WAEC · 2023 · May/June · General Maths · Paper 2

Question 1

(a)
A car travels a distance of $112\text{ km}$ at an average speed of $70\text{ km/h}$ . It then travels further for $60\text{ km}$ at an average speed of $50\text{ km/h}$ . Calculate, for the entire journey, the total time taken (in hours).
(b)
If $\dfrac xy = 2$ and $\dfrac yz = 3$ , find the value of $\dfrac{x + y}{y + z}$ .

Worked solution (try it first)

(a)

Time $= \frac{\text{distance}}{\text{speed}}$ for each part of the journey.
First part: $\frac{112}{70} = 1.6$ hours.
Second part: $\frac{60}{50} = 1.2$ hours.
Total time $= 1.6 + 1.2 = 2.8$ hours (2 hours 48 minutes).

(b)

From $\frac yz = 3$ , $y = 3z$ .
From $\frac xy = 2$ , $x = 2y = 6z$ .
So $\frac{x + y}{y + z} = \frac{6z + 3z}{3z + z}$
$= \frac{9z}{4z}$
$= \frac94$
$= 2\frac14$ .

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Question 2

In a football match, the tickets for children and adults were sold at D3.00 and D5.00 respectively. 400 people attended the match and D1,700.00 was collected in ticket sales.

(a)
How many tickets were sold to adults?
(b)
Mr. Sonko sold 250 tickets. If 175 of the tickets were for adults, how much sales (in D) did he make altogether?

Worked solution (try it first)

(a)

Let $c$ children's tickets and $a$ adults' tickets be sold.
How many: $c + a = 400$ (1).
How much: $3c + 5a = 1700$ (2).
Multiply (1) by 3: $3c + 3a = 1200$ (3).
Take (3) from (2): $2a = 500$ , so $a = 250$ . 250 tickets were sold to adults.

(b)

Of his 250 tickets, 175 were for adults, so $250 - 175 = 75$ were for children.
Sales $= 75 \times 3 + 175 \times 5$
$= 225 + 875$
$=$ D1,100.00.

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In the diagram, $PQR$ is an equilateral triangle of side $18\text{ cm}$ . $M$ is the midpoint of $\overline{QR}$ . An arc of a circle with centre $P$ touches $\overline{QR}$ at $M$ and meets $\overline{PQ}$ at $A$ and $\overline{PR}$ at $B$ . Calculate, correct to two decimal places, the area of the shaded region. $\left[\text{Take } \pi = \frac{22}{7}\right]$

(a)
Area of the shaded region (cm²)

Worked solution (try it first)

(a)

The shaded region is the triangle minus the sector $PAMB$ .
The arc touches $QR$ at $M$ , so the radius is $PM$ , the height of the triangle: $PM = \sqrt{18^2 - 9^2}$
$= \sqrt{243}$
$= 9\sqrt3$
$\approx 15.59$ cm.
The angle at $P$ is $60^\circ$ (equilateral triangle).
Sector $= \frac{60}{360} \times \frac{22}{7} \times 243$
$\approx 127.29\text{ cm}^2$ .
Triangle $PQR = \frac12 \times 18 \times 9\sqrt3$
$\approx 140.30\text{ cm}^2$ .
Shaded area $\approx 140.30 - 127.29$
$= 13.01\text{ cm}^2$ .

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Question 4

Circle geometry

In the diagram, $P$ , $Q$ , $R$ and $S$ are points on the circle centre $K$ . $\overline{KR}$ is a bisector of $\angle SRQ$ , $\angle KSP = 41^\circ$ and $\angle SKR = 80^\circ$ . Find:

(a)
$\angle RQP$ ;
(b)
$\angle SPQ$ .

Worked solution (try it first)

(a)

$KR = KS$ (radii), so triangle $KRS$ is isosceles and $\angle KRS = \angle KSR$
$= \frac{180^\circ - 80^\circ}{2}$
$= 50^\circ$ .
So $\angle RSP = \angle RSK + \angle KSP$
$= 50^\circ + 41^\circ$
$= 91^\circ$ .
$PQRS$ is a cyclic quadrilateral, so $\angle RQP = 180^\circ - 91^\circ$
$= 89^\circ$ .

(b)

$KR$ bisects $\angle SRQ$ , so $\angle SRQ = 2 \times 50^\circ$
$= 100^\circ$ .
Opposite angles of the cyclic quadrilateral: $\angle SPQ = 180^\circ - 100^\circ$
$= 80^\circ$ .

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Question 5

Elevation, depression & bearings

A boy stands at a point $M$ on the same horizontal level as the foot, $T$ , of a vertical building. He observes an object on the top, $P$ , of the building at an angle of elevation of $66^\circ$ . He moves directly backwards to a new point $C$ and observes the same object at an angle of elevation of $53^\circ$ . If $|MT| = 50\text{ m}$ :

(a)
Illustrate the information in a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: draw the building $PT$ standing vertically on level ground, with a right angle at $T$ . $M$ is 50 m from $T$ , with the angle of elevation of $P$ equal to $66^\circ$ . $C$ is further back on the same line, with elevation $53^\circ$ (smaller, because it is further away). Mark the height $h$ and the unknown $|MC|$ .
(b)(i)
Calculate, correct to one decimal place, the height of the building;
(b)(ii)
$|MC|$ .

Worked solution (try it first)

(a)

Draw the building $TP$ upright, with $M$ 50 m from its foot and $C$ further back on the same line.
The angles of elevation of $P$ are $66^\circ$ from $M$ and $53^\circ$ from $C$ .

(b)(i)

In triangle $MTP$ : $|PT| = 50\tan 66^\circ$
$\approx 50 \times 2.2460$
$\approx 112.3$ m.

(ii)

In triangle $CTP$ : $\tan 53^\circ = \frac{112.3}{|CT|}$ , so $|CT| = \frac{112.3}{1.3270} \approx 84.63$ m.
So $|MC| = 84.63 - 50 \approx 34.6$ m.

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Question 6 ✱

(a)(i)
$M = \{n : 2n - 3 \le 37\}$ , where $n$ is a counting number. Write down all the elements in $M$ .
Show the answer
$\{1, 2, 3, \dots, 20\}$
(a)(ii)
If a number is selected at random from $M$ , what is the probability that it is a: ( $\alpha$ ) multiple of 3; ( $\beta$ ) factor of 10? Enter both.
Separate values with commas, e.g. 3, −2
(b)
A shop owner gave an end-of-year bonus to two of his attendants, Kontor and Gapson, in the ratio of their ages. Kontor’s age is one and a half times that of Gapson, who is 20 years old. If Kontor received Le 200,000.00, find: (i) the total amount shared; (ii) Gapson’s share.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Solve the inequality: $2n - 3 \le 37$ , so $2n \le 40$ and $n \le 20$ .
Counting numbers start at 1, so $M = \{1, 2, 3, \ldots, 20\}$ .

(ii)

$M$ has 20 members.
( $\alpha$ ) Multiples of 3: $3, 6, 9, 12, 15, 18$ , which is 6 numbers, so $P = \frac{6}{20} = \frac{3}{10}$ .
( $\beta$ ) Factors of 10: $1, 2, 5, 10$ , which is 4 numbers, so $P = \frac{4}{20} = \frac15$ .

(b)(i)

Kontor is $1\frac12 \times 20 = 30$ years old, so the ratio of their ages is $30 : 20 = 3 : 2$ .
Kontor's 3 parts are Le 200,000, so 1 part is $\frac{200\,000}{3}$ and the 5 parts total $\frac{200\,000}{3} \times 5 \approx \text{Le } 333,333.33$ .

(ii)

Gapson's 2 parts: $\frac{200\,000}{3} \times 2 \approx \text{Le } 133,333.33$ .

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Question 7

(a)
The sum of three numbers is 81. The second number is twice the first. Given that the third number is 6 more than the second, find the numbers.
Separate values with commas, e.g. 3, −2
(b)
Given the points $P(3, 5)$ and $Q(-5, 7)$ on the Cartesian plane such that $R$ is the midpoint of $\overline{PQ}$ , find the equation of the line that passes through $R$ and is perpendicular to $\overline{PQ}$ .

Worked solution (try it first)

(a)

Let the first number be $x$ .
The second is twice the first, $2x$ , and the third is 6 more than the second, $2x + 6$ .
They add up to 81: $x + 2x + (2x + 6) = 81$ .
So $5x + 6 = 81$ , $5x = 75$ and $x = 15$ .
The numbers are 15, 30 and 36.

(b)

Midpoint $R = \left(\frac{3 + (-5)}{2}, \frac{5 + 7}{2}\right)$
$= (-1, 6)$ .
Gradient of $PQ = \frac{7 - 5}{-5 - 3}$
$= \frac{2}{-8}$
$= -\frac14$ .
A perpendicular line has gradient $m$ with $m \times \left(-\frac14\right) = -1$ , so $m = 4$ .
The line through $R(-1, 6)$ with gradient 4: $y - 6 = 4(x + 1)$ .
So $y = 4x + 10$ .

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Question 8

Quadratics & their graphs

(a)
Copy and complete the table of values for $y = 2x^2 - x - 4$ for $-3 \le x \le 3$ .

$x$ −3 −2 −1 0 1 2 3

$y$ 17 −4

Model answer

$x$ −3 −2 −1 0 1 2 3

$y$ 17 6 −1 −4 −3 2 11

For example, at $x = -2$ : $y = 2(4) + 2 - 4 = 6$ .
(b)
Using a scale of 2 cm to 1 unit on the $x$ -axis and 2 cm to 2 units on the $y$ -axis, draw the graph of $y = 2x^2 - x - 4$ for $-3 \le x \le 3$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 1 unit across, 2 cm to 2 units up.

For (c): (i) roots $x \approx −1.2 and 1.7$ ; (ii) $y$ increases as $x$ increases to the right of the lowest point, $x > 0.25$ ; (iii) the minimum point is $(0.25, -4.125)$ .
(c)
Use the graph to find the: (i) roots of the equation $2x^2 - x - 4 = 0$ ; (ii) values of $x$ for which $y$ increases as $x$ increases; (iii) minimum point of $y$ . Enter the two roots for (i).
Separate values with commas, e.g. 3, −2

$x$	−3	−2	−1	0	1	2	3
$y$	17	6	−1	−4	−3	2	11

Try it on a graph

Plot the curves, move them, and read values off the graph.

Worked solution (try it first)

(a)

Put each $x$ into $y = 2x^2 - x - 4$ .
For $x = -2$ : $8 + 2 - 4 = 6$ .
For $x = -1$ : $2 + 1 - 4 = -1$ .
For $x = 1$ : $2 - 1 - 4 = -3$ .
For $x = 2$ : $8 - 2 - 4 = 2$ .
For $x = 3$ : $18 - 3 - 4 = 11$ .
The row is $17, 6, -1, -4, -3, 2, 11$ .

(b)

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

(c)(i)

The roots are where the curve crosses the $x$ -axis: $x \approx -1.2$ and $x \approx 1.7$ .
(Exactly, $\frac{1 \pm \sqrt{33}}{4}$ , about $-1.19$ and $1.69$ .)

(ii)

The lowest point of the curve is halfway between the roots, at $x = 0.25$ .
To its right the curve rises, so $y$ increases as $x$ increases for $x > 0.25$ (on this graph, $0.25 < x \le 3$ ).

(iii)

At $x = 0.25$ , $y = 2(0.0625) - 0.25 - 4 = -4.125$ .
So the minimum point is about $(0.25, -4.1)$ .

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Question 9

Height (m)	3	4	5	6	7	8
Number of trees	4	6	4	5	6	2

The table shows the heights of teak trees harvested by a farmer.

(a)
Find the median height.
(b)
Calculate, correct to one decimal place, the (i) mean; (ii) standard deviation.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

There are $4 + 6 + 4 + 5 + 6 + 2 = 27$ trees, so the median is the $\frac{27 + 1}{2} = 14$ th height.
Running totals: $4, 10, 14, \ldots$ .
The 11th to 14th trees are 5 m, so the median height is 5 m.

(b)(i)

$\sum fx = 12 + 24 + 20 + 30 + 42 + 16$
$= 144$ , so the mean is $\frac{144}{27} \approx 5.3$ m.

(ii)

$\sum fx^2 = 4(9) + 6(16) + 4(25) + 5(36) + 6(49) + 2(64)$
$= 36 + 96 + 100 + 180 + 294 + 128$
$= 834$ .
Standard deviation $= \sqrt{\frac{\sum fx^2}{\sum f} - \bar x^2}$
$= \sqrt{\frac{834}{27} - \left(\frac{144}{27}\right)^2}$
$= \sqrt{30.889 - 28.444}$
$= \sqrt{2.444}$
$\approx 1.6$ m.

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Question 10

In a town, Chief $X$ resides $60\text{ m}$ away on a bearing of $057^\circ$ from the palace $P$ , while Chief $Y$ resides on a bearing of $150^\circ$ from the same palace $P$ . The residences of $X$ and $Y$ are $180\text{ m}$ apart.

(a)
Illustrate the information in a diagram.
Model answer
A clear sketch is enough (it need not be to scale), but it must show every given fact: draw a north line at $P$ , with $X$ on bearing $057^\circ$ (60 m) and $Y$ on bearing $150^\circ$ , and $XY = 180$ m. The angle $XPY = 150^\circ - 57^\circ = 93^\circ$ . Add a north line at $Y$ for the bearing of $X$ from $Y$ in (b). $PY$ is unknown (it works out to about 167 m).
(b)
Find, correct to three significant figures, the: (i) bearing of $X$ from $Y$ ; (ii) distance between $P$ and $Y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw north at the palace $P$ .
Chief $X$ is 60 m on $057^\circ$ and Chief $Y$ on $150^\circ$ .
Join $X$ and $Y$ (180 m).
The angle at $P$ is $\angle XPY = 150^\circ - 57^\circ$
$= 93^\circ$ .

(b)(i)

Sine rule for the angle at $Y$ : $\sin\angle PYX = \frac{60\sin 93^\circ}{180}$
$\approx 0.3329$ , so $\angle PYX \approx 19.4^\circ$ .
At $Y$ , the direction back to $P$ is $150^\circ + 180^\circ = 330^\circ$ , and $X$ is $19.4^\circ$ further round clockwise.
Bearing of $X$ from $Y$ $= 330^\circ + 19.4^\circ$
$\approx 349^\circ$ .

(ii)

The third angle is $\angle PXY = 180^\circ - 93^\circ - 19.4^\circ$
$= 67.6^\circ$ .
Sine rule: $|PY| = \frac{180\sin 67.6^\circ}{\sin 93^\circ}$
$\approx 167$ m.

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Question 11

(a)
Two regular polygons $P$ and $Q$ are such that the number of sides of $P$ is twice the number of sides of $Q$ . The difference between the exterior angles of $Q$ and $P$ is $45^\circ$ . Find the number of sides of $P$ .
(b)
The area of a semi-circle is $32\pi\text{ cm}^2$ . Find, in terms of $\pi$ , the circumference of the semi-circle.

Worked solution (try it first)

(a)

Let $Q$ have $n$ sides, so $P$ has $2n$ .
The exterior angle of a regular polygon is $\frac{360^\circ}{\text{number of sides}}$ : $Q$ 's is $\frac{360}{n}$ and $P$ 's is $\frac{360}{2n} = \frac{180}{n}$ .
Their difference is $45^\circ$ : $\frac{360}{n} - \frac{180}{n} = \frac{180}{n}$
$= 45$ , so $n = 4$ .
$P$ has $2 \times 4 = 8$ sides.

(b)

$\frac12\pi r^2 = 32\pi$ , so $r^2 = 64$ and $r = 8$ cm.
The boundary of the semicircle is the curved half, $\pi r = 8\pi$ , plus the diameter, $2r = 16$ : $(8\pi + 16)$ cm.

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Question 12

(a)
In the diagram, $P$ , $Q$ , $R$ and $S$ are points on the circle with centre $O$ ; $QR \parallel OS$ , $\angle QOR = 2m$ , $\angle QPR = n$ and $\angle SOR = 54^\circ$ . Find the values of $m$ and $n$ .
Separate values with commas, e.g. 3, −2
(b)
The length of a rectangle is $4\text{ cm}$ more than its width. If the perimeter is $40\text{ cm}$ , find its area.

Worked solution (try it first)

(a)

$QR \parallel OS$ , so $\angle ORQ = \angle SOR = 54^\circ$ (alternate angles).
$|OQ| = |OR|$ (radii), so triangle $OQR$ is isosceles and $\angle OQR = \angle ORQ = 54^\circ$ .
Angles in triangle $OQR$ : $\angle QOR = 180^\circ - 54^\circ - 54^\circ$
$= 72^\circ$ .
So $2m = 72^\circ$ and $m = 36^\circ$ .
The angle at the centre is twice the angle at the circumference on the same arc $QR$ , so $n = \frac12 \times 72^\circ = 36^\circ$ .

(b)

Let the width be $w$ cm.
The length is then $(w + 4)$ cm.
Perimeter: $2(w + w + 4) = 40$ , so $4w + 8 = 40$ and $w = 8$ .
The length is $8 + 4 = 12$ cm, so the area is $12 \times 8 = 96$ .
The area is $96\text{ cm}^2$ .

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Question 13

(a)
In the diagram, the radius of the sector of the circle centre $O$ is $7\text{ cm}$ and $\angle MON = 60^\circ$ ; $NT \perp OM$ . Find, correct to one decimal place, the area of the shaded portion. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
The $x$ and $y$ intercepts of a straight line are $-\frac34$ and $\frac27$ respectively. Find the equation of the line.

Worked solution (try it first)

(a)

The shaded portion is sector $MON$ minus the right-angled triangle $ONT$ .
Sector $= \frac{60}{360} \times \frac{22}{7} \times 7^2$
$\approx 25.67\text{ cm}^2$ .
In triangle $ONT$ (right angle at $T$ ), $OT = 7\cos 60^\circ = 3.5$ cm and $NT = 7\sin 60^\circ \approx 6.062$ cm, so its area is $\frac12 \times 3.5 \times 6.062 \approx 10.61\text{ cm}^2$ .
Shaded area $\approx 25.67 - 10.61 = 15.1\text{ cm}^2$ .

(b)

The line passes through $\left(-\frac34, 0\right)$ and $\left(0, \frac27\right)$ .
Gradient $= \frac{2/7 - 0}{0 - (-3/4)}$
$= \frac{2}{7} \times \frac43$
$= \frac{8}{21}$ , so $y = \frac{8}{21}x + \frac27$ .
Multiply by 21: $21y = 8x + 6$ .

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