \;\Rightarrow\; b = 400\]So tank B holds \(400\) litres and tank A holds \(400 + 600 = 1000\) litres.
Check: after pumping, A = \(900\), B = \(300\), and \(900 = 3\times 300\). Correct.
Tank A has a capacity of \(1000\) litres and tank B has a capacity of \(400\) litres.
Détails de la réponse
(a) Finding x
From the diagram, \(RT\) and \(RS\) are tangents drawn from the external point R, touching the circle (centre O) at T and S. U lies on the circle, with the inscribed angle \(\angle TUS = 68^\circ\), and \(\angle SRT = x\).
The inscribed angle \(\angle TUS\) stands on the arc \(TS\) that faces R (the arc not containing U). By the inscribed-angle theorem, the central angle on that same arc is
\[\angle TOS = 2\times \angle TUS = 2\times 68^\circ = 136^\circ\]In quadrilateral \(OTRS\), the radii meet the tangents at right angles, so \(\angle OTR = \angle OSR = 90^\circ\). The angles of the quadrilateral sum to \(360^\circ\):
\[\angle TOS + \angle OTR + \angle SRT + \angle OSR = 360^\circ\]\[136^\circ + 90^\circ + x + 90^\circ = 360^\circ\]\[x = 360^\circ - 316^\circ = 44^\circ\]Therefore \(x = 44^\circ\).
(b) Capacity of the two diesel tanks
Let the capacity of tank B be \(b\) litres. Then tank A holds \(b + 600\) litres.
After pumping out 100 litres from each, tank A contains 3 times as much as tank B:
\[(b + 600) - 100 = 3\big(b - 100\big)\]\[b + 500 = 3b - 300\]\[500 + 300 = 3b - b\]\[800 = 2b \;\Rightarrow\; b = 400\]So tank B holds \(400\) litres and tank A holds \(400 + 600 = 1000\) litres.
Check: after pumping, A = \(900\), B = \(300\), and \(900 = 3\times 300\). Correct.
Tank A has a capacity of \(1000\) litres and tank B has a capacity of \(400\) litres.
Question 26 Rapport
(a) Find the equation of the line which passes through the points A(-2, 7) and B(2, -3)
(b) Given that \(\frac{5b - a}{8b + 3a} = \frac{1}{5}\) = find, correct to two decimal places, the value \(\frac{a}{b}\)
(a) Equation of the line through \(A(-2,7)\) and \(B(2,-3)\).
Gradient \(m=\dfrac{-3-7}{2-(-2)}=\dfrac{-10}{4}=-\dfrac{5}{2}\).
\(y-7=-\dfrac{5}{2}(x+2)\Rightarrow y=-\dfrac{5}{2}x-5+7=-\dfrac{5}{2}x+2\).
\(y=-\dfrac{5}{2}x+2\), i.e. \(5x+2y-4=0\).
(b) \(\dfrac{5b-a}{8b+3a}=\dfrac{1}{5}\).
Cross-multiply: \(5(5b-a)=8b+3a \Rightarrow 25b-5a=8b+3a\).
\(25b-8b=3a+5a \Rightarrow 17b=8a \Rightarrow \dfrac{a}{b}=\dfrac{17}{8}=2.125\).
\(\dfrac{a}{b}\approx 2.13\) (2 d.p.).
Détails de la réponse
(a) Equation of the line through \(A(-2,7)\) and \(B(2,-3)\).
Gradient \(m=\dfrac{-3-7}{2-(-2)}=\dfrac{-10}{4}=-\dfrac{5}{2}\).
\(y-7=-\dfrac{5}{2}(x+2)\Rightarrow y=-\dfrac{5}{2}x-5+7=-\dfrac{5}{2}x+2\).
\(y=-\dfrac{5}{2}x+2\), i.e. \(5x+2y-4=0\).
(b) \(\dfrac{5b-a}{8b+3a}=\dfrac{1}{5}\).
Cross-multiply: \(5(5b-a)=8b+3a \Rightarrow 25b-5a=8b+3a\).
\(25b-8b=3a+5a \Rightarrow 17b=8a \Rightarrow \dfrac{a}{b}=\dfrac{17}{8}=2.125\).
\(\dfrac{a}{b}\approx 2.13\) (2 d.p.).
Question 27 Rapport
a) A twenty - kilogram bag of rice is consumed by m number of boys in 10 days. When four more boys joined them, the same quantity of rice lasted only 8 days. If the rate of consumption is the same, find the value of m.
(b) If \(\frac{5}{6}\) of a number is 10 greater than \(\frac{1}{3}\) of it. find the number
(c) Find the equation of the line which passes through the points (2, \(\frac{1}{2}\)) and (-1, -\(\frac{1}{2}\)
(a) Value of m. The same quantity of rice is fixed, so boy-days are equal:
\(m\times 10 = (m+4)\times 8 \Rightarrow 10m = 8m+32 \Rightarrow 2m=32 \Rightarrow m=\mathbf{16}\).
(b) The number. Let the number be \(n\):
\(\dfrac{5}{6}n = \dfrac{1}{3}n + 10 \Rightarrow \left(\dfrac{5}{6}-\dfrac{2}{6}\right)n = 10 \Rightarrow \dfrac{1}{2}n = 10 \Rightarrow n=\mathbf{20}\).
(c) Equation of the line through \(\left(2,\tfrac{1}{2}\right)\) and \(\left(-1,-\tfrac{1}{2}\right)\).
Gradient \(m=\dfrac{\frac{1}{2}-(-\frac{1}{2})}{2-(-1)}=\dfrac{1}{3}\).
\(y-\dfrac{1}{2}=\dfrac{1}{3}(x-2)\Rightarrow y=\dfrac{1}{3}x-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{1}{3}x-\dfrac{1}{6}\).
\(y=\dfrac{1}{3}x-\dfrac{1}{6}\), i.e. \(2x-6y-1=0\).
Détails de la réponse
(a) Value of m. The same quantity of rice is fixed, so boy-days are equal:
\(m\times 10 = (m+4)\times 8 \Rightarrow 10m = 8m+32 \Rightarrow 2m=32 \Rightarrow m=\mathbf{16}\).
(b) The number. Let the number be \(n\):
\(\dfrac{5}{6}n = \dfrac{1}{3}n + 10 \Rightarrow \left(\dfrac{5}{6}-\dfrac{2}{6}\right)n = 10 \Rightarrow \dfrac{1}{2}n = 10 \Rightarrow n=\mathbf{20}\).
(c) Equation of the line through \(\left(2,\tfrac{1}{2}\right)\) and \(\left(-1,-\tfrac{1}{2}\right)\).
Gradient \(m=\dfrac{\frac{1}{2}-(-\frac{1}{2})}{2-(-1)}=\dfrac{1}{3}\).
\(y-\dfrac{1}{2}=\dfrac{1}{3}(x-2)\Rightarrow y=\dfrac{1}{3}x-\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{1}{3}x-\dfrac{1}{6}\).
\(y=\dfrac{1}{3}x-\dfrac{1}{6}\), i.e. \(2x-6y-1=0\).
