WAEC · 2020 · Private, 2nd series · General Maths · Paper 2

(a)(i)

1. Let the universal set be the teachers in the school, $G$ the graduate teachers and $H$ the teachers who received National honours. "Some teachers are graduates": draw $G$ inside the rectangle, leaving room outside it. "Only graduate teachers received National honours": draw $H$ inside $G$.

(ii)

1. (I)** Mr.
2. Sowah could be in $G$ but outside $H$: a graduate with no honour.
3. So "graduate $\Rightarrow$ honours" is not valid.
4. (II) $H$ lies wholly inside $G$, so anyone in $H$ is in $G$. "Honours $\Rightarrow$ graduate" is valid.
5. (III) A teacher outside $H$ can still be inside $G$.
6. So "no honour $\Rightarrow$ not a graduate" is not valid.

(b)

1. Put $n = 1, 2, 3$ into $2^{2n + 1}$: $T_1 = 2^3 = 8$, $T_2 = 2^5 = 32$ and $T_3 = 2^7 = 128$.

(a)

1. Let a cutlass cost GH₵ $c$ and a hoe GH₵ $h$.
2. Then $3c + 2h = 85$ (1) and $2c + 3h = 90$ (2).
3. Add the two equations: $5c + 5h = 175$, so $c + h = 35$.
4. A cutlass and a hoe together cost GH₵ 35.00.
5. (Solving fully gives $c = 15$ and $h = 20$.)

(b)

1. The line $x = 0$ is the $y$-axis, so the graph passes through $(0, 24)$.
2. Substitute: $24 = k(0 + 4)(0 - 1) = -4k$.
3. So $k = -6$.

(a)(i)

1. Draw north at $P$ and $PQ$, 500 km on $053^\circ$.
2. Draw north at $Q$ and $QR$, 700 km on $165^\circ$.
3. Join $R$ to $P$.

(ii)

1. At $Q$, the direction back to $P$ is $053^\circ + 180^\circ = 233^\circ$ and the direction to $R$ is $165^\circ$, so $\angle PQR = 233^\circ - 165^\circ$
   $= 68^\circ$.
2. Cosine rule: $|PR|^2 = 500^2 + 700^2 - 2(500)(700)\cos 68^\circ$
   $= 740\,000 - 700\,000 \times 0.3746$
   $\approx 477\,780$.
3. So $|PR| \approx 691$ km (3 significant figures).

(b)

1. $XY$ is a diameter, so $\angle XZY = 90^\circ$ (angle in a semicircle).
2. In triangle $XYZ$: $\angle XYZ = 180^\circ - 90^\circ - 28^\circ$
   $= 62^\circ$.
3. $WXYZ$ is a cyclic quadrilateral, so $\angle XWZ = 180^\circ - \angle XYZ$
   $= 118^\circ$ (opposite angles).

(a)

1. Drop perpendiculars from $A$ and $B$ to $DC$, meeting it at $N$ and $M$.
2. Both are the height, 6 cm.
3. In the right-angled triangle $ADN$: $|DN| = \sqrt{7.5^2 - 6^2}$
   $= \sqrt{20.25}$
   $= 4.5$ cm.
4. In triangle $BMC$: $|MC| = \sqrt{6.8^2 - 6^2}$
   $= \sqrt{10.24}$
   $= 3.2$ cm.
5. $NM = AB = 8$ cm, so $|DC| = 4.5 + 8 + 3.2 = 15.7$ cm.
6. Area $= \frac12(8 + 15.7) \times 6$
   $= 71.1\text{ cm}^2$.

(b)

1. The point lies on the line, so its coordinates satisfy the equation: $6(-4) + k(-3) - 12 = 0$, so $-36 - 3k = 0$ and $k = -12$.

1. There are 25 equally likely numbers.

(a)

1. The even numbers are $2, 4, \ldots, 24$: 12 of them.
2. $P(\text{even}) = \frac{12}{25}$.

(b)

1. The primes up to 25 are $2, 3, 5, 7, 11, 13, 17, 19, 23$: 9 of them.
2. $P(\text{prime}) = \frac{9}{25}$.

(c)

1. Only 2 is both even and prime, so it's in both lists.
2. Add the two lists and take it away once: $\frac{12}{25} + \frac{9}{25} - \frac{1}{25} = \frac{20}{25}$
   $= \frac45$.

(d)

1. Both even and prime: only 2.
2. $P = \frac{1}{25}$.

(a)(i)

1. The price went up by $16\%$, so he paid $116\%$ of GH₵ 27,200: $1.16 \times 27\,200 =$ GH₵ 31,552.00.

(ii)

1. To make a $15\%$ profit on what he paid, he must sell at $115\%$ of it: $1.15 \times 31\,552 =$ GH₵ 36,284.80.

(b)

1. Let Morgan receive GH₵ $m$.
2. Aku received three times as much, $3m$, and Danso GH₵ 14,000 more than Morgan, $m + 14\,000$.
3. The three shares make up the estate: $3m + (m + 14\,000) + m = 64\,000$.
4. So $5m = 50\,000$ and $m = 10\,000$.
5. Morgan received GH₵ 10,000.00, Aku GH₵ 30,000.00 and Danso GH₵ 24,000.00.
6. Check: $10\,000 + 30\,000 + 24\,000 = 64\,000$ ✓.

(a)

1. Put each $x$ into $y = 2x^2 - 5x - 3$.
2. For $x = -2$: $8 + 10 - 3 = 15$.
3. For $x = -1$: $2 + 5 - 3 = 4$.
4. For $x = 1$: $2 - 5 - 3 = -6$.
5. For $x = 2$: $8 - 10 - 3 = -5$.
6. For $x = 3$: $18 - 15 - 3 = 0$.
7. For $x = 5$: $50 - 25 - 3 = 22$.
8. The row is $30, 15, 4, -3, -6, -5, 0, 9, 22$.

(b)

1. With 2 cm to 1 unit across and 2 cm to 5 units up, plot the nine points and join them with a smooth U-shaped curve.

(c)(i)

1. Take $x - 4$ from both sides of $2x^2 - 4x - 7 = x - 4$: $2x^2 - 5x - 3 = 0$, which is $y = 0$.
2. The curve crosses the $x$-axis at $x = -0.5$ and $x = 3$, so the truth set is $\{-0.5, 3\}$.

(ii)

1. The line of symmetry is halfway between the roots: $x = \frac{-0.5 + 3}{2} = 1.25$.
2. Its equation is $x = 1.25$.

(iii)

1. Draw the line $y = 7$.
2. It meets the curve at $x \approx -1.3$ and $x \approx 3.8$.
3. (Exactly, $2x^2 - 5x - 10 = 0$ gives $x = \frac{5 \pm \sqrt{105}}{4}$.)

(a)(i)

1. Two sides and the angle between them: cosine rule.
2. $|BC|^2 = 6^2 + 6^2 - 2(6)(6)\cos 130^\circ$
   $= 72 + 72(0.6428)$
   $\approx 118.28$, so $|BC| \approx 10.88$, which is 11 cm to two significant figures.

(ii)

1. Area $= \frac12 \times 6 \times 6 \times \sin 130^\circ$
   $\approx 18 \times 0.7660$
   $\approx 13.79$, which is 14 cm² to two significant figures.

(b)

1. The park has radius 25 m: area $= \frac{22}{7} \times 25^2$
   $\approx 1964.29\text{ m}^2$.
2. Each lamp base has radius 0.5 m: 10 bases cover $10 \times \frac{22}{7} \times 0.5^2 \approx 7.86\text{ m}^2$.
3. Grass $= 1964.29 - 7.86$
   $\approx 1956.43\text{ m}^2$.

(a)(i)

1. Draw north at $A$ and $AB$, 100 km on $330^\circ$ ($30^\circ$ west of north).
2. From $B$, draw $BC$, 300 km due west.

(ii)

1. On the first leg, the westward part is the side opposite the $30^\circ$ angle between $AB$ and north: $100\sin 30^\circ = 50$ km.
2. The second leg adds 300 km more to the west.
3. So $C$ is $50 + 300 = 350$ km west of $A$.

(b)

1. The odd numbers from 11 form an A.P. with $a = 11$ and $d = 2$.
2. $S_n = \frac n2[2a + (n - 1)d]$
   $= \frac n2[22 + 2(n - 1)]$
   $= n(n + 10)$.
3. So $n^2 + 10n = 551$, which is $n^2 + 10n - 551 = 0$.
4. Factorise: $(n - 19)(n + 29) = 0$.
5. The number of terms is positive, so $n = 19$.

1. Put the ages in order first: $1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10, 11, 12, 13$ (14 ages).

(a)(i)

1. Range $= 13 - 1 = 12$.

(ii)

1. Number the positions: 1st $= 1$, 2nd $= 2$, 3rd $= 3$, 4th $= 4$, 5th $= 5$, 6th $= 5$, 7th $= 6$, 8th $= 7$, 9th $= 8$, 10th $= 9$, 11th $= 10$, 12th $= 11$, 13th $= 12$, 14th $= 13$.
2. $Q_1$ is at position $\frac{14}{4} = 3.5$: halfway between the 3rd and 4th ages, $\frac{3 + 4}{2} = 3.5$.
3. $Q_3$ is at position $\frac{3 \times 14}{4} = 10.5$: halfway between the 10th and 11th ages, $\frac{9 + 10}{2} = 9.5$.
4. Semi-interquartile range $= \frac{Q_3 - Q_1}{2}$
   $= \frac{9.5 - 3.5}{2}$
   $= 3$.

(iii)

1. The ages add up to 96, so the mean is $\frac{96}{14} \approx 6.86$.

(b)

1. The squares of the ages add up to $1 + 4 + 9 + 16 + 25 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 + 169 = 844$.
2. Standard deviation $= \sqrt{\frac{\sum x^2}{n} - \bar x^2}$
   $= \sqrt{\frac{844}{14} - \left(\frac{96}{14}\right)^2}$
   $= \sqrt{60.286 - 47.020}$
   $= \sqrt{13.265}$
   $\approx 3.64$.

(a)(i)

1. $\angle ADB$ stands on the same arc $AB$ as the angle $AOB$ at the centre, so $\angle ADB = \frac12 \times 98^\circ$
   $= 49^\circ$.
2. Then $\angle ADC = \angle ADB + \angle BDC$
   $= 49^\circ + 47^\circ$
   $= 96^\circ$.
3. $ABCD$ is cyclic, so $\angle ABC = 180^\circ - 96^\circ$
   $= 84^\circ$ (opposite angles).
4. So $\angle CBD = \angle ABC - \angle ABD$
   $= 84^\circ - 68^\circ$
   $= 16^\circ$.

(ii)

1. In triangle $BCD$: $\angle DCB = 180^\circ - 47^\circ - 16^\circ$
   $= 117^\circ$.

(b)

1. $10\cos(x + 17^\circ) = 2$, so $\cos(x + 17^\circ) = 0.2$.
2. Then $x + 17^\circ = \cos^{-1} 0.2$
   $\approx 78.46^\circ$, and $x \approx 61.46^\circ \approx 61^\circ$.

(c)

1. The interior angles of an $n$-sided polygon add up to $(n - 2) \times 180^\circ$.
2. So $(n - 2) \times 180 = 120n$.
3. Then $180n - 360 = 120n$, $60n = 360$ and $n = 6$.

(a)

1. $|MN| = |NQ|$, so triangle $MNQ$ is isosceles and $\angle NQM = \angle NMQ = 34^\circ$.
2. The exterior angle at $N$ equals the sum of the two opposite interior angles: $\angle QNO = 34^\circ + 34^\circ$
   $= 68^\circ$.
3. $|NQ| = |QO|$, so triangle $NQO$ is isosceles and $\angle QON = \angle QNO = 68^\circ$.
4. $N$, $O$ and $P$ are on a straight line, so $x = \angle QOP$
   $= 180^\circ - 68^\circ$
   $= 112^\circ$.

(b)

1. There are 7 balls.
2. Without replacement, the second ball is picked from the 6 left.

(i)

1. Both red: $\frac37 \times \frac26 = \frac{6}{42}$
   $= \frac17$.

(ii)

1. Both white: $\frac47 \times \frac36 = \frac{12}{42}$
   $= \frac27$.

(iii)

1. Same colour means both red or both white: $\frac17 + \frac27 = \frac37$.

(iv)

1. Different colours is everything else: $1 - \frac37 = \frac47$.
2. (Check: red then white $\frac37 \times \frac46 = \frac{12}{42}$, plus white then red $\frac47 \times \frac36 = \frac{12}{42}$, gives $\frac{24}{42} = \frac47$.)

(a)(i)

1. With 1 cm to 10 m: $AB = 8.5$ cm and $BC = 11$ cm.
2. Draw a base line from $A$, construct $60^\circ$ at $A$, and mark $AB = 8.5$ cm on the arm.
3. With centre $B$ and radius 11 cm, cut the base line at $C$ (so $|AC| \approx 12.4$ cm).

(ii)

1. Equidistant from $AC$ and $BC$: construct the bisector of $\angle ACB$.
2. On the perpendicular from $B$ to $AC$: construct that perpendicular with arcs.
3. $X$ is where the two lines cross.

(b)

1. Measure $|XC| \approx 8.8$ cm and change back with the scale: about 88 m.
2. Check: by calculation $|XC| \approx 87.6$ m.