WAEC · 2010 · May/June · General Maths · Paper 2

1. List $C$ first: the whole numbers in $U$ between 3 and 9 are $C = \{4, 5, 6, 7, 8\}$.
2. Complements are the members of $U$ not in the set: $B' = \{4, 5, 6, 8\}$ and $C' = \{2, 3, 9\}$.

(a)

1. $B'$ and $C'$ have no member in common, so $B' \cap C' = \varnothing$.
2. Any set intersected with the empty set is empty: $A \cap (B' \cap C') = \varnothing$.

(b)

1. Join the sets: $A \cup B = \{2, 3, 4, 6, 7, 8, 9\}$ and $B \cup C = \{2, 3, 4, 5, 6, 7, 8, 9\}$.
2. Keep the members in both: $(A \cup B) \cap (B \cup C) = \{2, 3, 4, 6, 7, 8, 9\}$.

(a)

1. Draw the cliff with its mid-point $M$ at height $\frac h2$ above the foot $F$, and the boat $B$ $120\text{ m}$ from $F$.
2. The angle of depression at $M$ equals the angle of elevation at $B$: $\angle MBF = 35^\circ$.
3. In the right-angled triangle $MFB$: $\tan 35^\circ = \dfrac{h/2}{120}$.
4. So $\frac h2 = 120 \tan 35^\circ$
   $= 120 \times 0.7002$
   $= 84.02\text{ m}$.
5. Double it for the whole cliff: $h = 2 \times 84.02 = 168.0\text{ m}$.
6. The cliff is about $168\text{ m}$ high.

(b)

1. Time is distance over speed: the slower motorist takes $\frac{x}{60}$ hours and the faster one $\frac{x}{80}$ hours.
2. 30 minutes is $\frac12$ hour, so $\dfrac{x}{60} - \dfrac{x}{80} = \dfrac12$.
3. Multiply by 240 (the LCM of 60, 80 and 2): $4x - 3x = 120$.
4. So $x = 120$: the towns are $120\text{ km}$ apart.

(a)

1. Draw a line through $R$ and a line through $S$, both parallel to $PQ$ and $UT$.
2. They split $r$ and $\angle RST$ into pieces you can find.
3. At $Q$: the angle between $QR$ and the line through $R$ is $180^\circ - 125^\circ = 55^\circ$ (co-interior angles add up to $180^\circ$).
4. At $S$: the lower piece of $\angle RST$ is $44^\circ$ (alternate to $\angle STU$).
5. So the upper piece of $\angle RST$ is $80^\circ - 44^\circ = 36^\circ$, and the piece of $r$ below the line through $R$ is also $36^\circ$ (alternate angles).
6. Add the two pieces: $r = 55^\circ + 36^\circ = 91^\circ$.

(b)

1. Angle in the alternate segment: $\angle BAS = \angle ADB = 60^\circ$.
2. $AB \parallel CE$, so $\angle BAE$ and $\angle AEC$ are co-interior: $\angle BAE = 180^\circ - 5x$.
3. The angles at $A$ on the straight line $SAT$ add up to $180^\circ$: $60 + (180 - 5x) + x = 180$.
4. Simplify: $240 - 4x = 180$, so $4x = 60$ and $x = 15$.

(a)

1. The curved surface area of a cone is $\pi r l$.
2. Put in the values: $\frac{22}{7} \times r \times 10.5 = 115.5$.
3. $\frac{22}{7} \times 10.5 = 33$, so $33r = 115.5$.
4. Divide: $r = 3.50\text{ cm}$.

(b)

1. The radius, height and slant height form a right-angled triangle: $h^2 = 10.5^2 - 3.5^2$
   $= 110.25 - 12.25$
   $= 98$.
2. So $h = \sqrt{98}$
   $= 9.899$
   $\approx 9.90\text{ cm}$.

(c)

1. The volume of a cone is $\frac13 \pi r^2 h = \frac13 \times \frac{22}{7} \times 3.5^2 \times 9.899$.
2. That is $\frac13 \times 38.5 \times 9.899 = 127.0$, so the volume is $127\text{ cm}^3$ to 3 significant figures.

(a)

1. Write each outcome as (first die, second die).
2. A sum of 10 comes from $M = \{(4,6), (5,5), (6,4)\}$.
3. A difference of 3 comes from $N = \{(1,4), (4,1), (2,5), (5,2), (3,6), (6,3)\}$.

(b)

1. There are $6 \times 6 = 36$ equally likely outcomes, so $P(M) = \frac{3}{36} = \frac{1}{12}$ and $P(N) = \frac{6}{36} = \frac16$.
2. No outcome is in both, so add: $P(M \text{ or } N) = \frac{1}{12} + \frac{2}{12}$
   $= \frac{3}{12}$
   $= \frac14$.

(c)

1. Yes, $M$ and $N$ are mutually exclusive: no outcome is in both lists ($M \cap N = \varnothing$), so they cannot happen together.

(a)

1. On the map, $1\text{ cm}$ stands for $20\,000\text{ cm} = 200\text{ m}$
   $= 0.2\text{ km}$.
2. Square it for areas: $1\text{ cm}^2$ on the map stands for $0.2 \times 0.2 = 0.04\text{ km}^2$.
3. Divide: $85 \div 0.04 = 2125$.
4. The forest covers $2125\text{ cm}^2$ on the map.

(b)

1. Let the length of the field be $a\text{ m}$.
2. With the path, the outer rectangle is $(a + 12)\text{ m}$ by $18 + 12 = 30\text{ m}$.
3. Area of the path = outer area − field area $= 30(a + 12) - 18a = 12a + 360$.
4. The path and the field have equal areas: $12a + 360 = 18a$.
5. So $6a = 360$ and $a = 60$.
6. The field is $60\text{ m}$ long.

(c)

1. The shaded sector has angle $360^\circ - x$.
2. Its area is $\dfrac{360 - x}{360} \times \dfrac{22}{7} \times 3.5^2 = 27.5$.
3. $\frac{22}{7} \times 3.5^2 = 38.5$, so $\dfrac{360 - x}{360} = \dfrac{27.5}{38.5}$
   $= \dfrac57$.
4. Then $360 - x = \frac57 \times 360 = 257.14$.
5. So $x = 102.86^\circ$, which is $103^\circ$ to the nearest degree.

(a)(i)

1. Draw $PS = 6\text{ cm}$.
2. At $S$ construct $90^\circ$ (bisect a straight angle) and mark $R$ on that arm with $SR = 9\text{ cm}$.
3. With centre $R$, radius $8.4\text{ cm}$, and centre $P$, radius $5.4\text{ cm}$, draw two arcs on the side away from $S$.
4. They cross at $Q$.
5. Join $RQ$ and $QP$.

(ii)

1. Bisect $\angle RSP$ (this gives a $45^\circ$ line from $S$) and bisect $\angle SPQ$.
2. Label their meeting point $X$.

(iii)

1. From $X$, construct the perpendicular to $PS$: arcs from $X$ cut $PS$ twice, then bisect that chord.
2. The foot is $T$.

(b)

1. Measure $XT$ with the ruler: $|XT| \approx 3.4\text{ cm}$.
2. Check by calculation: $\angle SPQ \approx 106^\circ$, so in triangle $SXP$ the angles at $S$ and $P$ are $45^\circ$ and $53^\circ$, which gives $XT = \dfrac{6}{1 + \cot 53^\circ}$
   $\approx 3.4\text{ cm}$.

(a)

1. Find $\angle ABC$.
2. From $B$, $A$ is on $310^\circ$ and $C$ is on $230^\circ - 180^\circ = 050^\circ$ (the back bearing of $230^\circ$).
3. Going round from $050^\circ$ to $310^\circ$ is $260^\circ$, so the angle inside the triangle is $360^\circ - 260^\circ = 100^\circ$.
4. Cosine rule: $AC^2 = 8^2 + 13^2 - 2(8)(13)\cos 100^\circ$
   $= 233 + 36.12$
   $= 269.12$.
5. So $AC = \sqrt{269.12} = 16.4\text{ km}$ to 3 significant figures.

(b)

1. Sine rule: $\sin \angle CAB = \dfrac{13 \sin 100^\circ}{16.40}$
   $= 0.7805$, so $\angle CAB = 51.3^\circ$.
2. The bearing of $B$ from $A$ is $310^\circ - 180^\circ = 130^\circ$.
3. $C$ is $51.3^\circ$ further round towards north: $130^\circ - 51.3^\circ = 78.7^\circ$.
4. The bearing of $C$ from $A$ is $079^\circ$ to the nearest degree.

(c)

1. $BC$ makes $50^\circ$ with the north line at $B$ (bearing $050^\circ$), so the eastward distance is $13 \sin 50^\circ = 13 \times 0.7660$
   $= 9.96\text{ km}$.

(a)

1. Substitute each $x$.
2. For example, $x = -3$: $-9 - 3 + 2 = -10$.
3. $x = 1$: $-1 + 1 + 2 = 2$.
4. The row is $-10, -4, 0, 2, 2, 0, -4$.

(b)

1. Plot the seven points with the scales given and join them with one smooth curve.
2. It opens downwards, with its top at $x = 0.5$, $y = 2.25$.

(c)(i)

1. For $-3 \le x \le 3$ the lowest point of the curve is at the left end, $x = -3$: the minimum value of $y$ is $-10$.

(ii)

1. $x^2 - x - 2 = 0$ is the same as $-x^2 + x + 2 = 0$, so read where the curve crosses the $x$-axis: $x = -1$ and $x = 2$.

(iii)

1. Draw the tangent to the curve at $(-0.5, 1.25)$.
2. It passes through about $(-1.5, -0.75)$ and $(0.5, 3.25)$: a rise of 4 over a run of 2, so the gradient is $2$.

(a)

1. Pythagoras in triangle $PTQ$: $|PT| = \sqrt{10^2 - 6^2}$
   $= \sqrt{64}$
   $= 8\text{ cm}$.
2. $TQ \parallel SR$ (both are perpendicular to $PS$), so triangles $PTQ$ and $PSR$ are similar: $\dfrac{|SR|}{|TQ|} = \dfrac{|PS|}{|PT|}$.
3. So $|SR| = 6 \times \dfrac{14.4}{8}$
   $= 10.8\text{ cm}$.
4. $QRST$ is a trapezium with parallel sides $6$ and $10.8$, and height $|TS| = 14.4 - 8 = 6.4\text{ cm}$.
5. Area $= \frac12(6 + 10.8) \times 6.4$
   $= 53.76\text{ cm}^2$.

(b)

1. Let the side of the square be $y$.
2. The rectangle is $0.9y$ by $1.15y$.
3. Its area is $0.9 \times 1.15 \times y^2 = 1.035y^2$.
4. The ratio is $1.035y^2 : y^2 = 1.035 : 1$, which is $207 : 200$.

(a)

1. Add the frequencies as you go: $10, 28, 50, 75, 91, 100$.
2. Pair each total with the upper class boundary: $29.5, 39.5, 49.5, 59.5, 69.5, 79.5$.

(b)

1. Plot cumulative frequency against upper class boundary, starting at $(19.5, 0)$, and join the points with a smooth curve.

(c)(i)

1. The median is at $\frac{100}{2} = 50$ on the cumulative frequency axis.
2. Reading across and down gives about $49.5\text{ kg}$.

(ii)

1. Read the quartiles at $\frac{100}{4} = 25$ and $\frac{3 \times 100}{4} = 75$: $Q_1 \approx 37.8\text{ kg}$ and $Q_3 \approx 59.5\text{ kg}$.
2. The semi-interquartile range is $\frac12(Q_3 - Q_1) = \frac12(59.5 - 37.8)$
   $\approx 10.8\text{ kg}$.

(iii)

1. Weights of at least $60\text{ kg}$ start at the boundary $59.5$, where the curve reads 75.
2. So $100 - 75 = 25$ participants weigh at least $60\text{ kg}$.
3. The probability is $\frac{25}{100} = \frac14$.

(a)

1. The $n$th term of a G.P. is $ar^{n-1}$, so $ar^2 = 24$ and $ar^6 = 4\frac{20}{27} = \frac{128}{27}$.
2. Divide the second equation by the first: $r^4 = \dfrac{128}{27} \div 24$
   $= \dfrac{128}{648}$
   $= \dfrac{16}{81}$.
3. So $r^2 = \frac49$ (and $r = \pm\frac23$).
4. Then $a = \dfrac{24}{r^2}$
   $= 24 \times \dfrac94$
   $= 54$.
5. The first term is 54.

(b)

1. Write the variation: $y = \dfrac{kx}{z^2}$.
2. Put in $y = 4$, $x = 3$, $z = 1$: $4 = 3k$, so $k = \frac43$.
3. Now $x = 3$, $z = 2$: $y = \dfrac{\frac43 \times 3}{2^2}$, which is $\dfrac44 = 1$.