WAEC SSCE - General Mathematics - 2001 (Essay)

Detalhes da Resposta

Given \(\log_{10}2=0.3010,\ \log_{10}3=0.4771\).

(a) \(54=2\times 27=2\times 3^{3}\).

\(\log_{10}54=\log_{10}2+3\log_{10}3=0.3010+3(0.4771)=0.3010+1.4313=\mathbf{1.7323}\).

(b) \(0.24=\dfrac{24}{100}=\dfrac{2^{3}\times 3}{100}\).

\(\log_{10}24=3\log_{10}2+\log_{10}3=0.9030+0.4771=1.3801\).

\(\log_{10}0.24=\log_{10}24-\log_{10}100=1.3801-2=\mathbf{-0.6199}\) (i.e. \(\bar{1}.3801\)).

Detalhes da Resposta

From the diagram: \(AB=15\text{ m}\) on a bearing of \(053^\circ\), \(BC=18\text{ m}\) on a bearing of \(161^\circ\), the pole BT is vertical at B, and \(\angle ATB=58^\circ\).

(a) Length of AC. First find \(\angle ABC\). The bearing of A from B is the back-bearing of \(053^\circ\):

\[053^\circ+180^\circ=233^\circ\]

The bearing of C from B is \(161^\circ\), so

\[\angle ABC=233^\circ-161^\circ=72^\circ\]

Apply the cosine rule to triangle ABC:

\[AC^2=AB^2+BC^2-2\,(AB)(BC)\cos(\angle ABC)\]\[AC^2=15^2+18^2-2(15)(18)\cos72^\circ\]\[AC^2=225+324-540(0.30902)=382.13\]\[AC=19.5\text{ m (3 s.f.)}\]

(b) Bearing of C from A. Use the sine rule to find \(\angle BAC\):

\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\]\[\sin\angle BAC=\frac{18\sin72^\circ}{19.548}=\frac{17.119}{19.548}=0.87575\]\[\angle BAC=61.1^\circ\]

C lies clockwise of B as seen from A, so the bearing of C from A is

\[053^\circ+61.1^\circ=114^\circ\ (\text{3 s.f.})\]

(c) Height of the pole BT. The pole is vertical and BA is horizontal, so triangle ATB is right-angled at B. With \(\angle ATB=58^\circ\) and \(AB=15\text{ m}\) opposite that angle:

\[\tan(\angle ATB)=\frac{AB}{BT}\Rightarrow BT=\frac{AB}{\tan58^\circ}=\frac{15}{1.6003}\]\[BT=9.37\text{ m (3 s.f.)}\]

Detalhes da Resposta

Detalhes da Resposta

(a) Volume of the balloon. Diameter \(=15\text{ cm}\Rightarrow r=7.5\text{ cm}\).

\(V=\dfrac{4}{3}\pi r^{3}=\dfrac{4}{3}\times\dfrac{22}{7}\times 7.5^{3}=\dfrac{88}{21}\times 421.875=\dfrac{37125}{21}=1767.86\).

Volume \(\approx \mathbf{1768\ cm^{3}}\).

(b) X: \(28^{\circ}\)N, \(105^{\circ}\)W; Y: \(28^{\circ}\)N, \(35^{\circ}\)E. Difference in longitude \(=105+35=140^{\circ}\).

(i) Distance X to Y along latitude \(28^{\circ}\)N:

\(d=\dfrac{140}{360}\times 2\pi R\cos 28^{\circ}=\dfrac{140}{360}\times 2\times\dfrac{22}{7}\times 6400\times\cos 28^{\circ}\).

\(=\dfrac{140}{360}\times 40228.57\times 0.8829\approx 13{,}800\text{ km}\ (\mathbf{1.38\times 10^{4}\ km})\).

(ii) Distance of X from the equator (along its meridian):

\(d=\dfrac{28}{360}\times 2\pi R=\dfrac{28}{360}\times 40228.57\approx \mathbf{3130\ km}\).

Detalhes da Resposta

(a) Apples \(A=40\), Oranges \(O=46\), Mangoes \(M=50\). \(A\cap O=14\), \(A\cap M=15\), all three \(=10\); every trader sells at least one, total \(=100\).

(i) Venn diagram. Three intersecting circles; centre (all three) \(=10\).

(ii) Let \(O\cap M\) (total) \(=y\). By the inclusion-exclusion principle:

\(|A|+|O|+|M|-|A\cap O|-|A\cap M|-|O\cap M|+|A\cap O\cap M|=100\).

\(40+46+50-14-15-y+10=100 \Rightarrow 117-y=100 \Rightarrow y=17\).

Oranges and mangoes only \(=y-10=17-10=\mathbf{7}\).

(b) \(312_{four}+52_x=96_{ten}\).

\(312_{four}=3(16)+1(4)+2=54\). So \(52_x=96-54=42\).

\(52_x=5x+2=42 \Rightarrow 5x=40 \Rightarrow x=\mathbf{8}\).

Detalhes da Resposta

Read the height of every bar against the vertical scale, which is marked in steps of \(5\) vehicles. Match each shading to the key: fine dots are cars, solid black is lorries, diagonal cross-hatching is buses. Reading each of the three bars in each month group gives the table below.

Month	Cars	Lorries	Buses	Monthly total
January	10	15	20	45
February	40	25	30	95
March	5	35	50	90

The same data drawn to scale:

(a) Vehicles produced in February. Add the three February bars:

\[ 40 + 25 + 30 = 95 \text{ vehicles.} \]

(b) Fraction of February vehicles that were cars. Cars in February over the February total:

\[ \frac{40}{95} = \frac{8}{19}. \]

(c) Buses produced altogether, January to March. Add the three bus readings:

\[ 20 + 30 + 50 = 100 \text{ buses.} \]

(d) Ratio of lorries in February to lorries in March. The lorry bars read \(25\) and \(35\). Divide both by their highest common factor, \(5\):

\[ 25 : 35 = 5 : 7. \]

The key point when reading a grouped bar chart is to fix the scale first (each square here is \(5\) vehicles) and to match every bar to the correct shading in the key before adding. Miscounting the scale or reading the wrong shading is what leads to a wrong total such as treating February as \(75\) instead of \(95\).

Detalhes da Resposta

(a) Pass probabilities \(\tfrac{2}{3},\tfrac{5}{8},\tfrac{3}{4}\); fail probabilities \(\tfrac{1}{3},\tfrac{3}{8},\tfrac{1}{4}\).

(i) All three pass \(=\dfrac{2}{3}\times\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{30}{96}=\dfrac{5}{16}\).

(ii) None passes \(=\dfrac{1}{3}\times\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{96}=\dfrac{1}{32}\).

(iii) Exactly two pass:

• 1&2 pass, 3 fails: \(\tfrac{2}{3}\cdot\tfrac{5}{8}\cdot\tfrac{1}{4}=\tfrac{10}{96}\)
• 1&3 pass, 2 fails: \(\tfrac{2}{3}\cdot\tfrac{3}{8}\cdot\tfrac{3}{4}=\tfrac{18}{96}\)
• 2&3 pass, 1 fails: \(\tfrac{1}{3}\cdot\tfrac{5}{8}\cdot\tfrac{3}{4}=\tfrac{15}{96}\)

Total \(=\dfrac{10+18+15}{96}=\dfrac{43}{96}\).

(b) Marked price \(=\text{N}36{,}000\) gives \(20\%\) profit, so cost price \(=\dfrac{36000}{1.20}=\text{N}30{,}000\).

Discount of \(5\%\): selling price \(=36000\times 0.95=\text{N}34{,}200\).

Profit \(=34200-30000=\text{N}4{,}200\); actual percentage profit \(=\dfrac{4200}{30000}\times 100=\mathbf{14\%}\).

Detalhes da Resposta

(a) \(\dfrac{1}{3^{5n}}\times 9^{n-1}\times 27^{n+1}\). Convert everything to base 3:

\(9^{n-1}=3^{2(n-1)}=3^{2n-2}\); \(27^{n+1}=3^{3(n+1)}=3^{3n+3}\); \(\dfrac{1}{3^{5n}}=3^{-5n}\).

Add exponents: \(-5n+(2n-2)+(3n+3)= (-5n+2n+3n)+(-2+3)=0+1=1\).

Result \(=3^{1}=\mathbf{3}\).

(b) Let woman \(=w\), daughter \(=d\). \(w+d=46\).

In 4 years: \(\dfrac{w+4}{d+4}=\dfrac{7}{2}\Rightarrow 2(w+4)=7(d+4)\Rightarrow 2w+8=7d+28\Rightarrow 2w-7d=20\).

Substitute \(w=46-d\): \(2(46-d)-7d=20\Rightarrow 92-9d=20\Rightarrow 9d=72\Rightarrow d=8\), \(w=38\).

Woman is 38 years, daughter is 8 years. Check: in 4 years \(42:12=7:2\). \(\checkmark\)

Detalhes da Resposta

(a) Find \(x\).

From the diagram, \(MN\parallel ST\) and \(NP\parallel QT\), with \(\angle STQ=70^{\circ}\) at \(T\), and \(x\) is the angle at \(N\) between \(NM\) and \(NP\).

Extend \(QT\) to cut \(MN\) at a point \(W\). Because \(MN\parallel ST\) and \(QT\) is a transversal, corresponding angles give

\[\angle MWQ=\angle STQ=70^{\circ}\]

Now \(MN\) is a transversal of the parallel lines \(NP\) and \(QT\). The ray \(NP\) points to the same side of \(MN\) as \(P\) (above), while \(WQ\) points to the opposite side (below), so \(x\) and \(\angle MWQ\) are co-interior (allied) angles:

\[x+\angle MWQ=180^{\circ}\]\[x=180^{\circ}-70^{\circ}=\mathbf{110^{\circ}}\]

(b) Find \(\angle BDA\).

From the diagram \(A,B,C\) lie on the straight line \(AC\), with \(D\) above it, \(|BC|=|BD|\), \(\angle BCD=50^{\circ}\) and \(\angle BAD=55^{\circ}\).

Triangle \(BCD\) is isosceles with \(|BC|=|BD|\), so the base angles are equal:

\[\angle BDC=\angle BCD=50^{\circ}\]\[\angle DBC=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]

Since \(A,B,C\) are collinear, \(\angle DBA\) and \(\angle DBC\) are angles on a straight line at \(B\):

\[\angle DBA=180^{\circ}-80^{\circ}=100^{\circ}\]

In triangle \(ABD\):

\[\angle BDA=180^{\circ}-\angle BAD-\angle DBA=180^{\circ}-55^{\circ}-100^{\circ}=\mathbf{25^{\circ}}\]

Detalhes da Resposta

Sides \(x\) and \((x+7)\), diagonal \((x+8)\). By Pythagoras:

\(x^{2}+(x+7)^{2}=(x+8)^{2}\).

\(x^{2}+x^{2}+14x+49=x^{2}+16x+64\).

\(2x^{2}+14x+49-x^{2}-16x-64=0 \Rightarrow x^{2}-2x-15=0\).

\((x-5)(x+3)=0 \Rightarrow x=5\) (reject \(x=-3\), a length must be positive).

(a) \(x=5\ \text{m}\).

(b) Area \(=x(x+7)=5\times 12=\mathbf{60\ m^{2}}\).

Detalhes da Resposta

(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).

Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).

Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).

\(p=-9,\ q=6\).

(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):

\(k=80\times 6400^{2}\).

At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):

\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).

Weight \(\approx \mathbf{63\ kg}\).

Detalhes da Resposta