WAEC · 2009 · Nov/Dec · General Maths · Paper 2

Question 1

(a)
Simplify, without using tables or a calculator, $3\frac13 \div 1\frac14 - \frac49$ .
(b)
Simplify, without using tables or a calculator, $2 + \sqrt{96} - 4(\sqrt6 - 1)^2$ , and express your answer in the form $m + n\sqrt6$ , where $m$ and $n$ are real numbers.

Worked solution (try it first)

(a)

Change to improper fractions: $3\frac13 = \frac{10}{3}$ and $1\frac14 = \frac54$ .
Divide first (BODMAS): $\frac{10}{3} \div \frac54 = \frac{10}{3} \times \frac45$
$= \frac83$ .
Subtract with denominator 9: $\frac{24}{9} - \frac49 = \frac{20}{9}$
$= 2\frac29$ .

(b)

Simplify the surd: $\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt6$ .
Expand the square: $(\sqrt6 - 1)^2 = 6 - 2\sqrt6 + 1$
$= 7 - 2\sqrt6$ .
Multiply by 4: $4(7 - 2\sqrt6) = 28 - 8\sqrt6$ .
Combine: $2 + 4\sqrt6 - 28 + 8\sqrt6 = -26 + 12\sqrt6$ .

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Question 2

Linear & simultaneous equations

(a)
Solve the equation $\dfrac{4x - 1}{3} - \dfrac{3x - 1}{2} = \dfrac{5 - 2x}{4}$ .
(b)
From a shop, Kofi bought 2 singlets and 3 shirts for GH¢31.00, while Kwasi bought 3 singlets and 2 shirts for GH¢29.00. How much will Yaw pay for one singlet and one shirt bought from the same shop?

Worked solution (try it first)

(a)

Multiply every term by the LCM 12: $4(4x - 1) - 6(3x - 1) = 3(5 - 2x)$ .
Expand: $16x - 4 - 18x + 6 = 15 - 6x$ .
Simplify the left: $-2x + 2 = 15 - 6x$ .
Collect terms: $4x = 13$ , so $x = \frac{13}{4} = 3\frac14$ .

(b)

Let a singlet cost GH¢ $x$ and a shirt GH¢ $y$ : $2x + 3y = 31$ and $3x + 2y = 29$ .
Add the equations: $5x + 5y = 60$ , so $x + y = 12$ .
(Solving fully gives $x = 5$ and $y = 7$ .) Yaw pays GH¢12.00.

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The probability that a malaria patient $M$ survives when given a newly discovered drug is $0.27$ , and the probability that a typhoid patient $T$ survives when injected with another newly discovered drug is $0.85$ . Give your answers correct to 2 significant figures.

(a)
What is the probability that either of the two patients survives?
(b)
What is the probability that neither of the two patients survives?
(c)
What is the probability that at least one of the two patients survives?

Worked solution (try it first)

Write down the probabilities of not surviving: $P(M') = 1 - 0.27 = 0.73$ and $P(T') = 1 - 0.85 = 0.15$ .

(a)

"Either survives" means exactly one of them survives: $M$ survives and $T$ does not, or $T$ survives and $M$ does not.
$P = 0.27 \times 0.15 + 0.85 \times 0.73$
$= 0.0405 + 0.6205$
$= 0.661$ , which is $0.66$ to 2 s.f.

(b)

Neither survives: $0.73 \times 0.15 = 0.1095$ , which is $0.11$ to 2 s.f.

(c)

At least one survives is everything except "neither": $1 - 0.1095 = 0.8905$ , which is $0.89$ to 2 s.f.

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Question 4 ✱

Solid mensuration

A sector of angle $135^\circ$ is cut from a thin circular metal sheet of radius 40 cm. The sector is then folded, with its straight edges coinciding, to form a right circular cone. [Take $\pi = \frac{22}{7}$ ]

(a)
Calculate the base radius of the cone, correct to two decimal places.
(b)
Calculate the greatest volume of liquid which the cone can hold, correct to the nearest $\text{cm}^3$ .

Worked solution (try it first)

(a)

The arc of the sector becomes the circumference of the base, and the radius of the sheet becomes the slant height $l = 40$ cm.
Arc length $= \frac{135}{360} \times 2\pi \times 40$
$= 30\pi$ cm.
Set it equal to $2\pi r$ : $2\pi r = 30\pi$ , so $r = 15.00$ cm.

(b)

Height by Pythagoras: $h = \sqrt{40^2 - 15^2}$
$= \sqrt{1375}$
$= 37.081$ cm.
Volume $= \frac13\pi r^2 h$
$= \frac13 \times \frac{22}{7} \times 15^2 \times 37.081$ .
$= 8740.5$ , so the cone holds $8741\text{ cm}^3$ to the nearest $\text{cm}^3$ .

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Question 5

(a)
In the diagram, the two circles intersect at $X$ and $Y$ . The centre $O$ of the smaller circle is on the circumference of the bigger circle. $A$ and $B$ are any two points on the major arcs, one on each circle. Find an equation connecting $a$ and $b$ .
Model answer
Join $OX$ and $OY$ . In the small circle, $\angle XOY = 2b^\circ$ (angle at the centre). $AXOY$ is a cyclic quadrilateral of the big circle, so $a + 2b = 180$ .
(b)
In the diagram, $\angle QPR = \angle PTR = 90^\circ$ , $|PR| = 8$ cm and $|QP| = 6$ cm. Find $|TR|$ .

Worked solution (try it first)

(a)

Join $OX$ and $OY$ .
In the smaller circle (centre $O$ ), the angle at the centre is twice the angle at the circumference: $\angle XOY = 2b^\circ$ .
$A$ , $X$ , $O$ and $Y$ all lie on the bigger circle, so $AXOY$ is a cyclic quadrilateral and $\angle XAY + \angle XOY = 180^\circ$ .
So $a + 2b = 180$ .

(b)

In triangle $PQR$ , $\tan R = \frac{6}{8} = 0.75$ , so $R = 36.87^\circ$ .
In the right-angled triangle $PTR$ , $|TR| = 8\cos R = 8 \times 0.8 = 6.4$ cm.

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Question 6 ✱

(a)
By how much is $11000_2$ greater than or less than $111_2 \times 11_2$ ? Give your answer in base two.
Model answer
$111_2 \times 11_2 = 10101_2$ and $11000_2 - 10101_2 = 11_2$ , so $11000_2$ is greater by $11_2$ (3 in base ten).
(b)
A shopkeeper has 20 television sets in stock. He sells 18 of them at a profit of 15% and the remaining two at a loss of 5%. Find his percentage profit on the 20 sets.

Worked solution (try it first)

(a)

Multiply in base two: $111_2 \times 11_2 = 111_2 \times 10_2 + 111_2$
$= 1110_2 + 111_2$
$= 10101_2$ .
Subtract: $11000_2 - 10101_2 = 11_2$ (check in base ten: $24 - 21 = 3$ ).
So $11000_2$ is greater by $11_2$ .

(b)

Let each set cost $p$ .
The 18 sets sell for $18 \times 1.15p = 20.7p$ .
The other 2 sell for $2 \times 0.95p = 1.9p$ .
Total sales $= 22.6p$ .
Total cost $= 20p$ .
Profit $= 2.6p$ .
Percentage profit $= \frac{2.6p}{20p} \times 100 = 13\%$ .

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Question 7

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 boundaries 20.5–30.5 30.5–40.5 40.5–50.5 50.5–60.5 60.5–70.5 70.5–80.5

Frequency 2 10 12 15 8 3
(b)
Draw a histogram to represent the distribution.
Model answer
Use the class boundaries 20.5, 30.5, …, 80.5 on the horizontal axis and draw bars with no gaps, of heights 2, 10, 12, 15, 8 and 3.
(c)
Use your histogram to estimate the modal mark.
(d)
If a student is selected at random, find the probability that he or she obtains a mark greater than 63.

Marks	21–30	31–40	41–50	51–60	61–70	71–80
Class boundaries	20.5–30.5	30.5–40.5	40.5–50.5	50.5–60.5	60.5–70.5	70.5–80.5
Frequency	2	10	12	15	8	3

Worked solution (try it first)

(a)

Tally the marks into the classes: the frequencies are 2, 10, 12, 15, 8 and 3 (total 50).

(b)

Draw the bars on the class boundaries $20.5, 30.5, \dots, 80.5$ , touching each other, with heights equal to the frequencies.

(c)

On the tallest bar (50.5–60.5), join its top-left corner to the top of the next bar (height 8) and its top-right corner to the top of the previous bar (height 12).
Read the mark where the lines cross: about $50.5 + \frac{15 - 12}{(15 - 12) + (15 - 8)} \times 10 = 53.5$ .

(d)

Count the marks greater than 63 in the list: 67, 73, 69, 67, 65, 74, 68, 71 and 70, which is 9 students.
Probability $= \frac{9}{50} = 0.18$ .

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Question 8

The area of a rectangular football field is $7200\text{ m}^2$ and its perimeter is 360 m.

(a)
Calculate the dimensions of the field.
Separate values with commas, e.g. 3, −2
(b)
Calculate the cost of clearing the field at ₦6.50 per square metre, leaving a margin 2 m wide along the longer sides.
(c)
Calculate the percentage of the field not cleared.

Worked solution (try it first)

(a)

Let the length be $x$ m and the breadth $y$ m: $xy = 7200$ and $2(x + y) = 360$ , so $x + y = 180$ .
Substitute $y = 180 - x$ : $x(180 - x) = 7200$ , so $x^2 - 180x + 7200 = 0$ .
Factorise: $(x - 120)(x - 60) = 0$ .
The field is 120 m long and 60 m wide.

(b)

A 2 m margin along each longer side takes 4 m off the breadth: the cleared part is $120 \times 56 = 6720\text{ m}^2$ .
Cost $= 6720 \times 6.50 = 43\,680$ , so ₦43,680.00.

(c)

Area not cleared $= 7200 - 6720 = 480\text{ m}^2$ .
Percentage $= \frac{480}{7200} \times 100$
$= 6\frac23\%$
$\approx 6.67\%$ .

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Question 9

(a)(i)
$A$ (lat $43^\circ$ N, long $77^\circ$ E), $B$ (lat $43^\circ$ N, long $103^\circ$ W) and $C$ (lat $57^\circ$ S, long $77^\circ$ E) are three points on the surface of the earth. Find the distance from $A$ to $B$ along latitude $43^\circ$ N. [Take $\pi = 3.142$ and $R = 6400$ km]
(a)(ii)
Find the distance from $A$ to $C$ along the great circle joining the two points.
(b)
In triangle $XYZ$ , $|XY| = 9$ cm, $|XZ| = 10$ cm and $\angle YXZ = 75^\circ$ . Find $|YZ|$ .

Worked solution (try it first)

(a)(i)

$A$ and $B$ are on opposite sides of the Greenwich meridian: the longitude difference is $77^\circ + 103^\circ = 180^\circ$ .
The radius of latitude $43^\circ$ is $R\cos 43^\circ$ .
Distance $= \frac{180}{360} \times 2 \times 3.142 \times 6400\cos 43^\circ$ .
$= 3.142 \times 6400 \times 0.7314$
$= 14\,706.65$ km.

(ii)

$A$ and $C$ are on the same meridian ( $77^\circ$ E), on opposite sides of the equator: the latitude difference is $43^\circ + 57^\circ = 100^\circ$ .
Distance $= \frac{100}{360} \times 2 \times 3.142 \times 6400$
$= 11\,171.56$ km.

(b)

Cosine rule: $|YZ|^2 = 9^2 + 10^2 - 2 \times 9 \times 10 \cos 75^\circ$ .
$= 181 - 180 \times 0.2588 = 134.41$ .
So $|YZ| = \sqrt{134.41} = 11.59$ cm.

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Question 10

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct a triangle $ABC$ such that $|AB| = 7.1$ cm, $|AC| = 7$ cm and $\angle BAC = 105^\circ$ . Construct the bisector of $\angle BAC$ to meet $BC$ at $X$ , and the perpendicular bisector of $AC$ to meet $AX$ produced at $Y$ .
Model answer
Construct $105^\circ$ at $A$ as $90^\circ + 15^\circ$ (bisect the $30^\circ$ between $90^\circ$ and $120^\circ$ ). Mark $B$ and $C$ , join $BC$ . Bisect angle $BAC$ and extend the bisector beyond $BC$ . Construct the perpendicular bisector of $AC$ ; where it meets $AX$ produced is $Y$ . Leave all arcs visible.
(b)
Measure: (i) $|XY|$ ; (ii) $|BC|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Draw $AB = 7.1$ cm.
At $A$ construct $90^\circ$ , then $120^\circ$ , and bisect the angle between them to get $105^\circ$ .
Mark $C$ on this arm with $AC = 7$ cm and join $BC$ .
Bisect angle $BAC$ .
The bisector meets $BC$ at $X$ .
Extend it beyond $X$ .
Construct the perpendicular bisector of $AC$ .
It meets $AX$ produced at $Y$ .

(b)

Measure: $|XY| \approx 1.5$ cm and $|BC| \approx 11.2$ cm.

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Question 11

Quadratics & their graphs

(a)
Copy and complete the table of values for $y = x^2 - 2$ for $-3 \le x \le 4$ .

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

$y$ $2$ $-2$ $2$

Model answer

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

$y$ $7$ $2$ $-1$ $-2$ $-1$ $2$ $7$ $14$

The row is symmetric about $x = 0$ : $x = 3$ and $x = -3$ both give 7.
(b)
Using a scale of 2 cm to 1 unit on the $x$ -axis and 2 cm to 2 units on the $y$ -axis, draw the graph of $y = x^2 - 2$ .
Model answer
Plot the eight points and join them with a smooth U-shaped curve through $(0, -2)$ . For (c): it cuts the $x$ -axis at $x \approx \pm 1.4$ ; the line $y = 1$ meets it at $x \approx \pm 1.7$ ; the tangent at $x = -1$ has gradient $-2$ .
(c)(i)
Use your graph to find the roots of the equation $x^2 - 2 = 0$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Use your graph to find the values of $x$ for which $x^2 - 3 = 0$ .
Separate values with commas, e.g. 3, −2
(c)(iii)
Use your graph to find the gradient of the curve at the point where $x = -1$ .

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y$	$7$	$2$	$-1$	$-2$	$-1$	$2$	$7$	$14$

Try it on a graph

The x-axis gives (c)(i); the line y = 1 gives (c)(ii).

Worked solution (try it first)

(a)

Square each $x$ and subtract 2: for example, $x = -3$ gives $9 - 2 = 7$ and $x = 4$ gives $16 - 2 = 14$ .
The row is $7, 2, -1, -2, -1, 2, 7, 14$ .

(b)

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

(c)(i)

The roots are where the curve crosses the $x$ -axis: $x \approx -1.4$ and $x \approx 1.4$ ( $\pm\sqrt2 = \pm 1.41$ ).

(ii)

Write $x^2 - 3 = 0$ as $x^2 - 2 = 1$ .
Draw the line $y = 1$ and read down: $x \approx -1.7$ and $x \approx 1.7$ ( $\pm\sqrt3 = \pm 1.73$ ).

(iii)

Draw the tangent to the curve at $(-1, -1)$ and pick two points on it, such as $(-2, 1)$ and $(0, -3)$ .
Gradient $= \frac{-3 - 1}{0 - (-2)}$
$= \frac{-4}{2}$
$= -2$ .

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Question 12

(a)
Three positive numbers are in arithmetic progression (A.P.). The sum of the squares of the three numbers is 155, while the sum of the numbers is 21. If the common difference is positive, find the numbers.
Separate values with commas, e.g. 3, −2
(b)
If the total surface area of a sphere is $154\text{ cm}^2$ , find its radius. [Take $\pi = \frac{22}{7}$ ]

Worked solution (try it first)

(a)

Write the numbers as $a - d$ , $a$ , $a + d$ .
Their sum is $3a = 21$ , so $a = 7$ .
Sum of squares: $(7 - d)^2 + 49 + (7 + d)^2 = 155$ .
Expand: $49 - 14d + d^2 + 49 + 49 + 14d + d^2 = 147 + 2d^2$
$= 155$ .
So $2d^2 = 8$ , $d^2 = 4$ and $d = 2$ (positive).
The numbers are 5, 7 and 9 (check: $25 + 49 + 81 = 155$ ).

(b)

Surface area of a sphere $= 4\pi r^2$ : $4 \times \frac{22}{7} \times r^2 = 154$ .
$r^2 = \frac{154 \times 7}{88} = 12.25$ , so $r = 3.5$ cm.

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Question 13 ✱

Elevation, depression & bearings

(b)
The pilot of an aircraft 2000 metres above sea level observes at an instant that the angles of depression of two boats, which are in a direct straight line with the aircraft, are $58^\circ$ and $72^\circ$ . Find, correct to the nearest metre, the distance between the two boats.

Worked solution (try it first)

(b)

Let $X$ be the point on the sea directly below the aircraft $P$ , with $|PX| = 2000$ m, and let $M$ and $N$ be the nearer and farther boats.
The angle of depression of $M$ is $72^\circ$ , so $\angle XPM = 90^\circ - 72^\circ$
$= 18^\circ$ .
For $N$ , $\angle XPN = 90^\circ - 58^\circ$
$= 32^\circ$ .
$|XM| = 2000\tan 18^\circ = 649.84$ m and $|XN| = 2000\tan 32^\circ = 1249.74$ m.
Distance between the boats $= 1249.74 - 649.84 = 599.9$ m, which is 600 m to the nearest metre.

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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