WAEC · 2013 · Nov/Dec · General Maths · Paper 2

Question 1 ✱✱

(a)
Evaluate $\left(1 - \frac{63}{64}\right)^{-\frac12} \times 2^{-1}$ .
(b)
The interior angle of a regular polygon is $108^\circ$ greater than the exterior angle. How many sides has the polygon?

Worked solution (try it first)

(a)

Work out the bracket first: $1 - \frac{63}{64} = \frac{1}{64}$ .
A negative power means turn the fraction upside down: $\left(\frac{1}{64}\right)^{-\frac12} = 64^{\frac12}$ .
A power of $\frac12$ is a square root: $64^{\frac12} = 8$ .
In the same way, $2^{-1} = \frac12$ .
Multiply: $8 \times \frac12 = 4$ .
The value is 4.

(b)

Let the exterior angle be $e$ .
The interior angle is then $e + 108^\circ$ .
An interior angle and its exterior angle make a straight line: $e + (e + 108^\circ) = 180^\circ$ .
So $2e = 72^\circ$ , which gives $e = 36^\circ$ .
The exterior angles of any polygon add up to $360^\circ$ , so the number of sides is $360^\circ \div 36^\circ = 10$ .
The polygon has 10 sides.

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Question 2

(a)
The diagram shows a triangular prism with $|QR| = |MN| = |OP| = 10\text{ cm}$ and $|NR| = |QM| = 8\text{ cm}$ . If $\angle RON = 90^\circ$ and $\angle RNO = 30^\circ$ , calculate, correct to 3 significant figures, the volume of the prism.
(b)
A bird on top of a tree sights a prey $18\text{ m}$ away and on the same horizontal ground as the foot of the tree. If the height of the tree is $8\text{ m}$ , calculate, correct to the nearest degree, the angle of depression through which the bird sights the prey.

Worked solution (try it first)

(a)

The cross-section is the right-angled triangle $RON$ : the hypotenuse is $|NR| = 8\text{ cm}$ and the right angle is at $O$ .
The side opposite the $30^\circ$ angle: $|RO| = 8\sin30^\circ = 4\text{ cm}$ .
The side next to it: $|NO| = 8\cos30^\circ = 4\sqrt3\text{ cm}$ .
Area of the cross-section: $\frac12 \times 4 \times 4\sqrt3 = 8\sqrt3\text{ cm}^2$ .
Volume = area of cross-section × length: $8\sqrt3 \times 10 = 80\sqrt3$
$\approx 138.56$ .
The volume is $139\text{ cm}^3$ to 3 significant figures.

(b)

Sketch the tree, 8 m tall, and the prey on the ground 18 m from its foot.
The angle of depression $x$ at the top of the tree equals the angle of elevation at the prey (alternate angles).
In the right-angled triangle, 8 m is opposite $x$ and 18 m is adjacent, so $\tan x = \frac{8}{18}$ .
$x = \tan^{-1}(0.4444) \approx 23.96^\circ$ .
The angle of depression is $24^\circ$ to the nearest degree.

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Question 5 ✱

(a)
The pie chart shows the result of a survey conducted on the maintenance of five small and medium scale industries in a state. If the cost of maintenance of the plywood industry was ₦34,500.00, calculate the cost of maintenance of the textile industry.
(b)
In the diagram, $TYZ$ and $UWZ$ are tangents to the circle $WXY$ , $\angle XYZ = 100^\circ$ , $\angle YZW = 42^\circ$ and $\angle UWX = m$ . Find the value of $m$ .

Worked solution (try it first)

(a)

The angles at the centre add up to $360^\circ$ : $x + x + 82^\circ + 68^\circ + 72^\circ = 360^\circ$ .
So $2x + 222^\circ = 360^\circ$ , which gives $2x = 138^\circ$ and $x = 69^\circ$ .
The plywood sector is $69^\circ$ .
$69^\circ$ stands for ₦34,500, so the whole $360^\circ$ stands for $\frac{360}{69} \times 34\,500 = 180\,000$ .
The total cost is ₦180,000.
The textile sector is $68^\circ$ : $\frac{68}{360} \times 180\,000 = 34\,000$ .
The maintenance of the textile industry cost ₦34,000.00.

(b)

Tangents from the same point are equal, so $|ZY| = |ZW|$ and triangle $ZYW$ is isosceles with $\angle ZYW = \angle ZWY$ .
Angles in triangle $ZYW$ : $\angle ZYW = \frac{180^\circ - 42^\circ}{2}$
$= 69^\circ$ .
Alternate segment theorem: the angle between the tangent $WU$ and the chord $WX$ equals the angle in the alternate segment, so $\angle XYW = \angle UWX = m$ .
At $Y$ , $\angle XYZ = \angle XYW + \angle WYZ$ , so $100^\circ = m + 69^\circ$ .
So $m = 31^\circ$ .

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Question 6

(a)
If $y^2 - x^2 = 5(y - x)^2$ , find $x : y$ .
Separate values with commas, e.g. 3, −2
(b)
In a partnership, Ajayi contributed ₦500,000.00 more than Kunle. The total profit made was $15\%$ of their total contribution. If Kunle received $\frac25$ of the total profit, which amounted to ₦84,000.00, how much was: (i) Ajayi's share of the profit? (ii) Kunle's contribution to the partnership?
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Expand the right-hand side: $5(y - x)^2 = 5y^2 - 10xy + 5x^2$ .
So $y^2 - x^2 = 5y^2 - 10xy + 5x^2$ .
Collect every term on one side: $4y^2 - 10xy + 6x^2 = 0$ .
Divide by 2: $2y^2 - 5xy + 3x^2 = 0$ .
Factorise: $(2y - 3x)(y - x) = 0$ .
Either $y = x$ , which gives $x : y = 1 : 1$ , or $2y = 3x$ , which gives $x : y = 2 : 3$ .

(b)(i)

$\frac25$ of the profit is ₦84,000, so the total profit is $84\,000 \times \frac52 = 210\,000$ , that is ₦210,000.
Ajayi gets the other $\frac35$ : $\frac35 \times 210\,000 = 126\,000$ .
Ajayi's share is ₦126,000.00.

(ii)

Let Kunle's contribution be ₦ $x$ .
Ajayi's is then ₦ $(x + 500\,000)$ , so together they put in ₦ $(2x + 500\,000)$ .
The profit is $15\%$ of the total: $\frac{15}{100}(2x + 500\,000) = 210\,000$ .
Multiply both sides by $\frac{100}{15}$ : $2x + 500\,000 = 1\,400\,000$ .
So $2x = 900\,000$ and $x = 450\,000$ .
Kunle contributed ₦450,000.00.

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In the diagram, $PQS$ is a right-angled triangle, $\angle PSQ = 90^\circ$ , $|PS| = 3\text{ cm}$ , $|PQ| = 5\text{ cm}$ , $\angle PRS = 61^\circ$ and $\angle QPR = x^\circ$ . Calculate, correct to one decimal place, the:

(a)
value of $x$ ;
(b)
area of $\triangle PQR$ .

Worked solution (try it first)

(a)

In triangle $PQS$ , $PS$ is next to $\angle QPS$ and $PQ$ is the hypotenuse, so $\cos\angle QPS = \frac35$ .
So $\angle QPS = \cos^{-1}(0.6) = 53.13^\circ$ .
In triangle $PRS$ , the angles add up to $180^\circ$ : $\angle RPS = 180^\circ - 90^\circ - 61^\circ$
$= 29^\circ$ .
$x^\circ = \angle QPS - \angle RPS$
$= 53.13^\circ - 29^\circ$
$= 24.13^\circ$ .
So $x = 24.1$ to one decimal place.

(b)

Pythagoras in triangle $PQS$ : $|QS| = \sqrt{5^2 - 3^2} = 4\text{ cm}$ .
Area of $\triangle PQS = \frac12 \times 3 \times 4$
$= 6\text{ cm}^2$ .
In triangle $PRS$ , $|RS| = 3\tan29^\circ = 1.6629\text{ cm}$ .
Area of $\triangle PRS = \frac12 \times 3 \times 1.6629$
$= 2.4944\text{ cm}^2$ .
Area of $\triangle PQR$ is the difference: $6 - 2.4944 = 3.5056$ .
The area is $3.5\text{ cm}^2$ to one decimal place.

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Question 8

(a)
Given the Cartesian coordinates $A(1, 2)$ , $B(0, 4)$ , $C(-2, -2)$ and $D(-3, 0)$ , on a graph sheet and using a scale of 2 cm to represent 1 unit on both axes, plot the points $A$ , $B$ , $C$ and $D$ .
Model answer
Draw both axes with 2 cm to 1 unit ( $x$ from $-5$ to $2$ and $y$ from $-3$ to $5$ is enough) and mark the four points. The drawing under (c) shows them.
(b)
(i) Join the points to form a quadrilateral. (ii) What type of quadrilateral is formed?
(c)
Using a ruler and a pair of compasses only, construct: (i) the locus $l_1$ of points equidistant from $A$ and $C$ ; (ii) the locus $l_2$ of points equidistant from $\overline{AC}$ and $\overline{BA}$ ; (iii) locate $M$ , the point of intersection of $l_1$ and $l_2$ .
Model answer
$l_1$ is the perpendicular bisector of $AC$ ; $l_2$ is the bisector of $\angle CAB$ . They meet at $M$ , about $(-3.7, 2.4)$ .
(d)
Measure: (i) $\angle MAB$ ; (ii) $|MA|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw the axes with 2 cm to 1 unit on both, and plot $A(1, 2)$ , $B(0, 4)$ , $C(-2, -2)$ and $D(-3, 0)$ .

(b)(i)

Join $A$ to $B$ , $B$ to $D$ , $D$ to $C$ and $C$ to $A$ .

(ii)

Compare opposite sides.
From $A$ to $B$ is 1 left and 2 up, and from $C$ to $D$ is also 1 left and 2 up.
From $B$ to $D$ is 3 left and 4 down, and from $A$ to $C$ is also 3 left and 4 down.
Both pairs of opposite sides are equal and parallel, and the corners are not right angles, so $ABDC$ is a parallelogram.

(c)(i)

Points equidistant from $A$ and $C$ lie on the perpendicular bisector of $AC$ .
With centres $A$ and $C$ and the same radius (more than half of $AC$ ), draw arcs on both sides of $AC$ and join the two crossings.
This is $l_1$ .
It passes through the midpoint $(-0.5, 0)$ .

(ii)

Points equidistant from the lines $AC$ and $AB$ lie on the bisector of $\angle CAB$ .
With centre $A$ , draw an arc cutting $AC$ and $AB$ .
From those two points draw equal arcs that cross, and join $A$ to the crossing.
This is $l_2$ .

(iii)

Extend $l_1$ and $l_2$ until they meet, and label the point $M$ .
It is near $(-3.7, 2.4)$ .

(d)(i)

$\angle CAB \approx 116.6^\circ$ and $l_2$ halves it, so $\angle MAB \approx 58^\circ$ .

(ii)

Measure $MA$ with the ruler: $|MA| \approx 9.5\text{ cm}$ (about 4.76 units at 2 cm to 1 unit).

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Question 9

(a)
Solve: $\dfrac{x}{2} + \dfrac{1}{x} = 1\frac12$ .
Separate values with commas, e.g. 3, −2
(b)
The range of the numbers $p$ , 6, 8, 11 and $q$ , arranged in ascending order, is 16. If the mean is 9, find the value of $(2p + q)$ .

Worked solution (try it first)

(a)

Write $1\frac12$ as $\frac32$ and multiply every term by $2x$ , the LCM of the denominators: $x^2 + 2 = 3x$ .
Rearrange: $x^2 - 3x + 2 = 0$ .
Factorise: $(x - 1)(x - 2) = 0$ , so $x = 1$ or $x = 2$ .

(b)

The numbers are in ascending order, so $p$ is the smallest and $q$ the largest.
Range: $q - p = 16$ .
Mean: $p + 6 + 8 + 11 + q = 9 \times 5 = 45$ , so $p + q = 20$ .
Add the two equations: $2q = 36$ , so $q = 18$ .
Then $p = 20 - 18 = 2$ .
$2p + q = 2(2) + 18 = 22$ .

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Question 10

Town $Q$ is $20\text{ km}$ due north of $P$ . The bearing of town $R$ from $Q$ is $140^\circ$ . If $R$ is $8\text{ km}$ from $Q$ , calculate:

(a)
the bearing of $R$ from $P$ , to the nearest degree;
(b)
how far north of $P$ $R$ is, correct to 2 significant figures.

Worked solution (try it first)

(a)

Sketch it: $Q$ is 20 km due north of $P$ , and $R$ is 8 km from $Q$ on a bearing of $140^\circ$ (to the south-east of $Q$ ).
At $Q$ , the direction back to $P$ is due south ( $180^\circ$ ), so $\angle PQR = 180^\circ - 140^\circ$
$= 40^\circ$ .
Cosine rule: $|PR|^2 = 20^2 + 8^2 - 2(20)(8)\cos40^\circ$
$= 218.87$ , so $|PR| = 14.79\text{ km}$ .
Sine rule for $\theta = \angle QPR$ : $\frac{\sin\theta}{8} = \frac{\sin40^\circ}{14.79}$ , so $\sin\theta = 0.3476$ and $\theta = 20.34^\circ$ .
$\theta$ is measured clockwise from north at $P$ , so the bearing of $R$ from $P$ is $020^\circ$ to the nearest degree.

(b)

$R$ is $8\cos40^\circ = 6.13\text{ km}$ south of $Q$ .
So $R$ is $20 - 6.13 = 13.87\text{ km}$ north of $P$ (the same as $|PR|\cos20.34^\circ$ ).
To 2 significant figures, $R$ is $14\text{ km}$ north of $P$ .

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Question 11

Statistics: data & averages

The histogram shows the height of a group of students in a school.

(a)

Use the histogram to construct the frequency distribution table.

Model answer

Height (cm)	150–154	155–159	160–164	165–169	170–174	175–179
Class boundaries	149.5–154.5	154.5–159.5	159.5–164.5	164.5–169.5	169.5–174.5	174.5–179.5
Class mark $x$	152	157	162	167	172	177
Frequency $f$	4	7	9	13	5	2
$fx$	608	1099	1458	2171	860	354

$\Sigma f = 40$ and $\Sigma fx = 6550$ .

(b)
What percentage of the students have their heights between $159.5\text{ cm}$ and $164.5\text{ cm}$ ?
(c)
Calculate the mean height.

Worked solution (try it first)

(a)

The height of each bar is the frequency of the class between its two boundaries.
The class mark is the middle of the class, for example $\frac{149.5 + 154.5}{2} = 152$ .

Height (cm)	150–154	155–159	160–164	165–169	170–174	175–179
Class boundaries	149.5–154.5	154.5–159.5	159.5–164.5	164.5–169.5	169.5–174.5	174.5–179.5
Class mark $x$	152	157	162	167	172	177
Frequency $f$	4	7	9	13	5	2
$fx$	608	1099	1458	2171	860	354

Total frequency: $4 + 7 + 9 + 13 + 5 + 2 = 40$ students.

(b)

The bar from 159.5 to 164.5 has height 9, so 9 of the 40 students: $\frac{9}{40} \times 100\% = 22.5\%$ .

(c)

Add the $fx$ row: $\Sigma fx = 6550$ .
Mean $= \frac{\Sigma fx}{\Sigma f}$
$= \frac{6550}{40}$
$= 163.75$ .
The mean height is $163.75\text{ cm}$ .

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Question 12

(a)
An open rectangular tank is made from a steel plate of area $1440\text{ m}^2$ . Its length is twice its width. If the depth of the tank is $4\text{ m}$ less than the width, find the length of the tank.
(b)
In the diagram, $|XU| = 10\text{ cm}$ , $|WU| = 8\text{ cm}$ , $|UV| = 3.5\text{ cm}$ and $XV$ is perpendicular to $WZ$ . If the radius of the circle is $7\text{ cm}$ , calculate, correct to 2 decimal places, the length of the arc $WV$ . $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

Let the width be $w$ m.
Then the length is $2w$ m and the depth is $(w - 4)$ m.
The tank is open, so the plate makes the base and four sides but no top.
Base: $2w \times w = 2w^2$ .
Two long sides: $2 \times 2w(w - 4) = 4w^2 - 16w$ .
Two short sides: $2 \times w(w - 4) = 2w^2 - 8w$ .
Add them: the total area is $8w^2 - 24w$ .
Set it equal to the plate: $8w^2 - 24w = 1440$ .
Divide by 8: $w^2 - 3w - 180 = 0$ .
Factorise: $(w - 15)(w + 12) = 0$ .
A width can't be negative, so $w = 15$ .
The length is $2 \times 15 = 30$ .
The tank is $30\text{ m}$ long.

(b)

In the right-angled triangle $WUX$ , $\tan\angle WXU = \frac{|WU|}{|XU|}$
$= \frac{8}{10}$ , so $\angle WXV = 38.66^\circ$ .
The arc $WV$ subtends $\angle WXV$ at the circumference, so it subtends twice as much at the centre: $2 \times 38.66^\circ = 77.32^\circ$ .
Arc length $= \frac{77.32}{360} \times 2 \times \frac{22}{7} \times 7$
$= 9.450$ .
The arc $WV$ is $9.45\text{ cm}$ long.

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Question 13

(a)
Given that $Y = \{x : -2 \le x \le 5\}$ and $W = \{x : 1 < x < 6\}$ , illustrate $Y \cap W$ on the number line.
Model answer
Open circle at 1, filled dot at 5, joined by a line.
(b)(i)
$x$ varies jointly as the square of $m$ and the cube of $n$ . When $x = 9$ , $m = \frac34$ and $n = \frac12$ . Determine the relationship between $x$ , $m$ and $n$ .
(b)(ii)
Calculate, correct to 3 significant figures, the value of: ( $\alpha$ ) $x$ when $m = \frac23$ and $n = \frac15$ ; ( $\beta$ ) $m$ when $x = 5$ and $n = \frac18$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$Y \cap W$ holds the numbers in both sets: greater than 1 (from $W$ ) and at most 5 (from $Y$ ).
So $Y \cap W = \{x : 1 < x \le 5\}$ .
On the number line, draw an open circle at 1 (1 is left out), a filled dot at 5 (5 is included), and join them.

(b)(i)

Joint variation: $x = km^2n^3$ for some constant $k$ .
Put in $x = 9$ , $m = \frac34$ and $n = \frac12$ : $9 = k \times \frac{9}{16} \times \frac18$ , which is $9 = \frac{9k}{128}$ .
So $k = 128$ , and the relationship is $x = 128m^2n^3$ .

(ii)

( $\alpha$ )** Put in $m = \frac23$ and $n = \frac15$ : $x = 128 \times \frac49 \times \frac{1}{125}$
$= \frac{512}{1125}$ .
$\frac{512}{1125} = 0.4551$ , so $x = 0.455$ to 3 significant figures.
( $\beta$ ) Make $m^2$ the subject: $m^2 = \frac{x}{128n^3}$ .
Put in $x = 5$ and $n = \frac18$ : $128 \times \frac{1}{512} = \frac14$ , so $m^2 = 5 \div \frac14 = 20$ .
$m = \sqrt{20} = 4.47$ to 3 significant figures.

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