WAEC SSCE - General Mathematics - 2008 (Essay)

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(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:

\[\cos C=\frac{CA^{2}+CB^{2}-AB^{2}}{2\,(CA)(CB)}=\frac{10^{2}+12^{2}-17^{2}}{2(10)(12)}=\frac{100+144-289}{240}=\frac{-45}{240}=-0.1875\]

\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).

Largest angle \(\approx 100.8^{\circ}\).

(ii) Area.

\[\text{Area}=\frac{1}{2}(CA)(CB)\sin C=\frac{1}{2}(10)(12)\sin 100.8^{\circ}=60\times0.9823\approx 58.9\text{ m}^{2}\]

(b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.

\[\tan 63^{\circ}=\frac{RS}{TS}=\frac{38}{TS}\Rightarrow TS=\frac{38}{\tan 63^{\circ}}=\frac{38}{1.9626}\approx 19.4\text{ m}\]

Distance \(TS\approx 19\) m (nearest metre).

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(a) The exterior angles of any polygon add up to \(360^\circ\). The pentagon has five exterior angles: one is \(60^\circ\), two are \((90-m)^\circ\) each, and the remaining two are \((30+2m)^\circ\) each. Therefore:

\[60+2(90-m)+2(30+2m)=360\]\[60+180-2m+60+4m=360\]\[300+2m=360\]\[2m=60\Rightarrow m=30\]

(b) From the diagram PQR is a straight line, \(\overline{QR}=\sqrt{3}\text{ cm}\), \(\overline{SQ}=2\text{ cm}\), and the right-angle mark at R shows \(SR\perp QR\). So triangle SQR is right-angled at R, with SQ the hypotenuse.

In right-angled triangle SQR, using \(\angle SQR\):

\[\cos(\angle SQR)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\overline{QR}}{\overline{SQ}}=\frac{\sqrt{3}}{2}\]\[\angle SQR=\cos^{-1}\!\left(\tfrac{\sqrt{3}}{2}\right)=30^\circ\]

Since PQR is a straight line, \(\angle PQS\) and \(\angle SQR\) are angles on a straight line at Q:

\[\angle PQS=180^\circ-\angle SQR=180^\circ-30^\circ=150.0^\circ\]

Akọwa Nkọwa

(a) Express both equations to base 2.

\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)

Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)

Solve the pair:

\[x+y=4,\qquad x-y=-2.5\]

Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).

\(x=0.75,\ y=3.25\)

(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).

Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).

When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).

\(Q=5\)

Akọwa Nkọwa

(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).

Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):

\[\text{angle between } DE \text{ and the } BC\text{-family}=\widehat{DEF}=26^\circ.\]

Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is

\[75^\circ-26^\circ=49^\circ.\]

Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:

\[\boxed{x=49^\circ.}\]

By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):

\[y=360^\circ-49^\circ=\boxed{311^\circ.}\]

(b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).

(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:

\[\widehat{ABC}=\tfrac12(190^\circ)=95^\circ.\]

(ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):

\[\widehat{ADC}=\tfrac12(170^\circ)=85^\circ.\]

(Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)

(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):

\[\text{arc}=\frac{170}{360}\times 2\times 3.142\times 7=\frac{170}{360}\times 43.988=20.77\approx 20.8\text{ cm}.\]

(c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).

\[AD=2r\sin\!\left(\tfrac{110^\circ}{2}\right)=2(7)\sin 55^\circ=14\times 0.8192=11.47\approx 11.5\text{ cm}.\]

Akọwa Nkọwa

(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):

\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]

\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]

(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes

\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]

\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]

Converting back to base two, \(y = 1_2\).

Akọwa Nkọwa

Akọwa Nkọwa

(a) \(\dfrac25(x - 2) - \dfrac16(x + 5) \le 0\). Multiply through by the LCM \(30\):

\[12(x - 2) - 5(x + 5) \le 0\]

\[12x - 24 - 5x - 25 \le 0 \;\Rightarrow\; 7x - 49 \le 0 \;\Rightarrow\; 7x \le 49 \;\Rightarrow\; x \le 7\]

(b)(i) Factorise numerator and denominator of \(P = \dfrac{x^2 - y^2}{x^2 + xy}\):

\[P = \frac{(x - y)(x + y)}{x(x + y)} = \frac{x - y}{x}\]

(ii) With \(x = -4,\, y = -6\):

\[P = \frac{-4 - (-6)}{-4} = \frac{2}{-4} = -\frac12\]

Akọwa Nkọwa

(a) \(3,\ x,\ y,\ 18\) are consecutive terms of an A.P. The first term is \(3\) and the fourth term is \(18\):

\[18=3+(4-1)d=3+3d\Rightarrow d=5\]

So \(x=3+5=8\) and \(y=8+5=13\).

\(x=8,\ y=13\)

(b)(i) For a G.P. with first term \(a\) and ratio \(r\), terms are \(a,\ ar,\ ar^{2},\ ar^{3},\ldots\)

"Sum of second and third terms \(=6\times\) fourth term":

\[ar+ar^{2}=6ar^{3}\]

Divide through by \(ar\ (a,r\neq0)\): \(1+r=6r^{2}\Rightarrow 6r^{2}-r-1=0\).

\[r=\frac{1\pm\sqrt{1+24}}{12}=\frac{1\pm5}{12}\]

\(r=\tfrac{1}{2}\) or \(r=-\tfrac{1}{3}\).

(ii) The ratio is positive, so \(r=\tfrac{1}{2}\). The second term is \(ar=8\):

\[a\times\tfrac{1}{2}=8\Rightarrow a=16\]

First six terms: \(16,\ 8,\ 4,\ 2,\ 1,\ \tfrac{1}{2}\).

Akọwa Nkọwa

(a) Let the length be \(l\) and the breadth \(b\). The perimeter is \(280\) m:

\[2(l + b) = 280 \;\Rightarrow\; l + b = 140\]

The length is two and a half times the breadth: \(l = 2.5b\). Substitute:

\[2.5b + b = 140 \;\Rightarrow\; 3.5b = 140 \;\Rightarrow\; b = 40\text{ m}, \quad l = 2.5(40) = 100\text{ m}\]

(b) Taking the base as a square of side \(4.5\) m and height \(4\) m, the volume of a pyramid is one-third of base area times height:

\[V = \frac13 \times (\text{base area}) \times h = \frac13 \times (4.5 \times 4.5) \times 4\]

\[= \frac13 \times 20.25 \times 4 = \frac{81}{3} = 27\text{ m}^3\]

Akọwa Nkọwa

(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).

Using the formula with \(a=4,\ b=-11,\ c=-21\):

\[x=\frac{11\pm\sqrt{(-11)^{2}-4(4)(-21)}}{2(4)}=\frac{11\pm\sqrt{121+336}}{8}=\frac{11\pm\sqrt{457}}{8}\]

\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).

\(x\approx 4.05\) or \(x\approx -1.30\)

(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).

(i) First-year interest \(=1560-1500=\pounds60\).

\[\text{Rate}=\frac{60}{1500}\times100\%=4\%\]

(ii) Interest for the second year is charged on \(\pounds1560\):

\[\frac{4}{100}\times1560=\pounds62.40\]

(c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).

After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).

After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).

Value after 2 years \(=N180{,}000.00\)

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