WAEC SSCE - General Mathematics - 1995 (Essay)

Antwoorddetails

(a) Cumulative frequency table

Marks (%)	Number of pupils, \(f\)	Cumulative frequency
11 - 20	68	68
21 - 30	184	252
31 - 40	294	546
41 - 50	402	948
51 - 60	480	1428
61 - 70	310	1738
71 - 80	164	1902
81 - 90	98	2000

Using upper class boundaries, plot the cumulative frequencies and join the points with a smooth increasing curve.

(b)(i) Cut-off mark

The 300 candidates admitted are the highest scorers. Hence the number below the cut-off mark is

\[2000-300=1700.\]

From the ogive, at cumulative frequency \(1700\), the corresponding mark is approximately \(68\).

Therefore, the cut-off mark is 68 marks (68%).

(b)(ii) Probability of scoring at least \(45\%\)

From the ogive, the cumulative frequency at \(45\%\) is approximately \(720\).

\[\text{Number scoring at least }45\%=2000-720=1280.\]

\[P(\text{score at least }45\%)=\frac{1280}{2000}=0.64.\]

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(a) \(p = x + ym^3\). Make \(m\) the subject.

\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)

(b) \(x^2 - 6x + 7 = 0\) by completing the square.

\(x^2 - 6x = -7\)

Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:

\((x-3)^2 = -7 + 9 = 2\)

\(x - 3 = \pm\sqrt{2} = \pm 1.414\)

\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)

(c) Let the consecutive odd numbers be \(n\) and \(n+2\).

\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)

\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)

Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).

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(a) "Partly constant and partly varies inversely as the square of \(x\)" means \[y = a + \frac{b}{x^2},\] where \(a\) and \(b\) are constants.

(b) Using the given values:

When \(x = 1, y = 11\): \(a + b = 11.\quad(1)\)

When \(x = 2, y = 5\): \(a + \dfrac{b}{4} = 5.\quad(2)\)

Subtract \((2)\) from \((1)\): \(b - \dfrac{b}{4} = 6 \Rightarrow \dfrac{3b}{4} = 6 \Rightarrow b = 8.\) Then \(a = 11 - 8 = 3.\)

So \(y = 3 + \dfrac{8}{x^2}.\) When \(x = 4\): \[y = 3 + \frac{8}{16} = 3 + 0.5 = \mathbf{3.5}.\]

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(a) Finding /PZ/

The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.

In a parallelogram opposite sides are equal, so:

• In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
• In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).

Since \(P, Q, Y, Z\) are collinear in that order:

\[ PZ = PQ + QY + YZ \]

\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]

/PZ/ = 8 cm.

(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N

Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:

\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]

The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:

\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]

Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):

\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]

\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]

\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]

Distance PQ \(\approx\) 4374 km (to the nearest km).

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(a) Grouping the four terms \(px - 2qx - 4qy + 2py\) in pairs: \[(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q) = (p - 2q)(x + 2y).\] So the expression factorises as \(\mathbf{(p - 2q)(x + 2y)}.\)

(b) \(U = \{1,2,3,4,5,6,7,8,9,10\},\ P = \{1,2,4,6,10\},\ Q = \{2,3,6,9\}.\)

Left side: \(P \cup Q = \{1,2,3,4,6,9,10\}\), so \[(P \cup Q)' = \{5, 7, 8\}.\]

Right side: \(P' = \{3,5,7,8,9\}\) and \(Q' = \{1,4,5,7,8,10\}\), so \[P' \cap Q' = \{5, 7, 8\}.\]

Since \((P \cup Q)' = \{5,7,8\} = P' \cap Q'\), the identity is verified (De Morgan's law).

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Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).

(a)(i) Mean. \(\sum fx = 167\).

\[\bar{x}=\frac{167}{32}\approx 5.22.\]

(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.

\[\text{Median}=\frac{5+5}{2}=5.\]

(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.

(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.

\[\frac{6}{32}\times100\% = 18.75\%.\]

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(a) For each value of x, evaluate \(y=2\cos 2x-1\).

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(y=2\cos2x-1\)	1.0	0	-2.0	-3.0	-2.0	0	1.0

(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.

For the straight line, the plotting values are:

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(y=\dfrac{x-360}{180}\)	-2.00	-1.83	-1.67	-1.50	-1.33	-1.17	-1.00

(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]

(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]

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(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.

\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)

(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).

• \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
• \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
• \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)

Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.

\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)

Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).

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Construction and measurement

1. Draw \(BC=8.1\text{ cm}\).
2. At \(B\), construct an angle of \(105^\circ\). Construct \(90^\circ\) and \(120^\circ\), then bisect the angle between them to obtain \(105^\circ\).
3. On the \(105^\circ\) ray, use compasses to mark \(BA=7.5\text{ cm}\). Join \(A\) to \(C\).
4. Divide \(BC\) in the ratio \(3:2\). Since there are \(3+2=5\) equal parts, \[ BD=\frac{3}{5}\times 8.1=4.86\text{ cm}. \] Mark \(D\) \(4.86\text{ cm}\) from \(B\) along \(BC\).
5. Construct the perpendicular to \(BC\) through \(D\). Where it meets \(AC\) is \(P\).
6. Measure \(BP\).

The measurement from \(B\) to \(P\) is approximately

Examination reminder: The ratio \(3:2\) means that \(BC\) is split into \(5\) equal parts altogether, with \(D\) located \(3\) of those parts from \(B\), not \(2\) parts from \(B\).

Antwoorddetails

(a)(i) From tables, \(\sin 63.35° = 0.8937\).

\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)

(ii) \(\cos 44.74° = 0.7103\).

\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)

(b) \(\log K - \log(K-2) = \log 5\)

\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)

\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)

(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.

\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)

Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).

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Total balls \(= 5 + 3 + 2 = 10\). Since each ball is replaced, the two draws are independent and the probabilities on each draw stay the same.

\(P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2},\quad P(\text{black}) = \tfrac{3}{10},\quad P(\text{red}) = \tfrac{2}{10} = \tfrac{1}{5}\)

(a) Two red balls:

\(P = \tfrac{1}{5}\times\tfrac{1}{5} = \dfrac{1}{25}\)

(b) Two blue OR two black (mutually exclusive, so add):

\(P = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{3}{10}\right)^2 = \tfrac{1}{4} + \tfrac{9}{100} = \tfrac{25}{100} + \tfrac{9}{100} = \dfrac{34}{100} = \dfrac{17}{50}\)

(c) One black and one red in any order (black-then-red or red-then-black):

\(P = 2\times\tfrac{3}{10}\times\tfrac{2}{10} = 2\times\tfrac{6}{100} = \dfrac{12}{100} = \dfrac{3}{25}\)

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(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).

\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)

From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).

Z is due East of X, so its North component is \(0\):

\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)

(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).

East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).

(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).

(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).

(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)

\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).

(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).