WAEC · 2012 · Nov/Dec · General Maths · Paper 2

Question 1

(a)
Without using tables or a calculator, simplify $\dfrac{0.25 \times 3.3 \times 4200}{2.1 \times 1.65 \times 2}$ , leaving your answer in standard form.
(b)
Solve for $x$ if $\log_{10}(3x + 1) + \log_{10}\left(\frac12\right) - \log_{10}(2x - 5) = 0$ .

Worked solution (try it first)

(a)

Multiply out the decimals as whole numbers and track the powers of 10: $0.25 \times 3.3 \times 4200 = 3465$ and $2.1 \times 1.65 \times 2 = 6.93$ .
Then $\frac{3465}{6.93} = 500 = 5.0 \times 10^2$ .
(Cancelling first is quicker: $\frac{0.25 \times 3.3 \times 4200}{2.1 \times 1.65 \times 2}$ .
$\frac{3.3}{1.65} = 2$ and $\frac{4200}{2.1} = 2000$ , so it is $\frac{0.25 \times 2 \times 2000}{2} = 500$ .)

(b)

Combine into one log: $\log_{10}\frac{(3x + 1) \times \frac12}{2x - 5} = 0$ .
A log is 0 when the number is 1, so $\frac{3x + 1}{2(2x - 5)} = 1$ .
Then $3x + 1 = 4x - 10$ , so $x = 11$ .
Check: $3x + 1 = 34$ and $2x - 5 = 17$ are both positive.

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Question 2

(a)
Given that $U = \{1, 2, 3, 4, 5, 6, 7, 8\}$ , $X = \{1, 3, 5, 7\}$ and $Y = \{1, 5, 8\}$ , find: (i) $X' \cap Y$ ; (ii) $(X' \cup Y)'$ .
Show the answer
(i) $\{8\}$ ; (ii) $\{3, 7\}$
(b)
The probabilities that Kwakye and Sekyere will pass an examination are $\frac23$ and $\frac34$ respectively. If both of them take the examination, find the probability that exactly one of them would pass.

Worked solution (try it first)

(a)(i)

$X'$ is everything in $U$ that is not in $X$ : $X' = \{2, 4, 6, 8\}$ .
The elements in both $X'$ and $Y = \{1, 5, 8\}$ : $X' \cap Y = \{8\}$ .

(ii)

$X' \cup Y = \{1, 2, 4, 5, 6, 8\}$ .
Its complement is what's left in $U$ : $(X' \cup Y)' = \{3, 7\}$ .

(b)

$P(\text{Kwakye fails}) = 1 - \frac23$
$= \frac13$ and $P(\text{Sekyere fails}) = 1 - \frac34$
$= \frac14$ .
Exactly one passes in two ways: Kwakye passes and Sekyere fails, $\frac23 \times \frac14 = \frac16$ .
Or Kwakye fails and Sekyere passes, $\frac13 \times \frac34 = \frac14$ .
Add them: $\frac16 + \frac14 = \frac{5}{12}$ .

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Question 3

(a)
In the diagram, $ABCD$ is part of a right-angled triangle $ODC$ (with $A$ on $OD$ and $B$ on $OC$ ). If $|AB| = 6\text{ cm}$ , $|CD| = 15\text{ cm}$ , $|BC| = 8\text{ cm}$ , $\angle BCD = 90^\circ$ and $AB \parallel DC$ , calculate, correct to 1 decimal place, the: (i) height; (ii) perimeter of the triangle $ODC$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

$AB \parallel DC$ , so triangles $OBA$ and $OCD$ are similar and their sides are in proportion: $\frac{OB}{OC} = \frac{AB}{DC}$ .
With $OC = OB + 8$ : $\frac{OB}{OB + 8} = \frac{6}{15}$ .
Cross-multiply: $15\,OB = 6\,OB + 48$ , so $9\,OB = 48$ and $OB = 5\frac13$ cm.
The height is $OC = 5\frac13 + 8$
$= 13\frac13$
$\approx 13.3$ cm.

(ii)

$\angle OCD = 90^\circ$ , so $OD = \sqrt{15^2 + (13\frac13)^2}$
$= \sqrt{225 + 177.78}$
$\approx 20.07$ cm.
Perimeter $= 13.33 + 15 + 20.07 \approx 48.4$ cm.

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Question 4

(a)
In the diagram, $\overline{SR}$ is parallel to $\overline{UW}$ , $SVT$ and $UVW$ are straight lines. If $\angle RSP = 45^\circ$ (i.e. $\angle RST = 45^\circ$ ) and $\angle VTU = 20^\circ$ , calculate the values of $x = \angle WVT$ and $y = \angle VUT$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $OPR$ is a right-angled triangle with $\angle ORP = 45^\circ$ , $\angle OQP = 60^\circ$ ( $Q$ on $RP$ ) and $\angle OPR = 90^\circ$ . If $|OR| = 5\text{ m}$ , find, correct to 4 significant figures, $|QP|$ .

Worked solution (try it first)

(a)

$SR \parallel UW$ and $SVT$ crosses both, so $\angle WVT = \angle RST = 45^\circ$ (corresponding angles).
So $x = 45^\circ$ .
In $\triangle VUT$ , $\angle WVT$ is an exterior angle, so it equals the sum of the two opposite interior angles: $x = y + 20^\circ$ .
So $y = 45^\circ - 20^\circ = 25^\circ$ .

(b)

In $\triangle OPR$ the right angle is at $P$ , so $OR$ is the hypotenuse.
$OP$ is opposite the $45^\circ$ angle at $R$ : $|OP| = 5\sin 45^\circ \approx 3.536$ m.
In $\triangle OPQ$ , also right-angled at $P$ , $OP$ is opposite the $60^\circ$ angle at $Q$ and $QP$ is adjacent: $\tan 60^\circ = \frac{|OP|}{|QP|}$ .
So $|QP| = \frac{3.536}{\tan 60^\circ}$
$= \frac{3.536}{1.732}$
$\approx 2.041\text{ m}$ (4 significant figures).

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Question 5

(a)
A trader bought an article in Sierra Leone at Le 360,000. He paid $25\%$ of the cost price as import duty on entering Nigeria. If the rate of exchange is ₦1.00 = Le 16.00, how much, in naira, did he pay as import duty?
(b)
The mean of marks scored by 3 students in a test is 9. If the modal mark is 11, find the lowest mark.

Worked solution (try it first)

(a)

Import duty $= 25\%$ of Le 360,000 $=$ Le 90,000.
At ₦1.00 = Le 16.00, divide by 16 to change leones to naira: $90\,000 \div 16 = 5625$ , so the duty is ₦5,625.

(b)

The mean of 3 marks is 9, so the marks add up to $3 \times 9 = 27$ .
The mode is 11, so 11 appears more than once.
It can't appear three times ( $3 \times 11 = 33$ , not 27), so exactly two marks are 11.
The third mark is $27 - 11 - 11 = 5$ , and it's the lowest.

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Question 6 ✱✱

(a)
Simplify: $\left[\dfrac{3m + 1}{6} - \dfrac{m + 2}{5}\right] \div \dfrac{9m - 7}{15}$ .
(b)
The external diameter of a tyre is $52\text{ cm}$ . How many revolutions, correct to four significant figures, will the tyre make to cover a distance of $2.05\text{ km}$ ? $\left[\text{Take }\pi = \frac{22}{7}\right]$
(c)
The diagram is a square cardboard $PRSU$ of side $10\text{ cm}$ , with $Q$ on $PR$ , $T$ on $US$ , $V$ on $PU$ and $QT \parallel PU$ . $|PQ| = 4\text{ cm}$ , $|PV| = 8\text{ cm}$ and $|UT| = 4\text{ cm}$ . If a hole is made at random on the cardboard, what is the probability that it is made on $\triangle QVT$ ?

Worked solution (try it first)

(a)

Combine the bracket over the common denominator 30: $\frac{5(3m + 1) - 6(m + 2)}{30} = \frac{15m + 5 - 6m - 12}{30}$
$= \frac{9m - 7}{30}$ .
Dividing by a fraction means multiplying by it upside down: $\frac{9m - 7}{30} \times \frac{15}{9m - 7}$ .
The $(9m - 7)$ cancels, leaving $\frac{15}{30} = \frac12$ .

(b)

Radius $= 26\text{ cm}$ , so one revolution covers the circumference: $2 \times \frac{22}{7} \times 26 = \frac{1144}{7}\text{ cm}$ .
The distance is $2.05\text{ km} = 205\,000\text{ cm}$ .
Revolutions $= 205\,000 \div \frac{1144}{7}$
$= \frac{1\,435\,000}{1144}$
$\approx 1254.4$ .
Correct to 4 significant figures: 1254 revolutions.

(c)

The square has area $10 \times 10 = 100\text{ cm}^2$ .
In $\triangle QVT$ , take $QT$ as the base: it runs straight across the square, so $|QT| = 10\text{ cm}$ .
The height is the distance from $V$ to the line $QT$ , which is $|PQ| = 4\text{ cm}$ .
Area $= \frac12 \times 10 \times 4$
$= 20\text{ cm}^2$ .
Probability $= \frac{20}{100} = \frac15$ .

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Question 7

(a)
If $x = -\frac14$ , $y = \frac12$ and $z = -\frac13$ , evaluate $\dfrac{x^2 - y}{z - x}$ .
(b)
In one month, a man spent ₦650 on newspapers and ₦240 on soap. The next month, he reduced his newspaper spending by $40\%$ . If the ratio of the amount spent on soap in the second month to that of the first month is $5 : 3$ , what is the difference in the amount spent on both items in the two months?

Worked solution (try it first)

(a)

Top: $x^2 - y = \frac{1}{16} - \frac12$
$= -\frac{7}{16}$ .
Bottom: $z - x = -\frac13 - \left(-\frac14\right)$
$= -\frac13 + \frac14$
$= -\frac{1}{12}$ .
Divide: $-\frac{7}{16} \div \left(-\frac{1}{12}\right) = \frac{7}{16} \times 12$
$= \frac{84}{16}$
$= 5\frac14$ .

(b)

First month: $650 + 240 = ₦890$ .
Second month: newspapers $0.6 \times 650 = ₦390$ .
Soap is $\frac53$ of 240, which is ₦400.
Total ₦790.
Difference $= 890 - 790 = ₦100$ .

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Question 8

Construction & loci

(a)
Using a ruler and a pair of compasses only: (i) construct a parallelogram $PQRS$ with diagonals $|PR| = 8\text{ cm}$ and $|QS| = 10\text{ cm}$ , which intersect at $O$ with $\angle POS = 75^\circ$ ; (ii) construct the perpendicular from $R$ to $\overline{PQ}$ produced, meeting it at $C$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the marks are for the construction. The diagonals of a parallelogram bisect each other: draw $PR = 8$ cm and mark its midpoint $O$ . Construct $75^\circ$ at $O$ ( $60^\circ + 15^\circ$ , bisecting a $30^\circ$ angle) and mark $OS = OQ = 5$ cm on the line through $O$ . Join $PQRS$ . Then drop the perpendicular from $R$ to $PQ$ (produced if necessary): arcs from $R$ cut the line twice, and bisecting between those points gives $C$ . With these measurements the angle at $Q$ is about $77^\circ$ (acute), so $C$ actually lands on $PQ$ itself, about 1.2 cm from $Q$ . Measured: $|PQ| \approx 7.2$ cm, $|PS| \approx 5.5$ cm, $\angle QRS \approx 103^\circ$ .
(b)
Measure: (i) $|PQ|$ ; (ii) $|PS|$ ; (iii) $\angle QRS$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The diagonals of a parallelogram bisect each other, so $|OP| = |OR| = 4$ cm and $|OQ| = |OS| = 5$ cm.
Draw $PR = 8$ cm and mark its midpoint $O$ .
At $O$ construct $75^\circ$ ( $60^\circ$ plus half of the $30^\circ$ between $60^\circ$ and $90^\circ$ ), and draw the line $SOQ$ through $O$ at that angle.
Mark $|OS| = 5$ cm on one side and $|OQ| = 5$ cm on the other, with $\angle POS = 75^\circ$ .
Join $PQRS$ .

(ii)

Produce $PQ$ beyond $Q$ .
With centre $R$ , draw an arc cutting the produced line twice.
From those points draw equal arcs crossing on the other side.
Join $R$ to the crossing.
It meets the line at $C$ .

(b)

Measure: (i) $|PQ| \approx 7.2$ cm.

(ii)

$|PS| \approx 5.5$ cm.

(iii)

$\angle QRS \approx 103^\circ$ .
Check with the cosine rule in triangles $POS$ and $POQ$ : $|PS|^2 = 4^2 + 5^2 - 40\cos 75^\circ$ , so $|PS| \approx 5.5$ cm.
$|PQ|^2 = 4^2 + 5^2 - 40\cos 105^\circ$ , so $|PQ| \approx 7.2$ cm.

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Question 9

Quadratics & their graphs

(a)
Copy and complete the table of values for the relation $y = x^2 + 2x - 2$ .

$x$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$

$y$ $-3$ $-2$

Model answer

$x$ $-4$ $-3$ $-2$ $-1$ $0$ $1$ $2$

$y$ $6$ $1$ $-2$ $-3$ $-2$ $1$ $6$

For example $x = -4$ : $16 - 8 - 2 = 6$ . The values are symmetrical about $x = -1$ , where $y$ is least.
(b)
Using a scale of 2 cm to 1 unit on both axes, draw the graph of $y = x^2 + 2x - 2$ for $-4 \le x \le 2$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The lowest point is $(-1, -3)$ .

For (c): (i) the roots are where the curve crosses the $x$ -axis: $x \approx −2.7$ and $0.7$ . (ii) $x^2 + 2x < 0.5$ means $x^2 + 2x - 2 < -1.5$ : draw $y = -1.5$ and take the part of the curve below it, $−2.2 < x < 0.2$ .
(c)
Use your graph to: (i) find the roots of the equation $x^2 + 2x - 2 = 0$ ; (ii) indicate the region where $x^2 + 2x < 0.5$ .
Separate values with commas, e.g. 3, −2

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$
$y$	$6$	$1$	$-2$	$-3$	$-2$	$1$	$6$

Try it on a graph

The curve and the line y = −1.5.

Worked solution (try it first)

(a)

Put each $x$ into $y = x^2 + 2x - 2$ .
For $x = -4$ : $16 - 8 - 2 = 6$ .
For $x = -3$ : $9 - 6 - 2 = 1$ .
For $x = -2$ : $4 - 4 - 2 = -2$ .
For $x = 1$ : $1 + 2 - 2 = 1$ .
For $x = 2$ : $4 + 4 - 2 = 6$ .
The row is $6, 1, -2, -3, -2, 1, 6$ .

(b)

With 2 cm to 1 unit on both axes, plot the seven points and join them with a smooth U-shaped curve.
Its lowest point is at $(-1, -3)$ .

(c)(i)

The roots of $x^2 + 2x - 2 = 0$ are where the curve crosses the $x$ -axis ( $y = 0$ ): $x \approx -2.7$ and $x \approx 0.7$ .
(Exactly, $-1 \pm \sqrt3$ .)

(ii)

Take 2 from both sides: $x^2 + 2x < 0.5$ becomes $x^2 + 2x - 2 < -1.5$ , that is $y < -1.5$ .
Draw the line $y = -1.5$ and shade the part of the curve below it.
The line meets the curve at $x \approx -2.2$ and $x \approx 0.2$ , so the region is $-2.2 < x < 0.2$ .

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Question 10

(a)
If $3^x + 3^{x + 1} = 36$ , find $x$ .
(b)
The bearing of $P$ from $X$ , 10 km away, is $025^\circ$ . Another point $Q$ is 6 km from $X$ on a bearing of $162^\circ$ . Calculate the: (i) distance $PQ$ ; (ii) bearing of $P$ from $Q$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$3^{x + 1} = 3 \times 3^x$ , so $3^x + 3 \times 3^x = 36$ , which is $4 \times 3^x = 36$ .
So $3^x = 9 = 3^2$ and $x = 2$ .

(b)

Draw north at $X$ .
$P$ is 10 km away on $025^\circ$ and $Q$ is 6 km away on $162^\circ$ .
The angle between the two lines at $X$ is $\angle PXQ = 162^\circ - 25^\circ$
$= 137^\circ$ .

(i)

Cosine rule: $|PQ|^2 = 10^2 + 6^2 - 2(10)(6)\cos 137^\circ$
$= 136 + 120(0.7314)$
$\approx 223.76$ .
So $|PQ| \approx 14.96\text{ km}$ .

(ii)

Sine rule for the angle at $Q$ : $\sin\angle PQX = \frac{10\sin 137^\circ}{14.96}$
$\approx \frac{6.820}{14.96}$
$\approx 0.4559$ , so $\angle PQX \approx 27.1^\circ$ .
At $Q$ , the direction to $X$ is the back bearing of $162^\circ$ , which is $342^\circ$ .
$P$ is $27.1^\circ$ further round clockwise: $342^\circ + 27.1^\circ = 369.1^\circ$ , which is $009^\circ$ (take away $360^\circ$ ).

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Question 11

A sector of a circle with radius $20\text{ cm}$ has an area of $396\text{ cm}^2$ . Calculate, correct to 1 decimal place, the: $\left[\text{Take }\pi = \frac{22}{7}\right]$

(a)
sectoral angle;
(b)
perimeter of the sector;
(c)
volume of the cone formed when the sector is bent such that its straight edges coincide.

Worked solution (try it first)

(a)

$\frac{\theta}{360} \times \frac{22}{7} \times 20^2 = 396$ , so $\theta = \frac{396 \times 360 \times 7}{22 \times 400}$
$= 113.4^\circ$ .

(b)

Arc $= \frac{2 \times \text{area}}{r}$
$= \frac{2 \times 396}{20}$
$= 39.6$ cm.
Perimeter $= 39.6 + 20 + 20 = 79.6$ cm.

(c)

Bent into a cone, the arc becomes the base circumference: $2 \times \frac{22}{7} \times r = 39.6$ , so $r = 6.3$ cm.
The sector's radius, 20 cm, becomes the slant height.
Height: $h = \sqrt{20^2 - 6.3^2}$
$= \sqrt{400 - 39.69}$
$= \sqrt{360.31}$
$\approx 18.98$ cm.
Volume $= \frac13 \times \frac{22}{7} \times 6.3^2 \times 18.98$
$\approx 789.3\text{ cm}^3$ .

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Question 12

The ages (in years) of men selected from a community are as follows:

44 28 58 50 93 35 34 52 57 61 40 63 90 67 64 56 51 82 73 43 73 73 44 71 95 52 71 25 35 79 28 40 72 88 82 63 53 48 98 65 63 44 73 70 68 46 54 62 41 70

(a)
Prepare a grouped frequency distribution table with the intervals $20$ – $29$ , $30$ – $39$ , …
Show the answer
Frequencies $3, 3, 9, 9, 9, 10, 3, 4$ (total 50)
(b)
Calculate the mean deviation of the distribution.

Worked solution (try it first)

(a)

Go through the list once, making a tally mark for each age in its class, then count:
Ages 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–99

Frequency 3 3 9 9 9 10 3 4
The frequencies add up to 50, the number of ages in the list.

Ages	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	3	3	9	9	9	10	3	4

(b)

Class marks: $24.5, 34.5, \ldots, 94.5$ .
$\sum fx = 3(24.5) + 3(34.5) + 9(44.5) + 9(54.5) + 9(64.5) + 10(74.5) + 3(84.5) + 4(94.5)$
$= 3025$ , so the mean is $\frac{3025}{50} = 60.5$ .
The distances of the class marks from 60.5 are $36, 26, 16, 6, 4, 14, 24, 34$ .
Multiply by the frequencies: $108, 78, 144, 54, 36, 140, 72, 136$ , which add up to 768.
Mean deviation $= \frac{\sum f|x - \bar x|}{\sum f}$
$= \frac{768}{50}$
$= 15.36$ .

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Question 13

(a)
The internal diameter of a spherical bowl, half full of water, is $20\text{ cm}$ . The content is poured into an empty cylindrical vessel with internal diameter $10\text{ cm}$ . Calculate, correct to one decimal place, the depth of water in the vessel. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
The fourth term of an Arithmetic Progression (A.P.) is 1 less than twice the second term. If the sixth term is 7, find the first term.

Worked solution (try it first)

(a)

The bowl is a sphere of radius 10 cm, half full, so the water is half a sphere: $\frac12 \times \frac43\pi \times 10^3 = \frac{2000}{3}\pi\text{ cm}^3$ .
In the cylinder (radius 5 cm) the same volume stands to a depth $h$ : $\pi \times 5^2 \times h = \frac{2000}{3}\pi$ .
The $\pi$ cancels: $25h = \frac{2000}{3}$ , so $h = \frac{80}{3} \approx 26.7$ cm.

(b)

$T_4 = a + 3d$ and $T_2 = a + d$ , so $a + 3d = 2(a + d) - 1$ , which gives $a - d = 1$ .
$T_6 = a + 5d = 7$ .
Subtract: $6d = 6$ , so $d = 1$ and $a = 2$ .
The first term is 2.

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