WAEC · 2020 · May/June · General Maths · Paper 2

Question 1

(a)
If $A = \{\text{multiples of }2\}$ , $B = \{\text{multiples of }3\}$ and $C = \{\text{factors of }6\}$ are subsets of $\mu = \{x : 1 \le x \le 10\}$ , find $A' \cap B' \cap C'$ .
Show the answer
$\{5, 7\}$
(b)
Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?

Worked solution (try it first)

(a)

List the sets in $\mu = \{1, 2, \ldots, 10\}$ : $A = \{2, 4, 6, 8, 10\}$ , $B = \{3, 6, 9\}$ , $C = \{1, 2, 3, 6\}$ .
Complements: $A' = \{1, 3, 5, 7, 9\}$ , $B' = \{1, 2, 4, 5, 7, 8, 10\}$ , $C' = \{4, 5, 7, 8, 9, 10\}$ .
The numbers in all three: $A' \cap B' \cap C' = \{5, 7\}$ .

(b)

Bought in bulk, each ticket costs $\frac{80}{5} = 16$ dollars.
Each person saves $18.50 - 16.00 = 2.50$ dollars, that is $2.50.

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Question 2

(a)
Given that $P = \left(\dfrac{rk}{Q} - ms\right)^{\frac23}$ , (i) make $Q$ the subject of the relation; (ii) find, correct to two decimal places, the value of $Q$ when $P = 3$ , $m = 15$ , $s = 0.2$ , $k = 4$ and $r = 10$ .
(b)
Given that $\dfrac{x + 2y}{5} = x - 2y$ , find $x : y$ .
Show the answer
$3 : 1$

Worked solution (try it first)

(a)(i)

$Q$ is inside a bracket raised to the power $\frac23$ .
Undo it by raising both sides to the power $\frac32$ : $P^{\frac32} = \frac{rk}{Q} - ms$ .
Add $ms$ to both sides: $P^{\frac32} + ms = \frac{rk}{Q}$ .
Multiply both sides by $Q$ and divide by $\left(P^{\frac32} + ms\right)$ : $Q = \frac{rk}{P^{\frac32} + ms}$ .

(ii)

$P^{\frac32} = 3^{1.5}$
$= \sqrt{27}$
$\approx 5.196$ and $ms = 15 \times 0.2 = 3$ .
So $Q = \frac{10 \times 4}{5.196 + 3}$
$= \frac{40}{8.196}$
$\approx 4.880$ .
Correct to two decimal places: $Q = 4.88$ .

(b)

Multiply both sides by 5: $x + 2y = 5(x - 2y) = 5x - 10y$ .
Collect terms: $12y = 4x$ , so $x = 3y$ .
Divide both sides by $y$ : $\frac xy = 3$ .
So $x : y = 3 : 1$ .

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Question 3

(a)
In the diagram, $O$ is the centre of the circle $ABCDE$ , $BE$ and $AD$ are diameters, $|BC| = |CD|$ and $\angle BCD = 108^\circ$ . Find $\angle CDE$ .
(b)
Given that $\tan x = \sqrt3$ , $0^\circ \le x \le 90^\circ$ , evaluate $\dfrac{(\cos x)^2 - \sin x}{(\sin x)^2 + \cos x}$ .

Worked solution (try it first)

(a)

In triangle $BCD$ , $|BC| = |CD|$ , so the base angles are equal: $\angle CDB = \frac{180^\circ - 108^\circ}{2}$
$= 36^\circ$ .
$BE$ is a diameter, so $\angle BDE = 90^\circ$ (angle in a semicircle).
So $\angle CDE = \angle CDB + \angle BDE$
$= 36^\circ + 90^\circ$
$= 126^\circ$ .

(b)

$\tan x = \sqrt3$ with $x$ acute, so $x = 60^\circ$ , $\sin x = \frac{\sqrt3}{2}$ and $\cos x = \frac12$ .
Top: $\left(\frac12\right)^2 - \frac{\sqrt3}{2} = \frac14 - \frac{\sqrt3}{2}$
$= \frac{1 - 2\sqrt3}{4}$ .
Bottom: $\left(\frac{\sqrt3}{2}\right)^2 + \frac12 = \frac34 + \frac12$
$= \frac54$ .
So the value is $\frac{1 - 2\sqrt3}{4} \times \frac45 = \frac{1 - 2\sqrt3}{5}$ .

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Question 4

Solid mensuration

The total surface area of a cone of slant height $l\text{ cm}$ and base radius $r\text{ cm}$ is $224\pi\text{ cm}^2$ . If $r : l = 2 : 5$ , find: $\left[\text{Take }\pi = \frac{22}{7}\right]$

(a)
correct to one decimal place, the value of $r$ ;
(b)
correct to the nearest whole number, the volume of the cone.

Worked solution (try it first)

(a)

$r : l = 2 : 5$ , so $l = \frac52 r$ .
Total surface area $= \pi r(r + l)$
$= \pi r\left(r + \frac52 r\right)$
$= \frac72\pi r^2$ .
So $\frac72\pi r^2 = 224\pi$ .
The $\pi$ cancels: $r^2 = 64$ and $r = 8.0$ cm.

(b)

$l = \frac52 \times 8 = 20$ cm, so $h = \sqrt{20^2 - 8^2}$
$= \sqrt{336}$
$\approx 18.33$ cm.
Volume $= \frac13 \times \frac{22}{7} \times 8^2 \times 18.33$
$\approx 1229\text{ cm}^3$ .

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Question 5

Probability

Number	1	2	3	4	5	6
Outcomes	32	$m$	25	40	28	45

A die was rolled a number of times. The outcomes are as shown in the table. If the probability of obtaining 2 is 0.15, find the:

(a)
value of $m$ ;
(b)
number of times the die was rolled;
(c)
probability of obtaining an even number.

Worked solution (try it first)

(a)

The total number of rolls is $32 + m + 25 + 40 + 28 + 45 = 170 + m$ .
So $P(2) = \frac{m}{170 + m} = 0.15$ .
Then $m = 0.15(170 + m) = 25.5 + 0.15m$ , so $0.85m = 25.5$ and $m = 30$ .

(b)

The die was rolled $170 + 30 = 200$ times.

(c)

Even numbers are 2, 4 and 6: $30 + 40 + 45 = 115$ outcomes.
$P(\text{even}) = \frac{115}{200}$
$= \frac{23}{40}$ .

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Question 6

Trigonometric graphs

(a)

Copy and complete the table of values for the relation $y = 3\sin2x$ .

$x$	$0^\circ$	$15^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$75^\circ$	$90^\circ$	$105^\circ$	$120^\circ$	$135^\circ$	$150^\circ$
$y$	$0.0$					$1.5$					$-2.6$

Model answer

$x$	0°	15°	30°	45°	60°	75°	90°	105°	120°	135°	150°
$y$	0.0	1.5	2.6	3.0	2.6	1.5	0.0	−1.5	−2.6	−3.0	−2.6

For example, at $x = 15^\circ$ : $y = 3\sin30^\circ = 1.5$ .

(b)
Using a scale of 2 cm to $15^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of $y = 3\sin2x$ for $0^\circ \le x \le 150^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The curve reaches 3 at $45^\circ$ , crosses the axis at $90^\circ$ and reaches $-3$ at $135^\circ$ .

For (c): (i) $3\sin 2x + 2 = 0$ is $y = -2$ : $x \approx 111^\circ$ . (ii) $\frac32\sin 2x = 0.25$ is $3\sin 2x = 0.5$ , so draw $y = 0.5$ : $x \approx 5^\circ$ and $85^\circ$ .
(c)(i)
Use the graph to find the truth set of $3\sin2x + 2 = 0$ .
(c)(ii)
Use the graph to find the truth set of $\frac32\sin2x = 0.25$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

x in degrees. The lines y = −2 and y = 0.5 give the truth sets.

Worked solution (try it first)

(a)

Double the angle first, then take the sine.
For example, $x = 30^\circ$ : $3\sin 60^\circ = 3 \times 0.866$
$\approx 2.6$ .
$x = 135^\circ$ : $3\sin 270^\circ = -3.0$ .
The full row is $0.0, 1.5, 2.6, 3.0, 2.6, 1.5, 0.0, -1.5, -2.6, -3.0, -2.6$ .

(b)

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

(c)(i)

$3\sin 2x + 2 = 0$ is $3\sin 2x = -2$ .
Draw the line $y = -2$ and read where it cuts the curve in this range: $x \approx 111^\circ$ .
The truth set is $\{x : x \approx 111^\circ\}$ .

(ii)

Multiply both sides of $\frac32\sin 2x = 0.25$ by 2 to match the graph: $3\sin 2x = 0.5$ .
Draw $y = 0.5$ and read where it crosses the curve: $x \approx 5^\circ$ and $x \approx 85^\circ$ .
The truth set is $\{x : x \approx 5^\circ, 85^\circ\}$ .

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Question 7

(a)
The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is $48\text{ m}$ and the base radius is $14\text{ m}$ . Calculate, correct to three significant figures, the surface area of the structure. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.

Worked solution (try it first)

(a)

The surface of the structure is the curved surface of the cone plus the curved surface of the hemisphere (the flat circle where they join is hidden).
Slant height of the cone: $l = \sqrt{48^2 + 14^2}$
$= \sqrt{2304 + 196}$
$= \sqrt{2500}$
$= 50\text{ m}$ .
Curved surface of the cone $= \pi r l$
$= \frac{22}{7} \times 14 \times 50$
$= 2200\text{ m}^2$ .
Curved surface of the hemisphere $= 2\pi r^2$
$= 2 \times \frac{22}{7} \times 196$
$= 1232\text{ m}^2$ .
Total $= 2200 + 1232 = 3432\text{ m}^2$ .
Correct to three significant figures: $3430\text{ m}^2$ .

(b)

Let Sesay's age now be $x$ years.
Since the ages add up to 100, Musah is $100 - x$ .
Five years ago they were $x - 5$ and $95 - x$ , and Musah was twice as old: $95 - x = 2(x - 5)$ .
Expand: $95 - x = 2x - 10$ .
So $3x = 105$ and $x = 35$ .
Sesay is 35 years old.

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Question 8

(a)
Ms. Maureen spent $\frac14$ of her monthly income at a shopping mall, $\frac13$ at an open market and $\frac25$ of the remaining amount at a mechanic workshop. If she had ₦225,000.00 left, find: (i) her monthly income; (ii) the amount spent at the open market.
Separate values with commas, e.g. 3, −2
(b)
The third term of an Arithmetic Progression (A.P.) is $4m - 2n$ . If the ninth term of the progression is $2m - 8n$ , find the common difference in terms of $m$ and $n$ .

Worked solution (try it first)

(a)

The mall and market take $\frac14 + \frac13 = \frac{7}{12}$ of her income, leaving $\frac{5}{12}$ .
The workshop takes $\frac25$ of that remainder: $\frac25 \times \frac{5}{12} = \frac16$ .
What's left is $\frac{5}{12} - \frac16 = \frac14$ of her income.

(i)

$\frac14$ of her income is ₦225,000, so her income is $4 \times 225\,000 = ₦900,000$ .

(ii)

Open market: $\frac13 \times 900\,000 = ₦300,000$ .

(b)

$T_3 = a + 2d = 4m - 2n$ and $T_9 = a + 8d = 2m - 8n$ .
Subtract: $6d = -2m - 6n$ , so $d = -\frac{2m + 6n}{6} = -\frac13(m + 3n)$ .

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Two cyclists $X$ and $Y$ leave town $Q$ at the same time. Cyclist $X$ travels at the rate of $5\text{ km/h}$ on a bearing of $049^\circ$ and cyclist $Y$ travels at the rate of $9\text{ km/h}$ on a bearing of $319^\circ$ .

(a)
Illustrate the information on a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: after 2 hours, $QX = 5 \times 2 = 10$ km on bearing $049^\circ$ and $QY = 9 \times 2 = 18$ km on bearing $319^\circ$ ( $41^\circ$ west of north). The angle between them at $Q$ is $49^\circ + 41^\circ = 90^\circ$ . Draw a north line at $Y$ as well, since part (b) asks for a bearing measured from $Y$ .
(b)
After travelling for two hours, calculate, correct to the nearest whole number, the: (i) distance between cyclists $X$ and $Y$ ; (ii) bearing of cyclist $X$ from $Y$ .
Separate values with commas, e.g. 3, −2
(c)
Find the average speed at which cyclist $X$ would cover the distance to $Y$ in 4 hours.

Worked solution (try it first)

(a)

Draw north at $Q$ .
After 2 hours, $X$ is $5 \times 2 = 10$ km from $Q$ on $049^\circ$ and $Y$ is $9 \times 2 = 18$ km from $Q$ on $319^\circ$ .
Join $X$ and $Y$ .

(b)(i)

The angle between the two directions at $Q$ is $049^\circ + (360^\circ - 319^\circ) = 49^\circ + 41^\circ$
$= 90^\circ$ .
So triangle $XQY$ is right-angled at $Q$ : $|XY| = \sqrt{10^2 + 18^2}$
$= \sqrt{424}$
$\approx 20.59$ km, which is 21 km to the nearest whole number.

(ii)

At $Y$ : $\tan\angle QYX = \frac{10}{18}$ , so $\angle QYX \approx 29.1^\circ$ .
At $Y$ , the direction back to $Q$ is $319^\circ - 180^\circ = 139^\circ$ , and $X$ is $29.1^\circ$ further round anticlockwise.
Bearing of $X$ from $Y$ $= 139^\circ - 29.1^\circ$
$\approx 110^\circ$ .

(c)

Average speed $= \frac{\text{distance}}{\text{time}}$
$= \frac{20.59}{4}$
$\approx 5.15$ km/h.

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Question 10

Dispersion & cumulative frequency

Marks (%)	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	7	11	17	20	29	34	30	25	21	6

The table shows the distribution of marks obtained by students in an examination.

(a)

Construct a cumulative frequency table for the distribution.

Model answer

Marks (%)	Frequency	Upper class boundary	Cumulative frequency
0–9	7	9.5	7
10–19	11	19.5	18
20–29	17	29.5	35
30–39	20	39.5	55
40–49	29	49.5	84
50–59	34	59.5	118
60–69	30	69.5	148
70–79	25	79.5	173
80–89	21	89.5	194
90–99	6	99.5	200

Each cumulative frequency is the running total of the frequencies; the last one equals the total, 200.

(b)
Draw the cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary of its class, starting from $(-0.5, 0)$ where the cumulative frequency is 0 and ending at $(99.5, 200)$ . Join the points with a smooth rising S-shaped curve (an ogive), not straight lines. Label both axes. Readings from a hand-drawn curve differ a little from person to person; examiners accept a small range, usually about ±1.

For (c): the median is the 100th mark. Read across from 100: about 54.2. The top 5% are the last 10 of 200, so read across from 190: the lowest distinction mark is about 87.1.
(c)
Using the curve, find, correct to one decimal place, the: (i) median mark; (ii) lowest mark for distinction if $5\%$ of the students passed with distinction.
Separate values with commas, e.g. 3, −2

Try it on a graph

The ogive, with the median (purple) and distinction (red) readings.

Worked solution (try it first)

(a)

The running totals of the frequencies, with the upper class boundaries:

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Upper boundary	9.5	19.5	29.5	39.5	49.5	59.5	69.5	79.5	89.5	99.5
Cumulative frequency	7	18	35	55	84	118	148	173	194	200

(b)

Plot each cumulative frequency at its upper boundary, starting from $(-0.5, 0)$ , and join the points with a smooth S-shaped curve.

(c)(i)

The median is at $\frac{200}{2} = 100$ .
Go across from 100 to the curve and down: about 54.5.
(Check: 100 lies between 84 at 49.5 and 118 at 59.5, and $49.5 + \frac{100 - 84}{34} \times 10 \approx 54.2$ .)

(ii)

The top $5\%$ got distinctions, so $95\%$ are below the lowest distinction mark: $0.95 \times 200 = 190$ .
Go across from 190: about 88.
(Check: $79.5 + \frac{190 - 173}{21} \times 10 \approx 87.6$ .)

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Question 11

(a)
In the diagram, $MNPQ$ is a circle with centre $O$ , and $MOQ$ is a diameter. $|MN| = |NP|$ and $\angle OMN = 50^\circ$ . Find: (i) $\angle MNP$ ; (ii) $\angle POQ$ .
Separate values with commas, e.g. 3, −2
(b)
Find the equation of the line which has the same gradient as $8y + 4x = 24$ and passes through the point $(-8, 12)$ .

Worked solution (try it first)

(a)(i)

$MNPQ$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $\angle NPQ = 180^\circ - \angle NMQ$
$= 180^\circ - 50^\circ$
$= 130^\circ$ .
$MQ$ is a diameter, so $\angle MPQ = 90^\circ$ (angle in a semicircle).
So $\angle MPN = 130^\circ - 90^\circ$
$= 40^\circ$ .
$|MN| = |NP|$ , so triangle $MNP$ is isosceles and $\angle NMP = \angle NPM = 40^\circ$ .
So $\angle MNP = 180^\circ - 40^\circ - 40^\circ$
$= 100^\circ$ .

(ii)

$\angle PMQ = \angle NMQ - \angle NMP$
$= 50^\circ - 40^\circ$
$= 10^\circ$ .
The angle at the centre is twice the angle at the circumference on the same arc $PQ$ : $\angle POQ = 2 \times 10^\circ = 20^\circ$ .

(b)

$8y + 4x = 24$ gives $y = -\frac12x + 3$ , so the gradient is $-\frac12$ .
Through $(-8, 12)$ : $y - 12 = -\frac12(x + 8)$ , so $y = -\frac12x - 4 + 12 = -\frac12x + 8$ .

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Question 12

(a)
In the diagram, $\overline{AB}$ is a tangent to the circle with centre $O$ and $COEB$ is a straight line. If $\overline{CD} \parallel \overline{AB}$ and $\angle ABE = 40^\circ$ , find $\angle ODE$ .
(b)
$ABCD$ is a parallelogram in which $|CD| = 7\text{ cm}$ , $|AD| = 5\text{ cm}$ and $\angle ADC = 125^\circ$ . (i) Illustrate the information in a diagram. (ii) Find, correct to one decimal place, the area of the parallelogram.
(c)
If $x = \frac12(1 - \sqrt2)$ , evaluate $(2x^2 - 2x)$ .

Worked solution (try it first)

(a)

$CD \parallel AB$ , so alternate angles are equal: $\angle OCD = \angle ABE = 40^\circ$ .
$OC = OD$ (radii), so $\angle ODC = \angle OCD = 40^\circ$ .
$CE$ is a diameter, so the angle in a semicircle is a right angle: $\angle CDE = 90^\circ$ .
Then $\angle ODE = 90^\circ - 40^\circ$
$= 50^\circ$ .

(b)(i)

Draw $DC$ (7 cm) along the bottom and $DA$ (5 cm) at $125^\circ$ to it, then complete the parallelogram.

(ii)

Area $= DC \times DA \times \sin\angle ADC$
$= 7 \times 5 \times \sin 125^\circ$
$\approx 35 \times 0.8192$
$\approx 28.7\text{ cm}^2$ .

(c)

$2x^2 - 2x = 2x(x - 1)$ .
Here $2x = 1 - \sqrt2$ and $x - 1 = \frac{1 - \sqrt2}{2} - 1$
$= \frac{-1 - \sqrt2}{2}$ .
So $2x(x - 1) = \frac{(1 - \sqrt2)(-1 - \sqrt2)}{2}$
$= \frac{-(1 - \sqrt2)(1 + \sqrt2)}{2}$
$= \frac{-(1 - 2)}{2}$
$= \frac12$ .

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Question 13

(a)
Using a ruler and a pair of compasses only: (i) construct (α) $\triangle ABC$ with $|AB| = 7\text{ cm}$ , $|AC| = 13.5\text{ cm}$ and $\angle ABC = 120^\circ$ ; (β) the locus $l_1$ of points equidistant from $A$ and $B$ ; (γ) the locus $l_2$ of points equidistant from $B$ and $C$ . (ii) Using $N$ , the point of intersection of $l_1$ and $l_2$ , as centre, draw a circle to pass through points $A$ , $B$ and $C$ .
Model answer
Draw $AB = 7$ cm, construct $120^\circ$ at $B$ , and with centre $A$ and radius $13.5$ cm cut the arm at $C$ ( $|BC| \approx 8.6$ cm). The perpendicular bisectors $l_1$ of $AB$ and $l_2$ of $BC$ meet at $N$ , outside the triangle beyond $AC$ , because the angle at $B$ is obtuse. The circle with centre $N$ and radius $|NA| \approx 7.8$ cm passes through $A$ , $B$ and $C$ .
(b)
Using the method of completing the square, solve $4x^2 - 4\sqrt3x + 3 = 0$ , leaving the answer in surd form.

Worked solution (try it first)

(a)(i)

(α)** Draw $AB = 7$ cm and construct $120^\circ$ at $B$ (two $60^\circ$ angles).
With centre $A$ and radius 13.5 cm, cut the arm at $C$ .
Join $AC$ .

(β)

$l_1$ : the perpendicular bisector of $AB$ .

(γ)

$l_2$ : the perpendicular bisector of $BC$ .

(ii)

$N$ , where $l_1$ and $l_2$ cross, is the same distance from $A$ , $B$ and $C$ .
With centre $N$ and radius $NA$ , draw the circle: it passes through $A$ , $B$ and $C$ .
(Its radius is $\frac{13.5}{2\sin 120^\circ} \approx 7.8$ cm.)

(b)

Divide by 4: $x^2 - \sqrt3x + \frac34 = 0$ , so $x^2 - \sqrt3x = -\frac34$ .
Add $\left(\frac{\sqrt3}{2}\right)^2 = \frac34$ to both sides: $\left(x - \frac{\sqrt3}{2}\right)^2 = 0$ .
So $x - \frac{\sqrt3}{2} = 0$ and $x = \frac{\sqrt3}{2}$ (a repeated root).

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