WAEC · 2016 · Private · General Maths · Paper 2

Question 1

(a)
Simplify: $\dfrac{6\frac14 - 3\frac25}{2\frac12 - 1\frac35}$ .
(b)
The average age of students in a class is 15 years. When the teacher's age of 45 years is added, the average age becomes 18. How many students are in the class?

Worked solution (try it first)

(a)

Work out the top and the bottom separately.
Top: $6\frac14 - 3\frac25 = \frac{25}{4} - \frac{17}{5}$
$= \frac{125 - 68}{20}$
$= \frac{57}{20}$ .
Bottom: $2\frac12 - 1\frac35 = \frac52 - \frac85$
$= \frac{25 - 16}{10}$
$= \frac{9}{10}$ .
Then divide: $\frac{57}{20} \div \frac{9}{10} = \frac{57}{20} \times \frac{10}{9}$
$= \frac{570}{180}$
$= \frac{19}{6}$
$= 3\frac16$ .

(b)

Let there be $n$ students.
Their ages add up to $15n$ (mean × number).
Adding the teacher makes $n + 1$ people with a total age of $15n + 45$ and a mean of 18: $\frac{15n + 45}{n + 1} = 18$ .
So $15n + 45 = 18n + 18$ , which gives $3n = 27$ and $n = 9$ .
There are 9 students.
Check: 9 students total 135 years.
With the teacher, $\frac{180}{10} = 18$ .

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Question 2

(a)
If $\dfrac{m}{m - y + 2} = \dfrac{r}{y + r - 1}$ , make $y$ the subject of the equation.
(b)
In a school, $60\%$ of the girls and $70\%$ of the boys own a bicycle. If a boy and a girl are selected at random from the school, find the probability that only one of them owns a bicycle.

Worked solution (try it first)

(a)

Cross-multiply: $m(y + r - 1) = r(m - y + 2)$ .
Expand: $my + mr - m = mr - ry + 2r$ .
The $mr$ terms cancel: $my - m = -ry + 2r$ .
Collect the $y$ terms on the left: $my + ry = m + 2r$ .
Take out $y$ : $y(m + r) = m + 2r$ .
So $y = \frac{m + 2r}{m + r}$ .

(b)

A girl owns a bicycle with probability $0.6$ (so doesn't with $0.4$ ).
A boy with $0.7$ (doesn't with $0.3$ ). "Only one of them" happens in two ways.
The girl owns one and the boy doesn't: $0.6 \times 0.3 = 0.18$ .
The boy owns one and the girl doesn't: $0.7 \times 0.4 = 0.28$ .
Add the two ways: $0.18 + 0.28 = 0.46 = \frac{23}{50}$ .

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Question 3

(a)
If $x = -1$ , $y = -3$ , $z = -4$ and $w = -7$ , evaluate $\dfrac{x^3 - y^2}{2w - z}$ .
(b)
In the diagram, $\overline{MN}$ and $\overline{MQ}$ are tangents to the circle centre $O$ , and $PQ$ is a diameter. If $\angle MNQ = x$ , $\angle NMQ = y$ and $\angle NQP = 46^\circ$ , find the value of: (i) $x$ ; (ii) $y$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Substitute, with brackets round each negative number: $\frac{(-1)^3 - (-3)^2}{2(-7) - (-4)} = \frac{-1 - 9}{-14 + 4}$
$= \frac{-10}{-10}$
$= 1$ .

(b)(i)

The tangent $MQ$ meets the radius $OQ$ at $90^\circ$ , and $PQ$ is a diameter, so $\angle MQP = 90^\circ$ .
Then $\angle MQN = 90^\circ - 46^\circ$
$= 44^\circ$ .
Tangents from $M$ are equal ( $MN = MQ$ ), so triangle $MNQ$ is isosceles and $x = \angle MNQ = \angle MQN = 44^\circ$ .

(ii)

In triangle $MNQ$ : $y = 180^\circ - 44^\circ - 44^\circ$
$= 92^\circ$ .

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Question 4

(a)
Solve the inequality $\frac23(1 - 4x) - \frac12(5 - 3x) \le \frac14(7 + 9x) - \frac13$ .
Show the answer
$x \ge -\frac{39}{41}$
(b)
A man standing $3\text{ m}$ away from a tree observed that the angle of elevation of the top of the tree and the angle of depression of the bottom of the tree are $65^\circ$ and $20^\circ$ respectively. Find, correct to 3 significant figures, the height of the tree.

Worked solution (try it first)

(a)

Multiply every term by 12: $8(1 - 4x) - 6(5 - 3x) \le 3(7 + 9x) - 4$ .
Expand: $8 - 32x - 30 + 18x \le 21 + 27x - 4$ .
So $-22 - 14x \le 17 + 27x$ .
Collect terms: $-39 \le 41x$ , so $x \ge -\frac{39}{41}$ .

(b)

The man's eye is level with some point on the tree.
Up from his eye level to the top: $3\tan 65^\circ \approx 6.434$ m.
Down from his eye level to the bottom: $3\tan 20^\circ \approx 1.092$ m.
The tree's height is the sum: $6.434 + 1.092 \approx 7.53\text{ m}$ .

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Question 5

(a)
The diagram shows a semicircular arc on a school gate with centre $O$ and diameter $AE$ , with a shaded segment $BCD$ . If $|AE| = 14$ metres and $\angle AOB = \angle DOE = 45^\circ$ , calculate the cost of painting the shaded portion at the rate of ₦750.00 per square metre. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

The radius is half of $AE$ : $r = 7$ m.
The angles on the diameter add up to $180^\circ$ , so $\angle BOD = 180^\circ - 45^\circ - 45^\circ$
$= 90^\circ$ .
The shaded segment $BCD$ is sector $BOD$ minus triangle $BOD$ .
Sector $= \frac{90}{360} \times \frac{22}{7} \times 7^2$
$= 38.5\text{ m}^2$ .
Triangle $= \frac12 \times 7 \times 7$
$= 24.5\text{ m}^2$ .
Segment $= 38.5 - 24.5 = 14\text{ m}^2$ .
Cost $= 14 \times ₦750 = ₦10,500$ .

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Question 6

(a)
A worker's annual income was $20\%$ tax-free and the remainder was taxed at a rate of 25 kobo in the naira. If he paid a tax of ₦14,000.00 in one year, how much was his income?
(b)
Every month a family spends $\frac25$ of its monthly income on education, $\frac16$ on clothes, $\frac38$ on food and saves the rest. If its monthly income is Le 36,000.00, how many years would it take to save Le 63,000.00?

Worked solution (try it first)

(a)

Let the income be ₦ $x$ .
The taxed part is $80\%$ of it, $0.8x$ .
Tax at 25 kobo in the naira is $25\%$ : $0.25 \times 0.8x = 0.2x = 14\,000$ , so $x = ₦70,000$ .

(b)

The fractions spent add up to $\frac25 + \frac16 + \frac38 = \frac{48 + 20 + 45}{120}$
$= \frac{113}{120}$ , so the family saves $\frac{7}{120}$ of its income: $\frac{7}{120} \times 36\,000 =$ Le 2,100 a month.
To save Le 63,000 takes $\frac{63\,000}{2100} = 30$ months $= 2\frac12$ years.

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Question 7

Longitude & latitude

$X(60^\circ\text{N}, 12^\circ\text{E})$ and $Y(60^\circ\text{N}, 42^\circ\text{E})$ are points on the earth's surface.

(a)
Illustrate this 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 earth with centre $O$ , the equator, and the circle of latitude $60^\circ$ N, with $X$ and $Y$ on it. The $60^\circ$ angle is at $O$ , from the equator plane up to the latitude. A second sketch of the $60^\circ$ N circle seen from above helps: its radius is $6400\cos 60^\circ = 3200$ km, and $X$ and $Y$ are $42^\circ - 12^\circ = 30^\circ$ apart at its centre. The chord $XY$ is a straight line; the distance "along the common latitude" is the arc.
(b)
Taking $\pi = 3.142$ and the radius of the earth $= 6400\text{ km}$ , calculate the: (i) length of the chord $XY$ , correct to the nearest 10 km; (ii) angle that the chord $XY$ subtends at the centre of the earth, correct to one decimal place; (iii) distance between $X$ and $Y$ along their common latitude.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw the earth with the equator and the circle of latitude $60^\circ$ N.
Mark $X$ at $12^\circ$ E and $Y$ at $42^\circ$ E on that circle, and the angle $60^\circ$ at the centre of the earth up to the latitude.
The circle of latitude $60^\circ$ N has radius $6400\cos 60^\circ = 3200$ km, and the difference in longitude is $42^\circ - 12^\circ = 30^\circ$ .

(b)(i)

In the circle of latitude, the chord $XY$ subtends $30^\circ$ at its centre: $XY = 2 \times 3200 \times \sin 15^\circ$
$\approx 1656.4$ km, which is 1660 km to the nearest 10 km.

(ii)

The same chord seen from the centre of the earth (radius 6400 km): $2 \times 6400 \times \sin\frac{\alpha}{2} = 1656.4$ , so $\sin\frac{\alpha}{2} \approx 0.1294$ , $\frac{\alpha}{2} \approx 7.44^\circ$ and $\alpha \approx 14.9^\circ$ .

(iii)

Along the latitude: $\frac{30}{360} \times 2 \times 3.142 \times 3200 \approx 1675.73$ km.

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An eagle flies from point $X$ to $Y$ , a distance of $320\text{ km}$ , on a bearing of N $35^\circ$ W. It then changes direction and flies $550\text{ km}$ to point $Z$ on a bearing of S $55^\circ$ W. It then returns to its starting point $X$ from $Z$ .

(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 each point before measuring a bearing. From $X$ , $Y$ is 320 km on N $35^\circ$ W ( $35^\circ$ west of north). From $Y$ , $Z$ is 550 km on S $55^\circ$ W ( $55^\circ$ west of south). The angle at $Y$ is $35^\circ + 55^\circ = 90^\circ$ , which makes the calculation easy. Join $Z$ back to $X$ .
(b)
Calculate the: (i) total distance covered, correct to 3 significant figures; (ii) bearing of $X$ from $Z$ , correct to the nearest degree.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

N $35^\circ$ W is the bearing $325^\circ$ , and S $55^\circ$ W is $235^\circ$ .
Draw north at $X$ and draw $XY$ (320 km) on $325^\circ$ .
Draw north at $Y$ and draw $YZ$ (550 km) on $235^\circ$ .
Join $Z$ to $X$ .

(b)(i)

At $Y$ , the direction back to $X$ is $325^\circ - 180^\circ = 145^\circ$ and the direction to $Z$ is $235^\circ$ , so $\angle XYZ = 235^\circ - 145^\circ$
$= 90^\circ$ .
Pythagoras: $|XZ| = \sqrt{320^2 + 550^2}$
$= \sqrt{404\,900}$
$\approx 636.3$ km.
The total distance flown is $320 + 550 + 636.3 = 1506.3 \approx 1510$ km (3 significant figures).

(ii)

In the right-angled triangle, at $Z$ : $\tan\angle YZX = \frac{320}{550}$ , so $\angle YZX \approx 30.2^\circ$ .
At $Z$ , the direction to $Y$ is the back bearing of $235^\circ$ , which is $055^\circ$ .
$X$ is $30.2^\circ$ further round clockwise: $055^\circ + 30.2^\circ \approx 085^\circ$ (N $85^\circ$ E).

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Question 9 ✱

(a)
Solve: $3\log_{10}2 - 2\log_{10}3 = 1 + \log_{10}\left(\frac1x\right)$ .
(b)
The height of a cylindrical water container is $8\text{ m}$ . It took an athlete, running with a speed of $3\text{ km/h}$ , 3 minutes to run round the container once, keeping a constant distance of one metre from the container. Calculate, correct to the nearest whole number, the: (i) radius of the container; (ii) volume of the container. $\left[\text{Take }\pi = \frac{22}{7}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$3\log_{10}2 - 2\log_{10}3 = \log_{10}\frac{8}{9}$ , and $1 + \log_{10}\frac1x = \log_{10}\frac{10}{x}$ .
So $\frac89 = \frac{10}{x}$ and $x = \frac{90}{8} = 11\frac14$ .

(b)(i)

In 3 minutes at 3 km/h the athlete runs $3000 \times \frac{3}{60} = 150$ m.
He runs 1 m outside the container, on a circle of radius $r + 1$ : $2 \times \frac{22}{7} \times (r + 1) = 150$ , so $r + 1 \approx 23.86$ and $r \approx 22.86$ , which is 23 m to the nearest whole number.

(ii)

Volume $= \pi r^2 h$
$= \frac{22}{7} \times 23^2 \times 8$
$\approx 13\,300.6$ , which is 13,301 m³ to the nearest whole number.

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Question 10

(a)
Copy and complete the table of values for the relation $y = 2x^2 - 3x - 1$ .

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

$y$ $13$ $-1$

Model answer

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

$y$ $26$ $13$ $4$ $-1$ $-2$ $1$ $8$ $19$

For example $x = -3$ : $18 + 9 - 1 = 26$ , and $x = 1$ : $2 - 3 - 1 = -2$ .
(b)
Using scales of 2 cm to 1 unit on the $x$ -axis and 2 cm to 5 units on the $y$ -axis, draw the graph of $y = 2x^2 - 3x - 1$ for $-3 \le x \le 4$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points).

For (c): (i) $2x^2 - 3x = 1$ is $y = 0$ , so read where the curve crosses the $x$ -axis: $x \approx −0.3 and 1.8$ . (ii) The gradient is 0 at the lowest point, $(0.75, -2.125)$ . (iii) Draw $y = 3x - 1$ (through $(0, -1)$ and $(3, 8)$ ); the curve is below the line between $x = 0$ and $x = 3$ , so shade the area between them there.
(c)
Using the graph: (i) solve the equation $2x^2 - 3x = 1$ ; (ii) find the coordinates of the point where the gradient of the curve is $0$ ; (iii) shade the area for which $y \le 3x - 1$ .
Separate values with commas, e.g. 3, −2

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$26$	$13$	$4$	$-1$	$-2$	$1$	$8$	$19$

Try it on a graph

The curve, the line y = 3x − 1, and the turning point.

Worked solution (try it first)

(a)

Put each $x$ into $y = 2x^2 - 3x - 1$ .
For example $x = -3$ : $2(9) + 9 - 1 = 26$ .
$x = 1$ : $2 - 3 - 1 = -2$ .
The row is $26, 13, 4, -1, -2, 1, 8, 19$ .

(b)

Plot the eight points with the given scales and join them with a smooth U-shaped curve (not straight lines).

(c)(i)

$2x^2 - 3x = 1$ is the same as $2x^2 - 3x - 1 = 0$ , that is $y = 0$ .
Read where the curve crosses the $x$ -axis: $x \approx -0.3$ and $x \approx 1.8$ .

(ii)

The gradient is 0 at the lowest point of the curve: about $(0.75, -2.1)$ .
(Exactly: halfway between the roots, $x = \frac34$ , where $y = -2.125$ .)

(iii)

Draw the line $y = 3x - 1$ through $(0, -1)$ and $(3, 8)$ .
It meets the curve at $x = 0$ and $x = 3$ .
The points of the curve with $y \le 3x - 1$ are where the curve is on or below the line: shade the area between the line (above) and the curve (below), from $x = 0$ to $x = 3$ .

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Question 11

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct: (i) a parallelogram $ABCD$ with diagonals $|AC| = 10\text{ cm}$ and $|BD| = 7\text{ cm}$ intersecting at $K$ , with $\angle BKC = 60^\circ$ ; (ii) the locus $l_1$ of points equidistant from $B$ and $C$ ; (iii) the locus $l_2$ of points $5\text{ cm}$ from $B$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. The diagonals bisect each other: draw $AC = 10$ cm and mark its midpoint $K$ . Construct $60^\circ$ at $K$ and mark $KB = KD = 3.5$ cm on the line through $K$ . Join $ABCD$ . Then (ii) bisect $BC$ perpendicularly for $l_1$ , and (iii) draw the circle centre $B$ , radius 5 cm, for $l_2$ .
(b)
Locate the points of intersection $M$ and $N$ of loci $l_1$ and $l_2$ . Measure: (i) $|MN|$ ; (ii) $|BC|$ .
Separate values with commas, e.g. 3, −2
Model answer
$M$ and $N$ are where the perpendicular bisector cuts the circle. Measured: $|MN| \approx$ 9.0 cm and $|BC| \approx$ 4.4 cm.

Worked solution (try it first)

(a)(i)

The diagonals bisect each other: $|AK| = |KC| = 5$ cm and $|BK| = |KD| = 3.5$ cm.
Draw $AC = 10$ cm and mark its midpoint $K$ .
At $K$ construct $60^\circ$ and draw the line $BKD$ at that angle, with $\angle BKC = 60^\circ$ .
Mark 3.5 cm each side of $K$ for $B$ and $D$ .
Join $ABCD$ .

(ii)

$l_1$ , equidistant from $B$ and $C$ : construct the perpendicular bisector of $BC$ .

(iii)

$l_2$ , 5 cm from $B$ : draw the circle with centre $B$ and radius 5 cm.

(b)

$M$ and $N$ are where the circle cuts the bisector.
Measure: (i) $|MN| \approx 9.0$ cm.

(ii)

$|BC| \approx 4.4$ cm.
Check: by the cosine rule, $|BC|^2 = 5^2 + 3.5^2 - 2(5)(3.5)\cos 60^\circ$
$= 19.75$ , so $|BC| \approx 4.44$ cm.
$M$ and $N$ are $\sqrt{5^2 - 2.22^2} \approx 4.48$ cm either side of the midpoint of $BC$ , so $|MN| \approx 9.0$ cm.

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Question 12

Marks	3	4	5	6	7	8	9
Frequency	1	4	3	5	2	$x$	2

The table shows the distribution of marks of students in a Mathematics test. If the mean of the distribution is 6, calculate the:

(a)
value of $x$ ;
(b)
standard deviation, correct to 2 decimal places.

Worked solution (try it first)

(a)

$\sum f = 1 + 4 + 3 + 5 + 2 + x + 2$
$= 17 + x$ and $\sum fx = 3 + 16 + 15 + 30 + 14 + 8x + 18$
$= 96 + 8x$ .
The mean is 6, so $\frac{96 + 8x}{17 + x} = 6$ .
Then $96 + 8x = 102 + 6x$ , so $2x = 6$ and $x = 3$ .
Check: $\sum f = 20$ , $\sum fx = 120$ , and $\frac{120}{20} = 6$ .

(b)

Set out the table with the mean 6:
$x$ 3 4 5 6 7 8 9 Total

$f$ 1 4 3 5 2 3 2 20

$(x - 6)^2$ 9 4 1 0 1 4 9

$f(x - 6)^2$ 9 16 3 0 2 12 18 60
Standard deviation $= \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}$
$= \sqrt{\frac{60}{20}}$
$= \sqrt 3$
$\approx 1.73$ .

$x$	3	4	5	6	7	8	9	Total
$f$	1	4	3	5	2	3	2	20
$(x - 6)^2$	9	4	1	0	1	4	9
$f(x - 6)^2$	9	16	3	0	2	12	18	60

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Question 13

(a)
Using a scale of 2 cm to 2 units on each axis, draw on a sheet of graph paper two perpendicular axes $Ox$ and $Oy$ for $-10 \le x \le 10$ and $-12 \le y \le 12$ .
Model answer
Draw both axes with 2 cm to 2 units, from $-10$ to $10$ across and $-12$ to $12$ up, and label them. Then plot and join each shape:
- $PQRS$ : $P(-5, -4)$ , $Q(2, -1)$ , $R(0, 3)$ , $S(-8, 4)$ .
- Translation by $\begin{pmatrix} 3 \\ -8 \end{pmatrix}$ (add 3 to $x$ , subtract 8 from $y$ ): $P_1(-2, -12)$ , $Q_1(5, -9)$ , $R_1(3, -5)$ , $S_1(-5, -4)$ .
- Enlargement, scale factor $-\frac12$ from $O$ (halve each coordinate and change its sign): $P_2(2.5, 2)$ , $Q_2(-1, 0.5)$ , $R_2(0, -1.5)$ , $S_2(4, -2)$ . The image is upside down and on the other side of $O$ .
Notice $S_1$ lands exactly on $P$ .
(b)
Draw on this graph, indicating the coordinates of all vertices: (i) the quadrilateral $PQRS$ with vertices $P(-5, -4)$ , $Q(2, -1)$ , $R(0, 3)$ and $S(-8, 4)$ ; (ii) the image $P_1Q_1R_1S_1$ of $PQRS$ under a translation by the vector $\begin{pmatrix} 3 \\ -8 \end{pmatrix}$ ; (iii) the image $P_2Q_2R_2S_2$ of $PQRS$ under an enlargement from the origin with scale factor $-\frac12$ .
Model answer
Plot and join each shape, labelling every vertex with its coordinates: $PQRS$ with $P(-5, -4)$ , $Q(2, -1)$ , $R(0, 3)$ , $S(-8, 4)$ ; the translation $P_1(-2, -12)$ , $Q_1(5, -9)$ , $R_1(3, -5)$ , $S_1(-5, -4)$ (dashed; $S_1$ lands on $P$ ); and the enlargement with scale factor $-\frac12$ from $O$ , $P_2(2.5, 2)$ , $Q_2(-1, 0.5)$ , $R_2(0, -1.5)$ , $S_2(4, -2)$ (highlighted), which is upside down on the other side of $O$ .
(c)
Find the equation of the line $P_1S$ .
Show the answer
$8x + 3y + 52 = 0$

Try it on a graph

PQRS (blue), the translated image (red), the enlarged image (green).

Worked solution (try it first)

(a)

With 2 cm to 2 units, each 1 cm is 1 unit.
Draw the $x$ -axis from $-10$ to $10$ and the $y$ -axis from $-12$ to $12$ , crossing at $O$ , and number them.

(b)(i)

Plot $P(-5, -4)$ , $Q(2, -1)$ , $R(0, 3)$ and $S(-8, 4)$ and join them in order.

(ii)

A translation by $\begin{pmatrix} 3 \\ -8 \end{pmatrix}$ adds 3 to each $x$ and takes 8 from each $y$ : $P_1(-2, -12)$ , $Q_1(5, -9)$ , $R_1(3, -5)$ , $S_1(-5, -4)$ .
Plot and join them.

(iii)

An enlargement from the origin with scale factor $-\frac12$ multiplies each coordinate by $-\frac12$ : $P_2(2.5, 2)$ , $Q_2(-1, 0.5)$ , $R_2(0, -1.5)$ , $S_2(4, -2)$ .
Plot and join them.

(c)

$P_1(-2, -12)$ and $S(-8, 4)$ : gradient $= \frac{4 - (-12)}{-8 - (-2)}$
$= \frac{16}{-6}$
$= -\frac83$ .
Through $S(-8, 4)$ : $y - 4 = -\frac83(x + 8)$ .
Multiply by 3: $3y - 12 = -8x - 64$ , so $8x + 3y + 52 = 0$ .
(Check with $P_1$ : $-16 - 36 + 52 = 0$ ✓.)

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