WAEC · 2012 · May/June · General Maths · Paper 2

(a)

1. Work out the top and bottom separately.
2. Top: $1\frac14 + \frac79 = \frac54 + \frac79$
   $= \frac{45 + 28}{36}$
   $= \frac{73}{36}$.
3. Bottom: multiply first, $2\frac23 \times \frac{9}{64} = \frac83 \times \frac{9}{64}$
   $= \frac38$.
4. Then $1\frac49 - \frac38 = \frac{13}{9} - \frac38$
   $= \frac{104 - 27}{72}$
   $= \frac{77}{72}$.
5. Divide: $\frac{73}{36} \times \frac{72}{77} = \frac{146}{77}$
   $= 1\frac{69}{77}$.

(b)

1. $\sin x = \frac23$: draw a right-angled triangle with opposite 2 and hypotenuse 3.
2. The adjacent side is $\sqrt{9 - 4} = \sqrt5$.
3. So $\cos x = \frac{\sqrt5}{3}$ and $\tan x = \frac{2}{\sqrt5}$.
4. Rationalise: $\tan x = \frac{2}{\sqrt5} \times \frac{\sqrt5}{\sqrt5}$
   $= \frac{2\sqrt5}{5}$.
5. Then $\tan x - \cos x = \frac{2\sqrt5}{5} - \frac{\sqrt5}{3}$
   $= \frac{6\sqrt5 - 5\sqrt5}{15}$
   $= \frac{\sqrt5}{15}$.

(a)

1. Let Wale's age now be $w$ years.
2. Sonny is twice as old: $2w$ years.
3. Four years ago they were $w - 4$ and $2w - 4$, and Sonny was four times as old as Wale: $2w - 4 = 4(w - 4)$.
4. Expand: $2w - 4 = 4w - 16$, so $2w = 12$ and $w = 6$.
5. Wale is 6 and Sonny is 12.
6. In $n$ years they will be $6 + n$ and $12 + n$, and their sum is 66: $(6 + n) + (12 + n) = 66$.
7. So $18 + 2n = 66$, $2n = 48$ and $n = 24$.
8. The sum of their ages will be 66 in 24 years' time.

(a)

1. $RVUS$ is a cyclic quadrilateral, and $RST$ is a straight line.
2. The exterior angle of a cyclic quadrilateral equals the interior angle opposite it: $\angle UST = \angle RVU = 100^\circ$.
3. $TU$ is a tangent at $U$, so by the alternate segment theorem the angle between the tangent and the chord $US$ equals the angle in the other segment: $\angle SUT = \angle URS = 36^\circ$.
4. In triangle $SUT$: $\angle STU = 180^\circ - 100^\circ - 36^\circ$
   $= 44^\circ$.

(b)

1. Draw triangle $XYZ$ with $P$ on $XY$, $|XP| = 2$ cm, and $PQ \parallel YZ$ meeting $XZ$ at $Q$.
2. Triangles $XPQ$ and $XYZ$ have the same angles (corresponding angles), so they are similar, with $XQ$ matching $XZ$ and $XP$ matching $XY$.
3. $\frac{|XQ|}{|XZ|} = \frac{|XP|}{|XY|}$: $\frac{|XQ|}{6} = \frac25$, so $|XQ| = 2.4$ cm.
4. Then $|QZ| = 6 - 2.4 = 3.6$ cm.

(a)(i)

1. $P(\text{red}) = \frac{\text{red discs}}{40}$
   $= \frac14$, so there are $\frac14 \times 40 = 10$ red discs and $40 - 10 = 30$ white discs.

(ii)

1. Add $r$ red discs.
2. Then there are $10 + r$ red discs out of $40 + r$ altogether, and we want $\frac{10 + r}{40 + r} = \frac13$.
3. Cross-multiply: $3(10 + r) = 40 + r$, so $30 + 3r = 40 + r$, $2r = 10$ and $r = 5$.
4. Check: $\frac{15}{45} = \frac13$.

(b)

1. A $20\%$ profit means the selling price is $120\%$ of the cost price.
2. So $1.2 \times$ cost $=$ ₦600, and the cost is ₦600 $\div 1.2 =$ ₦500.
3. At ₦50.00 each, he bought $\frac{500}{50} = 10$ plates.

(a)

1. Join $O$ to $X$ and $Y$.
2. The triangle $OXY$ is isosceles ($OX = OY = 5$), so the perpendicular from $O$ bisects the chord: each half is 3 cm, and it cuts the angle $\theta$ at $O$ in half.
3. In one right-angled half, $\sin\frac\theta2 = \frac35 = 0.6$.
4. So $\frac\theta2 \approx 36.87^\circ$ and $\theta \approx 73.74^\circ$.

(b)

1. Area of the sector $OXY = \frac{\theta}{360} \times \pi r^2$
   $= \frac{73.74}{360} \times \frac{22}{7} \times 25$
   $\approx 16.09\text{ cm}^2$.
2. The perpendicular from $O$ to $XY$ is $\sqrt{5^2 - 3^2} = 4$ cm, so the area of triangle $OXY = \frac12 \times 6 \times 4$
   $= 12\text{ cm}^2$.
3. Minor segment $=$ sector $-$ triangle $\approx 16.09 - 12 = 4.09\text{ cm}^2$.

(a)

1. After spending D15 the boy had $M - 15$.
2. He shared it equally, so his sister got $\frac{M - 15}{2}$.
3. This equals $\frac13$ of $M$: $\frac{M - 15}{2} = \frac{M}{3}$.
4. Multiply both sides by 6: $3(M - 15) = 2M$.
5. So $3M - 45 = 2M$ and $M = 45$.

(b)

1. Let the number of tourists be $x$.
2. Road only $= \frac23x - \frac4{15}x = \frac{6}{15}x$.
3. Air only $= \frac{13}{30}x - \frac4{15}x$
   $= \frac{5}{30}x$
   $= \frac16x$.
4. Both $= \frac4{15}x$, and neither $= 20$.
5. The four regions of the Venn diagram make up everyone: $\frac{6}{15}x + \frac16x + \frac4{15}x + 20 = x$.
6. With denominator 30: $\frac{12 + 5 + 8}{30}x + 20 = x$, so $\frac{25}{30}x + 20 = x$.
7. Then $\frac{5}{30}x = 20$ and $x = 120$ tourists.
8. Air only $= \frac16 \times 120 = 20$ tourists.

(a)(i)

1. Rearrange each equation for $y$: $y = \frac34x + 3$ and $y = 6 - 2x$.
2. Plot points for each.
3. For $y = \frac34x + 3$: $(-4, 0)$, $(0, 3)$, $(4, 6)$.
4. For $y = 6 - 2x$: $(0, 6)$, $(1, 4)$, $(3, 0)$.
5. Join each set with a straight line.

(ii)

1. Read where the lines cross: about $(1.1, 3.8)$.
2. (Check by algebra: $\frac34x + 3 = 6 - 2x$ gives $\frac{11}{4}x = 3$, so $x = \frac{12}{11} \approx 1.1$ and $y = 6 - \frac{24}{11}$
   $= \frac{42}{11}$
   $\approx 3.8$.)

(iii)

1. $y - \frac34x \ge 3$ is $y \ge \frac34x + 3$.
2. Test the origin: $0 - 0 \ge 3$ is false, so the region is the side of the line away from the origin: on and above the line $y = \frac34x + 3$.
3. Draw the line solid (it is included) and label the region.

(b)

1. Expand: $(x + 2)(x + c) = x^2 + (2 + c)x + 2c$.
2. Compare with $x^2 + bx + 18$: the numbers give $2c = 18$, so $c = 9$.
3. The $x$ terms give $b = 2 + c = 11$.

1. Draw the tower $FT$ vertical, with $P$ on it, and $H$ on the ground 20 m from $F$.
2. $\angle FHP = 30^\circ$ and $\angle FHT = 50^\circ$, both measured up from the ground.

(a)

1. In triangle $HFT$: $|FT| = 20\tan 50^\circ \approx 23.84$ m.
2. In triangle $HFP$: $|FP| = 20\tan 30^\circ \approx 11.55$ m.
3. So $|PT| = |FT| - |FP|$
   $\approx 23.84 - 11.55$
   $= 12.29$
   $\approx 12.3\text{ m}$.

(b)

1. $HT$ is the hypotenuse of triangle $HFT$: $\cos 50^\circ = \frac{20}{|HT|}$, so $|HT| = \frac{20}{\cos 50^\circ}$
   $\approx 31.1\text{ m}$.

(c)

1. An angle of depression of $40^\circ$ from $T$ equals an angle of elevation of $40^\circ$ from the new point (alternate angles).
2. So $\tan 40^\circ = \frac{|FT|}{d}$, and $d = \frac{23.84}{\tan 40^\circ}$
   $\approx 28.4\text{ m}$.

(a)

1. With 1 cm to 2 km: $XY = 10$ cm, $YZ = 11$ cm, $XZ = 9$ cm, and 10 km is 5 cm.
2. Draw $XY = 10$ cm.
3. With centre $X$ and radius 9 cm, and centre $Y$ and radius 11 cm, draw arcs meeting at $Z$.
4. Join $XZ$ and $YZ$.
5. Equal distances from $X$ and $Y$: construct the perpendicular bisector of $XY$.
6. Exactly 10 km from $Z$: draw the circle with centre $Z$ and radius 5 cm.
7. The Health Centre is where they cross.

(b)(i)

1. The circle cuts the bisector in 2 places, so there are 2 possible locations.

(ii)

1. Measure each from $X$ and change back to km: about 14.0 cm $= 28$ km and 6.3 cm $= 12.7$ km.

(iii)

1. The location 12.7 km from $X$, inside the triangle, is convenient for all three towns.

1. Use the class marks $62, 67, 72, \ldots, 97$.
2. The numbers are large, so an assumed mean $A = 77$ keeps the arithmetic small: $d = x - 77$.

3. Marks	$f$	$x$	$d$	$fd$	$fd^2$
   60–64	2	62	$-15$	$-30$	450
   65–69	3	67	$-10$	$-30$	300
   70–74	6	72	$-5$	$-30$	150
   75–79	11	77	0	0	0
   80–84	8	82	5	40	200
   85–89	7	87	10	70	700
   90–94	2	92	15	30	450
   95–99	1	97	20	20	400
   Total	40			70	2650

(a)

1. Mean $= A + \frac{\sum fd}{\sum f}$
   $= 77 + \frac{70}{40}$
   $= 77 + 1.75$
   $= 78.75$.

(b)

1. Standard deviation $= \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}$
   $= \sqrt{\frac{2650}{40} - 1.75^2}$
   $= \sqrt{66.25 - 3.0625}$
   $= \sqrt{63.1875}$
   $\approx 7.95$.

(a)

1. Read the diagram carefully: the mesh runs along the length 3 times (the two long sides and one line between them) and across the width 5 times (the two short sides and three lines between them).
2. So the total length of mesh is $3(3n - 1) + 5(2n + 1) = 135$.
3. Expand: $9n - 3 + 10n + 5 = 135$.
4. So $19n + 2 = 135$, $19n = 133$ and $n = 7$.

(b)

1. Volume of the cube $= 22^3 = 10\,648\text{ cm}^3$.
2. The cylinder has the same volume: $\frac{22}{7} \times 14^2 \times h = 10\,648$, so $616h = 10\,648$ and $h = \frac{121}{7}\text{ cm}$.
3. Total surface area of the cylinder $= 2\pi r(r + h)$
   $= 2 \times \frac{22}{7} \times 14 \times \left(14 + \frac{121}{7}\right)$
   $= 88 \times \frac{219}{7}$
   $= \frac{19\,272}{7}\text{ cm}^2$.
4. Total surface area of the cube $= 6 \times 22^2 = 2904\text{ cm}^2$.
5. Ratio $= \frac{19\,272}{7} : 2904$
   $= 19\,272 : 20\,328$
   $= 73 : 77$.

(a)

1. Use a calculator in degree mode and round to 1 decimal place.
2. For example, $x = 30^\circ$: $1 - 4(0.8660) = -2.46 \approx -2.5$.
3. $x = 150^\circ$: $1 - 4(-0.8660) = 4.46 \approx 4.5$.
4. The full row is $-3.0, -2.5, -1.0, 1.0, 3.0, 4.5, 5.0, 4.5, 3.0, 1.0, -1.0$.

(b)

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

(c)(i)

1. $1 - 4\cos x = 0$ is where the curve crosses the line $y = 0$ (the $x$-axis): $x \approx 76^\circ$ and $x \approx 284^\circ$.

(ii)

1. Read up from $x = 105^\circ$ to the curve and across: $y \approx 2.0$.

(iii)

1. Draw the line $y = 1.5$ and read down from where it meets the curve: $x \approx 97^\circ$ and $x \approx 263^\circ$.