WAEC SSCE - General Mathematics - 2015 (Essay)

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(a) Area. The perpendicular distance between the parallel sides is 40 cm.

\[A = \tfrac{1}{2}(PQ + SR) \times h = \tfrac{1}{2}(20 + 60) \times 40 = 1600 \text{ cm}^2\]

(b) Angle QRS. Place S at \((0, 0)\) and R at \((60, 0)\). P is 40 cm above RS; since \(SP = 50\), its horizontal offset is \(\sqrt{50^2 - 40^2} = 30\), so P is at \((30, 40)\). As \(PQ \parallel SR\) with \(PQ = 20\), Q is at \((50, 40)\).

At R, \(\vec{RQ} = (-10, 40)\) and \(\vec{RS} = (-60, 0)\):

\[\cos(\angle QRS) = \frac{(-10)(-60) + 0}{\sqrt{10^2 + 40^2} \times 60} = \frac{600}{2474} = 0.2425\] \[\angle QRS = 76^{\circ} \text{ (2 s.f.)}\]

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(a)(i) Venn diagram: Draw a rectangle for the school. Inside it draw a circle for the General Arts class \((A)\). Entirely within that circle draw a smaller circle for good Literature students \((L)\). This shows \(L \subset A\): every good Literature student is in the General Arts class, but not every General Arts student is a good Literature student.

(ii) Testing the conclusions:

1. Vivian is in General Arts, therefore she is a good Literature student. NOT valid: being in \(A\) does not place her inside the smaller circle \(L\).
2. Audu is not a good Literature student, therefore he is not in General Arts. NOT valid: he could be in \(A\) but outside \(L\).
3. Kweku is not in General Arts, therefore he is not a good Literature student. VALID: since \(L \subset A\), anyone outside \(A\) must be outside \(L\).

(b) Cost \(c = h + y\) where \(y = kn\) (\(y\) varies directly as \(n\)), so \(c = h + kn\).

\(950 = h + 600k \quad(1)\)

\(1030 = h + 1000k \quad(2)\)

\((2) - (1):\ 80 = 400k \ \Rightarrow\ k = 0.2\). Then \(h = 950 - 600(0.2) = 830\).

(i) \(\mathbf{c = h + 0.2n}\), that is \(c = 830 + 0.2n\) (with \(h = \text{GH¢}830\)).

(ii) For \(n = 500\): \(c = 830 + 0.2(500) = 830 + 100 = \mathbf{\text{GH¢}930.00}\).

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(a) Brackets first:

\[5\tfrac{1}{3} - 2\tfrac{3}{4} = \tfrac{16}{3} - \tfrac{11}{4} = \tfrac{64 - 33}{12} = \tfrac{31}{12}\] \[3\tfrac{4}{9} \div \tfrac{31}{12} = \tfrac{31}{9} \times \tfrac{12}{31} = \tfrac{12}{9} = \tfrac{4}{3}\] \[\tfrac{4}{3} + 5\tfrac{9}{10} = \tfrac{4}{3} + \tfrac{59}{10} = \tfrac{40 + 177}{30} = \tfrac{217}{30} = 7\tfrac{7}{30}\]

(b) One number from \(\{2, 3, 4\}\) and one from \(\{1, 3, 5\}\): \(3 \times 3 = 9\) equally likely pairs.

Pairs whose sum satisfies \(3 < \text{sum} < 7\): \((2,3) = 5,\ (3,1) = 4,\ (3,3) = 6,\ (4,1) = 5\), giving 4 outcomes.

\[P = \frac{4}{9}\]

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(a) For an AP, the \(n\)th term is \(a + (n-1)d\), with \(a = 8\).

7th term \(= 8 + 6d\); 9th term \(= 8 + 8d\). Given ratio \(5:8\):

\(\dfrac{8 + 6d}{8 + 8d} = \dfrac{5}{8}\)

\(8(8 + 6d) = 5(8 + 8d)\)

\(64 + 48d = 40 + 40d\)

\(8d = -24 \ \Rightarrow\ \mathbf{d = -3}\)

(b) Let cost price be \(N\,p\) per basket of pawpaw and \(N\,m\) per basket of mango.

\(30p + 100m = 2450 \quad(1)\)

Profit: \(0.40(30p) + 0.30(100m) = 855 \ \Rightarrow\ 12p + 30m = 855 \quad(2)\)

Simplify: \((1)\div 10: 3p + 10m = 245\); \((2)\div 6: 2p + 5m = 142.5\).

Multiply the last by 2: \(4p + 10m = 285\). Subtract \(3p + 10m = 245\): \(p = 40\).

Then \(2(40) + 5m = 142.5 \ \Rightarrow\ 5m = 62.5 \ \Rightarrow\ m = 12.5\).

(i) Cost price of a basket of pawpaw \(= \mathbf{N40.00}\).

(ii) Selling price of 100 baskets of mango \(= 100m \times 1.30 = 1250 \times 1.30 = \mathbf{N1{,}625.00}\).

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Construction and measurements

(a)(i) Constructing \(WXYZ\)

1. Draw \(WX=8\text{ cm}\).
2. At \(X\), construct an angle of \(60^\circ\) with \(XW\). On this ray, mark \(Y\) so that \(XY=5.5\text{ cm}\).
3. Construct a line through \(Y\) parallel to \(WX\): construct a perpendicular to \(WX\), then construct a perpendicular to that new line through \(Y\).
4. On this parallel line, on the side towards \(W\), mark \(Z\) such that \(YZ=8.3\text{ cm}\). Join \(Z\) to \(W\).

(a)(ii) Constructing rectangle \(PQYZ\)

Construct perpendiculars from \(Y\) and \(Z\) to \(WX\). Their intersections with \(WX\) are \(Q\) and \(P\), respectively. Since \(PQ\) lies on \(WX\), \(PQ\parallel YZ\), and the other two sides are perpendicular to \(WX\), \(PQYZ\) is a rectangle.

(b)(i) Length \(QX\)

\(Q\) is the perpendicular projection of \(Y\) onto \(WX\). Therefore \(QX\) is the horizontal component of \(XY\):

\[ QX=XY\cos60^\circ=5.5\times0.5=2.75\text{ cm}. \]

Measured to one decimal place, \(QX\approx\mathbf{2.8\text{ cm}}\). A ruler measurement of \(2.7\text{ cm}\) may result from drawing accuracy or truncation.

(b)(ii) Angle \(XWZ\)

The height of the trapezium is

\[ YQ=5.5\sin60^\circ\approx4.763\text{ cm}. \]

Also, \(WQ=8-2.75=5.25\text{ cm}\), whereas \(PQ=YZ=8.3\text{ cm}\). Thus \(P\), and therefore \(Z\), is \(8.3-5.25=3.05\text{ cm}\) to the left of \(W\).

The acute angle between \(WZ\) and the leftward extension of \(WX\) is

\[ \tan^{-1}\left(\frac{4.763}{3.05}\right)\approx57.4^\circ. \]

However, \(\angle XWZ\) is measured from the ray \(WX\), which points to the right. Hence

\[ \angle XWZ=180^\circ-57.4^\circ\approx\mathbf{122.6^\circ}, \]

so the measured answer is about \(\mathbf{123^\circ}\).

The stated value \(75^\circ\) is not consistent with the given side lengths, the \(60^\circ\) angle at \(X\), and \(WX\parallel ZY\). A useful check is that \(Z\) lies to the left of \(W\), so \(\angle XWZ\) must be obtuse, not \(75^\circ\).

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(a) Diagram: A cone sitting on top of a hemisphere, sharing the same circular base of radius \(r\). The flat face of the hemisphere coincides with the base of the cone, and the cone height is \(h = 6r\).

(b) Let the common radius be \(r\), so \(h = 6r\).

\(V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi r^2(6r) = 2\pi r^3\)

\(V_{\text{hemisphere}} = \tfrac{2}{3}\pi r^3\)

Total: \(2\pi r^3 + \tfrac{2}{3}\pi r^3 = \tfrac{8}{3}\pi r^3 = 333\tfrac{1}{3}\pi = \tfrac{1000}{3}\pi\).

\(8r^3 = 1000 \ \Rightarrow\ r^3 = 125 \ \Rightarrow\ r = 5\text{ m}\) (so \(h = 30\text{ m}\)).

(I) Volume of hemisphere: \(\tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\times\tfrac{22}{7}\times 125 = \dfrac{5500}{21} = 261.9 \approx \mathbf{262\text{ m}^3}\).

(II) Total surface area \(=\) curved surface of cone \(+\) curved surface of hemisphere (the joining faces are internal).

Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{25 + 900} = \sqrt{925} = 30.41\text{ m}\).

\(\pi r l = \tfrac{22}{7}\times 5\times 30.41 = 477.9\text{ m}^2\); \(2\pi r^2 = 2\times\tfrac{22}{7}\times 25 = 157.1\text{ m}^2\).

Total \(= 477.9 + 157.1 = 635.0 \approx \mathbf{635\text{ m}^2}\).

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(a) Sum \(= 3\tfrac{2}{3} + 2\tfrac{1}{5} = \tfrac{11}{3} + \tfrac{11}{5} = \tfrac{55 + 33}{15} = \tfrac{88}{15}\).

It is less than 7 by \(7 - \tfrac{88}{15} = \tfrac{105 - 88}{15} = \tfrac{17}{15} = 1\tfrac{2}{15}\).

(b)(i) With \(p = 4\): \(15p = 60^{\circ}\) and \(\cos 60^{\circ} = 0.5\).

\[h = 6 + 4(0.5) = 8 \text{ m}\]

(b)(ii) With \(h = 9\):

\[9 = 6 + 4\cos(15p)^{\circ} \Rightarrow \cos(15p)^{\circ} = 0.75\] \[15p = \cos^{-1} 0.75 = 41.4^{\circ} \Rightarrow p = \frac{41.4}{15} = 2.8 \text{ (2 s.f.)}\]

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(a) Solving the inequality

\[4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1\]

Multiply every term by 8 (the LCM of the denominators) to clear fractions:

\[8(4) + 8\cdot\frac{3}{4}(x+2) \le 8\cdot\frac{3}{8}x + 8(1)\]

\[32 + 6(x+2) \le 3x + 8\]

\[32 + 6x + 12 \le 3x + 8\]

\[6x + 44 \le 3x + 8\]

Collect like terms:

\[6x - 3x \le 8 - 44\]

\[3x \le -36\]

\[x \le -12\]

(b) Finding x from the rectangle

From the diagram, rectangle PQRS has height \(|PS| = 20\ \text{cm}\). Along the base SR a square of side x cm is cut out, leaving 10 cm on the left and 10 cm on the right. So the full length of the rectangle is

\[|SR| = 10 + x + 10 = (20 + x)\ \text{cm}\]

Area of the whole rectangle:

\[(20 + x)\times 20 = 400 + 20x\]

Area of the square removed:

\[x \times x = x^2\]

The shaded (remaining) area is 484 cm\(^2\):

\[(400 + 20x) - x^2 = 484\]

\[-x^2 + 20x + 400 - 484 = 0\]

\[-x^2 + 20x - 84 = 0\]

\[x^2 - 20x + 84 = 0\]

Solve using factorisation (\(-6\) and \(-14\) multiply to 84 and add to \(-20\)):

\[(x - 6)(x - 14) = 0\]

\[x = 6 \quad\text{or}\quad x = 14\]

Both values are geometrically valid (the cut height stays within the 20 cm rectangle), so

\[x = 6\ \text{cm} \quad\text{or}\quad x = 14\ \text{cm}\]

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(a) Use exact surd values: \(\tan 60° = \sqrt{3}\), \(\cos 30° = \dfrac{\sqrt{3}}{2}\), \(\sin 60° = \dfrac{\sqrt{3}}{2}\).

Numerator: \(2\sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{4\sqrt{3} + \sqrt{3}}{2} = \dfrac{5\sqrt{3}}{2}\).

\(\dfrac{2\tan 60° + \cos 30°}{\sin 60°} = \dfrac{5\sqrt{3}/2}{\sqrt{3}/2} = \mathbf{5}\)

(b) Horizontal distance to the tower \(= 1050\text{ m}\). The top has depression \(36°\) and the base \(41°\). The vertical drop from the aeroplane's level to each point is \(1050\tan(\text{angle})\).

(i) Height of tower \(=\) (drop to base) \(-\) (drop to top):

\(= 1050\tan 41° - 1050\tan 36° = 1050(0.8693 - 0.7265)\)

\(= 1050 \times 0.1428 = 149.9 \approx \mathbf{150\text{ m}}\).

(ii) Shortest distance (line of sight to the base):

\(= \dfrac{1050}{\cos 41°} = \dfrac{1050}{0.7547} = 1391.3 \approx \mathbf{1391\text{ m}}\).

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(a) Number of sides of the polygon

At each vertex the interior and exterior angles are supplementary, so they add up to \(180^\circ\). The ratio interior : exterior is \(5:2\), giving \(5+2 = 7\) equal parts.

\[\text{Exterior angle} = \frac{2}{7}\times 180^\circ = \frac{360^\circ}{7}\]

For any regular polygon the exterior angles sum to \(360^\circ\), so the number of sides is

\[n = \frac{360^\circ}{\text{exterior angle}} = \frac{360^\circ}{\tfrac{360^\circ}{7}} = 7\]

The polygon has 7 sides (a regular heptagon).

(b) Finding \(\angle PRS\)

From the diagram, \(TPQU\) is a straight line through the two points \(P\) and \(Q\) on the circle, and \(PQRS\) is a cyclic quadrilateral. The given angles are \(\angle TPS = 74^\circ\), \(\angle UQR = 68^\circ\) and \(\angle QSR = 40^\circ\).

Step 1: Interior angles of the cyclic quadrilateral. Since \(TPQ\) is a straight line,

\[\angle SPQ = 180^\circ - \angle TPS = 180^\circ - 74^\circ = 106^\circ\]

Since \(PQU\) is a straight line,

\[\angle PQR = 180^\circ - \angle UQR = 180^\circ - 68^\circ = 112^\circ\]

Step 2: Use the opposite-angles property of a cyclic quadrilateral.

\[\angle SRQ = 180^\circ - \angle SPQ = 180^\circ - 106^\circ = 74^\circ\]

Step 3: Split \(\angle SRQ\) using angles in the same segment. The angle \(\angle PRQ\) stands on chord \(PQ\), as does \(\angle PSQ\); angles in the same segment are equal, so \(\angle PRQ = \angle PSQ\).

Now find \(\angle PSQ\). We have \(\angle PSR = 180^\circ - \angle PQR = 180^\circ - 112^\circ = 68^\circ\), and this splits as \(\angle PSR = \angle PSQ + \angle QSR\). Therefore

\[\angle PSQ = 68^\circ - 40^\circ = 28^\circ \;\Rightarrow\; \angle PRQ = 28^\circ\]

Step 4: Obtain \(\angle PRS\). Since \(\angle SRQ = \angle PRS + \angle PRQ\),

\[\angle PRS = 74^\circ - 28^\circ = \boxed{46^\circ}\]

Hence \(\angle PRS = 46^\circ\).

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