WAEC · 2008 · May/June · General Maths · Paper 2

(a)

1. A ratio of $2 : 3$ means $x = 2k$ and $y = 3k$ for some number $k$.
2. Put these in: $x^2 - y^2 = 4k^2 - 9k^2 = -5k^2$ and $y^2 + x^2 = 9k^2 + 4k^2 = 13k^2$.
3. Divide, and $k^2$ cancels: the value is $-\frac{5}{13}$.

(b)

1. Draw the right-angled triangle from the man's eye: the horizontal side is $12.82\text{ m}$ and the angle at his eye is $52^\circ$.
2. The height of the tree above his eye is $h = 12.82 \tan 52^\circ$
   $= 12.82 \times 1.2799$
   $= 16.409\text{ m}$.
3. Add the man's height, because the angle is measured from his eye: $16.409 + 1.24 = 17.649$.
4. The tree is $17.65\text{ m}$ tall, to two decimal places.

(a)

1. Factorise the top: $x^2 - 8x + 16 = (x - 4)^2$.
2. Factorise the bottom: $x^2 - 7x + 12 = (x - 3)(x - 4)$.
3. Cancel the common factor $(x - 4)$: the fraction simplifies to $\frac{x - 4}{x - 3}$.

(b)

1. In an A.P. the common difference is the same between neighbouring terms, so $\frac1x - \frac12 = \frac13 - \frac1x$.
2. Write each side as one fraction: $\frac{2 - x}{2x} = \frac{x - 3}{3x}$.
3. Multiply both sides by $6x$: $3(2 - x) = 2(x - 3)$.
4. Divide both sides by $3(x - 3)$: $\frac{2 - x}{x - 3} = \frac23$, which is what we had to show.

1. The bucket is a frustum: a cone with its tip cut off.
2. The radii are $R = 10\text{ cm}$ at the top and $r = 6\text{ cm}$ at the bottom, and the depth is $h = 16\text{ cm}$.
3. Use the frustum volume $V = \frac13\pi h(R^2 + Rr + r^2)$.
4. Put in the numbers: $V = \frac13 \times \frac{22}{7} \times 16 \times (100 + 60 + 36)$
   $= \frac13 \times \frac{22}{7} \times 16 \times 196$.
5. Simplify: $V = \frac{68\,992}{21}$
   $= 3285\frac13\text{ cm}^3$.
6. The tin has radius $14\text{ cm}$, so its base area is $\frac{22}{7} \times 14^2 = 616\text{ cm}^2$.
7. Depth of water $= \frac{\text{volume}}{\text{base area}}$
   $= \frac{3285\frac13}{616}$
   $= 5\frac13$.
8. The water is $5\frac13\text{ cm}$ (about $5.33\text{ cm}$) deep.

(a)

1. Use the power law: $2\log x = \log x^2$.
2. Use the division law on the left: $\log x^2 - \log(1 - x) = \log\frac{x^2}{1 - x}$.
3. The logs are equal, so the numbers are equal: $\frac{x^2}{1 - x} = 2 - x$.
4. Multiply out: $x^2 = (2 - x)(1 - x) = 2 - 3x + x^2$.
5. The $x^2$ terms cancel: $3x = 2$, so $x = \frac23$ (and $1 - x$, $2 - x$ are positive, so the logs exist).

(b)

1. Let Ade have ₦$x$ and Chidi ₦$y$.
2. First condition: $x - 5 = y + 5$, so $x - y = 10$.
3. Second condition: $x + 5 = 2(y - 5)$, so $x - 2y = -15$.
4. Subtract the second equation from the first: $y = 25$.
5. Then $x = 25 + 10 = 35$.
6. Ade had ₦35 and Chidi had ₦25.

(a)

1. The perimeter of a rectangle is $2(l + b)$, so $2(l + b) = 280$ and $l + b = 140$.
2. The length is two and a half times the breadth: $l = 2.5b$.
3. Substitute: $2.5b + b = 140$, so $3.5b = 140$ and $b = 40$.
4. Then $l = 2.5 \times 40 = 100$.
5. The field is $100\text{ m}$ long and $40\text{ m}$ wide.

(b)

1. The volume of a pyramid is $\frac13 \times \text{base area} \times \text{height}$.
2. The base area is $4.5 \times 2.5 = 11.25\text{ m}^2$.
3. $V = \frac13 \times 11.25 \times 4 = 15$.
4. The volume is $15\text{ m}^3$.

(a)

1. Write both sides as powers of 2: $2^{x + y} = 2^4$, so $x + y = 4$.
2. $4^{x - y} = 2^{2(x - y)}$ and $\frac{1}{32} = 2^{-5}$, so $2(x - y) = -5$ and $x - y = -\frac52$.
3. Add the two equations: $2x = \frac32$, so $x = \frac34$.
4. Then $y = 4 - \frac34 = 3\frac14$.

(b)

1. Write the variation with a constant: $P = \frac{kQ^2}{R}$.
2. Find $k$ from $P = 36$, $Q = 3$, $R = 4$: $36 = \frac{9k}{4}$, so $k = 16$.
3. So $P = \frac{16Q^2}{R}$.
4. Put in $P = 200$, $R = 2$: $200 = \frac{16Q^2}{2} = 8Q^2$.
5. So $Q^2 = 25$ and $Q = 5$.

(a)

1. Rearrange to the standard form: $4x^2 - 11x - 21 = 0$, so $a = 4$, $b = -11$, $c = -21$.
2. Use the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
3. The discriminant is $121 + 336 = 457$, and $\sqrt{457} = 21.378$.
4. So $x = \frac{11 \pm 21.378}{8}$.
5. $x = \frac{32.378}{8} = 4.05$ or $x = \frac{-10.378}{8} = -1.30$, to two decimal places.

(b)(i)

1. The interest in the first year is £1560 − £1500 = £60.
2. Rate $= \frac{60}{1500} \times 100\% = 4\%$.

(ii)

1. Compound interest: the second year's interest is on the new amount, £1560.
2. Interest $= \frac{4}{100} \times 1560 = 62.40$, so £62.40.

(c)

1. After the first year the car is worth $75\%$ of ₦300,000: $0.75 \times 300\,000 = 225\,000$.
2. The second year's $20\%$ comes off the new value: $0.80 \times 225\,000 = 180\,000$.
3. The car is worth ₦180,000.00 after 2 years.

(a)

1. Tally each age into its class.
2. The frequencies are: $21 - 25$: 6, $26 - 30$: 9, $31 - 35$: 9, $36 - 40$: 9, $41 - 45$: 6, $46 - 50$: 6.
3. They add up to 45.

(b)

1. A histogram uses the class boundaries: $20.5, 25.5, 30.5, 35.5, 40.5, 45.5, 50.5$.
2. Draw touching bars of heights 6, 9, 9, 9, 6, 6.

(c)

1. The three bars from $25.5$ to $40.5$ are equally tall, so take them together as the modal block.
2. Join the top-left corner of the block to the top of the next bar on the right, $(40.5, 6)$, and the top-right corner to the top of the bar on the left, $(25.5, 6)$.
3. The two lines cross above $33$, so the mode is about $33$ years.

(d)

1. Use the class midpoints $x = 23, 28, 33, 38, 43, 48$ and find $fx$: $138, 252, 297, 342, 258, 288$.
2. Add them: $\sum fx = 1575$, and $\sum f = 45$.
3. Mean $= \frac{\sum fx}{\sum f}$
   $= \frac{1575}{45}$
   $= 35$.
4. The mean age is $35$ years.

(a)(i)

1. The largest angle is opposite the longest side, $AB = 17$, so it is angle $C$.
2. Use the cosine rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
   $= \frac{12^2 + 10^2 - 17^2}{2 \times 12 \times 10}$
   $= \frac{-45}{240}$
   $= -0.1875$.
3. The cosine is negative, so $C$ is obtuse: $C = 180^\circ - 79.19^\circ$
   $= 100.81^\circ$.
4. The largest angle is $100.8^\circ$.

(ii)

1. Area $= \frac12 ab\sin C$
   $= \frac12 \times 12 \times 10 \times \sin 100.81^\circ$.
2. $\sin 100.81^\circ = 0.9823$, so the area is $60 \times 0.9823 = 58.94\text{ m}^2$.

(b)

1. In the right-angled triangle $RST$, $RS = 38$ is opposite the $63^\circ$ angle and $TS$ is adjacent.
2. $\tan 63^\circ = \frac{38}{TS}$, so $TS = \frac{38}{\tan 63^\circ}$
   $= \frac{38}{1.9626}$
   $= 19.36$.
3. The distance $TS$ is $19\text{ m}$ to the nearest metre.

(i)

1. Draw $PQ = 7\text{ cm}$ with a ruler.
2. At $P$, construct $60^\circ$: an arc centred at $P$, then the same radius from where it meets $PQ$.
3. Mark $S$ on this arm with $PS = 6.5\text{ cm}$.
4. At $Q$, construct $135^\circ$ with $QP$: construct $90^\circ$, then bisect the $90^\circ$ between the perpendicular and the extension of $PQ$.
5. Join $QS$.
6. With the compasses set to $|QS|$ (about $6.8\text{ cm}$ when measured), centre $Q$, cut the $135^\circ$ arm at $R$.
7. Join $RS$ to complete $PQRS$.

(ii)

1. The points equidistant from $P$ and $Q$ lie on the perpendicular bisector of $PQ$.
2. Construct it with equal arcs from $P$ and $Q$ and label it $l_1$.

(iii)

1. Likewise, construct the perpendicular bisector of $PS$ and label it $l_2$.

(a)

1. Extend $CD$ beyond $D$ to meet $EF$ at $G$.
2. On the straight line $CDG$, $\angle EDG = 180^\circ - 75^\circ$
   $= 105^\circ$.
3. In triangle $DEG$ the angles add up to $180^\circ$: $\angle DGE = 180^\circ - 105^\circ - 26^\circ$
   $= 49^\circ$.
4. $BC \parallel FE$, so the angle between $CD$ and $BC$ equals the angle between $CD$ and $FE$ (corresponding angles): $\angle BCD = 49^\circ$.
5. $AB \parallel CD$, so $x = \angle ABC = \angle BCD$ (alternate angles).
6. So $x = 49^\circ$.
7. $y$ is the reflex angle at $C$: $y = 360^\circ - 49^\circ = 311^\circ$.

(b)(i)

1. $B$ is on the arc opposite the reflex angle, and the angle at the centre is twice the angle at the circumference: $\angle ABC = \frac12 \times 190^\circ$
   $= 95^\circ$.

(ii)

1. The other angle $AOC$ is $360^\circ - 190^\circ = 170^\circ$, so $\angle ADC = \frac12 \times 170^\circ$
   $= 85^\circ$.
2. (Check: $95^\circ + 85^\circ = 180^\circ$, as in any cyclic quadrilateral.)

(c)(i)

1. Arc $ABC$ subtends the $170^\circ$ angle at $O$.
2. Its length is $\frac{170}{360} \times 2 \times \frac{22}{7} \times 7 = 20.78$, so $20.8\text{ cm}$.

(ii)

1. Triangle $AOD$ is isosceles ($OA = OD = 7$), so $\angle ODA = 35^\circ$ and $\angle AOD = 180^\circ - 70^\circ$
   $= 110^\circ$.
2. Split it into two right-angled triangles: $AD = 2 \times 7 \sin 55^\circ$
   $= 14 \times 0.8192$
   $= 11.47$, so $11.5\text{ cm}$.

(a)

1. Work out each value in degree mode.
2. For example, at $60^\circ$: $3(0.8660) + 2(0.5) = 3.60$.
3. At $120^\circ$: $3(0.8660) + 2(-0.5) = 1.60$.
4. At $180^\circ$: $3(0) + 2(-1) = -2.00$.
5. At $240^\circ$: $3(-0.8660) + 2(-0.5) = -3.60$.
6. At $300^\circ$: $3(-0.8660) + 2(0.5) = -1.60$.
7. The completed row is $2.00, 3.60, 1.60, -2.00, -3.60, -1.60, 2.00$.

(b)

1. Plot the seven points with the scales given and join them with a smooth curve.

(c)

1. Draw the line $y = 1.5$ and read down from where it meets the curve: $x \approx 122^\circ$ and $x \approx 351^\circ$.

(d)

1. Draw the line $y = -1$.
2. The curve is below it between the two crossing points, about $162^\circ$ and $310^\circ$.
3. So $162^\circ < x < 310^\circ$.

(a)

1. $18$ is the fourth term, so $3 + 3d = 18$ and the common difference is $d = 5$.
2. Then $x = 3 + 5 = 8$ and $y = 8 + 5 = 13$.

(b)(i)

1. With first term $a$ and ratio $r$: $ar + ar^2 = 6ar^3$.
2. Divide by $ar$ (both are non-zero): $1 + r = 6r^2$, so $6r^2 - r - 1 = 0$.
3. Factorise: $(3r + 1)(2r - 1) = 0$, so $r = \frac12$ or $r = -\frac13$.

(ii)

1. The positive ratio is $r = \frac12$.
2. The second term is $ar = 8$, so $a = 16$.
3. Halve each time: the first six terms are $16, 8, 4, 2, 1, \frac12$.