WAEC · 2014 · May/June · General Maths · Paper 2

(a)

1. Remove the decimals: top $= 6 \times 32 \times 4 \times 10^{-4}$
   $= 768 \times 10^{-4}$.
2. Bottom $= 12 \times 8 \times 16 \times 10^{-6}$
   $= 1536 \times 10^{-6}$.
3. Divide: $\frac{768}{1536} \times 10^{-4 - (-6)} = 0.5 \times 10^2$
   $= 50$
   $= 5 \times 10^1$.

(b)

1. Draw a line through $B$ parallel to $EF$ and $GH$.
2. It splits $\angle ABC$ into two parts: $3x$ (alternate angles with $\angle AEF$) and $180^\circ - 7x$ (co-interior with $\angle CHG$).
3. So $3x + 180 - 7x = 120$, $4x = 60$ and $x = 15$.
4. $\angle GHB$ and $\angle CHG$ are on the straight line $BHC$: $\angle GHB = 180^\circ - 7 \times 15^\circ$
   $= 75^\circ$.

(a)

1. Simplify each surd using its largest square factor: $\sqrt{75} = 5\sqrt3$, $\sqrt{12} = 2\sqrt3$ and $\sqrt{108} = 6\sqrt3$.
2. Then $3 \times 5\sqrt3 - 2\sqrt3 + 6\sqrt3 = 15\sqrt3 - 2\sqrt3 + 6\sqrt3$
   $= 19\sqrt3$.

(b)

1. Change both sides to base ten: $124_n = n^2 + 2n + 4$ and $232_{\text{five}} = 2 \times 25 + 3 \times 5 + 2$
   $= 67$.
2. So $n^2 + 2n + 4 = 67$, $n^2 + 2n - 63 = 0$ and $(n + 9)(n - 7) = 0$.
3. A base is a positive whole number, larger than every digit used, so $n = 7$.

(a)

1. Treat $\frac1x$ and $\frac1y$ as the unknowns.
2. Adding the equations, the $\frac1x$ terms cancel: $\frac2y = 6$, so $\frac1y = 3$ and $y = \frac13$.
3. Taking the second equation from the first: $\frac2x = 4$, so $\frac1x = 2$ and $x = \frac12$.

(b)

1. Let the good surface be $x$ km.
2. The bad surface is then $(48 - x)$ km.
3. Time $= \frac{\text{distance}}{\text{speed}}$, so the two times are $\frac{x}{72}$ and $\frac{48 - x}{48}$ hours.
4. The whole journey takes 45 minutes $= \frac34$ hour: $\frac{x}{72} + \frac{48 - x}{48} = \frac34$.
5. Multiply every term by 144: $2x + 3(48 - x) = 108$.
6. So $2x + 144 - 3x = 108$, $-x = -36$ and $x = 36$.
7. There are 36 km of good surface.

(a)

1. The shaded segment is the quarter-circle sector $XOY$ minus the right-angled triangle $XOY$.
2. Sector $= \frac{90}{360} \times \frac{22}{7} \times r^2$
   $= \frac{11r^2}{14}$.
3. Triangle $= \frac12 r^2$.
4. Segment $= \frac{11r^2}{14} - \frac{7r^2}{14}$
   $= \frac{4r^2}{14}$
   $= \frac{2r^2}{7}$.
5. So $\frac{2r^2}{7} = 504$, $r^2 = 1764$ and $r = 42$ cm.

(b)

1. Triangle $PQS$ is isosceles with the $109^\circ$ angle at $S$ between the equal sides, so its base angles are $\frac12(180^\circ - 109^\circ) = 35.5^\circ$: $\angle PQS = 35.5^\circ$.
2. The triangles are on opposite sides of $PQ$, so at $Q$ the angles add: $\angle RQS = \angle PQR + \angle PQS$
   $= 66^\circ + 35.5^\circ$
   $= 101.5^\circ$.

1. Write all four probabilities first.
2. $P(\text{wins } X) = 0.5$, so $P(\text{loses } X) = 0.5$.
3. $P(\text{loses } Y) = 0.3$, so $P(\text{wins } Y) = 1 - 0.3 = 0.7$.
4. The contracts are independent, so for "and" we multiply.

(a)

1. Wins $X$ and wins $Y$: $0.5 \times 0.7 = 0.35$.

(b)

1. Exactly one happens in two ways: wins $X$ but not $Y$, $0.5 \times 0.3 = 0.15$.
2. Or wins $Y$ but not $X$, $0.5 \times 0.7 = 0.35$.
3. Add them: $0.15 + 0.35 = 0.5$.

(c)

1. Loses both: $0.5 \times 0.3 = 0.15$.
2. Check: both, exactly one and neither cover everything, and $0.35 + 0.5 + 0.15 = 1$.

(a)

1. One fraction equals another, so cross-multiply: $3\left(\frac14p + 1\right) = \frac13\left(2p - \frac12\right)$.
2. Expand both sides: $\frac34p + 3 = \frac23p - \frac16$.
3. Multiply every term by 12 to clear the fractions: $9p + 36 = 8p - 2$.
4. So $p = -38$.

(b)

1. The marked price, GH¢ 760.00, includes a $20\%$ profit, so it is $120\%$ of the cost price.
2. Cost price $= \frac{760}{1.2}$
   $= \frac{1900}{3}$
   $\approx$ GH¢ 633.33.
3. With a $5\%$ discount, the selling price is $95\%$ of the marked price: $0.95 \times 760 =$ GH¢ 722.00.
4. Profit $= 722 - \frac{1900}{3}$
   $= \frac{266}{3}$
   $\approx$ GH¢ 88.67.
5. Percentage profit $= \frac{266/3}{1900/3} \times 100\%$
   $= \frac{266}{1900} \times 100\%$
   $= 14\%$.

(a)

1. Use a calculator in degree mode, to 1 decimal place.
2. For example, $x = 30^\circ$: $2(0.5) + 1 = 2.0$.
3. $x = 90^\circ$: $2(1) + 1 = 3.0$.
4. $x = 270^\circ$: $2(-1) + 1 = -1.0$.
5. The full row is $1.0, 2.0, 2.7, 3.0, 2.7, 2.0, 1.0, 0.0, -0.7, -1.0$.

(b)

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

(c)

1. The graph is of $2\sin x + 1$, so change the equation to match: $\sin x = \frac14$ gives $2\sin x + 1 = 2 \times \frac14 + 1 = 1.5$.
2. Draw the line $y = 1.5$ and read down from where it crosses the curve: $x \approx 15^\circ$ and $x \approx 165^\circ$.

(a)

1. Multiply, then take the remainder on dividing by 11.
2. For example $9 \otimes 10 = 90 = 8 \times 11 + 2$, so the entry is 2.

3. $\otimes$	1	5	9	10
   1	1	5	9	10
   5	5	3	1	6
   9	9	1	4	2
   10	10	6	2	1

(i)

1. From the table, $9 \otimes 5 = 1$ and $10 \otimes 10 = 1$, so $(9 \otimes 5) \otimes (10 \otimes 10) = 1 \otimes 1$
   $= 1$.

(ii)

1. In the row of 10, the entry 2 is in the column of 9: the truth set of $10 \otimes m = 2$ is $\{9\}$.

(iii)

1. On the diagonal, $n \otimes n = 4$ only for $n = 9$: the truth set is $\{9\}$.

(b)

1. Let the fraction be $\frac{x}{y}$.
2. It equals $\frac34$, so $4x = 3y$.
3. Doubling the numerator gives 34 more than the denominator: $2x = y + 34$, so $y = 2x - 34$.
4. Then $4x = 3(2x - 34) = 6x - 102$, so $2x = 102$, $x = 51$ and $y = 68$.
5. The fraction is $\frac{51}{68}$.

(a)(i)

1. The eight regions together hold all 125 elements: $(16 - 2x) + 5x + (6 + x) + 8x + 7x + 4x + (19 - 3x) + 4 = 125$.
2. Collect the numbers and the $x$ terms: $45 + 20x = 125$, so $20x = 80$ and $x = 4$.

(ii)

1. $(P \cup Q) \cap R'$ is the part of $P$ or $Q$ that is outside $R$: the regions $P$ only, $P \cap Q$ only and $Q$ only.
2. With $x = 4$ these hold $16 - 8 = 8$, $5 \times 4 = 20$ and $6 + 4 = 10$, so $n[(P \cup Q) \cap R'] = 8 + 20 + 10 = 38$.

(b)(i)

1. $XOW$ is a diameter, so $\angle XYW = 90^\circ$ (angle in a semicircle).
2. In $\triangle WXY$: $\angle XWY = 180^\circ - 90^\circ - 50^\circ$
   $= 40^\circ$.
3. $WX \parallel YZ$, so $\angle WYZ = \angle XWY = 40^\circ$ (alternate angles).
4. $\angle WXZ$ and $\angle WYZ$ stand on the same arc $WZ$, so $\angle WXZ = 40^\circ$ (angles in the same segment).

(ii)

1. The angle at the centre is twice the angle at the circumference on the same arc: $\angle WOZ = 2 \times \angle WYZ$
   $= 80^\circ$.
2. In $\triangle OEW$: $\angle OEW = 180^\circ - 80^\circ - 40^\circ$
   $= 60^\circ$.
3. $\angle YEZ$ and $\angle OEW$ are vertically opposite, so $\angle YEZ = 60^\circ$.

(a)

1. Expand and bring everything to one side: $x^2 - 5x + 6 = 12$, so $x^2 - 5x - 6 = 0$.
2. Factorise: $(x - 6)(x + 1) = 0$, so $x = 6$ or $x = -1$.

(b)

1. The overlap is two equal segments back to back, one in each circle, on the common chord $PQ$.
2. Each is a $60^\circ$ sector minus the triangle on the chord.
3. Sector $= \frac{60}{360} \times \frac{22}{7} \times 7^2$
   $\approx 25.667\text{ cm}^2$.
4. Triangle $= \frac12 \times 7 \times 7 \times \sin 60^\circ$
   $\approx 21.218\text{ cm}^2$.
5. One segment $\approx 4.449\text{ cm}^2$.
6. Shaded area $= 2 \times 4.449 \approx 8.9$, which is $9\text{ cm}^2$ to the nearest whole number.

(a)

1. First the mean: $\sum fx = 1(2) + 2(5) + 3(13) + 4(11) + 5(9) + 6(10)$
   $= 2 + 10 + 39 + 44 + 45 + 60$
   $= 200$, and $\sum f = 50$, so $\bar x = \frac{200}{50} = 4$.
2. The distances from 4 are $|x - 4| = 3, 2, 1, 0, 1, 2$.
3. Multiply by the frequencies: $2(3) + 5(2) + 13(1) + 11(0) + 9(1) + 10(2) = 6 + 10 + 13 + 0 + 9 + 20$
   $= 58$.
4. Mean deviation $= \frac{\sum f|x - \bar x|}{\sum f}$
   $= \frac{58}{50}$
   $= 1.16$.

(b)

1. "At least 4" means a score of 4, 5 or 6.
2. That happened $11 + 9 + 10 = 30$ times out of 50, so the probability is $\frac{30}{50} = \frac35$.

(a)

1. $5\cos(x + 8.5)^\circ = 1$, so $\cos(x + 8.5)^\circ = 0.2$.
2. Treat $(x + 8.5)$ as one angle: $x + 8.5 = \cos^{-1} 0.2 \approx 78.46$.
3. So $x \approx 69.96 \approx 70^\circ$.

(b)(i)

1. At $P$, draw north: $Q$ is on $150^\circ$, 24 km away.
2. The bearing of $P$ from $R$ is $015^\circ$, so the bearing of $R$ from $P$ is the back bearing, $015^\circ + 180^\circ = 195^\circ$: $R$ is 32 km from $P$ on $195^\circ$.
3. The angle between the two lines at $P$ is $\angle QPR = 195^\circ - 150^\circ$
   $= 45^\circ$.

(ii)

1. Cosine rule: $|QR|^2 = 32^2 + 24^2 - 2(32)(24)\cos 45^\circ$
   $= 1600 - 1086.1$
   $\approx 513.9$.
2. So $|QR| \approx 22.67\text{ km}$.

(iii)

1. Use the cosine rule for the angle at $Q$ (it avoids any doubt about an obtuse angle): $\cos\angle PQR = \frac{24^2 + 22.67^2 - 32^2}{2 \times 24 \times 22.67}$
   $\approx \frac{65.9}{1088.2}$
   $\approx 0.0606$, so $\angle PQR \approx 86.5^\circ$.
2. At $Q$, the direction back to $P$ is $150^\circ + 180^\circ = 330^\circ$, and $R$ is $86.5^\circ$ further round anticlockwise.
3. Bearing of $R$ from $Q$ $= 330^\circ - 86.5^\circ$
   $= 243.5^\circ$
   $\approx 243^\circ$.

(a)(i)

1. Replace every $x$ in $f(x) = 2x^2 - 1$ by $(x - 1)$: $f(x - 1) = 2(x - 1)^2 - 1$.
2. So $2(x - 1)^2 - 1 - 7 = 0$, which gives $(x - 1)^2 = 4$.
3. Then $x - 1 = 2$ or $x - 1 = -2$.
4. So $x = 3$ or $x = -1$.

(ii)

1. $f\left(-\frac12\right) = 2 \times \frac14 - 1$
   $= -\frac12$.
2. $g(3) = 9 + 2 = 11$.
3. $f(4) = 32 - 1 = 31$.
4. $g(5) = 15 + 2 = 17$.
5. So the value is $\frac{-\frac12 \times 11}{31 - 17} = \frac{-\frac{11}{2}}{14}$
   $= -\frac{11}{28}$.

(b)

1. $m * n = \frac{-(-10)}{(-3)^2 + 1}$
   $= \frac{10}{10}$
   $= 1$.
2. Swap them: $n * m = \frac{-(-3)}{(-10)^2 + 1}$
   $= \frac{3}{101}$.
3. The two answers are different, so $*$ is not commutative.