WAEC SSCE - General Mathematics - 1996 (Essay)

Detalles de la respuesta

(a) Express 0.72 using 2 and 3: \(0.72 = \dfrac{72}{100} = \dfrac{2^3 \times 3^2}{10^2}\).

\[ \log_{10} 0.72 = 3\log_{10}2 + 2\log_{10}3 - 2\log_{10}10. \] \[ = 3(0.3010) + 2(0.4771) - 2(1) = 0.9030 + 0.9542 - 2 = 1.8572 - 2. \] \[ \log_{10} 0.72 = -0.1428. \]

(b) The tree is 20 m high; the chick is on the ground at angle of depression \(39^\circ\) from the top. The angle of depression equals the angle of elevation from the chick, so in the right-angled triangle (height 20 m, horizontal distance \(d\)):

\[ \tan 39^\circ = \frac{20}{d} \implies d = \frac{20}{\tan 39^\circ} = \frac{20}{0.8098} = 24.7\ \text{m}. \]

The distance of the chick from the bottom of the tree is 25 m (to 2 significant figures).

Detalles de la respuesta

Detalles de la respuesta

(a) Make d the subject of \(S=\frac{n}{2}[2a+(n-1)d]\).

\[2S=n[2a+(n-1)d]\]\[\frac{2S}{n}=2a+(n-1)d\]\[\frac{2S}{n}-2a=(n-1)d\]\[d=\frac{\dfrac{2S}{n}-2a}{n-1}=\frac{2S-2an}{n(n-1)}=\frac{2(S-an)}{n(n-1)}.\]

(b)(i) Prove that \(\angle AOB=2\angle APB\).

Given: circle with centre O; A, B and P on the circumference.
To prove: \(\angle AOB=2\angle APB\).
Construction: Join P to O and produce it to a point Q.
Proof: \(OA=OP\) (radii), so triangle OAP is isosceles and its base angles are equal, \(x_1=x_2\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so

\[\angle AOQ=x_1+x_2=2x_2.\]

Similarly, \(OB=OP\) (radii), giving \(\angle BOQ=2y_2\) where \(y_2=\angle OPB\). Adding the two central angles:

\[\angle AOB=\angle AOQ+\angle BOQ=2x_2+2y_2=2(x_2+y_2)=2\angle APB.\]\[\therefore\ \angle AOB=2\angle APB.\]

(b)(ii) Find x, y and z. The inscribed angle standing on the same arc is \(126^\circ\). The reflex angle at the centre is twice the angle at the circumference:

\[z=2\times126^\circ=252^\circ.\]

The angle at the centre going the other way completes the full turn:

\[x=360^\circ-252^\circ=108^\circ.\]

The angle at the circumference on this arc is half the central angle:

\[y=\tfrac{1}{2}\times108^\circ=54^\circ.\]\[\therefore\ x=108^\circ,\quad y=54^\circ,\quad z=252^\circ.\]

Detalles de la respuesta

Detalles de la respuesta

Detalles de la respuesta

(a) Solving the simultaneous equations

Label the equations:

\[5x + 6y = 12 \quad (1)\]\[-3x + 4y = 11 \quad (2)\]

Eliminate \(x\). Multiply (1) by 3 and (2) by 5 so the \(x\)-terms cancel:

\[15x + 18y = 36 \quad (3)\]\[-15x + 20y = 55 \quad (4)\]

Add (3) and (4):

\[38y = 91 \;\Rightarrow\; y = \frac{91}{38} = 2\tfrac{15}{38} \approx 2.39\]

Find \(x\). Multiply (1) by 2 and (2) by 3:

\[10x + 12y = 24, \qquad -9x + 12y = 33\]

Subtract the second from the first:

\[19x = 24 - 33 = -9 \;\Rightarrow\; x = -\frac{9}{19} \approx -0.47\]

Check in (2): \(-3\left(-\tfrac{9}{19}\right) + 4\left(\tfrac{91}{38}\right) = \tfrac{54}{38} + \tfrac{364}{38} = \tfrac{418}{38} = 11.\) Correct.

Hence \(x = -\dfrac{9}{19}\), \(y = \dfrac{91}{38}\).

(b) Finding \(y = |AD|\)

From the diagram, \(ADC\) is a straight line with \(BD \perp AC\), so \(BD\) is the altitude from the right angle \(B\) to the hypotenuse \(AC\) of the right-angled triangle \(ABC\) (\(\angle ABC = 90^\circ\)). Here \(|BD| = 36\text{ cm}\), \(|DC| = 48\text{ cm}\) and \(|AD| = y\).

For a right-angled triangle, the altitude to the hypotenuse is the geometric mean of the two segments it makes on the hypotenuse:

\[|BD|^{2} = |AD|\times|DC|\]\[36^{2} = y \times 48\]\[1296 = 48y\]\[y = \frac{1296}{48} = 27\]

Hence \(y = 27\text{ cm}\).

Detalles de la respuesta

(a) With U the set of integers: \(P = \{x : x \le 2\}\), \(Q = \{x : -7 < x < 15\}\), \(R = \{x : -2 \le x < 19\}\).

(i) \(P \cap Q\) contains integers that are both \(\le 2\) and strictly between \(-7\) and \(15\), i.e. \(-6 \le x \le 2\):

\[ P \cap Q = \{-6, -5, -4, -3, -2, -1, 0, 1, 2\} = \{x : -6 \le x \le 2\}. \]

(ii) First find \(R' = \{x : x < -2\ \text{or}\ x \ge 19\} = \{x : x \le -3\ \text{or}\ x \ge 19\}\).

Then \(Q \cup R'\): \(Q\) gives \(-6 \le x \le 14\) and \(R'\) gives all \(x \le -3\) together with \(x \ge 19\); their union is \(\{x : x \le 14\} \cup \{x : x \ge 19\}\).

Now intersect with \(P = \{x : x \le 2\}\). Every integer \(\le 2\) is already \(\le 14\), so

\[ P \cap (Q \cup R') = \{x : x \le 2\} = P. \]

(b)(i) Grouped frequency table (40 marks):

Class	Frequency \(f\)	Midpoint \(x\)
20 - 29	3	24.5
30 - 39	5	34.5
40 - 49	10	44.5
50 - 59	13	54.5
60 - 69	6	64.5
70 - 79	3	74.5
Total	40

(ii) Mean: \(\sum fx = 3(24.5)+5(34.5)+10(44.5)+13(54.5)+6(64.5)+3(74.5) = 2010\).

\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{2010}{40} = 50.25 \approx 50.3. \]

Detalles de la respuesta

Cost price (outlay) \(= \text{N}120.00\) for 40 kg. Note \(50\text{k} = \text{N}0.50\).

(a) Amount realised from sales:

Quantity	Price per kg	Amount
20 kg	N5.00	N100.00
10 kg	N3.00	N30.00
5 kg	N2.00	N10.00
5 kg	N0.50	N2.50

\[ \text{Total sales} = 100 + 30 + 10 + 2.50 = \text{N}142.50. \]

(b) Total profit:

\[ \text{Profit} = 142.50 - 120.00 = \text{N}22.50. \]

Since this is positive, it is a profit.

(c) Percentage profit on outlay of N120.00:

\[ \frac{22.50}{120} \times 100\% = 18.75\%. \]

Detalles de la respuesta

(a) Multiply numerator and denominator by \(10^4\) to remove decimals:

\[ \frac{0.45 \times 0.91}{0.0117} = \frac{4500 \times 91}{117} = \frac{409500}{11700}. \]

Since \(0.45 \times 0.91 = 0.4095\) and \(0.0117 \times 35 = 0.4095\),

\[ \frac{0.4095}{0.0117} = 35. \]

(b) The number halfway between two values is their average. Write both as improper fractions over 28:

\[ 1\tfrac{6}{7} = \frac{13}{7} = \frac{52}{28}, \qquad 2\tfrac{11}{28} = \frac{67}{28}. \] \[ \text{Halfway} = \frac{1}{2}\left(\frac{52}{28} + \frac{67}{28}\right) = \frac{1}{2}\times\frac{119}{28} = \frac{119}{56} = \frac{17}{8} = 2\tfrac{1}{8}. \]

(c) Each interior angle \(= 5 \times\) exterior angle, and interior \(+\) exterior \(= 180^\circ\):

\[ 5e + e = 180^\circ \implies 6e = 180^\circ \implies e = 30^\circ. \] \[ n = \frac{360^\circ}{e} = \frac{360^\circ}{30^\circ} = 12\ \text{sides}. \]

(d) Pipe length 20 cm, internal radius \(r = 3\) cm, external radius \(R = 4\) cm, \(\pi = \tfrac{22}{7}\). Volume of material (hollow cylinder):

\[ V = \pi(R^2 - r^2)\times \text{length} = \frac{22}{7}(4^2 - 3^2)(20) = \frac{22}{7}(7)(20) = 22 \times 20 = 440\ \text{cm}^3. \]

Detalles de la respuesta

(a) Finding \(\angle BDC\) in the trapezium

ABCD is a trapezium with \(AB \parallel DC\) and \(|AD| = |AB|\), and \(\angle BAD = 106^{\circ}\).

Since \(AB \parallel DC\) with \(AD\) as transversal, \(\angle BAD\) and \(\angle ADC\) are co-interior (allied) angles, so they add to \(180^{\circ}\):

\(\angle ADC = 180^{\circ} - 106^{\circ} = 74^{\circ}\).

Triangle ABD is isosceles because \(|AD| = |AB|\); the base angles are equal:

\(\angle ADB = \angle ABD = \dfrac{180^{\circ}-106^{\circ}}{2} = 37^{\circ}\).

Now \(\angle ADC = \angle ADB + \angle BDC\), so

\(\angle BDC = 74^{\circ} - 37^{\circ} = 37^{\circ}\).

(b) The cards

Card	A	B	C	D	E
Frequency	3	4	7	5	1

Total number of cards \(= 3+4+7+5+1 = 20\).

(i) Probability the card is E

\(P(E) = \dfrac{1}{20}\).

(ii) Probability that one of the two cards (drawn without replacement) is B

There are 4 B-cards and 16 non-B cards. It is easiest to find the probability of drawing no B, then subtract from 1.

\(P(\text{no B}) = \dfrac{16}{20}\times\dfrac{15}{19} = \dfrac{240}{380} = \dfrac{12}{19}\).

\(P(\text{at least one B}) = 1 - \dfrac{12}{19} = \dfrac{7}{19}\).

Detalles de la respuesta

Place base camp R at the origin. Convert each bearing to \((\text{East}, \text{North})\) components using East \(= r\sin\theta\), North \(= r\cos\theta\).

P: 20 km on bearing \(205^\circ\):

\[ P = (20\sin 205^\circ,\ 20\cos 205^\circ) = (-8.45,\ -18.13). \]

Q: 15 km on bearing \(060^\circ\):

\[ Q = (15\sin 60^\circ,\ 15\cos 60^\circ) = (12.99,\ 7.50). \]

(a) Distance of Q from P. The displacement is

\[ Q - P = (12.99 + 8.45,\ 7.50 + 18.13) = (21.44,\ 25.63). \] \[ PQ = \sqrt{21.44^2 + 25.63^2} = \sqrt{459.7 + 656.9} = \sqrt{1116.6} = 33.4 \approx 33\ \text{km}. \]

(Check by the cosine rule with included angle \(205^\circ - 60^\circ = 145^\circ\): \(PQ^2 = 20^2 + 15^2 - 2(20)(15)\cos 145^\circ = 1116.5\), giving \(PQ = 33\) km.)

(b) Bearing of Q from P. Both components of \(Q - P\) are positive, so Q is north-east of P:

\[ \tan\theta = \frac{\text{East}}{\text{North}} = \frac{21.44}{25.63} = 0.8366 \implies \theta = 39.9^\circ. \]

The bearing of Q from P is \(\approx 040^\circ\).

Detalles de la respuesta

(a) P and Q lie on the parallel \(68.7^\circ\)S, with longitudes \(124^\circ\)W and \(56^\circ\)E. The longitude difference is

\[ 124^\circ + 56^\circ = 180^\circ. \]

The radius of the parallel of latitude is \(R\cos 68.7^\circ\), so the distance along the parallel is

\[ \frac{180}{360} \times 2\pi R \cos 68.7^\circ = \frac{1}{2}\times 2 \times 3.142 \times 6400 \times \cos 68.7^\circ. \]

With \(\cos 68.7^\circ = 0.3633\):

\[ = 3.142 \times 6400 \times 0.3633 = 7304\ \text{km} \approx 7300\ \text{km (3 s.f.).} \]

(b) Bag: 4 red, 3 white, 5 green, total 12.

(i) \(P(\text{not green}) = 1 - P(\text{green}) = 1 - \dfrac{5}{12} = \dfrac{7}{12}.\)

(ii) Two balls without replacement, one red and one white (red-then-white or white-then-red):

\[ \frac{4}{12}\times\frac{3}{11} + \frac{3}{12}\times\frac{4}{11} = \frac{12}{132} + \frac{12}{132} = \frac{24}{132} = \frac{2}{11}. \]