WAEC SSCE - General Mathematics - 1991 (Essay)

Answer Details

(a) Solve \(2x^2 + 7x - 11 = 0\) using the quadratic formula with \(a = 2, b = 7, c = -11\).

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 88}}{4} = \frac{-7 \pm \sqrt{137}}{4}.\]

Now \(\sqrt{137} = 11.7047\), so

\[x = \frac{-7 + 11.7047}{4} = \frac{4.7047}{4} = 1.18,\qquad x = \frac{-7 - 11.7047}{4} = \frac{-18.7047}{4} = -4.68.\]

Hence \(x = 1.18\) or \(x = -4.68\) (2 d.p.).

(b) Let \(P = \dfrac{1}{x}\) and \(Q = \dfrac{1}{y}\). The equations become:

\[2P + Q = 3 \quad(1), \qquad P - 5Q = 7 \quad(2).\]

From (1): \(Q = 3 - 2P\). Substitute into (2):

\[P - 5(3 - 2P) = 7 \Rightarrow P - 15 + 10P = 7 \Rightarrow 11P = 22 \Rightarrow P = 2.\]

Then \(Q = 3 - 2(2) = -1\).

Returning to \(x\) and \(y\): \(x = \dfrac{1}{P} = \dfrac{1}{2}\) and \(y = \dfrac{1}{Q} = -1\).

Therefore \(x = \dfrac{1}{2},\; y = -1\).

Answer Details

Each ream is supposed to contain 480 sheets, so 5 reams should hold \(5 \times 480 = 2400\) sheets.

Actual sheets: \(435 + 420 + 405 + 415 + 440 = 2115\).

Total shortage \(= 2400 - 2115 = 285\) sheets.

(a) Percentage error.

\[\text{Percentage error} = \frac{\text{shortage}}{\text{supposed number}}\times 100 = \frac{285}{2400}\times 100 = 11.875\%.\]

To the nearest whole number, the percentage error \(= \mathbf{12\%}\).

(b) Amount cheated. A full ream of 480 sheets costs \(\text{N}35.00\), so one sheet is worth \(\dfrac{35}{480}\) naira.

The buyer paid for 2400 sheets but received only 2115, a shortage of 285 sheets:

\[\text{Amount cheated} = 285 \times \frac{35}{480} = \frac{9975}{480} = \text{N}20.78.\]

To the nearest naira, the buyer was cheated by \(\mathbf{\text{N}21}\).

Answer Details

Let the pilot be at a height \(H\) above the ground, with the horizontal distance to the tower equal to 8.5 km. The angle of depression to the base is \(33^\circ\) and to the top is \(30^\circ\) (the top is higher, so its depression is smaller).

Height of pilot above the ground, using the base:

\[\tan 33^\circ = \frac{H}{8.5} \Rightarrow H = 8.5\tan 33^\circ = 8.5(0.6494) = 5.520\text{ km}.\]

(a) Shortest distance from the pilot to the base. This is the straight line (line of sight):

\[d = \frac{8.5}{\cos 33^\circ} = \frac{8.5}{0.8387} = 10.135\text{ km} \approx \mathbf{10.1\text{ km}}.\]

(b) Height of the control tower. The height of the pilot above the top of the tower is

\[8.5\tan 30^\circ = 8.5(0.5774) = 4.908\text{ km}.\]

Therefore the tower height is

\[H - 8.5\tan 30^\circ = 5.520 - 4.908 = 0.612\text{ km}.\]

The height of the control tower \(\approx \mathbf{0.612\text{ km}}\) (about 612 m), correct to three significant figures.

Answer Details

Table of values

\(x\)	-3	-2	-1	0	1	2	3	4
\(y=3+2x-x^2\)	-12	-5	0	3	4	3	0	-5
\(y=2x-3\)	-9	-7	-5	-3	-1	1	3	5

Using the stated scales, the required graph is:

(i) At the points of intersection,

\[3+2x-x^2=2x-3\]

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

Reading the two intersection abscissae from the graph gives

\[x\approx -2.4\quad\text{or}\quad x\approx 2.4.\]

(ii) The highest point of the parabola is \((1,4)\). Hence, the maximum value of \(3+2x-x^2\) is

\[\boxed{4.0}.\]

(iii) The curve \(y=3+2x-x^2\) meets the line \(y=1\) at approximately \(x=-0.7\) and \(x=2.7\). The part of the parabola on or below \(y=1\) is therefore

\[\boxed{-3.0\leq x\leq -0.7\quad\text{or}\quad 2.7\leq x\leq4.0}.\]

Answer Details

Total balls \(= 9\): black \(= 3\), blue \(= 2\), red \(= 9 - 3 - 2 = 4\).

(a)(i) P(not blue) \(= \dfrac{9 - 2}{9} = \dfrac{7}{9}\).

(a)(ii) P(not red) \(= \dfrac{9 - 4}{9} = \dfrac{5}{9}\).

(b)(i) Both black, no replacement. First black: \(\dfrac{3}{9}\). After removing one black, 2 black remain out of 8: \(\dfrac{2}{8}\).

\[P = \frac{3}{9} \times \frac{2}{8} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}.\]

(b)(ii) Both blue, with replacement. The bag is restored, so each draw is \(\dfrac{2}{9}\).

\[P = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}.\]

Answer Details

(a)(i)

“Nobody takes Economics and Chemistry” is written as

\[n(E\cap C)=0\]

“4 pupils take Economics and Government” is written as

\[n(E\cap G)=4\]

(a)(ii) The Venn diagram is:

(b)(i) Let \(n(C\cap G)=x\). Since \(E\cap C=\varnothing\), there is no triple intersection.

Using the total number of pupils,

\[22=12+8+7-4-x\]

\[22=23-x\]

\[x=1\]

Therefore, the number who take both Chemistry and Government is \(\boxed{1}\).

(b)(ii)

\[\text{Government only}=8-4-1=3\]

Therefore, the number who take Government only is \(\boxed{3}\).

Answer Details

(a) Let \(N=\dfrac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\). Using logarithms:

\[\log N=0.04623-(-0.53873)=0.58496.\]

\[N=\text{antilog}(0.58496)=3.85\ (3\text{ s.f.}).\]

(b) Express in one base: \(81=3^{4}\) and \(\tfrac13=3^{-1}\).

\[\frac{\log81}{\log\frac13}=\frac{\log3^{4}}{\log3^{-1}}=\frac{4\log3}{-\log3}=-4.\]

Answer Details

(a) Proof: the angle at the centre is twice the angle at the circumference

Given: A circle centre \(O\), an arc \(AB\), with \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining part of the circumference, both standing on the same arc \(AB\).

To prove: \(\angle AOB = 2\,\angle ACB\).

Construction: Join \(C\) to \(O\) and produce it to a point \(D\).

Proof: In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles and \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the two opposite interior angles:

\[ \angle AOD = \angle OAC + \angle OCA = 2\,\angle OCA \]

Similarly, in triangle \(OBC\), \(OB = OC\) (radii), so \(\angle OBC = \angle OCB\) and

\[ \angle BOD = \angle OBC + \angle OCB = 2\,\angle OCB \]

Adding:

\[ \angle AOD + \angle BOD = 2\,\angle OCA + 2\,\angle OCB \]

\[ \angle AOB = 2(\angle OCA + \angle OCB) = 2\,\angle ACB \]

Hence the angle subtended at the centre is twice that subtended at the circumference. (The same argument holds when \(O\) lies outside triangle \(ACB\), using subtraction instead of addition.) \(\blacksquare\)

(b) Calculations

\(O\) is the centre, \(\angle OQR = 32^{\circ}\) and \(\angle MPQ = 15^{\circ}\).

(i) \(\angle QPR\)

In triangle \(OQR\), \(OQ = OR\) (radii), so it is isosceles and \(\angle ORQ = \angle OQR = 32^{\circ}\). Then

\[ \angle QOR = 180^{\circ} - 32^{\circ} - 32^{\circ} = 116^{\circ} \]

\(\angle QOR\) is the angle at the centre and \(\angle QPR\) is the angle at the circumference standing on the same arc \(QR\). By the theorem proved in (a):

\[ \angle QPR = \tfrac{1}{2}\,\angle QOR = \tfrac{1}{2}\times 116^{\circ} = 58^{\circ} \]

(ii) \(\angle MQO\)

\(\angle MPQ = 15^{\circ}\) is the angle at the circumference standing on arc \(MQ\). The angle at the centre on the same arc is

\[ \angle MOQ = 2 \times 15^{\circ} = 30^{\circ} \]

In triangle \(OMQ\), \(OM = OQ\) (radii), so it is isosceles with equal base angles:

\[ \angle MQO = \angle OMQ = \frac{180^{\circ} - 30^{\circ}}{2} = 75^{\circ} \]

\(\angle QPR = 58^{\circ}\) and \(\angle MQO = 75^{\circ}\).

Answer Details

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Answer Details

(a) Construction of parallelogram \(PQRS\)

1. Draw \(PR=9\text{ cm}\).
2. Bisect \(PR\) to obtain its midpoint \(K\), so that \(KP=KR=4.5\text{ cm}\).
3. At \(K\), construct a ray making \(60^\circ\) with \(KR\). With centre \(K\) and radius \(3\text{ cm}\), mark \(Q\) on this ray.
4. Produce \(QK\) through \(K\), and with the same radius \(3\text{ cm}\), mark \(S\) on the opposite ray. Thus \(QS=6\text{ cm}\) and \(K\) is the midpoint of \(QS\).
5. Join \(P\) to \(Q\), \(Q\) to \(R\), \(R\) to \(S\), and \(S\) to \(P\). Since the diagonals bisect each other at \(K\), \(PQRS\) is a parallelogram.

(b) Construction of the equal-area rectangle

1. From \(Q\), construct \(QD\perp PS\).
2. At \(P\), construct a perpendicular to \(PS\) on the side containing \(Q\). Transfer the length \(QD\) onto this perpendicular to locate \(A\), such that \(PA=QD\).
3. Through \(A\), construct a line parallel to \(PS\). Through \(S\), construct a line parallel to \(PA\); let the two lines meet at \(B\).
4. \(PABS\) is the required rectangle, since \(PA=QD\) and both figures have base \(PS\) and the same perpendicular height.

For the measurement,

\[|PS|=\sqrt{4.5^2+3^2-2(4.5)(3)\cos60^\circ}=\sqrt{15.75}\approx3.97\text{ cm}.\]

The area of the parallelogram is

\[\text{Area}=\frac12(9)(6)\sin60^\circ\approx23.38\text{ cm}^2.\]

Hence

\[|PA|=\frac{23.38}{3.97}\approx5.89\text{ cm}.\]

Therefore, \(|PA|\approx5.9\text{ cm}\).

Answer Details

(i) Interior angle of the pentagon. The sum of the interior angles of an \(n\)-sided polygon is \((n - 2)\times 180^\circ\). For a regular pentagon \(n = 5\):

\[\text{Each interior angle} = \frac{(5 - 2)\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.\]

(ii) < BXA. In the rectangle \(AXYE\), the side \(AE\) is a side of the pentagon and \(AX \perp AE\), so \(< XAE = 90^\circ\).

The interior angle of the pentagon at \(A\) is \(< BAE = 108^\circ\). Since \(X\) lies on \(BC\), the ray \(AX\) splits \(< BAE\):

\[< BAX = < BAE - < XAE = 108^\circ - 90^\circ = 18^\circ.\]

Also \(X\) lies on \(BC\), so in triangle \(ABX\) the angle at \(B\) equals the interior angle of the pentagon:

\[< ABX = 108^\circ.\]

The angles of triangle \(ABX\) sum to \(180^\circ\):

\[< BXA = 180^\circ - < ABX - < BAX = 180^\circ - 108^\circ - 18^\circ = 54^\circ.\]

Therefore \(< BXA = \mathbf{54^\circ}\).

Answer Details

(a) Class boundaries and cumulative frequencies

(b) Cumulative frequency curve

Plot cumulative frequency against the upper class boundaries, beginning with the point \((1.45,0)\).

The points to plot are:

\[ (1.45,0),\ (1.95,3),\ (2.45,13),\ (2.95,31),\ (3.45,41),\ (3.95,48),\ (4.45,50). \]

Join the points with a smooth cumulative frequency curve.

(c)(i) Interquartile range

Total number of customers, \(N=50\).

\[ Q_1=\frac{N}{4}=\frac{50}{4}=12.5 \]

\[ Q_3=\frac{3N}{4}=\frac{3(50)}{4}=37.5 \]