WAEC · 2010 · Nov/Dec · General Maths · Paper 2

Question 1

(a)
Two children shared an amount of money in the ratio $\frac34 : \frac25$ . If the smaller share was GH¢25.00, how much was shared between them?
(b)
A box contains 5 red, 3 green and 4 blue balls of the same size. If a boy picks two balls from the box one after the other without replacement, what is the probability that both balls are red?

Worked solution (try it first)

(a)

Clear the fractions: multiply both parts of $\frac34 : \frac25$ by 20 (the LCM of 4 and 5) to get $15 : 8$ .
The ratio has $15 + 8 = 23$ parts, and the smaller share is 8 of them.
So $\frac{8}{23}$ of the amount is GH¢25.00, and the amount is $25 \times \frac{23}{8} = 71.875$ .
They shared GH¢71.88 (to the nearest pesewa).

(b)

There are $5 + 3 + 4 = 12$ balls.
The first ball is red with probability $\frac{5}{12}$ .
One red ball has gone, so the second is red with probability $\frac{4}{11}$ .
Multiply: $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$
$= \frac{5}{33}$ .

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Question 2

(a)
(i) Solve the inequality $\frac12x - \frac56(x + 2) \le 1 + x$ . (ii) Illustrate the solution on a number line.
Model answer
A solid dot at $-2$ (it is included) with an arrow to the right.
(b)
When the price of an apple increased by ₦5.00, 18 apples cost ₦60.00 more than 20 apples cost before the increase. Find the new price of an apple.

Worked solution (try it first)

(a)(i)

Multiply every term by 6 (the LCM of 2 and 6): $3x - 5(x + 2) \le 6 + 6x$ .
Expand the bracket: $3x - 5x - 10 \le 6 + 6x$ , so $-2x - 10 \le 6 + 6x$ .
Collect terms: $-8x \le 16$ .
Divide by $-8$ and reverse the sign: $x \ge -2$ .

(ii)

On the number line, put a solid dot at $-2$ (because $-2$ is included) and draw an arrow from it to the right.

(b)

Let the old price be ₦ $x$ .
The new price is ₦ $(x + 5)$ .
18 apples at the new price cost ₦60 more than 20 at the old price: $18(x + 5) - 20x = 60$ .
Expand: $18x + 90 - 20x = 60$ , so $-2x = -30$ and $x = 15$ .
The new price is $15 + 5 =$ ₦20.00.

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Question 3 ✱

(a)
In a right-angled triangle, $\sin x = \frac35$ . Evaluate $5\cos^2 x - 3$ .
(b)
The angle of elevation of the top of a vertical pole from a point $63\text{ m}$ east of the base of the pole is $30^\circ$ . From another point due west of the pole, the angle of elevation of the top is $60^\circ$ . (i) Draw a sketch diagram to illustrate the information. (ii) Calculate, correct to three significant figures, the distance of the second point from the base of the pole.
Model answer
The pole $TB$ stands between the two points: $A$ is $63\text{ m}$ east of $B$ , $C$ is $x\text{ m}$ west of $B$ .

Worked solution (try it first)

(a)

Draw a right-angled triangle with opposite side 3 and hypotenuse 5.
By Pythagoras the adjacent side is $\sqrt{25 - 9} = 4$ .
So $\cos x = \frac45$ and $\cos^2 x = \frac{16}{25}$ .
Substitute: $5 \times \frac{16}{25} - 3 = \frac{16}{5} - 3$
$= \frac15$ .

(b)(i)

Sketch the pole $TB$ with $A$ $63\text{ m}$ east of the foot $B$ (elevation $30^\circ$ ) and $C$ west of $B$ (elevation $60^\circ$ ), both on level ground.

(ii)

From $A$ : the height is $|TB| = 63 \tan 30^\circ = 36.37\text{ m}$ .
From $C$ : $\tan 60^\circ = \dfrac{|TB|}{x}$ , so $x = \dfrac{63 \tan 30^\circ}{\tan 60^\circ}$ .
$\tan 30^\circ = \frac{1}{\sqrt3}$ and $\tan 60^\circ = \sqrt3$ , so $x = \frac{63}{3} = 21.0$ .
The second point is $21.0\text{ m}$ from the base of the pole.

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Question 4

(a)
Find the value of $x$ if $x3_{\text{five}} - 14_{\text{five}} = 2x_{\text{five}}$ .
(b)
The diagram is a circle passing through the points $A$ , $B$ , $C$ and $D$ such that $AC$ and $BD$ meet at a point $E$ inside the circle. If $\angle DAC = 27^\circ$ , $\angle ABD = 54^\circ$ and $\angle ACB = 63^\circ$ , find: (i) $\angle CAB$ ; (ii) $\angle AEB$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Write each number in base ten: $x3_{\text{five}} = 5x + 3$ , $14_{\text{five}} = 5 + 4 = 9$ and $2x_{\text{five}} = 10 + x$ .
So $(5x + 3) - 9 = 10 + x$ .
Simplify: $5x - 6 = 10 + x$ , so $4x = 16$ and $x = 4$ .

(b)(i)

Angles in the same segment are equal: $\angle DBC = \angle DAC = 27^\circ$ (both stand on arc $DC$ ).
The angles of triangle $ABC$ add up to $180^\circ$ , and $\angle ABC = 54^\circ + 27^\circ$
$= 81^\circ$ .
So $\angle CAB = 180^\circ - 81^\circ - 63^\circ$
$= 36^\circ$ .

(ii)

In triangle $ABE$ : $\angle AEB = 180^\circ - 36^\circ - 54^\circ$
$= 90^\circ$ .

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Question 5

(a)
The diagonals of a rhombus are $14\text{ cm}$ and $9\text{ cm}$ . Calculate, correct to the nearest centimetre, the perimeter of the rhombus.
(b)
The cross section of a rectangular tank measures $1.2\text{ m}$ by $0.9\text{ m}$ . It contains water to a depth of $0.4\text{ m}$ . If a cubical block of side $50\text{ cm}$ is lowered into the tank, calculate, correct to 2 significant figures, the rise in the water level (in metres).

Worked solution (try it first)

(a)

The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right-angled triangle with legs $7\text{ cm}$ and $4.5\text{ cm}$ .
Side $= \sqrt{7^2 + 4.5^2}$
$= \sqrt{69.25}$
$= 8.32\text{ cm}$ .
All four sides are equal: perimeter $= 4 \times 8.32 = 33.3$ , which is $33\text{ cm}$ to the nearest centimetre.

(b)

Work in metres: the block is $0.5\text{ m}$ on a side, so its volume is $0.5^3 = 0.125\text{ m}^3$ .
The block sinks below the surface (the water will not be $0.5\text{ m}$ deep), so it pushes up $0.125\text{ m}^3$ of water over the base area $1.2 \times 0.9 = 1.08\text{ m}^2$ .
Rise $= \dfrac{0.125}{1.08} = 0.1157\text{ m}$ .
The water level rises by $0.12\text{ m}$ to 2 significant figures.

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Question 6

(a)
In a class of 50 students, 30 offered History, 15 offered History and Geography while 3 did not offer any of the two subjects. (i) Represent the information on a Venn diagram. (ii) Find the number of students that offered: (A) History only; (B) Geography only.
Separate values with commas, e.g. 3, −2
(b)
A trader sold an article at a discount of $8\%$ for ₦828.00. If the article was initially marked to gain $25\%$ , find the: (i) cost price of the article; (ii) discount allowed.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw two overlapping circles $H$ and $G$ inside a rectangle $U$ with $n(U) = 50$ .
Put 15 in the overlap and 3 outside both circles.

(ii)

(A)** History only $= 30 - 15 = 15$ .
(B) The students who offered at least one subject number $50 - 3 = 47$ .
Geography only $= 47 - 30 = 17$ .

(b)

After an $8\%$ discount the price is $92\%$ of the marked price, so the marked price is $828 \times \frac{100}{92} =$ ₦900.00.

(i)

The marked price gives a $25\%$ gain, so it is $125\%$ of the cost price: cost price $= 900 \times \frac{100}{125} =$ ₦720.00.

(ii)

Discount allowed $= 900 - 828 =$ ₦72.00.

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Question 7

Quadratics & their graphs

(a)
Copy and complete the following table for the relation $y = \frac12x(x - 6)$ for $-2 \le x \le 8$ .

$x$ $-2$ $-1$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$y$ $8$ $0$ $-4$ $0$

Separate values with commas, e.g. 3, −2
(b)
Using scales of $2\text{ cm}$ to 1 unit on the $x$ -axis, and $2\text{ cm}$ to 2 units on the $y$ -axis, draw the graph of the relation $y = \frac12x(x - 6)$ for $-2 \le x \le 8$ .
Model answer
Plot all eleven points from the table (2 cm to 1 unit across, 2 cm to 2 units up) and join them with one smooth U-shaped curve, symmetrical about $x = 3$ , with its lowest point at $(3, -4.5)$ .

For (c): the curve is below the $x$ -axis for $0 < x < 6$ ; the minimum value is $-4.5$ ; the line $y = 5$ meets it at $x \approx -1.4$ and $x \approx 7.4$ .
(c)
Use the graph to find the: (i) range of values of $x$ for which $y$ is negative; (ii) minimum value of $y$ ; (iii) roots of the equation $\frac12x(x - 6) = 5$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

The x-axis gives (c)(i); the line y = 5 gives (c)(iii).

Worked solution (try it first)

(a)

Substitute each $x$ .
For example, $x = -1$ : $\frac12(-1)(-7) = 3.5$ .
$x = 3$ : $\frac12(3)(-3) = -4.5$ .
The row is $8.0, 3.5, 0.0, -2.5, -4.0, -4.5, -4.0, -2.5, 0.0, 3.5, 8.0$ .

(b)

Plot the eleven points with the scales given and join them with one smooth curve.
It is symmetrical about $x = 3$ .

(c)(i)

$y$ is negative where the curve is below the $x$ -axis: between the crossings at $x = 0$ and $x = 6$ , so $0 < x < 6$ .

(ii)

The lowest point of the curve is $(3, -4.5)$ , so the minimum value of $y$ is $-4.5$ .

(iii)

Draw the line $y = 5$ and read down from where it meets the curve: $x \approx -1.4$ and $x \approx 7.4$ .
Check: $\frac12x(x - 6) = 5$ gives $x^2 - 6x - 10 = 0$ , so $x = 3 \pm \sqrt{19} = -1.36$ or $7.36$ .

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Question 8

Using ruler and a pair of compasses only:

(a)
Construct a triangle $PQR$ with $|PQ| = 10\text{ cm}$ , $\angle QPR = 90^\circ$ and $\angle PQR = 30^\circ$ .
Model answer
Leave all the construction arcs showing. Draw $PQ = 10$ cm, construct $90^\circ$ at $P$ , and at $Q$ construct $60^\circ$ and bisect it to get $30^\circ$ . The two arms meet at $R$ , with $|PR| \approx 5.8$ cm and $\angle PRQ = 60^\circ$ .
(b)
(i) Construct $l$ , the locus of all points equidistant from $PR$ and $QR$ ; (ii) locate $M$ , the point where $l$ intersects $PQ$ .
Model answer
Points equidistant from the lines $PR$ and $QR$ lie on the bisector of the angle between them, so $l$ is the bisector of $\angle PRQ$ ( $60^\circ$ ), constructed with arcs from $R$ . It meets $PQ$ at $M$ , with $|PM| \approx 3.3$ cm (exactly $\frac{10}{3}$ cm).
(c)
(i) With $M$ as centre and radius $MP$ , draw a circle; (ii) calculate the area of the circle, correct to one decimal place. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

Draw $PQ = 10\text{ cm}$ .
At $P$ construct $90^\circ$ .
At $Q$ construct $60^\circ$ (an equilateral-triangle arc) and bisect it to get $30^\circ$ .
The two arms meet at $R$ .

(b)(i)

Points equidistant from the lines $PR$ and $QR$ lie on the bisector of the angle between them, so bisect $\angle PRQ$ (which is $60^\circ$ ).
This line is $l$ .

(ii)

$l$ cuts $PQ$ at $M$ .
Measure: $|PM| \approx 3.3\text{ cm}$ .
By calculation: $|PR| = 10 \tan 30^\circ = 5.77\text{ cm}$ , and in triangle $PRM$ , $|PM| = 5.77 \tan 30^\circ$
$= \frac{10}{3}$
$= 3.33\text{ cm}$ .

(c)(i)

With centre $M$ and radius $MP$ , draw the circle.
It touches $PR$ at $P$ and also touches $QR$ , because $M$ is the same distance from both lines.

(ii)

Area $= \pi r^2$
$= \frac{22}{7} \times \left(\frac{10}{3}\right)^2$
$= \frac{22}{7} \times \frac{100}{9}$
$= 34.9\text{ cm}^2$ .

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Question 9

$OABCD$ is a right pyramid with a rectangular base $ABCD$ . Its vertical height is $OG$ . If $|AB| = 6\text{ cm}$ , $|BC| = 8\text{ cm}$ and each slant edge is $13\text{ cm}$ ,

(a)
Calculate, correct to one decimal place, the: (i) vertical height $|OG|$ ; (ii) angle between a slant edge and the base $ABCD$ ; (iii) angle between the triangle $OAB$ and the base $ABCD$ .
Separate values with commas, e.g. 3, −2
(b)
Find the volume of the pyramid.

Worked solution (try it first)

(a)(i)

$G$ is the centre of the base, so $|AG|$ is half the diagonal: $|AC| = \sqrt{6^2 + 8^2} = 10$ , so $|AG| = 5\text{ cm}$ .
Pythagoras in triangle $OGA$ : $|OG|^2 = 13^2 - 5^2 = 144$ , so $|OG| = 12.0\text{ cm}$ .

(ii)

The angle between the edge $OA$ and the base is $\angle OAG$ : $\cos \angle OAG = \dfrac{5}{13}$ .
So $\angle OAG = 67.4^\circ$ .

(iii)

Let $X$ be the mid-point of $AB$ .
$GX$ is perpendicular to $AB$ and $|GX| = \frac12|BC| = 4\text{ cm}$ .
The angle between face $OAB$ and the base is $\angle OXG$ : $\tan \angle OXG = \dfrac{12}{4} = 3$ , so the angle is $71.6^\circ$ .

(b)

Volume $= \frac13 \times \text{base area} \times \text{height}$
$= \frac13 \times 6 \times 8 \times 12$
$= 192\text{ cm}^3$ .

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Question 10

Dispersion & cumulative frequency

The table gives the distribution of marks for 360 candidates who sat for an examination.

Marks (%)	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89
Number of candidates	20	48	60	72	80	40	25	10	5

(a)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary ( $9.5, 19.5, \ldots, 89.5$ ), starting from $(-0.5, 0)$ and ending at $(89.5, 360)$ , and join the points with a smooth S-shaped curve. Label both axes.

For (b) and (c): across from 90 and 270 the curve gives $Q_1 \approx 23.2$ and $Q_3 \approx 48.3$ ; up from 74.5 it reads about 350, so about 10 candidates scored $75\%$ or more.
(b)
Use your graph to estimate the semi-interquartile range.
(c)
If the minimum mark for distinction is $75\%$ , how many candidates passed with distinction?

Try it on a graph

The ogive: read the quartiles at 90 and 270, and the curve at 74.5.

Worked solution (try it first)

(a)

Add the frequencies as you go: $20, 68, 128, 200, 280, 320, 345, 355, 360$ .
Plot each total against its upper class boundary ( $9.5, 19.5, \ldots, 89.5$ ), start at $(-0.5, 0)$ , and join the points with a smooth curve.

(b)

Read the quartiles at $\frac{360}{4} = 90$ and $\frac{3 \times 360}{4} = 270$ : $Q_1 \approx 23.2$ and $Q_3 \approx 48.3$ .
Semi-interquartile range $= \frac12(Q_3 - Q_1)$
$= \frac12(48.3 - 23.2)$
$\approx 12.5$ marks.

(c)

A mark of $75\%$ or more starts at $74.5$ .
The curve reads about 350 there.
So about $360 - 350 = 10$ candidates passed with distinction.

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Question 11

(a)
A ship $P$ is $3\text{ km}$ due east of a harbour. Another ship $Q$ is also $3\text{ km}$ from the harbour but on a bearing of $042^\circ$ from the harbour. (i) Find the distance between the two ships. (ii) Find the bearing of ship $Q$ from ship $P$ .
Separate values with commas, e.g. 3, −2
Model answer
(b)
A motorist travelled $300\text{ km}$ at an average speed of $75\text{ km/h}$ and returned at an average speed of $v\text{ km/h}$ . If his average speed for the whole journey is $60\text{ km/h}$ , find $v$ .

Worked solution (try it first)

(a)(i)

Sketch the harbour $H$ with $P$ due east (bearing $090^\circ$ ) and $Q$ on $042^\circ$ .
The angle between them at $H$ is $\angle QHP = 90^\circ - 42^\circ$
$= 48^\circ$ .
$HP = HQ = 3\text{ km}$ , so triangle $HPQ$ is isosceles and the base angles are $\angle HQP = \angle HPQ$
$= \frac12(180^\circ - 48^\circ)$
$= 66^\circ$ .
Sine rule: $\dfrac{|PQ|}{\sin 48^\circ} = \dfrac{3}{\sin 66^\circ}$ , so $|PQ| = \dfrac{3 \sin 48^\circ}{\sin 66^\circ}$
$= 2.44\text{ km}$ .

(ii)

At $P$ , the direction to $H$ is due west ( $270^\circ$ ).
$Q$ is $66^\circ$ round from west towards north.
So the bearing of $Q$ from $P$ is $270^\circ + 66^\circ = 336^\circ$ .

(b)

Average speed is total distance over total time.
The whole journey is $600\text{ km}$ at $60\text{ km/h}$ , so it takes $\frac{600}{60} = 10$ hours.
The outward trip takes $\frac{300}{75} = 4$ hours, so the return takes $10 - 4 = 6$ hours.
So $v = \frac{300}{6} = 50$ .

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$P$ (lat $40^\circ$ N, long $18^\circ$ W) and $Q$ (lat $40^\circ$ N, long $78^\circ$ W) are two cities on the surface of the earth. Calculate the: $\left[\text{Take }\pi = \frac{22}{7}\text{ and the radius of the earth} = 6400\text{ km}\right]$

(a)
radius of the parallel of latitude on which $P$ and $Q$ lie, correct to the nearest $10\text{ km}$ ;
(b)
length of the minor arc $PQ$ , correct to the nearest $100\text{ km}$ ;
(c)
vertical distance between the centre of the earth and the centre of the small circle on which $P$ and $Q$ lie, correct to the nearest km.

Worked solution (try it first)

(a)

The radius of the parallel of latitude $\theta$ is $R\cos\theta$ : $r = 6400 \cos 40^\circ$
$= 6400 \times 0.7660$
$= 4902.7\text{ km}$ .
To the nearest $10\text{ km}$ , $r = 4900\text{ km}$ .

(b)

Both places are west, so the difference in longitude is $78^\circ - 18^\circ = 60^\circ$ .
Arc $PQ = \dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 4902.7$
$= 5136\text{ km}$ .
To the nearest $100\text{ km}$ , the arc is $5100\text{ km}$ .

(c)

In the cross-section, the centre $X$ of the small circle is directly above the centre $O$ of the earth, and $OP$ makes $40^\circ$ with the equator, so $|OX| = R \sin 40^\circ$ .
$|OX| = 6400 \times 0.6428$
$= 4114\text{ km}$ to the nearest km.

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Question 13

(a)
The second, fourth and sixth terms of an Arithmetic Progression (A.P.) are $x - 1$ , $x + 1$ and $7$ respectively. Find the: (i) common difference; (ii) first term; (iii) value of $x$ .
Separate values with commas, e.g. 3, −2
(b)
A spherical bowl of radius $r\text{ cm}$ is one-quarter full when 6 litres of water is poured into it. Calculate, correct to three significant figures, its diameter. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

The $n$ th term of an A.P. is $a + (n - 1)d$ .
So $a + d = x - 1$ , $a + 3d = x + 1$ and $a + 5d = 7$ .

(i)

Take the second term from the fourth: $2d = (x + 1) - (x - 1) = 2$ , so $d = 1$ .

(ii)

Put $d = 1$ into $a + 5d = 7$ : $a = 2$ .

(iii)

Put $a = 2$ , $d = 1$ into $a + d = x - 1$ : $3 = x - 1$ , so $x = 4$ .

(b)

6 litres is $6000\text{ cm}^3$ , and this is a quarter of the volume of the sphere: $\frac14 \times \frac43 \times \frac{22}{7} \times r^3 = 6000$ .
Simplify: $\frac{22}{21} r^3 = 6000$ , so $r^3 = 6000 \times \frac{21}{22} = 5727.3$ .
Take the cube root: $r = 17.89\text{ cm}$ .
The diameter is $2r = 35.8\text{ cm}$ to 3 significant figures.

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