WAEC · 2009 · May/June · General Maths · Paper 2

Question 1 ✱✱

(a)
Given that $(\sqrt3 - 5\sqrt2)(\sqrt3 + \sqrt2) = a + b\sqrt6$ , find $a$ and $b$ .
Separate values with commas, e.g. 3, −2
(b)
If $\dfrac{2^{1-y} \times 2^{y-1}}{2^{y+2}} = 8^{2-3y}$ , find $y$ .

Worked solution (try it first)

(a)

Expand the brackets term by term: $\sqrt3 \cdot \sqrt3 + \sqrt3 \cdot \sqrt2 - 5\sqrt2 \cdot \sqrt3 - 5\sqrt2 \cdot \sqrt2$ .
Simplify each term: $3 + \sqrt6 - 5\sqrt6 - 10$ .
Collect like terms: $-7 - 4\sqrt6$ .
Compare with $a + b\sqrt6$ : $a = -7$ and $b = -4$ .

(b)

Add the powers on top (same base 2): $(1 - y) + (y - 1) = 0$ , so the top is $2^0$ .
Divide by subtracting powers: the left side is $2^{0 - (y + 2)} = 2^{-y-2}$ .
Write the right side in base 2: $8^{2-3y} = (2^3)^{2-3y} = 2^{6-9y}$ .
Equate the powers: $-y - 2 = 6 - 9y$ , so $8y = 8$ .
So $y = 1$ .

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Question 2

(a)
If $9\cos x - 7 = 1$ and $0^\circ \le x \le 90^\circ$ , find $x$ .
(b)
Given that $x$ is an integer, find the three greatest values of $x$ which satisfy the inequality $7x < 2x - 13$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Add 7 to both sides: $9\cos x = 8$ .
Divide by 9: $\cos x = \frac89 = 0.8889$ .
Use tables or a calculator in degree mode: $x = \cos^{-1}(0.8889) = 27.27^\circ$ .

(b)

Subtract $2x$ from both sides: $5x < -13$ .
Divide by 5 (a positive number, so the sign stays): $x < -2.6$ .
The integers less than $-2.6$ are $-3, -4, -5, \dots$ .
The three greatest are $-3$ , $-4$ and $-5$ .

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Question 3

The table shows the number of children per family in a community.

No. of children	0	1	2	3	4	5
No. of families	3	5	7	4	3	2

(a)
Find the: (i) mode; (ii) third quartile; (iii) probability that a family has at least 2 children.
Separate values with commas, e.g. 3, −2
(b)
If a pie chart were to be drawn for the data, what would be the sectoral angle representing families with one child?

Worked solution (try it first)

(a)(i)

The mode is the number of children with the highest frequency: 7 families have 2 children, so the mode is 2.

(ii)

Total number of families: $3 + 5 + 7 + 4 + 3 + 2 = 24$ .
The third quartile is at position $\frac34 \times 24 = 18$ in the ordered list.
Running totals: 3 (0 children), 8 (1 child), 15 (2 children), 19 (3 children).
The 16th to 19th families have 3 children, so the 18th has 3.
The third quartile is 3 children.

(iii)

Families with at least 2 children: $7 + 4 + 3 + 2 = 16$ .
Probability $= \frac{16}{24} = \frac23$ .

(b)

Angle for one child $= \frac{5}{24} \times 360^\circ$
$= 75^\circ$ .

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Question 4

(a)
Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 have neither a degree nor a diploma. How many of them have both?
(b)
In triangle $PQR$ , $M$ and $N$ are points on the sides $PQ$ and $PR$ respectively such that $MN$ is parallel to $QR$ . If $\angle PRQ = 75^\circ$ , $|PN| = |QN|$ and $\angle PNQ = 125^\circ$ , determine: (i) $\angle NQR$ ; (ii) $\angle NPM$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Let $x$ have both.
Degree only is $17 - x$ and diploma only is $15 - x$ .
Everyone is in exactly one region: $(17 - x) + x + (15 - x) + 4 = 30$ .
Simplify: $36 - x = 30$ , so $x = 6$ .
Six applicants have both.

(b)

Sketch the triangle: $N$ is on $PR$ , $M$ on $PQ$ , and $NQ$ is joined.

(i)

$MN \parallel QR$ , so $\angle PNM = \angle PRQ = 75^\circ$ (corresponding angles).
$\angle MNQ = \angle PNQ - \angle PNM$
$= 125^\circ - 75^\circ$
$= 50^\circ$ .
$\angle NQR = \angle MNQ = 50^\circ$ (alternate angles, $MN \parallel QR$ ).

(ii)

$|PN| = |QN|$ , so triangle $PNQ$ is isosceles with equal base angles at $P$ and $Q$ .
$\angle NPM = \frac{180^\circ - 125^\circ}{2}$
$= 27.5^\circ$ .

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Question 5

In the diagram, $ABCDEF$ is a triangular prism. $\angle ABC = \angle DEF = 90^\circ$ , $|AB| = 24$ cm, $|BC| = 7$ cm and $|CD| = 40$ cm.

(a)
Calculate $|AC|$ .
(b)
Calculate the total surface area of the prism.

Worked solution (try it first)

(a)

Triangle $ABC$ is right-angled at $B$ , so use Pythagoras: $|AC|^2 = 24^2 + 7^2$ .
$|AC|^2 = 576 + 49 = 625$ .
So $|AC| = 25$ cm.

(b)

The prism has two triangular ends and three rectangular faces.
Each triangle: $\frac12 \times 24 \times 7 = 84\text{ cm}^2$ .
Both ends give $168\text{ cm}^2$ .
The rectangles are $24 \times 40$ , $7 \times 40$ and $25 \times 40$ : $960 + 280 + 1000 = 2240\text{ cm}^2$ .
Total surface area $= 168 + 2240 = 2408\text{ cm}^2$ .

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Question 6

(a)
If $\log 5 = 0.6990$ , $\log 7 = 0.8451$ and $\log 8 = 0.9031$ , evaluate $\log\left(\dfrac{35 \times 49}{40 \div 56}\right)$ .
(b)(i)
For a musical show, $x$ children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the value of $x$ .
(b)(ii)
Calculate the ratio of the number of children to the number of adults.
(b)(iii)
Calculate the average amount paid per person.
(b)(iv)
Calculate the percentage profit if the organisers spent D720 on the show.

Worked solution (try it first)

(a)

Write each number with factors 5, 7 and 8: $35 = 5 \times 7$ , $49 = 7 \times 7$ , $40 = 5 \times 8$ , $56 = 7 \times 8$ .
Dividing by $40 \div 56$ is multiplying by $\frac{56}{40}$ : the number is $\frac{5 \times 7 \times 7 \times 7 \times 7 \times 8}{5 \times 8} = 7^4$ .
So the log is $4\log 7 = 4 \times 0.8451 = 3.3804$ .

(b)(i)

Adults $= x + 60$ .
Money collected: $2x + 5(x + 60) = 1280$ .
Expand: $7x + 300 = 1280$ , so $7x = 980$ and $x = 140$ .

(ii)

There were 140 children and 200 adults: the ratio is $140 : 200 = 7 : 10$ .

(iii)

Number of people $= 140 + 200 = 340$ .
Average $= \frac{1280}{340} = 3.76$ , so D3.76 per person.

(iv)

Profit $= 1280 - 720 = 560$ , so the profit is D560.
Percentage profit $= \frac{560}{720} \times 100 = 77.8\%$ .

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Question 7

(a)
A woman looking out from the window of a building at a height of 30 m observed that the angle of depression of the top of a flag pole was $44^\circ$ . If the foot of the pole is 25 m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the: (i) angle of depression of the foot of the pole from the woman; (ii) height of the flag pole.
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $O$ is the centre of the circle, $\angle OQR = 32^\circ$ and $\angle TPQ = 15^\circ$ . Calculate: (i) $\angle QPR$ ; (ii) $\angle TQO$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Sketch it: the window $W$ is 30 m above the foot $F$ of the building.
The pole stands 25 m away with foot $O$ and top $T$ .

(i)

The angle of depression $x$ of the foot equals the angle of elevation of $W$ from $O$ : $\tan x = \frac{30}{25} = 1.2$ .
So $x = \tan^{-1}(1.2) = 50.2^\circ$ , which is $50^\circ$ to the nearest degree.

(ii)

The horizontal line from the window meets the pole 25 m away.
The drop from the window to the top of the pole is $25\tan 44^\circ = 24.14$ m.
Height of pole $= 30 - 24.14 = 5.86$ m, which is 6 m to the nearest metre.

(b)(i)

$|OQ| = |OR|$ (radii), so $\angle ORQ = 32^\circ$ and $\angle QOR = 180^\circ - 2(32^\circ)$
$= 116^\circ$ .
The angle at the centre is twice the angle at the circumference: $\angle QPR = \frac{116^\circ}{2}$
$= 58^\circ$ .

(ii)

$\angle TPR = \angle TPQ + \angle QPR$
$= 15^\circ + 58^\circ$
$= 73^\circ$ .
$QTPR$ is a cyclic quadrilateral, so opposite angles add to $180^\circ$ : $\angle TQR = 180^\circ - 73^\circ$
$= 107^\circ$ .
$\angle TQO = \angle TQR - \angle OQR$
$= 107^\circ - 32^\circ$
$= 75^\circ$ .

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Question 8 ✱

Statistics: data & averages

The marks scored by 50 students in a Geography examination are as follows:


60	54	40	67	53	73	37	55	62	43
44	69	39	32	45	58	48	67	39	51
46	59	40	52	61	48	23	60	59	47
65	58	74	47	40	59	68	51	50	50
71	51	26	36	38	70	46	40	51	42

(a)
Using class intervals 21 – 30, 31 – 40, …, prepare a frequency distribution table.
Model answer

Marks 21–30 31–40 41–50 51–60 61–70 71–80

Class mark $x$ 25.5 35.5 45.5 55.5 65.5 75.5

Frequency $f$ 2 10 12 15 8 3

$fx$ 51 355 546 832.5 524 226.5

The frequencies add up to 50 and $\sum fx = 2535$ .
(b)
Calculate the mean mark of the distribution.
(c)
What percentage of the students scored more than 60 marks?

Marks	21–30	31–40	41–50	51–60	61–70	71–80
Class mark $x$	25.5	35.5	45.5	55.5	65.5	75.5
Frequency $f$	2	10	12	15	8	3
$fx$	51	355	546	832.5	524	226.5

Worked solution (try it first)

(a)

Tally each mark into its class.
The frequencies are 2, 10, 12, 15, 8 and 3.
Check they add up to 50.

(b)

Class marks are the midpoints: $25.5, 35.5, \dots, 75.5$ .
Multiply each by its frequency: $51, 355, 546, 832.5, 524, 226.5$ .
Add them: $\sum fx = 2535$ .
Mean $= \frac{\sum fx}{\sum f}$
$= \frac{2535}{50}$
$= 50.7$ .

(c)

More than 60 marks means the classes 61–70 and 71–80: $8 + 3 = 11$ students.
Percentage $= \frac{11}{50} \times 100 = 22\%$ .

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Question 9

(a)
Simplify $\dfrac{x + 2}{x - 2} - \dfrac{x + 3}{x - 1}$ .
(b)(i)
The graph of the equation $y = Ax^2 + Bx + C$ passes through the points $(0, 0)$ , $(1, 4)$ and $(2, 10)$ . Find the value of $C$ .
(b)(ii)
Find the values of $A$ and $B$ .
Separate values with commas, e.g. 3, −2
(b)(iii)
Find the coordinates of the other point where the graph cuts the $x$ -axis.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Use the common denominator $(x - 2)(x - 1)$ : the top is $(x + 2)(x - 1) - (x + 3)(x - 2)$ .
Expand: $(x^2 + x - 2) - (x^2 + x - 6)$ .
Subtract: the top is $4$ , so the answer is $\dfrac{4}{(x - 2)(x - 1)}$ .

(b)(i)

Put in $(0, 0)$ : $0 = A(0) + B(0) + C$ , so $C = 0$ .

(ii)

Put in $(1, 4)$ : $A + B = 4$ .
Put in $(2, 10)$ : $4A + 2B = 10$ , so $2A + B = 5$ .
Subtract the first equation from the second: $A = 1$ .
Then $B = 4 - 1 = 3$ .

(iii)

The curve is $y = x^2 + 3x$ .
On the $x$ -axis $y = 0$ : $x(x + 3) = 0$ , so $x = 0$ or $x = -3$ .
$(0, 0)$ is given, so the other point is $(-3, 0)$ .

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Question 10

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct: (i) quadrilateral $PQRS$ such that $|PQ| = 10$ cm, $|QR| = 8$ cm, $|PS| = 6$ cm, $\angle PQR = 60^\circ$ and $\angle QPS = 75^\circ$ ; (ii) the locus $l_1$ of points equidistant from $QR$ and $RS$ ; (iii) the locus $l_2$ of points equidistant from $R$ and $S$ .
Model answer
Draw $PQ = 10$ cm. At $Q$ construct $60^\circ$ and mark $R$ with $QR = 8$ cm. At $P$ construct $75^\circ$ ( $60^\circ$ plus half of the $30^\circ$ between $60^\circ$ and $90^\circ$ ) and mark $S$ with $PS = 6$ cm. Join $RS$ . $l_1$ is the bisector of angle $QRS$ ; $l_2$ is the perpendicular bisector of $RS$ . Leave every construction arc visible.
(b)
Measure $|RS|$ .

Worked solution (try it first)

(a)(i)

Draw $PQ = 10$ cm.
At $Q$ , construct $60^\circ$ (an arc, then the same radius from where it cuts $QP$ ) and mark $R$ on the arm with $QR = 8$ cm.
At $P$ , construct $90^\circ$ and $60^\circ$ , then bisect the angle between them to get $75^\circ$ .
Mark $S$ with $PS = 6$ cm and join $RS$ .

(ii)

Points equidistant from the lines $QR$ and $RS$ lie on the bisector of angle $QRS$ : draw it and label it $l_1$ .

(iii)

Points equidistant from $R$ and $S$ lie on the perpendicular bisector of $RS$ : draw it and label it $l_2$ .

(b)

Measure $RS$ with the ruler: $|RS| \approx 4.6$ cm.

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Question 11

Plane mensuration

(a)
A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take $\pi = \frac{22}{7}$ ]
(b)
A rope 60 cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the: (i) length; (ii) diagonal of the rectangle.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The circle touches all four sides, so the side of the square is the diameter $2r$ .
Perimeter of the square $= 4 \times 2r = 8r$ .
Circumference $= 2\pi r = \frac{44}{7}r$ .
Add them: $8r + \frac{44}{7}r = 100$ , so $\frac{100}{7}r = 100$ .
So $r = 7$ cm.

(b)(i)

Let the breadth be $b$ .
The length is $4b$ .
The rope is the perimeter: $2(4b + b) = 60$ .
So $10b = 60$ and $b = 6$ cm.
The length is $4 \times 6 = 24.0$ cm.

(ii)

Pythagoras: diagonal $= \sqrt{24^2 + 6^2} = \sqrt{612} = 24.7$ cm.

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Question 12

Trigonometric graphs

(a)

Copy and complete the table of values for $y = \sin x + 2\cos x$ , correct to one decimal place.

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$
$y$		$2.2$				$-1.2$	$-2.0$		$-1.9$

Model answer

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$
$y$	$2.0$	$2.2$	$1.9$	$1.0$	$-0.1$	$-1.2$	$-2.0$	$-2.2$	$-1.9$

(b)
Using a scale of 2 cm to $30^\circ$ on the $x$ -axis and 2 cm to 0.5 units on the $y$ -axis, draw the graph of $y = \sin x + 2\cos x$ for $0^\circ \le x \le 240^\circ$ .
Model answer
Plot the nine points from the table and join them with one smooth curve. Scale: 2 cm to $30^\circ$ , 2 cm to 0.5 units. The curve falls from 2 at $0^\circ$ (after a small rise to about 2.2 near $27^\circ$ ), crosses the $x$ -axis near $117^\circ$ and reaches about $-2.2$ near $207^\circ$ .

For (c) and (d): the curve crosses $y = 0$ at $x \approx 117^\circ$ ; the line $y = 2.1$ meets it at $x \approx 6^\circ$ and $47^\circ$ ; at $x = 171^\circ$ , $y \approx -1.8$ .
(c)(i)
Use your graph to solve the equation $\sin x + 2\cos x = 0$ .
(c)(ii)
Use your graph to solve the equation $\sin x = 2.1 - 2\cos x$ .
Separate values with commas, e.g. 3, −2
(d)
From the graph, find $y$ when $x = 171^\circ$ .

Try it on a graph

x in degrees. The x-axis gives (c)(i); the line y = 2.1 gives (c)(ii).

Worked solution (try it first)

(a)

Work out each value in degree mode and round to 1 decimal place.
For example, $x = 60^\circ$ : $0.866 + 2(0.5) = 1.866 \approx 1.9$ .
$x = 120^\circ$ : $0.866 + 2(-0.5) = -0.134 \approx -0.1$ .
The completed row is $2.0, 2.2, 1.9, 1.0, -0.1, -1.2, -2.0, -2.2, -1.9$ .

(b)

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

(c)(i)

$\sin x + 2\cos x = 0$ where the curve crosses the $x$ -axis: $x \approx 117^\circ$ (exactly, $\tan x = -2$ gives $116.6^\circ$ ).

(ii)

Rearrange: $\sin x + 2\cos x = 2.1$ .
Draw the line $y = 2.1$ and read down where it meets the curve: $x \approx 6^\circ$ and $x \approx 47^\circ$ .

(d)

Read up from $x = 171^\circ$ to the curve and across: $y \approx -1.8$ .

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Question 13

Sequences & series (AP, GP)

(a)
How many numbers between 75 and 500 are divisible by 7?
(b)
The 8th term of an arithmetic progression (A.P.) is 5 times the third term, while the 7th term is 9 greater than the 4th term. Write down the first five terms of the A.P.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The multiples of 7 between 75 and 500 form an A.P.: the first is $77$ ( $7 \times 11$ ) and the last is $497$ ( $7 \times 71$ ), with $d = 7$ .
Use $T_n = a + (n - 1)d$ : $497 = 77 + (n - 1) \times 7$ .
So $(n - 1) \times 7 = 420$ , $n - 1 = 60$ and $n = 61$ .
There are 61 such numbers.

(b)

"7th term is 9 greater than the 4th": $a + 6d = a + 3d + 9$ , so $3d = 9$ and $d = 3$ .
"8th term is 5 times the 3rd": $a + 7d = 5(a + 2d)$ , so $a + 7d = 5a + 10d$ and $4a = -3d$ .
Put in $d = 3$ : $4a = -9$ , so $a = -\frac94$ .
Add 3 each time: the first five terms are $-\frac94, \frac34, \frac{15}{4}, \frac{27}{4}, \frac{39}{4}$ .

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60	54	40	67	53	73	37	55	62	43
44	69	39	32	45	58	48	67	39	51
46	59	40	52	61	48	23	60	59	47
65	58	74	47	40	59	68	51	50	50
71	51	26	36	38	70	46	40	51	42


60	54	40	67	53	73	37	55	62	43
44	69	39	32	45	58	48	67	39	51
46	59	40	52	61	48	23	60	59	47
65	58	74	47	40	59	68	51	50	50
71	51	26	36	38	70	46	40	51	42


60	54	40	67	53	73	37	55	62	43
44	69	39	32	45	58	48	67	39	51
46	59	40	52	61	48	23	60	59	47
65	58	74	47	40	59	68	51	50	50
71	51	26	36	38	70	46	40	51	42