WAEC · 2018 · Private · General Maths · Paper 2

Question 1

(a)
Evaluate without using a calculator: $\left(\frac14 \times 9\frac17 + \frac25\left(\frac23 + \frac34\right)\right) \div \left(\frac25 - \frac14\right)$ .
(b)
A hunter walked $250\text{ m}$ from point $P$ to $Q$ on a bearing of $042^\circ$ . Calculate, correct to the nearest metre, the vertical distance he has moved (i.e. how far north).

Worked solution (try it first)

(a)

Work inside the big bracket first.
$\frac14 \times 9\frac17 = \frac14 \times \frac{64}{7}$
$= \frac{16}{7}$ .
$\frac23 + \frac34 = \frac{17}{12}$ , and $\frac25 \times \frac{17}{12} = \frac{17}{30}$ .
So the bracket is $\frac{16}{7} + \frac{17}{30} = \frac{480 + 119}{210}$
$= \frac{599}{210}$ .
The divisor is $\frac25 - \frac14 = \frac{3}{20}$ .
So the value is $\frac{599}{210} \times \frac{20}{3} = \frac{1198}{63}$
$= 19\frac{1}{63}$ .

(b)

Draw north at $P$ and the path $PQ$ , 250 m on $042^\circ$ .
Drop a perpendicular from $Q$ to the north line.
The distance moved north is the side next to the $42^\circ$ angle, with $PQ$ as the hypotenuse: $250\cos 42^\circ = 250 \times 0.7431$
$\approx 186$ m.

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Question 2

Linear & simultaneous equations

(a)
Musa is three years older than Manya. Seven years ago, Musa was twice as old as Manya. How old are they now?
Separate values with commas, e.g. 3, −2
(b)
In how many years will the sum of their ages be 45?

Worked solution (try it first)

(a)

Let Manya's age now be $x$ years.
Musa is three years older: $x + 3$ years.
Seven years ago they were $x - 7$ and $x + 3 - 7 = x - 4$ .
Musa was twice as old as Manya: $x - 4 = 2(x - 7)$ .
Expand: $x - 4 = 2x - 14$ .
So $x = 10$ .
Manya is 10 years old and Musa is 13.
Check: seven years ago they were 3 and 6, and 6 is twice 3 ✓.

(b)

In $n$ years they will be $10 + n$ and $13 + n$ .
Their sum is 45: $(10 + n) + (13 + n) = 45$ .
So $23 + 2n = 45$ , $2n = 22$ and $n = 11$ .
In 11 years.

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Question 3

(a)
The diagram shows an athletics track with two parallel sides and two semicircular ends. Each of the parallel sides is 60 metres long and the diameter of each semicircular end is 120 metres. Calculate the distance covered by an athlete who runs round the track two times. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
If the athlete spends 200 seconds for the race, calculate the speed in km/h.

Worked solution (try it first)

(a)

One lap is the two straights and the two semicircular ends.
The two semicircles make a whole circle of diameter 120 m: $\frac{22}{7} \times 120 \approx 377.14$ m.
One lap $= 2 \times 60 + 377.14 = 497.14$ m.
Two laps $= 2 \times 497.14 \approx 994.3$ m.

(b)

Speed $= \frac{994.29\text{ m}}{200\text{ s}}$
$\approx 4.97$ m/s.
To change m/s to km/h, multiply by 3.6 (3600 seconds in an hour, 1000 m in a km): $4.97 \times 3.6 \approx 17.9$ km/h.

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Question 4

(a)
A man was charged 2 kobo per month for every ₦1.00 he borrowed from a bank. At what rate per annum was the interest charged?
(b)
In the diagram, $W$ , $X$ , $Y$ , $Z$ are points on a circle with $|WX| = |XY| = |YZ|$ and $\angle WXY = 80^\circ$ . What is the size of $\angle XWZ$ ?

Worked solution (try it first)

(a)

2 kobo on every ₦1.00 is $2\%$ , and that's per month.
Per annum: $2\% \times 12 = 24\%$ .

(b)

$|WX| = |XY|$ , so triangle $WXY$ is isosceles: $\angle XWY = \angle XYW$
$= \frac{180^\circ - 80^\circ}{2}$
$= 50^\circ$ .
Angles in the same segment: $\angle XZY = \angle XWY = 50^\circ$ .
$|XY| = |YZ|$ , so $\angle ZXY = \angle XZY = 50^\circ$ , and $\angle XYZ = 180^\circ - 100^\circ$
$= 80^\circ$ .
$WXYZ$ is a cyclic quadrilateral, so $\angle XWZ = 180^\circ - \angle XYZ$
$= 100^\circ$ .

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Question 5

Statistics: data & averages

Score	1	2	3	4	5	6
Frequency	2	5	13	11	9	10

The table shows the distribution of scores obtained when a fair die was rolled 50 times.

(a)
Draw a bar chart for the distribution.
Model answer
Put the scores on the horizontal axis and the frequency on the vertical axis, with equal-width bars and equal gaps between them. The bar heights are $2, 5, 13, 11, 9, 10$ ; the tallest bar is at score 3.
(b)
Calculate the mean score of the distribution.

Worked solution (try it first)

(a)

The scores 1 to 6 are separate values, so draw a bar chart: six bars of equal width with equal gaps between them, of heights $2, 5, 13, 11, 9, 10$ .
Label the axes "Score" and "Frequency", and state the scale.

(b)

$\sum fx = 1(2) + 2(5) + 3(13) + 4(11) + 5(9) + 6(10)$
$= 2 + 10 + 39 + 44 + 45 + 60$
$= 200$ and $\sum f = 50$ , so the mean score is $\frac{200}{50} = 4$ .

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Question 6

(a)
If $\tan x = \frac{5}{12}$ , $0^\circ < x < 90^\circ$ , evaluate, without using mathematical tables or a calculator, $\dfrac{\sin x}{(\sin x)^2 + \cos x}$ .
(b)
A rectangular lawn measures $14\text{ m}$ by $11\text{ m}$ . A path of uniform width $x\text{ m}$ surrounds it. If the total area of the path is $186\text{ m}^2$ , how wide is the path?

Worked solution (try it first)

(a)

$\tan x = \frac{5}{12}$ gives a right-angled triangle with sides 5, 12 and 13, so $\sin x = \frac{5}{13}$ and $\cos x = \frac{12}{13}$ .
The bottom is $\left(\frac{5}{13}\right)^2 + \frac{12}{13} = \frac{25}{169} + \frac{156}{169}$
$= \frac{181}{169}$ .
So the value is $\frac{5}{13} \div \frac{181}{169} = \frac{5}{13} \times \frac{169}{181}$
$= \frac{65}{181}$ .

(b)

The lawn and path together make a rectangle $(14 + 2x)$ m by $(11 + 2x)$ m (the path adds $x$ on each side).
Area of the path $=$ big rectangle $-$ lawn: $(14 + 2x)(11 + 2x) - 14 \times 11 = 186$ .
Expand: $154 + 50x + 4x^2 - 154 = 186$ , so $4x^2 + 50x - 186 = 0$ , which is $2x^2 + 25x - 93 = 0$ .
Factorise: $(2x + 31)(x - 3) = 0$ .
A width can't be negative, so $x = 3$ m.

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Question 7

Commercial arithmetic

(a)
A shop had two reduction sales during which prices of all items were reduced by $40\%$ in the first sale and $30\%$ in the second. If a shirt was sold at GH₵ 35.00 during the second reduction sale, find the price before the first sale.
(b)
If the price of an article before the first reduction was GH₵ 180.00, find the total: (i) reduction in the price due to the two sales; (ii) percentage reduction in the price of the article.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

A $40\%$ reduction multiplies the price by 0.6, and a $30\%$ reduction by 0.7.
Together: $0.6 \times 0.7 = 0.42$ .
So $0.42 \times \text{original} = 35$ and the original price was $\frac{35}{0.42} \approx$ GH₵ 83.33.

(b)(i)

From GH₵ 180: after the first sale, $0.6 \times 180 = 108$ .
After the second, $0.7 \times 108 = 75.60$ .
Total reduction $= 180 - 75.60 =$ GH₵ 104.40.

(ii)

$\frac{104.40}{180} \times 100\% = 58\%$ .

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Question 8

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct: (i) a trapezium $PQRS$ such that $|PQ| = 6.8\text{ cm}$ , $\angle PQR = 120^\circ$ , $QR \parallel PS$ , $|PS| = 10.6\text{ cm}$ and $|PR| = 9.3\text{ cm}$ ; (ii) the locus $l_1$ of points equidistant from $P$ and $R$ ; (iii) the locus $l_2$ of points equidistant from $Q$ and $R$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $PS = 10.6$ cm. $QR \parallel PS$ and $\angle PQR = 120^\circ$ , so $\angle QPS = 60^\circ$ : construct $60^\circ$ at $P$ and mark $Q$ with $PQ = 6.8$ cm. Draw the line through $Q$ parallel to $PS$ , and cut it with an arc of radius 9.3 cm from $P$ to get $R$ . Join $RS$ . Then bisect $PR$ ( $l_1$ ) and $QR$ ( $l_2$ ) perpendicularly; they meet at $Y$ , the centre of the circle through $P$ , $Q$ and $R$ . Measured: $|QR| \approx 3.8$ cm, $\angle PSR \approx 60^\circ$ , and $|QY| \approx$ 5.4 cm.
(b)
Measure: (i) $|QR|$ ; (ii) $\angle PSR$ ; (iii) $|QY|$ , where $Y$ is the point of intersection of $l_1$ and $l_2$ .
Show the answer
$|QR| \approx 3.8\text{ cm}$ , $\angle PSR \approx 60^\circ$ , $|QY| \approx 5.4\text{ cm}$

Try it on a graph

The accurate construction: P(0, 0), Q(6.8, 0), R(8.70, 3.29), S(5.3, 9.18); Y is the circumcentre of PQR.

Worked solution (try it first)

(a)(i)

Draw $PQ = 6.8$ cm and construct $120^\circ$ at $Q$ .
With centre $P$ and radius 9.3 cm, cut the arm at $R$ .
$PS \parallel QR$ : through $P$ construct a line parallel to $QR$ , and mark $PS = 10.6$ cm on it.
Join $RS$ .

(ii)

$l_1$ , equidistant from $P$ and $R$ : the perpendicular bisector of $PR$ .

(iii)

$l_2$ , equidistant from $Q$ and $R$ : the perpendicular bisector of $QR$ .
They cross at $Y$ , the centre of the circle through $P$ , $Q$ and $R$ .

(b)

Measure: (i) $|QR| \approx 3.8$ cm.

(ii)

$\angle PSR \approx 60^\circ$ .

(iii)

$|QY| \approx 5.4$ cm.
Check: the cosine rule gives $9.3^2 = 6.8^2 + |QR|^2 + 6.8|QR|$ , so $|QR| \approx 3.80$ cm.
$|QY|$ is the radius of the circle through $P$ , $Q$ , $R$ : $\frac{9.3}{2\sin 120^\circ} \approx 5.37$ cm.

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Question 9

(a)
A donkey is tied with a rope to a post which is $15\text{ m}$ from a fence. If the length of the rope between the donkey and the post is $17\text{ m}$ , calculate the length of the fence within the reach of the donkey.
(b)
The base of a right pyramid with vertex $V$ is a square $PQRS$ of side $15\text{ cm}$ . If the slant edge is $32\text{ cm}$ long, (i) represent the information in a diagram; (ii) calculate its: (I) height, correct to one decimal place; (II) volume, correct to the nearest $\text{cm}^3$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw the perpendicular from the post to the fence (15 m).
The rope (17 m) reaches the fence at two points, each the hypotenuse of a right-angled triangle: $x = \sqrt{17^2 - 15^2} = \sqrt{64} = 8$ m on each side of the foot.
The donkey reaches $2 \times 8 = 16$ m of fence.

(b)(i)

Draw the square base $PQRS$ with its diagonals crossing at the centre $M$ , and the vertex $V$ directly above $M$ , with the slant edges $VP$ , $VQ$ , $VR$ , $VS$ of 32 cm.

(ii)

(I)** The diagonal of the base is $15\sqrt2 \approx 21.21$ cm, so $MP \approx 10.61$ cm.
In the right-angled triangle $VMP$ : height $VM = \sqrt{32^2 - 10.61^2}$
$= \sqrt{1024 - 112.5}$
$\approx 30.2$ cm.
(II) Volume $= \frac13 \times 15^2 \times 30.19$
$\approx 2264\text{ cm}^3$ .

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Question 10

Trigonometric graphs

(a)

Copy and complete the table of values for $y = 2\cos x - \sin x$ , $0^\circ \le x \le 300^\circ$ .

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$	$270^\circ$	$300^\circ$
$y$	$2.00$		$0.13$		$-1.87$		$-2.00$		$-0.13$

Model answer

$x$	0°	30°	60°	90°	120°	150°	180°	210°	240°	270°	300°
$y$	2.00	1.23	0.13	−1.00	−1.87	−2.23	−2.00	−1.23	−0.13	1.00	1.87

For example, at $x = 30^\circ$ : $y = 2(0.866) - 0.5 = 1.23$ (2 d.p.).

(b)
Using scales of 2 cm to $30^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of $y = 2\cos x - \sin x$ for $0^\circ \le x \le 300^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to $30^\circ$ , 2 cm to 1 unit.

For (c): (i) draw $y = 1$ : it meets the curve at $x \approx 36.9^\circ$ and $270.0^\circ$ . (ii) $\tan x = 2$ means $\sin x = 2\cos x$ , i.e. $2\cos x - \sin x = 0$ : read where the curve crosses the $x$ -axis, $x \approx 63.4^\circ$ and $243.4^\circ$ .
(c)(i)
Use the graph to find the values of $x$ for which $2\cos x - \sin x = 1$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Use the graph to find the values of $x$ for which $\tan x = 2$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

x in degrees. The line y = 1 gives (c)(i); the x-axis gives (c)(ii).

Worked solution (try it first)

(a)

In degree mode, to 2 decimal places as in the table.
For example, $x = 30^\circ$ : $2(0.866) - 0.5 = 1.23$ .
$x = 150^\circ$ : $2(-0.866) - 0.5 = -2.23$ .
The full row is $2.00, 1.23, 0.13, -1.00, -1.87, -2.23, -2.00, -1.23, -0.13, 1.00, 1.87$ .

(b)

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

(c)(i)

Draw the line $y = 1$ and read down from where it crosses the curve: $x \approx 37^\circ$ and $x = 270^\circ$ .

(ii)

Rearrange so it matches the graph: $\tan x = 2$ means $\frac{\sin x}{\cos x} = 2$ , so $\sin x = 2\cos x$ , which is $2\cos x - \sin x = 0$ .
So read where the curve crosses the $x$ -axis: $x \approx 63^\circ$ and $x \approx 243^\circ$ .

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Question 11

(a)
If $\log_a(y + 2) = 1 + \log_a x$ , find $x$ in terms of $y$ .
(b)

Community Bibiani Amenfi Oda Wiawso Sankore

Timber production (tonnes) 600 900 1800 1500 2400

The table shows the distribution of timber production in five communities in a certain year. (i) Draw a pie chart to represent the information. (ii) What percentage of timber produced that year was from Amenfi? (iii) If a tonne of timber is sold at $560.00, how much more revenue would Oda community receive than Bibiani?
Separate values with commas, e.g. 3, −2

Community	Bibiani	Amenfi	Oda	Wiawso	Sankore
Timber production (tonnes)	600	900	1800	1500	2400

Worked solution (try it first)

(a)

Write 1 as $\log_a a$ : $\log_a(y + 2) = \log_a a + \log_a x$
$= \log_a(ax)$ .
The logs are equal, so $y + 2 = ax$ and $x = \frac{y + 2}{a}$ .

(b)(i)

The total is $600 + 900 + 1800 + 1500 + 2400 = 7200$ tonnes, and $360^\circ$ stands for 7200 tonnes, so each $1^\circ$ is 20 tonnes.
The sector angles: Bibiani $\frac{600}{20} = 30^\circ$ , Amenfi $45^\circ$ , Oda $90^\circ$ , Wiawso $75^\circ$ , Sankore $120^\circ$ (total $360^\circ$ ).
Draw them with a protractor and label each sector.

(ii)

Amenfi's share is $\frac{900}{7200} \times 100 = 12.5\%$ .

(iii)

Oda produced $1800 - 600 = 1200$ tonnes more than Bibiani, so it would receive $1200 \times \text{\textdollar}560 = \text{\textdollar}672,000$ more.

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Question 12 ✱✱

(a)
Mr John paid ₦8,400.00 for ₦1.00 ordinary shares of a company which sold at ₦2.50 per share. If a dividend was declared at 25 kobo per share, how much dividend did he get?
(b)
Using the method of completing the square, solve $\dfrac{1 - x}{x} + \dfrac{x}{1 - x} = \dfrac52$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The shares cost ₦2.50 each, so ₦8,400 buys $\frac{8400}{2.50} = 3360$ shares.
The dividend is 25 kobo (₦0.25) per share: $3360 \times 0.25 = 840$ , which is ₦840.00.

(b)

Multiply every term by $2x(1 - x)$ : $2(1 - x)^2 + 2x^2 = 5x(1 - x)$ .
Expand: $2 - 4x + 2x^2 + 2x^2 = 5x - 5x^2$ , so $9x^2 - 9x + 2 = 0$ .
Divide by 9: $x^2 - x = -\frac29$ .
Add $\left(\frac12\right)^2 = \frac14$ to both sides: $\left(x - \frac12\right)^2 = \frac14 - \frac29$
$= \frac{1}{36}$ .
So $x - \frac12 = \pm\frac16$ , giving $x = \frac23$ or $x = \frac13$ .

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Question 13

(a)
The points $R(3, -6)$ , $S(6, -2)$ and $T(p, q)$ are on the $xy$ -plane. If $\frac13\overrightarrow{OR} + \overrightarrow{OS} + \overrightarrow{OT} = \overrightarrow{RS}$ , find the coordinates of $T$ .
Separate values with commas, e.g. 3, −2
(b)
If $2\begin{pmatrix} 3 \\ m \end{pmatrix} + n\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ , find the values of $m$ and $n$ .
Separate values with commas, e.g. 3, −2
(c)
Given that $w * u = w + u + 1$ , if $(y * 4) * y = 12$ , find the value of $y$ .

Worked solution (try it first)

(a)

The position vectors are $\overrightarrow{OR} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}$ , $\overrightarrow{OS} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$ and $\overrightarrow{OT} = \begin{pmatrix} p \\ q \end{pmatrix}$ .
So $\frac13\overrightarrow{OR} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ .
$\overrightarrow{RS} = \overrightarrow{OS} - \overrightarrow{OR}$
$= \begin{pmatrix} 6 - 3 \\ -2 - (-6) \end{pmatrix}$
$= \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .
Add the left-hand side: $\begin{pmatrix} 1 + 6 + p \\ -2 - 2 + q \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$ .
So $7 + p = 3$ and $-4 + q = 4$ .
$p = -4$ and $q = 8$ : $T$ is the point $(-4, 8)$ .

(b)

Multiply out: $\begin{pmatrix} 6 \\ 2m \end{pmatrix} + \begin{pmatrix} n \\ 2n \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$ .
The top row gives $6 + n = 8$ , so $n = 2$ .
The bottom row gives $2m + 2n = 4$ , so $2m + 4 = 4$ and $m = 0$ .

(c)

Work out the bracket first: $y * 4 = y + 4 + 1 = y + 5$ .
Then $(y + 5) * y = (y + 5) + y + 1 = 2y + 6$ .
So $2y + 6 = 12$ , which gives $y = 3$ .

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