WAEC SSCE - General Mathematics - 1996 (Essay)
Question 1 Report
(a) Find the number N such that when \(\frac{1}{3}\) of it is added to 8, the result is the same as when \(\frac{1}{2}\) of it is subtracted from 18.
(b) Using a ruler and a pair of compasses only, construct a trapezium ABCD, in which the parallel sides AB and DC are 4 cm apart. < DAB = 60°, /AB/ = 8 cm and /BC/ = 5 cm. Measure /DC/.
(a) Let the number be \(N\).
\[\frac{N}{3}+8=18-\frac{N}{2}\]
\[\frac{N}{3}+\frac{N}{2}=18-8\]
\[\frac{5N}{6}=10\]
\[5N=60\]
\[\boxed{N=12}\]
(b) Construction of trapezium \(ABCD\)
- Draw \(AB=8\text{ cm}\).
- At \(A\), construct \(\angle DAB=60^\circ\), and draw the ray \(AD\).
- Construct a perpendicular to \(AB\) at \(A\). Mark a point \(4\text{ cm}\) from \(A\) on this perpendicular and draw through it a line parallel to \(AB\). This line meets the \(60^\circ\) ray at \(D\).
- With centre \(B\) and radius \(5\text{ cm}\), draw an arc to cut the parallel line at \(C\), taking the intersection to the left of \(B\).
- Join \(BC\) and \(DC\).
From the completed construction,
\[\boxed{DC\approx 2.7\text{ cm}}\]
Answer Details
(a) Let the number be \(N\).
\[\frac{N}{3}+8=18-\frac{N}{2}\]
\[\frac{N}{3}+\frac{N}{2}=18-8\]
\[\frac{5N}{6}=10\]
\[5N=60\]
\[\boxed{N=12}\]
(b) Construction of trapezium \(ABCD\)
- Draw \(AB=8\text{ cm}\).
- At \(A\), construct \(\angle DAB=60^\circ\), and draw the ray \(AD\).
- Construct a perpendicular to \(AB\) at \(A\). Mark a point \(4\text{ cm}\) from \(A\) on this perpendicular and draw through it a line parallel to \(AB\). This line meets the \(60^\circ\) ray at \(D\).
- With centre \(B\) and radius \(5\text{ cm}\), draw an arc to cut the parallel line at \(C\), taking the intersection to the left of \(B\).
- Join \(BC\) and \(DC\).
From the completed construction,
\[\boxed{DC\approx 2.7\text{ cm}}\]
Question 2 Report
(a) Without using tables, find the value of \(\frac{0.45 \times 0.91}{0.0117}\)
(b) Find the number which is exactly halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\).
(c) If each interior angle of a regular polygon is five times the exterior angle, how many sides has the polygon?
(d) Calculate the volume of the material used in making a pipe 20cm long, with an internal diameter 6cm and external diameter 8cm. [Take \(pi = \frac{22}{7}\)].
(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):
\[ V = \pi(R^2 - r^2)\times \text{length} = \frac{22}{7}(4^2 - 3^2)(20) = \frac{22}{7}(7)(20) = 22 \times 20 = 440\ \text{cm}^3. \]Answer Details
(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):
\[ V = \pi(R^2 - r^2)\times \text{length} = \frac{22}{7}(4^2 - 3^2)(20) = \frac{22}{7}(7)(20) = 22 \times 20 = 440\ \text{cm}^3. \]Question 3 Report
(a) ABCD is a trapezium with AB parallel to DC and /AD/ = /AB/. If < BAD = 106°, find < BDC.
(b) The table below shows the distribution of 20 cards labelled A - E.
| Card | A | B | C | D | E |
| Frequency | 3 | 4 | 7 | 5 | 1 |
(i) If a card is selected at random from the pack, what is the probability that the card is E? (ii) If two cards are selected at random one after the other without replacement from the pack, what is the probability that one of the two cards is B?
(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.
\(P(\text{no B}) = \dfrac{16}{20}\times\dfrac{15}{19} = \dfrac{240}{380} = \dfrac{12}{19}\).
\(P(\text{at least one B}) = 1 - \dfrac{12}{19} = \dfrac{7}{19}\).
Answer Details
(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.
\(P(\text{no B}) = \dfrac{16}{20}\times\dfrac{15}{19} = \dfrac{240}{380} = \dfrac{12}{19}\).
\(P(\text{at least one B}) = 1 - \dfrac{12}{19} = \dfrac{7}{19}\).
Question 4 Report
Two men P and Q set off from a base camp R, prospecting for oil. P moves 20km on a bearing of 205° and Q moves 15km on a bearing of 060°. Calculate the:
(a) distance of Q from P ;
(b) bearing of Q from P.
(Give your answer in each case to the nearest whole number)
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 \(\approx 040^\circ\).
Answer Details
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 \(\approx 040^\circ\).
Question 5 Report
The table below shows the marks obtained by forty pupils in a Mathematics test.
| Marks | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 |
| No of pupils | 4 | 5 | 6 | 12 | 8 | 5 |
(a) Draw a histogram for the mark distribution ;
(b) Use your histogram to estimate the mode ;
(c) Calculate the median of the distribution.
(a) Histogram
Use continuous class boundaries. Since all class widths are 10 marks, the bar heights are the frequencies.
| Marks | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 0 - 9 | -0.5 - 9.5 | 4 | 4 |
| 10 - 19 | 9.5 - 19.5 | 5 | 9 |
| 20 - 29 | 19.5 - 29.5 | 6 | 15 |
| 30 - 39 | 29.5 - 39.5 | 12 | 27 |
| 40 - 49 | 39.5 - 49.5 | 8 | 35 |
| 50 - 59 | 49.5 - 59.5 | 5 | 40 |
(b) Mode
The modal class is \(30-39\), with class boundaries \(29.5-39.5\).
\[\text{Mode}=L+\frac{d_1}{d_1+d_2}\times c\]
\[=29.5+\frac{12-6}{(12-6)+(12-8)}\times10\]
\[=29.5+\frac{6}{10}\times10=\boxed{35.5\text{ marks}}\]
(c) Median
\[N=40,\qquad \frac{N}{2}=20\]
The 20th value lies in the class \(30-39\), since the cumulative frequency rises from \(15\) to \(27\) in this class.
\[\text{Median}=L+\frac{\left(\frac{N}{2}-\text{cumulative frequency before median class}\right)}{f_m}\times c\]
\[=29.5+\frac{20-15}{12}\times10\]
\[=29.5+4.1667=\boxed{33.67\text{ marks}}\]
Answer Details
(a) Histogram
Use continuous class boundaries. Since all class widths are 10 marks, the bar heights are the frequencies.
| Marks | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 0 - 9 | -0.5 - 9.5 | 4 | 4 |
| 10 - 19 | 9.5 - 19.5 | 5 | 9 |
| 20 - 29 | 19.5 - 29.5 | 6 | 15 |
| 30 - 39 | 29.5 - 39.5 | 12 | 27 |
| 40 - 49 | 39.5 - 49.5 | 8 | 35 |
| 50 - 59 | 49.5 - 59.5 | 5 | 40 |
(b) Mode
The modal class is \(30-39\), with class boundaries \(29.5-39.5\).
\[\text{Mode}=L+\frac{d_1}{d_1+d_2}\times c\]
\[=29.5+\frac{12-6}{(12-6)+(12-8)}\times10\]
\[=29.5+\frac{6}{10}\times10=\boxed{35.5\text{ marks}}\]
(c) Median
\[N=40,\qquad \frac{N}{2}=20\]
The 20th value lies in the class \(30-39\), since the cumulative frequency rises from \(15\) to \(27\) in this class.
\[\text{Median}=L+\frac{\left(\frac{N}{2}-\text{cumulative frequency before median class}\right)}{f_m}\times c\]
\[=29.5+\frac{20-15}{12}\times10\]
\[=29.5+4.1667=\boxed{33.67\text{ marks}}\]
Question 6 Report
(a) If \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\), calculate without using tables, the value of \(\log_{10} 0.72\).
(b) A hawk on top of a tree, 20 metres high views a chick on the ground at an angle of depression of 39°. Find, correct to 2 significant figures, the distance of the chick from the bottom of the tree.
(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).
Answer Details
(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).
Question 7 Report
(a) Copy and complete the table for \(y = 3x^{2} - 5x - 7\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| \(y = 3x^{2} - 5x - 7\) | 35 | -7 | -9 | 5 |
(b) Using a scale of 2cm = 1 unit along the x- axis and 2cm = 5 units along the y- axis, draw the graph of \(y = 3x^{2} - 5x - 7\).
(c) On the same axis, draw the graph of \(y + 3x + 2 = 0\).
(d) From your graph, find the : (i) range of values of x for which \(3x^{2} - 5x - 7 < 0\) ; (ii) roots of the equation \(3x^{2} - 2x - 5 = 0\).
(a) For \(y=3x^{2}-5x-7\), the completed table is:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=3x^2-5x-7\) | 35 | 15 | 1 | −7 | −9 | −5 | 5 | 21 |
For example, when \(x=-2\),
\[ y=3(-2)^2-5(-2)-7=12+10-7=15. \](b) and (c) The required graphs, drawn on the same axes, are shown below. The curve is \(y=3x^2-5x-7\), and the straight line is \(y=-3x-2\).
(d)(i) \(3x^2-5x-7<0\) where the parabola lies below the \(x\)-axis. Reading the intercepts from the graph gives approximately \(-0.9\) and \(2.6\). Hence,
\[ -0.9<x<2.6. \](d)(ii) The roots of \(3x^2-2x-5=0\) are the \(x\)-coordinates of the points where the line and curve intersect:
\[ 3x^2-5x-7=-3x-2 \] \[ 3x^2-2x-5=0. \] From the graph, \[ x=-1.0\quad\text{or}\quad x\approx1.7. \]Indeed, \(3x^2-2x-5=(3x-5)(x+1)\), so the exact roots are \(x=-1\) and \(x=\frac53\).
Answer Details
(a) For \(y=3x^{2}-5x-7\), the completed table is:
| \(x\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|
| \(y=3x^2-5x-7\) | 35 | 15 | 1 | −7 | −9 | −5 | 5 | 21 |
For example, when \(x=-2\),
\[ y=3(-2)^2-5(-2)-7=12+10-7=15. \](b) and (c) The required graphs, drawn on the same axes, are shown below. The curve is \(y=3x^2-5x-7\), and the straight line is \(y=-3x-2\).
(d)(i) \(3x^2-5x-7<0\) where the parabola lies below the \(x\)-axis. Reading the intercepts from the graph gives approximately \(-0.9\) and \(2.6\). Hence,
\[ -0.9<x<2.6. \](d)(ii) The roots of \(3x^2-2x-5=0\) are the \(x\)-coordinates of the points where the line and curve intersect:
\[ 3x^2-5x-7=-3x-2 \] \[ 3x^2-2x-5=0. \] From the graph, \[ x=-1.0\quad\text{or}\quad x\approx1.7. \]Indeed, \(3x^2-2x-5=(3x-5)(x+1)\), so the exact roots are \(x=-1\) and \(x=\frac53\).
Question 8 Report
(a) P and Q are points on the parallel of latitude 68.7°S, their longitudes being 124°W and 56°E respectively. What is their distance apart measured along the parallel of latitude? [Take R = 6400km, \(\pi = 3.142\)]. (Give your answers to 3 significant figures).
(b) A bag contains four red, three white and five green balls. (i) If one ball is picked at random, what is the probability that it is not green? (ii) if two balls are picked at random without replacement, what is the probability that one is red and the other white?
(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):
\[ \frac{4}{12}\times\frac{3}{11} + \frac{3}{12}\times\frac{4}{11} = \frac{12}{132} + \frac{12}{132} = \frac{24}{132} = \frac{2}{11}. \]Answer Details
(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):
\[ \frac{4}{12}\times\frac{3}{11} + \frac{3}{12}\times\frac{4}{11} = \frac{12}{132} + \frac{12}{132} = \frac{24}{132} = \frac{2}{11}. \]Question 9 Report
A shopkeeper buys 40 kg of fruits for N120.00. He sells 20 kg at N5.00 per kg, 10 kg at N3.00 per kg, 5 kg at N2.00 per kg and the remaining 5 kg at 50k per kg. Calculate the :
(a) amount he realises from the sales ;
(b) total profit / loss ;
(c) percentage profit/ loss on his outlay of N120.00.
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 |
(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:
\[ \frac{22.50}{120} \times 100\% = 18.75\%. \]Answer Details
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 |
(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:
\[ \frac{22.50}{120} \times 100\% = 18.75\%. \]Question 10 Report
(a) The universal set U is the set of integers, P, Q and R are subsets of U defined as follows:
\(P = x : x \leq 2 \) ; \(Q = x : -7 < x < 15\) ; \(R = x : -2 \leq x < 19\).
Find (i) \(P \cap Q\) ; (ii) \(P \cap (Q \cup R')\), where R' is the complement of R with respect to U.
(b) The following data shows the marks of 40 students in a History examination.
41 52 37 56 63 48 65 46 54 32 51 66 74 23 35 61 58 44 49 53 45 57 56 38 59 28 50 49 67 56 36 45 79 68 43 56 26 47 55 71.
(i) Form a grouped frequency table with the class intervals 20 - 29, 30 - 39, 40 - 49 etc; (ii) Find the mean of the distribution.
(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: \(\sum fx = 3(24.5)+5(34.5)+10(44.5)+13(54.5)+6(64.5)+3(74.5) = 2010\).
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{2010}{40} = 50.25 \approx 50.3. \]Answer Details
(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: \(\sum fx = 3(24.5)+5(34.5)+10(44.5)+13(54.5)+6(64.5)+3(74.5) = 2010\).
\[ \bar{x} = \frac{\sum fx}{\sum f} = \frac{2010}{40} = 50.25 \approx 50.3. \]Question 11 Report
(a) Solve the simultaneous equation : \(6y + 5x = 12 ; 4y - 3x = 11\).
(b) 
In the diagram, ADC is a straight line. /CD/ = 48 cm, /BD/ = 36 cm and /AD/ = y cm. Find the value of y.
(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 \(y = 27\text{ cm}\).
Answer Details
(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 \(y = 27\text{ cm}\).
Question 12 Report
(a) Make d the subject of the formula \(S = \frac{n}{2}[2a + (n - 1) d]\).
(b) (i) 
In the diagram, O is the centre of the circle, A, B and P are points on the circumference. Prove that < AOB = 2 < APB.
(ii) 
Find the angles x, v and z in the diagram.
(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
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:
\[y=\tfrac{1}{2}\times108^\circ=54^\circ.\]\[\therefore\ x=108^\circ,\quad y=54^\circ,\quad z=252^\circ.\]Answer Details
(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
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:
\[y=\tfrac{1}{2}\times108^\circ=54^\circ.\]\[\therefore\ x=108^\circ,\quad y=54^\circ,\quad z=252^\circ.\]
