WAEC SSCE - General Mathematics - 2014 (Essay)

Maelezo ya Majibu

(a) Solving \(5\cos(x+8.5)^\circ - 1 = 0\).

\[\cos(x+8.5)^\circ = \frac{1}{5} = 0.2\]

\[x + 8.5 = \cos^{-1}(0.2) = 78.46^\circ\]

\[x = 78.46 - 8.5 = 69.96^\circ \approx 70^\circ\]

(b) Reading the bearings: the bearing of \(Q\) from \(P\) is \(150^\circ\) and the bearing of \(P\) from \(R\) is \(015^\circ\). Since the bearing of \(P\) from \(R\) is \(015^\circ\), the bearing of \(R\) from \(P\) is \(015^\circ + 180^\circ = 195^\circ\). With \(|PQ| = 24\) km and \(|PR| = 32\) km:

(i) Diagram. At \(P\), draw the north line. \(Q\) lies on a bearing of \(150^\circ\) at 24 km; \(R\) lies on a bearing of \(195^\circ\) at 32 km. The angle \(\angle QPR = 195^\circ - 150^\circ = 45^\circ\).

(ii) Distance \(|QR|\). By the cosine rule in \(\triangle PQR\):

\[|QR|^2 = 24^2 + 32^2 - 2(24)(32)\cos 45^\circ\]

\[= 576 + 1024 - 1536(0.7071) = 1600 - 1086.11 = 513.89\]

\[|QR| = \sqrt{513.89} = 22.67\text{ km (2 d.p.)}\]

(iii) Bearing of \(R\) from \(Q\). Taking \(P\) as origin with north as the \(y\)-axis: \(Q = (24\sin150^\circ,\,24\cos150^\circ) = (12,\,-20.78)\) and \(R = (32\sin195^\circ,\,32\cos195^\circ) = (-8.28,\,-30.91)\).

Displacement from \(Q\) to \(R\): \((-20.28,\,-10.13)\), which points south and west (third quadrant).

\[\theta = \tan^{-1}\!\left(\frac{20.28}{10.13}\right) = 63.5^\circ \text{ west of south}\]

\[\text{Bearing of }R\text{ from }Q = 180^\circ + 63.5^\circ = 243.5^\circ \approx 243^\circ\]

Maelezo ya Majibu

(a) Multiplication table modulo 11 (multiply, then take the remainder on division by 11).

For example \(5\otimes 5 = 25 = 2(11)+3 \equiv 3\), \(9\otimes 9 = 81 = 7(11)+4 \equiv 4\), \(10\otimes 10 = 100 = 9(11)+1 \equiv 1\).

\(\otimes\)	1	5	9	10
1	1	5	9	10
5	5	3	1	6
9	9	1	4	2
10	10	6	2	1

(i) \((9\otimes 5)\otimes(10\otimes 10) = 1 \otimes 1 = \mathbf{1}\).

(ii)(1) \(10\otimes m = 2\): from the row for 10, \(10\otimes 9 = 2\), so the truth set is \(\{9\}\).

(ii)(2) \(n\otimes n = 4\): scanning the leading diagonal, only \(9\otimes 9 = 4\), so the truth set is \(\{9\}\).

(b) The fraction. In lowest terms it equals \(\tfrac{3}{4}\), so let it be \(\dfrac{3k}{4k}\).

Numerator doubled is 34 more than the denominator:

\[2(3k) = 4k + 34 \;\Rightarrow\; 6k - 4k = 34 \;\Rightarrow\; 2k = 34 \;\Rightarrow\; k = 17\] \[\text{Fraction} = \frac{3(17)}{4(17)} = \frac{51}{68}\]

Check: \(2(51) = 102 = 68 + 34\). The fraction is \(\mathbf{\tfrac{51}{68}}\).

Maelezo ya Majibu

(a)(i) \(f(x) = 2x^2 - 1\), so \(f(x-1) = 2(x-1)^2 - 1\).

\[2(x-1)^2 - 1 - 7 = 0 \;\Rightarrow\; 2(x-1)^2 = 8 \;\Rightarrow\; (x-1)^2 = 4\]

\[x - 1 = \pm 2 \;\Rightarrow\; x = 3 \text{ or } x = -1\]

(ii) Evaluate each term:

• \(f\!\left(-\tfrac12\right) = 2\left(\tfrac14\right) - 1 = -\tfrac12\)
• \(g(3) = 3(3) + 2 = 11\)
• \(f(4) = 2(16) - 1 = 31\)
• \(g(5) = 3(5) + 2 = 17\)

\[\frac{f\!\left(-\tfrac12\right)\cdot g(3)}{f(4) - g(5)} = \frac{\left(-\tfrac12\right)(11)}{31 - 17} = \frac{-\tfrac{11}{2}}{14} = -\frac{11}{28}\]

(b) \(m \ast n = \dfrac{-n}{m^2 + 1}\). Test with \(m = -3,\, n = -10\):

\[(-3)\ast(-10) = \frac{-(-10)}{(-3)^2 + 1} = \frac{10}{10} = 1\]

\[(-10)\ast(-3) = \frac{-(-3)}{(-10)^2 + 1} = \frac{3}{101}\]

Since \(1 \neq \dfrac{3}{101}\), we have \((-3)\ast(-10) \neq (-10)\ast(-3)\). Therefore the operation \(\ast\) is not commutative.

Maelezo ya Majibu

(a) Mean deviation. First find the mean.

Score (x)	1	2	3	4	5	6	Total
Freq (f)	2	5	13	11	9	10	50
fx	2	10	39	44	45	60	200

\[\bar{x} = \frac{200}{50} = 4.0\]

x	\(|x-\bar{x}|\)	\(f|x-\bar{x}|\)
1	3	6
2	2	10
3	1	13
4	0	0
5	1	9
6	2	20
Total		58

\[\text{Mean deviation} = \frac{\sum f|x-\bar{x}|}{\sum f} = \frac{58}{50} = \mathbf{1.16}\]

(b) Probability of at least a 4. Scores 4, 5 or 6:

\[P(\ge 4) = \frac{11 + 9 + 10}{50} = \frac{30}{50} = \frac{3}{5} = 0.6\]

Maelezo ya Majibu

Setup. The contracts are independent. Given:

\[P(\text{win }X) = 0.5,\qquad P(\text{not win }Y) = 0.3 \Rightarrow P(\text{win }Y) = 0.7\]

Also \(P(\text{not win }X) = 0.5\).

(a) Wins both contracts

\[P(X\cap Y) = 0.5\times 0.7 = 0.35\]

(b) Wins exactly one contract

Either win \(X\) and lose \(Y\), or lose \(X\) and win \(Y\):

\[P = (0.5\times 0.3) + (0.5\times 0.7) = 0.15 + 0.35 = 0.50\]

(c) Wins neither contract

\[P(\text{not }X\cap\text{not }Y) = 0.5\times 0.3 = 0.15\]

Check: \(0.35 + 0.50 + 0.15 = 1.00\), confirming all outcomes are accounted for.

Maelezo ya Majibu

(a) Shaded segment of the circle.

From the diagram, \(O\) is the centre, \(OX=OY=r\), \(\angle XOY=90^\circ\), and the shaded part is the segment lying between chord \(XY\) and the arc \(XKY\).

\[\text{Shaded area}=\text{area of sector } XOY-\text{area of }\triangle XOY.\]

Sector \(XOY\) subtends \(90^\circ\), so it is a quarter of the circle:

\[\text{sector}=\frac{90}{360}\pi r^{2}=\frac{1}{4}\pi r^{2}.\]

Triangle \(XOY\) is right-angled at \(O\) with legs \(r\) and \(r\):

\[\triangle XOY=\frac{1}{2}\,r\times r=\frac{1}{2}r^{2}.\]

Hence

\[504=\frac{1}{4}\pi r^{2}-\frac{1}{2}r^{2}=r^{2}\left(\frac{1}{4}\times\frac{22}{7}-\frac{1}{2}\right)=r^{2}\left(\frac{22}{28}-\frac{14}{28}\right)=r^{2}\left(\frac{8}{28}\right)=\frac{2}{7}r^{2}.\]

\[r^{2}=504\times\frac{7}{2}=1764\quad\Rightarrow\quad r=\sqrt{1764}=\boxed{42\text{ cm}}.\]

(b) Two isosceles triangles on a common base PQ.

Triangles \(PQR\) and \(PQS\) stand on opposite sides of the common base \(PQ\).

In \(\triangle PQR\) (isosceles with \(RP=RQ\)), the base angles are equal, and \(\angle PQR=66^\circ\) is a base angle, so

\[\angle RPQ=\angle RQP=66^\circ.\]

In \(\triangle PQS\) (isosceles with \(SP=SQ\)), \(\angle PSQ=109^\circ\) is the apex angle, so the base angles are

\[\angle SPQ=\angle SQP=\frac{180^\circ-109^\circ}{2}=\frac{71^\circ}{2}=35.5^\circ.\]

Since \(R\) and \(S\) are on opposite sides of \(PQ\), the angle \(RQS\) is the sum of the two base angles at \(Q\):

\[\angle RQS=\angle RQP+\angle PQS=66^\circ+35.5^\circ=\boxed{101.5^\circ}.\]

Maelezo ya Majibu

(a) Solve \(\dfrac{3}{2p-\tfrac{1}{2}} = \dfrac{\tfrac{1}{3}}{\tfrac{1}{4}p + 1}\)

Cross multiply:

\[3\left(\tfrac{1}{4}p + 1\right) = \tfrac{1}{3}\left(2p - \tfrac{1}{2}\right)\]

Expand both sides:

\[\tfrac{3}{4}p + 3 = \tfrac{2}{3}p - \tfrac{1}{6}\]

Collect the \(p\) terms and constants:

\[\tfrac{3}{4}p - \tfrac{2}{3}p = -\tfrac{1}{6} - 3\]

\[\left(\tfrac{9-8}{12}\right)p = -\tfrac{19}{6} \;\Rightarrow\; \tfrac{1}{12}p = -\tfrac{19}{6}\]

\[p = -\tfrac{19}{6}\times 12 = -38\]

\(p = -38\).

(b) Actual percentage profit

The marked price of GH¢760.00 gives a 20% profit, so it is \(120\%\) of the cost price:

\[\text{Cost price} = \frac{760}{1.20} = \text{GH¢}633.33\]

Sold at a 5% discount off the marked price:

\[\text{Selling price} = 760\times 0.95 = \text{GH¢}722.00\]

Actual profit \(= 722.00 - 633.33 = \text{GH¢}88.67\).

\[\text{Percentage profit} = \frac{88.67}{633.33}\times 100 \approx 14\%\]

Actual percentage profit \(\approx 14\%\) (to 2 s.f.).

Maelezo ya Majibu

(a) Solve \((x-2)(x-3)=12\).

Expand the left side:

\[x^2-5x+6=12\]

Bring every term to one side:

\[x^2-5x-6=0\]

Factorise. Two numbers whose product is \(-6\) and whose sum is \(-5\) are \(-6\) and \(+1\):

\[(x-6)(x+1)=0\]

Hence \(x-6=0\) or \(x+1=0\), giving:

\[x=6 \quad\text{or}\quad x=-1\]

(b) Area of the shaded (overlap) portion.

From the diagram the two circles have centre \(M\) and \(N\), each of radius \(r=7\text{ cm}\), and \(\angle PMQ=\angle PNQ=60^{\circ}\). The shaded lens is the region common to both circles. It is made up of two equal circular segments cut off by the common chord \(PQ\): one from circle \(M\) and one from circle \(N\).

Area of one segment = area of sector \(-\) area of triangle.

Area of sector \(PMQ\):

\[\frac{60}{360}\times\frac{22}{7}\times 7^2=\frac{1}{6}\times\frac{22}{7}\times 49=\frac{154}{6}=25.667\text{ cm}^2\]

Area of triangle \(PMQ\) (two radii with included angle \(60^{\circ}\)):

\[\frac{1}{2}\times 7\times 7\times\sin 60^{\circ}=24.5\times 0.8660=21.217\text{ cm}^2\]

Area of one segment:

\[25.667-21.217=4.450\text{ cm}^2\]

The shaded lens is two such segments (one per circle):

\[2\times 4.450=8.90\text{ cm}^2\]

Correct to the nearest whole number, the area of the shaded portion is \(9\text{ cm}^2\).

Maelezo ya Majibu

(a) Simplify in standard form.

Write every number in standard form first, then combine powers of 10.

Numerator:

\[0.6\times32\times0.004=(6\times10^{-1})(3.2\times10^{1})(4\times10^{-3})\]\[=(6\times3.2\times4)\times10^{-1+1-3}=76.8\times10^{-3}=7.68\times10^{-2}\]

Denominator:

\[1.2\times0.008\times0.16=(1.2\times10^{0})(8\times10^{-3})(1.6\times10^{-1})\]\[=(1.2\times8\times1.6)\times10^{0-3-1}=15.36\times10^{-4}=1.536\times10^{-3}\]

Divide:

\[\frac{7.68\times10^{-2}}{1.536\times10^{-3}}=\frac{7.68}{1.536}\times10^{-2-(-3)}=5\times10^{1}\]

So the value is \(\mathbf{5.0\times10^{1}}\) (that is, \(50\)).

(b) Find \(\angle GHB\).

From the diagram, \(\overline{EF}\parallel\overline{GH}\). The line \(A\,E\,B\) is straight, it bends at \(B\) where \(\angle ABC=120^{\circ}\), and the line then continues straight through to \(H\). The marked angles are \(\angle AEF=3x^{\circ}\) and \(\angle CHG=7x^{\circ}\).

Draw a line \(BK\) through \(B\) parallel to both \(\overline{EF}\) and \(\overline{GH}\).

Since \(A,E,B\) are collinear, \(\angle FEB=180^{\circ}-3x\). With \(EF\parallel BK\), the co-interior angles give

\[\angle EBK=180^{\circ}-\angle FEB=3x^{\circ}\]

With \(BK\parallel GH\) and \(BH\) a transversal, alternate angles give

\[\angle KBH=\angle GHB=7x^{\circ}\]

The ray \(BK\) lies inside \(\angle EBH\), so

\[\angle EBH=\angle EBK+\angle KBH\]\[120=3x+7x=10x\]\[x=12\]

Therefore

\[\angle GHB=7x=7(12)=\mathbf{84^{\circ}}\]

Maelezo ya Majibu

(a)(i) Adding all the regions of the Venn diagram and equating to \( n(U) = 125 \):

\[ (16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125 \]

\[ 45 + 20x = 125 \implies 20x = 80 \implies x = 4 \]

(a)(ii) \( n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x) = 22 + 4x \)

\[ = 22 + 4(4) = 22 + 16 = \mathbf{38} \]

(b) XW is a diameter, so \( \angle XYW = 90^\circ \) (angle in a semicircle).

In triangle XYW: \( \angle XWY + 90^\circ + 50^\circ = 180^\circ \), so \( \angle XWY = 40^\circ \).

(i) Since \( WX \parallel YZ \), \( \angle WYZ = \angle XWY = \mathbf{40^\circ} \) (alternate angles).

(ii) \( \angle WOZ = 2 \times \angle WYZ = 2 \times 40^\circ = 80^\circ \) (angle at centre is twice angle at circumference). In triangle WEO, \( \angle WEO = 180^\circ - (40^\circ + 80^\circ) = 60^\circ \), therefore \( \angle YEZ = \mathbf{60^\circ} \) (vertically opposite angles).

Maelezo ya Majibu

Maelezo ya Majibu

(a) Simplify \(3\sqrt{75} - \sqrt{12} + \sqrt{108}\)

Reduce each surd to lowest terms by taking out perfect squares:

\[\sqrt{75} = \sqrt{25\times 3} = 5\sqrt{3} \;\Rightarrow\; 3\sqrt{75} = 15\sqrt{3}\]

\[\sqrt{12} = \sqrt{4\times 3} = 2\sqrt{3}\]

\[\sqrt{108} = \sqrt{36\times 3} = 6\sqrt{3}\]

Combine the like surds:

\[15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3} = 19\sqrt{3}\]

(b) If \(124_{n} = 232_{five}\), find \(n\).

Convert the right side to base ten:

\[232_{five} = 2(5^{2}) + 3(5) + 2 = 50 + 15 + 2 = 67\]

Express the left side in base ten:

\[124_{n} = 1(n^{2}) + 2(n) + 4 = n^{2} + 2n + 4\]

Set them equal and solve:

\[n^{2} + 2n + 4 = 67 \;\Rightarrow\; n^{2} + 2n - 63 = 0\]

\[(n+9)(n-7) = 0\]

Since a base must be positive, \(n = 7\).

Maelezo ya Majibu

(a) Solve \(\dfrac{1}{x}+\dfrac{1}{y}=5\) and \(\dfrac{1}{y}-\dfrac{1}{x}=1\)

Let \(u=\dfrac{1}{x}\) and \(v=\dfrac{1}{y}\). Then

\[u+v = 5 \quad\text{and}\quad v-u = 1\]

Add the two equations:

\[2v = 6 \;\Rightarrow\; v = 3 \;\Rightarrow\; \frac{1}{y}=3 \;\Rightarrow\; y = \frac{1}{3}\]

Then \(u = 5 - v = 2\), so \(\dfrac{1}{x}=2 \Rightarrow x = \dfrac{1}{2}\).

\(x = \tfrac{1}{2},\; y = \tfrac{1}{3}\).

(b) Length of good surface

Total distance \(48\) km in \(45\) minutes \(= 0.75\) h. Let good surface \(= g\) km at \(72\) km/h and bad surface \(= b\) km at \(48\) km/h.

Distance: \(g + b = 48\).

Time: \(\dfrac{g}{72} + \dfrac{b}{48} = 0.75\). Multiply through by \(144\):

\[2g + 3b = 108\]

Substitute \(b = 48 - g\):

\[2g + 3(48 - g) = 108 \;\Rightarrow\; 2g + 144 - 3g = 108 \;\Rightarrow\; -g = -36\]

\[g = 36\]

The good surface is 36 km.