WAEC · 2020 · Private · General Maths · Paper 2

(a)

1. Let the two fractions be $\frac x8$ and $\frac y8$ (a different letter for each).
2. Their sum is $\frac12$: $\frac x8 + \frac y8 = \frac12$, so $x + y = 4$ (1).
3. One added to 5 times the other is 2: $\frac x8 + \frac{5y}{8} = 2$, so $x + 5y = 16$ (2).
4. Take (1) from (2): $4y = 12$, so $y = 3$ and $x = 1$.
5. The fractions are $\frac18$ and $\frac38$.

(b)

1. First make $v$ the subject.
2. Multiply by $m$: $am = mg - kv^2$.
3. Rearrange: $kv^2 = mg - am = m(g - a)$.
4. So $v^2 = \frac{m(g - a)}{k}$ and $v = \sqrt{\frac{m(g - a)}{k}}$.
5. Substitute: $v = \sqrt{\frac{12 \times (9.8 - 2.8)}{\frac83}}$
   $= \sqrt{\frac{84 \times 3}{8}}$
   $= \sqrt{31.5}$
   $\approx 5.61$.
6. To the nearest whole number, $v = 6$.

(a)

1. $\tan 60^\circ = \frac{\sin 60^\circ}{\cos 60^\circ}$
   $= \sqrt3$ and $\tan 30^\circ = \frac{\sin 30^\circ}{\cos 30^\circ}$
   $= \frac{1}{\sqrt3}$.
2. So the value is $\frac{\sqrt3 - 1}{1 - \frac{1}{\sqrt3}}$.
3. Multiply the top and bottom by $\sqrt3$: $\frac{\sqrt3(\sqrt3 - 1)}{\sqrt3 - 1} = \sqrt3$.

(b)

1. Let $x$ students read both subjects.
2. Only Chemistry: $30 - x$.
3. Only Physics: $20 - x$.
4. Everyone reads at least one, so $(30 - x) + x + (20 - x) = 40$, which gives $50 - x = 40$ and $x = 10$.
5. Only Chemistry: $30 - 10 = 20$ students.
6. Probability $= \frac{20}{40} = \frac12$.

(a)

1. The ages add up to $42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48 + 48 = 480$, and there are 10, so the mean is $\frac{480}{10} = 48$ years.

(b)

1. Set out a table.
2. Each age appears twice, so $f = 2$ for each:

3. $x$	42	46	48	50	54	Total
   $f$	2	2	2	2	2	10
   $x - 48$	$-6$	$-2$	0	2	6
   $f(x - 48)^2$	72	8	0	8	72	160

4. Standard deviation $= \sqrt{\frac{160}{10}} = \sqrt{16} = 4$ years.

(a)

1. $RSTU$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$: $\angle URS = 180^\circ - 124^\circ$
   $= 56^\circ$.
2. $QRS$ is a straight line, so $\angle URQ = 180^\circ - 56^\circ$
   $= 124^\circ$.
3. In triangle $QUR$: $\angle RQU = 180^\circ - 124^\circ - 31^\circ$
   $= 25^\circ$.

(b)(i)

1. The radius is 7 cm.
2. The arc is $\frac37$ of the circumference: $\frac37 \times 2 \times \frac{22}{7} \times 7 \approx 18.86$ cm.
3. Perimeter of the sector $= 18.86 + 7 + 7 \approx 32.9$ cm.

(ii)

1. The sector is the same fraction $\frac37$ of the circle's area: $\frac37 \times \frac{22}{7} \times 7^2 = 66.0\text{ cm}^2$.

1. Draw the wall vertical and the ground horizontal, with the 10 m ladder at $70^\circ$ to the ground.
2. At first the top of the ladder is $10\sin 70^\circ \approx 9.397$ m up the wall, and the foot is $10\cos 70^\circ \approx 3.420$ m from the wall.

(a)

1. The top slips 4 m down, to $9.397 - 4 = 5.397$ m.
2. The ladder is still 10 m long: $\sin\theta = \frac{5.397}{10} = 0.5397$, so $\theta \approx 32.7^\circ$, which is $33^\circ$ to two significant figures.

(b)

1. The foot is now $10\cos 32.7^\circ \approx 8.419$ m from the wall.
2. It slipped back $8.419 - 3.420 \approx 5.0$ m.

(a)

1. Consecutive terms of an A.P. have equal gaps: $(y + 3) - (y + 2) = (2y^2 + 1) - (y + 3)$.
2. Simplify each side: $1 = 2y^2 - y - 2$.
3. Move everything to one side: $2y^2 - y - 3 = 0$.
4. Factorise: $(2y - 3)(y + 1) = 0$, so $y = \frac32$ or $y = -1$.
5. (Check: $y = -1$ gives $1, 2, 3$ and $y = \frac32$ gives $3\frac12, 4\frac12, 5\frac12$.)

(b)(i)

1. Multiply row by column: $XY = \begin{pmatrix} 2a + 4c & 2b + 4d \\ 8a + 6c & 8b + 6d \end{pmatrix}$
   $= \begin{pmatrix} 12 & 16 \\ 8 & 10 \end{pmatrix}$.
2. The first column gives $2a + 4c = 12$ and $8a + 6c = 8$.
3. Multiply the first by 4: $8a + 16c = 48$.
4. Take away the second: $10c = 40$, so $c = 4$, and $2a = 12 - 16 = -4$, so $a = -2$.
5. The second column gives $2b + 4d = 16$ and $8b + 6d = 10$.
6. Multiply the first by 4: $8b + 16d = 64$.
7. Take away the second: $10d = 54$, so $d = \frac{27}{5}$, and $2b = 16 - \frac{108}{5} = -\frac{28}{5}$, so $b = -\frac{14}{5}$.
8. So $Y = \begin{pmatrix} -2 & -2\frac45 \\ 4 & 5\frac25 \end{pmatrix}$.

(ii)

1. $|Y| = ad - bc$
   $= (-2) \times \frac{27}{5} - \left(-\frac{14}{5}\right) \times 4$
   $= -\frac{54}{5} + \frac{56}{5}$
   $= \frac25$.

(a)

1. In right-angled triangle $ADC$ (the right angle at $D$), $AC = 12.7$ cm is the hypotenuse: $|AD| = 12.7\cos 35^\circ \approx 10.403$ cm and $|CD| = 12.7\sin 35^\circ \approx 7.284$ cm.
2. The third angle of that triangle is $\angle ACD = 90^\circ - 35^\circ$
   $= 55^\circ$, so $\angle BCD = 120^\circ - 55^\circ$
   $= 65^\circ$.
3. In right-angled triangle $CDB$: $|DB| = |CD|\tan 65^\circ$
   $\approx 7.284 \times 2.1445$
   $\approx 15.621$ cm.
4. So $|AB| = |AD| + |DB|$
   $\approx 10.403 + 15.621$
   $= 26.0$ cm.

(b)

1. Integrate term by term: $\int (x^2 - 4)\,dx = \frac{x^3}{3} - 4x$.
2. Between the limits: $\left[\frac{x^3}{3} - 4x\right]_0^a = \frac{a^3}{3} - 4a - 0$
   $= \frac{a^3}{3} - 4a$.

(c)

1. Differentiate term by term: $\frac{dy}{dx} = 3x^2 - 2$.

(a)(i)

1. Fill the Venn diagram from the middle outwards.
2. All three: 5.
3. French and Mathematics only: $9 - 5 = 4$.
4. Mathematics and Economics only: $8 - 5 = 3$.
5. French and Economics only: $7 - 5 = 2$.
6. Mathematics only: $18 - 5 - 4 - 3 = 6$.
7. French only: $16 - 5 - 4 - 2 = 5$.
8. Economics only: $17 - 5 - 3 - 2 = 7$.
9. Outside the circles: 4.

(ii)

1. Total: $5 + 4 + 3 + 2 + 6 + 5 + 7 + 4 = 36$ students.

(b)

1. Let the pole be $h$ m high and its first shadow $x$ m long.
2. With the sun at $58^\circ$: $h = x\tan 58^\circ$.
3. With the sun at $36^\circ$ the shadow is $x + 90$: $h = (x + 90)\tan 36^\circ$.
4. Set them equal: $x\tan 58^\circ = x\tan 36^\circ + 90\tan 36^\circ$, so $x = \frac{90\tan 36^\circ}{\tan 58^\circ - \tan 36^\circ}$
   $= \frac{65.39}{0.8738}$
   $\approx 74.83$ m.
5. Then $h = 74.83\tan 58^\circ \approx 119.7$ m, which is 120 m to three significant figures.

(a)

1. Write $\frac1t$ as $t^{-1}$ and differentiate term by term: $\frac{dy}{dt} = 12t^3 - 12t^2 + 4t - (-1)t^{-2}$
   $= 12t^3 - 12t^2 + 4t + \frac{1}{t^2}$.
2. The constant $-5$ gives 0.

(b)

1. $P(\text{owns a car}) = \frac16$, so $P(\text{doesn't}) = \frac56$.

(i)

1. Each owns one: $\frac16 \times \frac16 = \frac{1}{36}$.

(ii)

1. Only one owns one: the first does and the second doesn't, $\frac16 \times \frac56$.
2. Or the other way round, $\frac56 \times \frac16$.
3. Together: $\frac{5}{36} + \frac{5}{36} = \frac{10}{36}$
   $= \frac{5}{18}$.

(c)

1. A difference of logs is the log of a quotient: $\log_{10}\frac{16x + 2}{4x + 2} = \log_{10}(2x + 1)$, so $\frac{16x + 2}{4x + 2} = 2x + 1$.
2. Multiply out: $16x + 2 = (2x + 1)(4x + 2)$
   $= 8x^2 + 8x + 2$, so $8x^2 - 8x = 0$, $8x(x - 1) = 0$, and $x = 0$ or $x = 1$.
3. Check both in the original: at $x = 0$, $\log 2 - \log 2 = 0 = \log 1$ ✓.
4. At $x = 1$, $\log 18 - \log 6 = \log 3$ ✓.
5. Both are solutions.

(a)

1. $r : h = 2 : 3$, so $r = \frac23 \times 8.1 = 5.4$ cm.
2. Volume $= \frac13 \times \frac{22}{7} \times 5.4^2 \times 8.1$
   $\approx 247.4\text{ cm}^3$.

(b)

1. Let the total be $y$.
2. The two parts are $\frac38 y$ at $8\%$ and $\frac58 y$ at $10\%$, so one year's interest is $\frac38 y \times 0.08 + \frac58 y \times 0.10 = 0.03y + 0.0625y$
   $= 0.0925y$.
3. $0.0925y = 92\,000$, so $y \approx ₦994,595$ to the nearest naira.

(a)

1. Arc length $= \frac{120}{360} \times 2\pi r$
   $= \frac13 \times 2 \times \frac{22}{7} \times 35$
   $= \frac{220}{3}$
   $\approx 73.33$ cm.
2. The sector's perimeter is the arc plus two radii: $73.33 + 70 = 143.3$ cm.

(b)

1. The perpendicular from $O$ bisects the chord and the $120^\circ$ angle.
2. Half the chord is $35\sin 60^\circ \approx 30.31$ cm, so $|MN| = 2 \times 30.31 \approx 60.6$ cm.

(c)

1. Area of the sector $= \frac13 \times \frac{22}{7} \times 35^2$
   $= \frac{3850}{3}$
   $\approx 1283.33\text{ cm}^2$.
2. Area of triangle $MON = \frac12 \times 35 \times 35 \times \sin 120^\circ$
   $\approx 530.44\text{ cm}^2$.
3. Minor segment $=$ sector $-$ triangle $\approx 1283.33 - 530.44$
   $= 752.9\text{ cm}^2$.

(a)

1. $\cos 60^\circ = \frac12$, so $2y - 16 = 60$.
2. Then $2y = 76$ and $y = 38$.

(b)(i)

1. (α) Draw $|BC| = 9$ cm, construct an angle of $75^\circ$ at $B$ (a $60^\circ$ angle plus half of the $30^\circ$ next to it), and mark $A$ on its arm with $|AB| = 8$ cm.
2. Join $AC$.

(β)

1. $|PA| = |PB|$, so $P$ lies on the perpendicular bisector of $AB$: construct it.
2. With centre $A$ and radius 4.5 cm, draw an arc to cut the bisector inside the triangle at $P$.

(ii)

1. Measure: $|CP| \approx 6.8$ cm and $\angle ACB \approx 48^\circ$.
2. (By calculation, $|AC| \approx 10.38$ cm, $\angle ACB \approx 48.1^\circ$ and $|CP| \approx 6.84$ cm.)

(a)

1. Let the base be $x$ cm, so the height is $(x + 3)$ cm.
2. The area is $\frac12 \times \text{base} \times \text{height}$: $\frac12 x(x + 3) = 54$.
3. Multiply by 2: $x^2 + 3x = 108$, so $x^2 + 3x - 108 = 0$.
4. Factorise: $(x + 12)(x - 9) = 0$.
5. A length can't be negative, so $x = 9$.
6. The base is $9$ cm (and the height is 12 cm: $\frac12 \times 9 \times 12 = 54$ ✓).

(b)(i)

1. The angle at the centre is twice the angle at the circumference: $\angle XZY = \frac12 \times 130^\circ$
   $= 65^\circ$.
2. $OX = OY$ (radii), so $\angle OXY = \angle OYX$
   $= \frac{180^\circ - 130^\circ}{2}$
   $= 25^\circ$.
3. In triangle $XYZ$: $\angle ZXY = 38^\circ + 25^\circ$
   $= 63^\circ$, so $\angle XYZ = 180^\circ - 63^\circ - 65^\circ$
   $= 52^\circ$.
4. Then $\angle OYZ = 52^\circ - 25^\circ$
   $= 27^\circ$.

(ii)

1. $OY = OZ$ (radii), so $\angle OZY = \angle OYZ = 27^\circ$, and $\angle ZOY = 180^\circ - 2 \times 27^\circ$
   $= 126^\circ$.