WAEC SSCE - General Mathematics - 2019 (Essay)

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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\(^2\).

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(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 \(x=222-71=151\).

\(x=151\)

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(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 \(x^{2}+x-6=0\)).

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(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 \(3^{-3}\cdot 9^{2}=81/27=3\). \(\checkmark\)

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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.

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(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: \(\dfrac{1}{8}=2^{-3},\ 32=2^{5}\).

\(2^{1}\cdot 2^{-3x}=2^{5(x-1)} \Rightarrow 2^{1-3x}=2^{5x-5}\).

\(1-3x=5x-5 \Rightarrow 6=8x \Rightarrow x=\dfrac{3}{4}\).

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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.).

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(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 \(x=342^\circ\):

\[y = 2\cos342^\circ + 3\sin342^\circ = 2(0.951)+3(-0.309)=1.90-0.93 \approx 1.0.\]

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(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.

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(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:

\[x + (x+20^\circ) + (x+40^\circ) + (x+80^\circ) + (x+60^\circ) = 540^\circ\]

\[5x + 200^\circ = 540^\circ\]

\[5x = 340^\circ\]

\[x = 68^\circ\]

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(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 \(=\sqrt{3}+6\sqrt{3}+12\sqrt{3}=19\sqrt{3}\).

\(a=19,\ b=3\).

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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.

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(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 \(=60000:120000:120000=\mathbf{1:2:2}\).

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(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.

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(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 \(=\dfrac{9256}{52000}\times 100\approx \mathbf{17.8\%}\).

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(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.).

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(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 \(BA\): \((8)(17)-(21)(7)=136-147=\mathbf{-11}\).

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(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:

\[-6+16\ge 36x-x\;\Rightarrow\;10\ge 35x.\]

\[x\le\frac{10}{35}=\frac{2}{7}.\]

\(\therefore x\le\dfrac{2}{7}.\)

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(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

\[35^{\circ}+41.87^{\circ}=76.87^{\circ}\text{ (2 d.p.).}\]

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(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 \(\angle BAC = 23^\circ\).

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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\(=7\).)

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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) \(P(\text{green})=\dfrac{5}{18}\).

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(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 \(=2.847-1.953=0.894\text{ m}\approx \mathbf{1\ m}\).

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(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 \(\mathbf{185^\circ}\).

Antwortdetails

(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. \(2x-6y-1=0\).

Antwortdetails

(a) Solve \(\dfrac{1+4x}{2}-\dfrac{5+2x}{7}\[7(1+4x)-2(5+2x)<14(x-2)\]\[7+28x-10-4x<14x-28\]\[24x-3<14x-28\]\[24x-14x<-28+3\]\[10x<-25\Rightarrow x<-2.5\]

\(x<-2.5\)

(b) \(x:y=3:5\), so let \(x=3k,\ y=5k\).

\[\frac{2x^{2}-y^{2}}{y^{2}-x^{2}}=\frac{2(3k)^{2}-(5k)^{2}}{(5k)^{2}-(3k)^{2}}=\frac{2(9k^{2})-25k^{2}}{25k^{2}-9k^{2}}=\frac{18k^{2}-25k^{2}}{16k^{2}}=\frac{-7k^{2}}{16k^{2}}=-\frac{7}{16}\]

\(\dfrac{2x^{2}-y^{2}}{y^{2}-x^{2}}=-\dfrac{7}{16}\)