WAEC SSCE - General Mathematics - 2003 (Essay)

Antwoorddetails

The percentages account for the whole salary, and they sum to

\[5+25+40+10+12.5+7.5 = 100.\]

(a) Pie chart

The whole circle is \(360^\circ\), representing \(100\%\). Each item's sector angle is its percentage multiplied by \(\dfrac{360^\circ}{100} = 3.6^\circ\) per unit:

Item	Percentage (%)	Sector angle
Pension	5	\(\frac{5}{100}\times360^\circ = 18^\circ\)
Income Tax	25	\(\frac{25}{100}\times360^\circ = 90^\circ\)
Food	40	\(\frac{40}{100}\times360^\circ = 144^\circ\)
Transport	10	\(\frac{10}{100}\times360^\circ = 36^\circ\)
Rent	12.5	\(\frac{12.5}{100}\times360^\circ = 45^\circ\)
Others	7.5	\(\frac{7.5}{100}\times360^\circ = 27^\circ\)
Total	100	\(360^\circ\)

Check: \(18^\circ+90^\circ+144^\circ+36^\circ+45^\circ+27^\circ = 360^\circ.\) The completed pie chart is:

(b) Salary calculations (gross monthly salary \(=\) N10,800.00)

(i) Pension contribution \(= 5\%\) of N10,800, i.e. the \(18^\circ\) sector:

\[\frac{5}{100}\times 10800 \;=\; \frac{18}{360}\times 10800 \;=\; \text{N}540.00.\]

(ii) Income tax \(= 25\%\) of N10,800, i.e. the \(90^\circ\) sector:

\[\frac{25}{100}\times 10800 \;=\; \frac{90}{360}\times 10800 \;=\; \text{N}2{,}700.00.\]

(c) Take-home pay

Deduct the pension and income tax from the gross salary. Together they are \((5+25)\% = 30\%\), so the take-home pay is \(100\% - 30\% = 70\%\):

\[10800 - 540 - 2700 \;=\; \frac{70}{100}\times 10800 \;=\; \text{N}7{,}560.00.\]

The employee's take-home pay is N7,560.00.

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(a) Copy and complete the table

Using \(y = 11 - 2x - 2x^{2}\), build the value from its parts \(x^{2}\), \(11\), \(-2x\) and \(-2x^{2}\) for each \(x\):

\(x=-3:\ 11-2(-3)-2(9)=11+6-18=-1\)
\(x=-2:\ 11+4-8=7\)
\(x=-1:\ 11+2-2=11\)
\(x=1:\ 11-2-2=7\)
\(x=2:\ 11-4-8=-1\)
\(x=3:\ 11-6-18=-13\)

\(x\)	-4	-3	-2	-1	0	1	2	3
\(x^{2}\)	16	9	4	1	0	1	4	9
\(11\)	11	11	11	11	11	11	11	11
\(-2x\)	8	6	4	2	0	-2	-4	-6
\(-2x^{2}\)	-32	-18	-8	-2	0	-2	-8	-18
\(y\)	-13	-1	7	11	11	7	-1	-13

(b) The graph of \(y = 11 - 2x - 2x^{2}\)

Plotting the eight points \((-4,-13),(-3,-1),(-2,7),(-1,11),(0,11),(1,7),(2,-1),(3,-13)\) with a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, and joining them with a smooth curve, gives a downward (maximum) parabola with its peak midway between \(x=-1\) and \(x=0\), at about \((-0.5,\ 11.5)\).

(c) Using the graph

(i) Roots of \(11 - 2x - 2x^{2} = 0\). These are the \(x\)-values where the curve crosses the \(x\)-axis (\(y=0\)). Reading off the graph:

\[x \approx -2.9 \quad \text{or} \quad x \approx 1.9\]

(ii) Values of \(x\) for which \(3 - 2x - 2x^{2} = 0\). Subtracting this equation from the curve:

\[y - (3 - 2x - 2x^{2}) = (11 - 2x - 2x^{2}) - (3 - 2x - 2x^{2}) = 8\]

So where \(3 - 2x - 2x^{2}=0\) the curve has \(y = 8\). Drawing the horizontal line \(y = 8\) (shown dashed) and reading where it meets the curve:

\[x \approx -1.8 \quad \text{or} \quad x \approx 0.8\]

(iii) Gradient of the curve at \(x = 1\). Draw the tangent to the curve at the point \((1,\ 7)\) (shown as the straight line touching the curve). Taking two points on this tangent, e.g. \((0,\ 13)\) and \((2,\ 1)\):

\[\text{gradient} = \frac{1 - 13}{2 - 0} = \frac{-12}{2} = -6\]

Hence the gradient of the curve at \(x = 1\) is \(-6\).

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(a) Write \(625=5^{4}\) and \(25=5^{2}\).

\[625^{\frac{3}{8}}\times5^{\frac{1}{2}}\div25=(5^{4})^{\frac{3}{8}}\times5^{\frac{1}{2}}\div5^{2}=5^{\frac{12}{8}}\times5^{\frac{1}{2}}\times5^{-2}\]

Add the indices: \(\dfrac{12}{8}+\dfrac{1}{2}-2=\dfrac{3}{2}+\dfrac{1}{2}-2=2-2=0\).

\[=5^{0}=1\]

\(=1\)

(b)(i) \(\tan(\theta+25)^{\circ}=5.145\).

\((\theta+25)^{\circ}=\tan^{-1}5.145=79.0^{\circ}\), so \(\theta=79.0-25=54.0^{\circ}\).

\(\theta\approx 54.0^{\circ}\)

(ii) \(5\cos\theta-1=0\Rightarrow \cos\theta=\dfrac{1}{5}=0.2\).

\(\theta=\cos^{-1}0.2=78.5^{\circ}\) (within \(0^{\circ}\le\theta\le90^{\circ}\)).

\(\theta\approx 78.5^{\circ}\)

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Reading the diagram. S is at the top and Q at the bottom, with centre O lying on SQ, so SQ is a vertical chord passing through the centre. P is on the left and R on the right, with PR a horizontal chord. The two chords SQ and PR cross at the interior point K, where \(\angle SKR = 90^\circ\). The radius is \(7\text{ cm}\).

Step 1: Apply the intersecting-chords angle rule. When two chords cross inside a circle, the angle between them equals half the sum of the two arcs it intercepts. The angle \(\angle SKR\) intercepts arc \(SR\), while its vertically opposite angle intercepts arc \(PQ\). Hence

\[\angle SKR = \tfrac{1}{2}\left(\text{arc } SR + \text{arc } PQ\right).\]

Step 2: Form an equation in the arc measures. Let arc \(PQ = x^\circ\). Since arc \(SR\) is four times arc \(PQ\), arc \(SR = 4x^\circ\). With \(\angle SKR = 90^\circ\):

\[90 = \tfrac{1}{2}(4x + x) \;\Rightarrow\; 180 = 5x \;\Rightarrow\; x = 36.\]

So the arc \(SR\) subtends an angle of \(4x = 4(36) = 144^\circ\) at the centre.

Step 3: Convert the arc angle to a length. Using \(\text{arc length} = \dfrac{\theta}{360}\times 2\pi r\) with \(\theta = 144^\circ\), \(r = 7\), \(\pi = \tfrac{22}{7}\):

\[\text{arc } SR = \frac{144}{360}\times 2\times\frac{22}{7}\times 7 = \frac{144}{360}\times 44 = 0.4\times 44 = 17.6\text{ cm}.\]

Answer: The length of arc \(SR\) is \(\mathbf{17.6\text{ cm}}\).

Antwoorddetails

(a) Simplify the compound fraction

\[\frac{\tfrac{1}{3}c^{2} - \tfrac{2}{3}cd}{\tfrac{1}{2}d^{2} - \tfrac{1}{4}cd}\]

Factorise numerator and denominator.

\[\text{Numerator} = \tfrac{1}{3}c^{2} - \tfrac{2}{3}cd = \tfrac{1}{3}c\,(c - 2d)\]\[\text{Denominator} = \tfrac{1}{2}d^{2} - \tfrac{1}{4}cd = \tfrac{1}{4}d\,(2d - c)\]

Write \((c - 2d) = -(2d - c)\):

\[\frac{\tfrac{1}{3}c\,(c-2d)}{\tfrac{1}{4}d\,(2d-c)} = \frac{\tfrac{1}{3}c\,[-(2d-c)]}{\tfrac{1}{4}d\,(2d-c)} = -\frac{\tfrac{1}{3}c}{\tfrac{1}{4}d}\]\[= -\frac{1}{3}c \times \frac{4}{d} = \boxed{-\frac{4c}{3d}}\]

(b) Two right-angled triangles on the straight line \(YPF\)

From the diagram, \(YPF\) is a straight line, \(\angle XYP = \angle MFP = 90^\circ\), \(\angle XPY = 44^\circ\), \(\angle MPF = 46^\circ\), \(|XY| = 7\text{ cm}\) and \(|MP| = 9\text{ cm}\).

(i) Calculate \(|XM|\) and \(|YF|\).

Triangle \(XYP\) (right-angled at \(Y\)):

\[\tan 44^\circ = \frac{|XY|}{|YP|} \Rightarrow |YP| = \frac{7}{\tan 44^\circ} = \frac{7}{0.9657} = 7.249\text{ cm}\]\[\sin 44^\circ = \frac{|XY|}{|XP|} \Rightarrow |XP| = \frac{7}{\sin 44^\circ} = \frac{7}{0.6947} = 10.08\text{ cm}\]

Triangle \(MPF\) (right-angled at \(F\), hypotenuse \(MP = 9\)):

\[|PF| = 9\cos 46^\circ = 9(0.6947) = 6.252\text{ cm}\]

Length \(YF\): since \(Y\), \(P\), \(F\) are collinear,

\[|YF| = |YP| + |PF| = 7.249 + 6.252 = 13.5\text{ cm (3 s.f.)}\]

Length \(XM\): at \(P\) the angles on the straight line give

\[\angle XPM = 180^\circ - 44^\circ - 46^\circ = 90^\circ\]

so triangle \(XPM\) is right-angled at \(P\) with \(|XP| = 10.08\) and \(|PM| = 9\):

\[|XM| = \sqrt{|XP|^{2} + |PM|^{2}} = \sqrt{10.08^{2} + 9^{2}} = \sqrt{101.5 + 81} = \sqrt{182.5} = 13.5\text{ cm (3 s.f.)}\]

(ii) Find \(\angle XMP\). In the right-angled triangle \(XPM\) (right angle at \(P\)),

\[\tan(\angle XMP) = \frac{|XP|}{|PM|} = \frac{10.08}{9} = 1.1196\]\[\angle XMP = \tan^{-1}(1.1196) = 48.2^\circ\]

Hence \(|XM| \approx 13.5\text{ cm}\), \(|YF| \approx 13.5\text{ cm}\) and \(\angle XMP \approx 48.2^\circ\).

Antwoorddetails

Given data.

No. of suitcases (x)	0	1	2	3	4	5
Travellers (f)	2	7	7	2	3	9

Total number of travellers: \(\sum f = 2+7+7+2+3+9 = 30\).

(a)(i) Median

With \(N = 30\) values (even), the median is the mean of the 15th and 16th values. Building the cumulative frequencies:

x	0	1	2	3	4	5
Cumulative f	2	9	16	18	21	30

Positions 10 to 16 correspond to \(x = 2\); hence both the 15th and 16th values equal 2.

\[\text{Median} = \frac{2+2}{2} = 2 \text{ suitcases.}\]

(a)(ii) Mean

\[\sum fx = (0)(2)+(1)(7)+(2)(7)+(3)(2)+(4)(3)+(5)(9) = 0+7+14+6+12+45 = 84.\]\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{84}{30} = 2.8 \approx 3 \text{ (nearest whole number).}\]

(b) Bar chart

The number of suitcases (0 to 5) is placed on the horizontal axis and the number of travellers on the vertical axis. Each bar has equal width with equal gaps, and the heights are 2, 7, 7, 2, 3 and 9 respectively.

The tallest bar occurs at 5 suitcases, with a height of 9 travellers.

Antwoorddetails

(a) Simplify \(\sqrt{50}-3\sqrt{2}(2\sqrt{2}-5)-5\sqrt{32}\).

\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), so \(5\sqrt{32}=20\sqrt{2}\).

Expand the bracket: \(3\sqrt{2}(2\sqrt{2}-5)=6(\sqrt{2}\cdot\sqrt{2})-15\sqrt{2}=6(2)-15\sqrt{2}=12-15\sqrt{2}\).

Combine:

\[5\sqrt{2}-(12-15\sqrt{2})-20\sqrt{2}=5\sqrt{2}-12+15\sqrt{2}-20\sqrt{2}=(5+15-20)\sqrt{2}-12=-12\]

\(=-12\)

(b) \(\tfrac{1}{2}\log_{10}\tfrac{25}{4}-2\log_{10}\tfrac{4}{5}+\log_{10}\tfrac{320}{125}\).

Bring the coefficients inside as powers: \(\tfrac{1}{2}\log\tfrac{25}{4}=\log\left(\tfrac{25}{4}\right)^{1/2}=\log\tfrac{5}{2}\); \(\;2\log\tfrac{4}{5}=\log\tfrac{16}{25}\).

\[\log\frac{5}{2}-\log\frac{16}{25}+\log\frac{320}{125}=\log\!\left(\frac{5}{2}\times\frac{25}{16}\times\frac{320}{125}\right)\]

\(\dfrac{5}{2}\times\dfrac{25}{16}=\dfrac{125}{32}\); then \(\dfrac{125}{32}\times\dfrac{320}{125}=\dfrac{320}{32}=10\).

\[=\log_{10}10=1\]

\(=1\)

Antwoorddetails

(a) P, Q, R, S lie on the circle, PQT is a straight line (T on PQ produced beyond Q), and \(SQ\parallel RT\). Join QR.

(i) Show that \(\angle RPS=\angle QRT\).

• \(\angle RPS\) and \(\angle RQS\) both stand on the same chord \(RS\) and P, Q are in the same segment, so \(\angle RPS=\angle RQS\) (angles in the same segment).
• \(SQ\parallel RT\) with QR as transversal gives \(\angle RQS=\angle QRT\) (alternate angles).

Therefore \(\angle RPS=\angle QRT\), as required.

(ii) Show that \(\angle PRS=\angle QTR\).

• \(\angle PRS\) and \(\angle PQS\) both stand on the same chord \(PS\) with R, Q in the same segment, so \(\angle PRS=\angle PQS\) (angles in the same segment).
• The straight line PQT is a transversal cutting the parallel lines \(SQ\) and \(RT\) at Q and T, so \(\angle PQS=\angle QTR\) (corresponding angles).

Therefore \(\angle PRS=\angle QTR\), as required.

(b) In triangle ABC, AB is produced to D and AC is produced to E, with \(\angle DBC=132^\circ\) and the exterior angle at C equal to \(96^\circ\).

\(\angle DBC\) is the exterior angle at B, and it is the supplement of the interior angle \(\angle ABC\) (angles on the straight line ABD):

\[\angle ABC=180^\circ-132^\circ=48^\circ\]

The exterior angle at C is the supplement of the interior angle \(\angle ACB\) (angles on the straight line ACE):

\[\angle ACB=180^\circ-96^\circ=84^\circ\]

The three interior angles of triangle ABC sum to \(180^\circ\):

\[\angle BAC=180^\circ-48^\circ-84^\circ=48^\circ\]

Hence \(\angle BAC=\angle ABC=48^\circ\). Since two angles are equal, the sides opposite them are equal, i.e. \(BC=AC\). Therefore triangle ABC is isosceles.

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Distance flown due west \(=\text{speed}\times\text{time}=600\times3=1800\) km, along the parallel of latitude \(50^{\circ}N\).

(a) Longitude of Q. Distance along a parallel \(=\dfrac{\theta}{360}\times2\pi R\cos(\text{lat})\), where \(\theta\) is the change in longitude.

\[1800=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\cos50^{\circ}\]

\(2\times\tfrac{22}{7}\times6400=40228.6\); \(\cos50^{\circ}=0.6428\); product \(=25859\).

\[\theta=\frac{1800\times360}{25859}\approx 25.1^{\circ}\]

Flying due west from long. \(60^{\circ}W\) increases the western longitude: \(60+25.1=85.1^{\circ}W\).

Longitude of Q \(\approx 85.1^{\circ}W\) (3 s.f.).

(b) Latitude of Y. Flying due south is along a meridian: distance \(=\dfrac{\phi}{360}\times2\pi R\).

\[500=\frac{\phi}{360}\times40228.6\Rightarrow \phi=\frac{500\times360}{40228.6}\approx 4.47^{\circ}\]

Moving south from \(50^{\circ}N\): latitude of \(Y=50-4.47\approx 45.5^{\circ}N\).

Latitude of Y \(\approx 45.5^{\circ}N\) (3 s.f.).

Antwoorddetails

(a) Completed frequency distribution

Sorting the 40 marks into the given class boundaries and tallying each mark:

Class Boundaries	Tally	Frequency
59.5 - 64.5	||	2
64.5 - 69.5	|||	3
69.5 - 74.5	|||| |	6
74.5 - 79.5	|||| |||| |	11
79.5 - 84.5	|||| |||	8
84.5 - 89.5	|||| |	6
89.5 - 94.5	||	2
94.5 - 99.5	||	2
Total		40

(Check: \(2+3+6+11+8+6+2+2 = 40\) students.)

(b) Histogram

Since all class widths are equal (5 marks), the bar heights equal the frequencies. The class boundaries are plotted on the horizontal axis and frequency on the vertical axis, with the bars drawn adjoining (no gaps):

(c) Modal mark from the histogram

The modal class is 74.5 - 79.5, the tallest bar (frequency 11). To read the mode, straight lines are drawn inside the modal bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. The vertical through their point of intersection meets the mark axis at the mode (shown dashed above), giving mode \(\approx 77.6\).

This agrees with the standard formula \[\text{Mode} = L + \left(\frac{f_1 - f_0}{(f_1-f_0)+(f_1-f_2)}\right)c,\] where \(L = 74.5,\ f_1 = 11,\ f_0 = 6,\ f_2 = 8,\ c = 5\): \[\text{Mode} = 74.5 + \left(\frac{11-6}{(11-6)+(11-8)}\right)\times 5 = 74.5 + \frac{5}{8}\times 5 = 74.5 + 3.125 \approx 77.6.\] So the modal mark is approximately 77.6 (about 78 marks).

(d) Probability of a mark greater than 79

Marks greater than 79 fall in the four highest classes (79.5 and above), giving \[8 + 6 + 2 + 2 = 18 \text{ students.}\] With 40 students in all, \[P(\text{mark} > 79) = \frac{18}{40} = \frac{9}{20} = 0.45.\]

Antwoorddetails

Antwoorddetails

(a) \(U=\{1,2,3,4,5,6,7,8,9,10\}\), \(A=\{1,2,5,7\}\), \(B=\{1,3,6,7\}\).

(i) \(A'=U\setminus A=\{3,4,6,8,9,10\}\).

(ii) \(A\cap B=\{1,7\}\), so \((A\cap B)'=\{2,3,4,5,6,8,9,10\}\).

(iii) \(A\cup B=\{1,2,3,5,6,7\}\), so \((A\cup B)'=\{4,8,9,10\}\).

(iv) The three-element subsets of \(B=\{1,3,6,7\}\) are \(\{1,3,6\},\ \{1,3,7\},\ \{1,6,7\},\ \{3,6,7\}\).

(b) The sequence \(\dfrac{2}{1\times3},\dfrac{3}{2\times4},\dfrac{4}{3\times5},\dfrac{5}{4\times6},\ldots\) has \(n\)th term \(\dfrac{n+1}{n(n+2)}\).

\[\text{15th term}=\frac{15+1}{15(15+2)}=\frac{16}{15\times17}=\frac{16}{255}\]

(c) A.P. with \(a=3,\ d=4\).

(i) \(n\)th term \(=a+(n-1)d=3+(n-1)4=4n-1\).

(ii) Least term greater than 100: \(4n-1>100\Rightarrow 4n>101\Rightarrow n>25.25\), so \(n=26\).

\[T_{26}=4(26)-1=103\]

Least term above 100 is \(103\).

Antwoorddetails

(a) \(p\propto r^{2}\Rightarrow p=kr^{2}\).

Using \(p=3.2,\ r=4\): \(3.2=k(16)\Rightarrow k=0.2\).

When \(r=6.5\): \(p=0.2(6.5)^{2}=0.2(42.25)=8.45\).

\(p=8.45\)

(b) Clear the fractions.

\(\dfrac{x}{2}+\dfrac{y}{4}=1\) \(\times4\Rightarrow 2x+y=4\) ... (1)

\(\dfrac{x}{3}-\dfrac{y}{4}=-\dfrac{1}{6}\) \(\times12\Rightarrow 4x-3y=-2\) ... (2)

From (1): \(y=4-2x\). Substitute into (2):

\[4x-3(4-2x)=-2\Rightarrow 4x-12+6x=-2\Rightarrow 10x=10\Rightarrow x=1\]

Then \(y=4-2(1)=2\).

\(x=1,\ y=2\)