WAEC · 2008 · Nov/Dec · General Maths · Paper 2

(b)

1. Each exterior angle is $180^\circ - (134 + n)^\circ = (46 - n)^\circ$.
2. The exterior angles of any polygon add up to $360^\circ$, and here there are $n$ equal ones: $n(46 - n) = 360$.
3. Expand and rearrange: $n^2 - 46n + 360 = 0$.
4. Factorise: $(n - 10)(n - 36) = 0$, so $n = 10$ or $n = 36$.
5. Both work: a 10-sided polygon has interior angles of $144^\circ$ ($= 134 + 10$) and a 36-sided one has $170^\circ$ ($= 134 + 36$).
6. So $n = 10$ or $36$.

1. The frequencies add up to 20: $1 + 3 + 1 + m + 4 + n + 2 + 1 = 20$, so $m + n = 8$.
2. The mean is $\frac{\sum fx}{\sum f} = 5.4$, so $\sum fx = 5.4 \times 20 = 108$.
3. Work out $\sum fx$: $2 + 9 + 4 + 5m + 24 + 7n + 16 + 9 = 64 + 5m + 7n$.
4. So $64 + 5m + 7n = 108$, which gives $5m + 7n = 44$.
5. From the first equation $m = 8 - n$.
6. Substitute: $5(8 - n) + 7n = 44$, so $40 + 2n = 44$ and $n = 2$.
7. Then $m = 8 - 2 = 6$.
8. So $m = 6$ and $n = 2$.

(a)(i)

1. Selling at the marked price gives $25\%$ profit, so D600 is $125\%$ of the cost price.
2. Cost price $= \frac{100}{125} \times 600 = 480$, so D480.

(ii)

1. With a $6\%$ discount she receives $94\%$ of the marked price: $\frac{94}{100} \times 600 = 564$, so D564.
2. Actual profit = D564 − D480 = D84.

(b)

1. Let her normal rate be $v\text{ km/h}$.
2. Her usual time is $\frac{1.5}{v}$ hours and her slow time is $\frac{1.5}{v - 0.5}$ hours.
3. Change 6 minutes to hours: $\frac{6}{60} = 0.1$ h.
4. So $\frac{1.5}{v - 0.5} - \frac{1.5}{v} = 0.1$.
5. Multiply both sides by $v(v - 0.5)$: $1.5v - 1.5(v - 0.5) = 0.1v(v - 0.5)$, so $0.75 = 0.1v^2 - 0.05v$.
6. Multiply both sides by 20: $15 = 2v^2 - v$, so $2v^2 - v - 15 = 0$.
7. Factorise: $(2v + 5)(v - 3) = 0$, so $v = 3$ (a speed can't be negative).
8. Isatu normally walks at $3\text{ km/h}$.

(a)

1. Substitute each $x$.
2. For example, $x = -1$: $-3(1) + 12(-1) = -15$.
3. $x = 2$: $-3(4) + 24 = 12$.
4. $x = 5$: $-75 + 60 = -15$.
5. The completed row is $-36, -15, 0, 9, 12, 9, 0, -15, -36$.

(b)

1. Plot the points with the scales given and draw a smooth curve through them.
2. It is symmetric about $x = 2$.

(c)

1. The line $y = 5x - 10$ passes through $(0, -10)$ and $(2, 0)$.
2. Plot a third point such as $(4, 10)$ and rule the line across the graph.

(d)

1. Take $5x$ from both sides of $7x + 10 > 3x^2$ and rearrange: $-3x^2 + 12x > 5x - 10$.
2. So the region is where the curve lies above the line.
3. The curve and line meet where $3x^2 - 7x - 10 = 0$, i.e. $(3x - 10)(x + 1) = 0$, so $x = -1$ and $x = 3\frac13$.
4. Shade the region between the line and the curve from $x = -1$ to $x = 3\frac13$.

(a)

1. The part of the ladder from the foot $L$ to the top of the wall $T$ is $10 - 2.5 = 7.5\text{ m}$.
2. In the right-angled triangle $LPT$, the wall $PT = 4.5$ is opposite $\theta$ and $LT = 7.5$ is the hypotenuse: $\sin\theta = \frac{4.5}{7.5} = 0.6$.
3. So $\theta = 36.87^\circ$, which is $36.9^\circ$ to one decimal place.

(b)

1. The top end $N$ is $10\text{ m}$ along the ladder: its height is $10\sin\theta = 10 \times 0.6 = 6\text{ m}$.

(c)

1. First find $LP$ by Pythagoras: $LP = \sqrt{7.5^2 - 4.5^2}$
   $= \sqrt{36}$
   $= 6\text{ m}$.
2. Moving the foot $2\text{ m}$ further makes $LP = 8\text{ m}$, with the ladder still resting on top of the wall.
3. Now $\tan\theta = \frac{4.5}{8} = 0.5625$.
4. So $\theta = 29.36^\circ$, which is $29^\circ$ to the nearest degree.

(a)

1. Start in the middle: 8 offered all three subjects.
2. Physics and Chemistry only: $23 - 8 = 15$.
3. Chemistry and Mathematics only: $41 - 8 = 33$.
4. Let $x$ offer Physics and Mathematics only.
5. Then Physics only is $70 - 15 - 8 - x = 47 - x$, Chemistry only is $90 - 15 - 8 - 33 = 34$ and Mathematics only is $100 - 33 - 8 - x = 59 - x$.
6. All the regions, with the 24 outside, add up to 200: $(47 - x) + 34 + (59 - x) + 15 + 33 + x + 8 + 24 = 200$, so $220 - x = 200$ and $x = 20$.
7. So Physics only is $27$, Mathematics only is $39$ and Physics and Mathematics only is $20$.
8. Fill these into the diagram.

(b)(i)

1. $P(\text{Physics only}) = \frac{27}{200}$.

(ii)

1. Exactly two subjects: $15 + 33 + 20 = 68$ students, so the probability is $\frac{68}{200} = \frac{17}{50}$.

(a)

1. Draw $QR = 6\text{ cm}$.
2. At $Q$, construct $90^\circ$ (bisect a straight angle).
3. At $R$, construct $120^\circ$ ($60^\circ$ twice) on the same side as the $90^\circ$ arm.
4. Mark $S$ on the $120^\circ$ arm with $RS = 8\text{ cm}$.
5. $|PQ| = |PS|$ means $P$ is equidistant from $Q$ and $S$, so $P$ lies on the perpendicular bisector of $QS$.
6. Join $QS$ and construct its perpendicular bisector.
7. $P$ is where this bisector cuts the $90^\circ$ arm at $Q$.
8. Join $PS$ to complete $PQRS$.

(b)

1. Measure $PQ$: it is about $10.7\text{ cm}$.

(a)

1. $|PR| = |RQ|$, so triangle $PQR$ is isosceles and $\angle QPR = \angle PQR = 30^\circ$.
2. $\angle PRS$ is an exterior angle of triangle $PQR$, so it equals the sum of the two opposite interior angles: $30^\circ + 30^\circ = 60^\circ$.
3. In triangle $PRS$: $\angle PSR = 180^\circ - 70^\circ - 60^\circ$
   $= 50^\circ$ (needed only as a check).
4. Sine rule in triangle $PRS$: $\frac{PS}{\sin 60^\circ} = \frac{RS}{\sin 70^\circ}$.
5. $PS = \frac{10 \sin 60^\circ}{\sin 70^\circ}$
   $= \frac{10 \times 0.8660}{0.9397}$
   $= 9.216$.
6. So $|PS| = 9.22\text{ cm}$.

(b)(i)

1. The radius of the parallel of latitude $60^\circ$ is $R\cos 60^\circ = 6400 \times 0.5$
   $= 3200\text{ km}$.

(ii)

1. Let $\theta$ be the difference in longitude.
2. The arc along the parallel is $\frac{\theta}{360} \times 2\pi r$, so $\frac{\theta}{360} \times 2 \times 3.142 \times 3200 = 1500$.
3. $\theta = \frac{1500 \times 360}{2 \times 3.142 \times 3200}$
   $= 26.85^\circ$.
4. $B$ is east of $A$, so add: $20^\circ + 26.85^\circ = 46.85^\circ$.
5. $B$ is on longitude $47^\circ\text{E}$, to the nearest degree.

(a)

1. The $n$th term of an A.P. is $a + (n - 1)d = 31 + 9(n - 1)$.
2. Expand: $31 + 9n - 9 = 9n + 22$, as required.
3. Hence the 20th term is $9(20) + 22 = 202$.

(b)(i)

1. The second term is $ar = 1$ and the fifth term is $ar^4 = \frac18$.
2. Divide the fifth term by the second: $r^3 = \frac18$, so the common ratio is $r = \frac12$.

(ii)

1. From $ar = 1$: $a = 1 \div \frac12 = 2$.
2. The first term is $2$.

(iii)

1. The eighth term is $ar^7 = 2 \times \left(\frac12\right)^7$
   $= \frac{2}{128}$
   $= \frac{1}{64}$.