WAEC SSCE - General Mathematics - 2009 (Essay)

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(a) Simplifying the algebraic fractions

Use the common denominator \((x-2)(x-1)\):

\[ \frac{x+2}{x-2}-\frac{x+3}{x-1} = \frac{(x+2)(x-1)-(x+3)(x-2)}{(x-2)(x-1)}. \] \[ = \frac{(x^2+x-2)-(x^2+x-6)}{(x-2)(x-1)} = \frac{x^2+x-2-x^2-x+6}{(x-2)(x-1)}. \] \[ \boxed{\frac{4}{(x-2)(x-1)}} \]

This is valid for \(x\ne2\) and \(x\ne1\), since these values would make an original denominator zero.

(b) Finding the quadratic equation

The equation is \(y=Ax^2+Bx+C\). Since the graph passes through \((0,0)\), substitute \(x=0\) and \(y=0\):

\[ 0=A(0)^2+B(0)+C, \] \[ \boxed{C=0}. \]

So the equation becomes:

\[ y=Ax^2+Bx. \]

Using \((1,4)\):

\[ 4=A(1)^2+B(1), \] \[ A+B=4. \]

Using \((2,10)\):

\[ 10=A(2)^2+B(2), \] \[ 4A+2B=10, \] \[ 2A+B=5. \]

Subtract \(A+B=4\) from \(2A+B=5\):

\[ A=1. \]

Then:

\[ 1+B=4 \quad\Rightarrow\quad B=3. \]

Therefore:

\[ \boxed{A=1,\quad B=3} \]

and the equation of the graph is:

\[ \boxed{y=x^2+3x}. \]

Other x-intercept

At an x-intercept, \(y=0\):

\[ x^2+3x=0, \] \[ x(x+3)=0. \] \[ x=0 \quad\text{or}\quad x=-3. \]

\((0,0)\) is the given intercept, so the other point where the graph cuts the x-axis is:

\[ \boxed{(-3,0)}. \]

Examination reminder: \((1,4)\) and \((2,10)\) are points on the graph, not roots. A root is an x-value for which \(y=0\). Factoring \(x^2+3x\) gives \(x=0\) and \(x=-3\), not \(x=3\).

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(a) The multiples of \(7\) between \(75\) and \(500\): the first is \(77 = 7\times 11\) and the last is \(497 = 7\times 71.\) \[\text{count} = 71 - 11 + 1 = \mathbf{61}.\]

(b) Let the first term be \(a\) and common difference \(d.\)

8th term is \(5\) times the 3rd: \(a + 7d = 5(a + 2d) \Rightarrow a + 7d = 5a + 10d \Rightarrow 4a + 3d = 0.\)

7th term is \(9\) greater than the 4th: \((a + 6d) - (a + 3d) = 9 \Rightarrow 3d = 9 \Rightarrow d = 3.\)

Then \(4a + 3(3) = 0 \Rightarrow 4a = -9 \Rightarrow a = -\tfrac{9}{4} = -2.25.\)

First five terms (adding \(d = 3\) each time): \[\mathbf{-2\tfrac{1}{4},\ \tfrac{3}{4},\ 3\tfrac{3}{4},\ 6\tfrac{3}{4},\ 9\tfrac{3}{4}}.\]

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(a) Expand \((\sqrt{3}-5\sqrt{2})(\sqrt{3}+\sqrt{2})\):

\(=\sqrt{3}\cdot\sqrt{3}+\sqrt{3}\cdot\sqrt{2}-5\sqrt{2}\cdot\sqrt{3}-5\sqrt{2}\cdot\sqrt{2}\)

\(=3+\sqrt{6}-5\sqrt{6}-10=-7-4\sqrt{6}\).

Comparing with \(a+b\sqrt{6}\): \(\mathbf{a=-7,\ b=-4}\).

(b) \(\dfrac{2^{1-y}\times 2^{y-1}}{2^{y+2}}=8^{2-3y}\).

Numerator: \(2^{(1-y)+(y-1)}=2^{0}=1\). Left side \(=2^{-(y+2)}\).

Right side: \(8^{2-3y}=2^{3(2-3y)}=2^{6-9y}\).

\(-(y+2)=6-9y \Rightarrow -y-2=6-9y \Rightarrow 8y=8 \Rightarrow y=\mathbf{1}\).

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(a) A circle inscribed in a square has diameter equal to the side, so side \(= 2r.\) Then \[\text{perimeter of square} + \text{circumference} = 4(2r) + 2\pi r = 100.\] \[8r + 2\times\frac{22}{7}r = 100 \Rightarrow 8r + \frac{44}{7}r = 100 \Rightarrow \frac{56r + 44r}{7} = 100.\] \[\frac{100r}{7} = 100 \Rightarrow r = \mathbf{7\,\text{cm}}.\]

(b) Perimeter of rectangle \(= 60\,\text{cm}\), with length \(= 4\times\)breadth. \[2(l + b) = 60 \Rightarrow l + b = 30,\quad l = 4b \Rightarrow 5b = 30 \Rightarrow b = 6.\]

(i) Length \(l = 4(6) = \mathbf{24\,\text{cm}}.\)

(ii) Diagonal \(= \sqrt{l^2 + b^2} = \sqrt{24^2 + 6^2} = \sqrt{576 + 36} = \sqrt{612} \approx \mathbf{24.7\,\text{cm}}.\)

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(a) \(n=30\); degrees \(=17\), diplomas \(=15\), neither \(=4\), so \(30-4=26\) have at least one.

\(|D\cup P|=|D|+|P|-|D\cap P| \Rightarrow 26=17+15-|D\cap P| \Rightarrow |D\cap P|=32-26=\mathbf{6}\).

Six candidates have both a degree and a diploma.

(b) In \(\triangle PQR\), \(M\) on \(PQ\), \(N\) on \(PR\), \(MN\parallel QR\); \(\angle PRQ=75^{\circ}\), \(PN=QN\), \(\angle PNQ=125^{\circ}\).

Triangle \(PNQ\) is isosceles (\(PN=QN\)), so \(\angle NPQ=\angle NQP\).

\(\angle NPQ+\angle NQP+\angle PNQ=180^{\circ}\Rightarrow 2\angle NPQ=180^{\circ}-125^{\circ}=55^{\circ}\Rightarrow \angle NPQ=\angle NQP=27.5^{\circ}\).

In \(\triangle PQR\): \(\angle P=27.5^{\circ}\), \(\angle R=75^{\circ}\), so \(\angle PQR=180^{\circ}-27.5^{\circ}-75^{\circ}=77.5^{\circ}\).

(i) \(\angle NQR=\angle PQR-\angle NQP=77.5^{\circ}-27.5^{\circ}=\mathbf{50^{\circ}}\).

(ii) \(\angle NPM\) is the angle at \(P\) between \(PM\) (on \(PQ\)) and \(PN\) (on \(PR\)), i.e. \(\angle QPR=\mathbf{27.5^{\circ}}\).

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Construction and loci

1. Draw \(PQ=10\text{ cm}\).
2. At \(Q\), construct an angle of \(60^\circ\) with \(QP\). On this ray, mark \(R\) so that \(QR=8\text{ cm}\).
3. At \(P\), construct an angle of \(75^\circ\) with \(PQ\). This can be made by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On this ray, mark \(S\) so that \(PS=6\text{ cm}\).
4. Join \(R\) to \(S\) to complete quadrilateral \(PQRS\).
5. The locus \(l_1\), equidistant from \(QR\) and \(RS\), is the internal angle bisector of \(\angle QRS\).
6. The locus \(l_2\), equidistant from points \(R\) and \(S\), is the perpendicular bisector of \(RS\).

Measurement of \(RS\): \(RS\approx 4.6\text{ cm}\), measured to the nearest \(0.1\text{ cm}\).

A check from the given lengths and angles gives \[ RS=\sqrt{(10-8\cos60^\circ-6\cos75^\circ)^2+(8\sin60^\circ-6\sin75^\circ)^2} \approx 4.59\text{ cm}. \] Therefore, \(4.6\text{ cm}\) is the appropriate measured value. The stated value \(4.4\text{ cm}\) is not consistent with an accurate construction using the dimensions given.

Examination reminder: “Equidistant from two intersecting sides” indicates an angle bisector; “equidistant from two points” indicates a perpendicular bisector.

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(a) The woman is at a window \(30\text{ m}\) above the ground. The foot of the flag pole is \(25\text{ m}\) horizontally from the foot of the building, on the same level ground.

(i) Angle of depression of the foot of the pole. From the window, the horizontal distance to the pole is \(25\text{ m}\) and the vertical drop to the foot of the pole is the full window height \(30\text{ m}\):

\[\tan\theta=\frac{30}{25}=1.2\]\[\theta=\tan^{-1}(1.2)\approx 50.19^\circ\approx\mathbf{50^\circ}.\]

(ii) Height of the flag pole. The angle of depression of the top of the pole is \(44^\circ\). Over the horizontal distance \(25\text{ m}\), the vertical drop from window level down to the top of the pole is

\[h_1=25\tan 44^\circ=25\times0.9657=24.14\text{ m}.\]

So the top of the pole is \(30-24.14=5.86\text{ m}\) above the ground. Since the pole stands on the same ground, its height is

\[|\text{pole}|=30-24.14\approx\mathbf{6\text{ m}}.\]

(b) \(O\) is the centre of the circle, \(\angle OQR=32^\circ\) and \(\angle TPQ=15^\circ\).

(i) \(\angle QPR\): \(OQ\) and \(OR\) are radii, so triangle \(OQR\) is isosceles and

\[\angle ORQ=\angle OQR=32^\circ.\]\[\angle QOR=180^\circ-32^\circ-32^\circ=116^\circ.\]

\(\angle QOR\) is the angle subtended by chord \(QR\) at the centre, and \(\angle QPR\) is subtended by the same chord at the circumference, so it is half:

\[\angle QPR=\tfrac{1}{2}\times116^\circ=\mathbf{58^\circ}.\]

(ii) \(\angle TQO\): \(\angle TPQ=15^\circ\) stands on chord \(TQ\), so the central angle on the same chord is

\[\angle TOQ=2\times15^\circ=30^\circ.\]

\(OT\) and \(OQ\) are radii, so triangle \(TOQ\) is isosceles:

\[\angle TQO=\angle OTQ=\frac{180^\circ-30^\circ}{2}=\mathbf{75^\circ}.\]

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The solid is a triangular prism with congruent right-angled triangular ends \(ABC\) and \(DEF\) (\(\angle ABC = \angle DEF = 90^\circ\)). The given lengths are \(|AB| = 24\) cm, \(|BC| = 7\) cm, and the prism length \(|CD| = 40\) cm.

(a) Length AC. Triangle \(ABC\) is right-angled at \(B\), so by Pythagoras:

\[ |AC| = \sqrt{|AB|^2 + |BC|^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\ \text{cm} \]

(b) Total surface area. The prism has two triangular ends and three rectangular faces. The prism length is \(40\) cm.

The two triangular ends:

\[ 2 \times \left(\tfrac{1}{2}\times 24 \times 7\right) = 2 \times 84 = 168\ \text{cm}^2 \]

The three rectangles have widths equal to the sides of the triangle \((24, 7, 25)\) and common length \(40\):

\[ (24 + 7 + 25)\times 40 = 56 \times 40 = 2240\ \text{cm}^2 \]

Total surface area:

\[ 168 + 2240 = 2408\ \text{cm}^2 \]

Answers: (a) \(|AC| = 25\) cm; (b) total surface area \(= 2408\ \text{cm}^2\).

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(a) Simplify inside first. \(40 \div 56 = \dfrac{40}{56} = \dfrac{5}{7}\), so \[\frac{35\times 49}{40\div 56} = \frac{35\times 49}{5/7} = 35\times 49\times\frac{7}{5} = 7\times 49\times 7 = 7^4 = 2401.\] \[\log 2401 = \log 7^4 = 4\log 7 = 4(0.8451) = \mathbf{3.3804}.\]

(b) Let there be \(x\) children; adults \(= x + 60\). Adult pays \(D5\), child \(D2\), total \(D1280\): \[5(x + 60) + 2x = 1280 \Rightarrow 7x + 300 = 1280 \Rightarrow 7x = 980 \Rightarrow x = 140.\]

(i) \(x = \mathbf{140}\) children (adults \(= 200\)).

(ii) children : adults \(= 140 : 200 = \mathbf{7 : 10}.\)

(iii) Total people \(= 340\); average \(= \dfrac{1280}{340} = \mathbf{D3.76}\) per person.

(iv) Gain \(= 1280 - 720 = D560\); percentage gain \(= \dfrac{560}{720}\times 100 = \mathbf{77.8\%}.\)

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(a) Frequency distribution table

Class interval	Frequency \(f\)	Midpoint \(x\)	\(fx\)
21 - 30	2	25.5	51.0
31 - 40	10	35.5	355.0
41 - 50	12	45.5	546.0
51 - 60	15	55.5	832.5
61 - 70	8	65.5	524.0
71 - 80	3	75.5	226.5
Total	50		2535.0

(b) Mean \[\bar{x} = \frac{\sum fx}{\sum f} = \frac{2535}{50} = \mathbf{50.7}.\]

(c) Marks more than \(60\): the values greater than \(60\) are \(61, 62, 65, 67, 67, 68, 69, 70, 71, 73, 74\) - that is \(11\) students. \[\text{percentage} = \frac{11}{50}\times 100 = \mathbf{22\%}.\]

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(a) \(9\cos x-7=1\Rightarrow 9\cos x=8\Rightarrow \cos x=\dfrac{8}{9}=0.8889\).

For \(0^{\circ}\le x\le 90^{\circ}\): \(x=\cos^{-1}(0.8889)\approx \mathbf{27.3^{\circ}}\).

(b) \(7x<2x-13 \Rightarrow 7x-2x<-13 \Rightarrow 5x<-13 \Rightarrow x<-2.6\).

Since \(x\) is an integer, the values are \(\ldots,-5,-4,-3\). The three greatest are \(\mathbf{-3,\ -4,\ -5}\).