WAEC · 2014 · Nov/Dec · General Maths · Paper 2

Question 1

(a)
Factorize completely: $m^2 - 2mn + n^2 - 9r^2$ .
Show the answer
$(m - n + 3r)(m - n - 3r)$
(b)
Solve simultaneously the equations $5x - 4y = 6$ and $3^{3(y - x)} = \frac{1}{27}$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The first three terms are a perfect square: $m^2 - 2mn + n^2 = (m - n)^2$ .
So the expression is $(m - n)^2 - (3r)^2$ , a difference of two squares.
Using $a^2 - b^2 = (a - b)(a + b)$ with $a = m - n$ and $b = 3r$ : $(m - n - 3r)(m - n + 3r)$ .

(b)

Write $\frac1{27}$ as a power of 3: $\frac1{27} = 3^{-3}$ .
So $3^{3(y - x)} = 3^{-3}$ , and the powers are equal: $3(y - x) = -3$ , so $y - x = -1$ and $y = x - 1$ .
Substitute into $5x - 4y = 6$ : $5x - 4(x - 1) = 6$ .
So $5x - 4x + 4 = 6$ and $x = 2$ .
Then $y = 2 - 1 = 1$ .

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Question 2

Commercial arithmetic

(a)
A man has a wife and 6 children and his total income in a year was GH¢ 850.00. He was given the following tax-free allowances: personal GH¢ 120.00; wife GH¢ 30.00; children GH¢ 25.00 per child, for a maximum of 4 children; medical GH¢ 40.00. The rest was taxed as follows: first GH¢ 200.00 at $10\%$ ; next GH¢ 200.00 at $15\%$ ; next GH¢ 200.00 at $20\%$ ; remainder at $25\%$ . Calculate his: (i) taxable income; (ii) monthly tax.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The children's allowance only counts 4 children.
Allowances $= 120 + 30 + 4 \times 25 + 40 =$ GH¢ 290.
Taxable income $= 850 - 290 =$ GH¢ 560.

(ii)

Tax the 560 in bands: the first 200 at $10\%$ is 20.
The next 200 at $15\%$ is 30.
The remaining 160 at $20\%$ is 32.
Total tax $= 20 + 30 + 32 =$ GH¢ 82 a year.
Monthly tax $= \frac{82}{12} \approx$ GH¢ 6.83.

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Question 3

(a)
In the diagram, $QSP$ and $QTR$ are straight lines, $|PS| = 6\text{ cm}$ , $|QS| = 4\text{ cm}$ , $|QT| = 5\text{ cm}$ and $\angle QTS = \angle RPQ$ . Calculate $|TR|$ .
(b)
If $\sin x = \frac{5}{13}$ , $0^\circ \le x \le 90^\circ$ , evaluate, without tables or a calculator, $\dfrac{\cos x - 2\sin x}{2\tan x}$ .

Worked solution (try it first)

(a)

Triangles $QTS$ and $QPR$ share the angle at $Q$ , and $\angle QTS = \angle QPR$ , so they are similar, with $T$ matching $P$ and $S$ matching $R$ .
Corresponding sides are in the same ratio: $\frac{|QT|}{|QP|} = \frac{|QS|}{|QR|}$ .
Here $|QP| = 4 + 6 = 10$ , so $\frac{5}{10} = \frac{4}{|QR|}$ and $|QR| = 8$ .
So $|TR| = 8 - 5 = 3\text{ cm}$ .

(b)

$\sin x = \frac{5}{13}$ : draw a right-angled triangle with opposite 5 and hypotenuse 13.
The adjacent side is $\sqrt{169 - 25} = 12$ .
So $\cos x = \frac{12}{13}$ and $\tan x = \frac{5}{12}$ .
Then $\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - \frac{10}{13}}{\frac{10}{12}}$
$= \frac{2}{13} \times \frac{12}{10}$
$= \frac{12}{65}$ .

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Question 4

(a)
Without using tables or a calculator, evaluate $\log_{10}\left(\frac{75}{10}\right) - 2\log_{10}\left(\frac59\right) + \log_{10}\left(\frac{100}{243}\right)$ .
(b)
Given that $X = \{x : 10 \le x < 15\}$ and $Y = \{\text{even numbers} < 18\}$ are subsets of $U = \{10, 11, 12, \ldots, 20\}$ , find: (i) $X \cap Y$ ; (ii) $n(X' \cap Y)$ .

Worked solution (try it first)

(a)

Move the 2 inside as a power: $2\log_{10}\frac59 = \log_{10}\frac{25}{81}$ .
Then combine: $\log_{10}\left(\frac{75}{10} \times \frac{100}{243} \div \frac{25}{81}\right)$ .
Simplify: $\frac{75}{10} \times \frac{100}{243} \times \frac{81}{25} = \frac{75 \times 100 \times 81}{10 \times 243 \times 25}$
$= 10$ .
So the answer is $\log_{10} 10 = 1$ .

(b)

$X = \{10, 11, 12, 13, 14\}$ and $Y$ (even numbers in $U$ less than 18) $= \{10, 12, 14, 16\}$ .

(i)

$X \cap Y = \{10, 12, 14\}$ .

(ii)

$X' = \{15, 16, 17, 18, 19, 20\}$ , so $X' \cap Y = \{16\}$ and $n(X' \cap Y) = 1$ .

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Question 5

The diagram shows a right pyramid with a rectangular base $WXYZ$ and vertex $O$ . If $|WX| = 8\text{ cm}$ , $|ZW| = 6\text{ cm}$ and $|OX| = 13\text{ cm}$ , calculate the:

Not to scale.

(a)
height of the pyramid;
(b)
value of $\angle OXZ$ , correct to the nearest degree;
(c)
volume of the pyramid.

Worked solution (try it first)

(a)

The vertex $O$ is directly above the centre $M$ of the base, where the diagonals cross.
The diagonal $|ZX| = \sqrt{8^2 + 6^2} = 10$ cm, so $|MX| = 5$ cm.
Triangle $OMX$ is right-angled at $M$ : height $|OM| = \sqrt{13^2 - 5^2}$
$= \sqrt{144}$
$= 12\text{ cm}$ .

(b)

$\angle OXZ$ is the angle at $X$ in triangle $OMX$ .
$MX = 5$ is adjacent and $OX = 13$ is the hypotenuse: $\cos\angle OXZ = \frac{5}{13}$
$\approx 0.3846$ .
So $\angle OXZ \approx 67^\circ$ .

(c)

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

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Question 6

(a)
In a class of 52 students, 16 are Science students. If $\frac13$ of the boys and $\frac14$ of the girls are Science students, how many boys are in the class?
(b)
The sum of the first and third terms of a Geometric Progression (G.P.) is 40 while the fourth and sixth terms are in the ratio $1 : 4$ . Find the: (i) common ratio; (ii) fifth term.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Let there be $x$ boys and $y$ girls.
The class has 52 students: $x + y = 52$ .
A third of the boys and a quarter of the girls do Science, making 16: $\frac x3 + \frac y4 = 16$ .
Multiply by 12: $4x + 3y = 192$ .
From the first equation $y = 52 - x$ , so $4x + 3(52 - x) = 192$ .
Then $4x + 156 - 3x = 192$ and $x = 36$ .
There are 36 boys.

(b)(i)

With first term $a$ and common ratio $r$ , the terms are $a, ar, ar^2, ar^3, \ldots$ The fourth and sixth terms are in the ratio $1 : 4$ : $\frac{ar^3}{ar^5} = \frac14$ , so $\frac1{r^2} = \frac14$ and $r^2 = 4$ .
So $r = 2$ or $r = -2$ .

(ii)

The first and third terms add up to 40: $a + ar^2 = 40$ , so $a + 4a = 40$ and $a = 8$ .
The fifth term is $ar^4 = 8 \times 16 = 128$ (the same for $r = 2$ or $r = -2$ ).

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Question 7

(a)
A spherical tank of diameter $3\text{ m}$ is filled with water from a pipe of radius $30\text{ cm}$ at $0.2\text{ m}$ per second. Calculate, correct to 3 significant figures, the time, in minutes, it takes to fill the tank. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
When $k$ is added to the expression $y^2 - 12y$ , the expression becomes $(y + p)^2$ . Find the values of $p$ and $k$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Work in metres.
The tank has radius 1.5 m: volume $= \frac43 \times \frac{22}{7} \times 1.5^3$
$= \frac{99}{7}$
$\approx 14.14\text{ m}^3$ .
Each second the water in the pipe moves 0.2 m, so it delivers a cylinder of radius 0.3 m and length 0.2 m: $\frac{22}{7} \times 0.3^2 \times 0.2 = \frac{0.396}{7}$
$\approx 0.0566\text{ m}^3$ per second.
Time $= \frac{99/7}{0.396/7}$
$= \frac{99}{0.396}$
$= 250$ seconds $= \frac{250}{60} \approx 4.17$ minutes.

(b)

$(y + p)^2 = y^2 + 2py + p^2$ must equal $y^2 - 12y + k$ .
Compare the $y$ terms: $2p = -12$ , so $p = -6$ .
Compare the constants: $k = p^2 = 36$ .

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Question 8

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct: (i) a parallelogram $PQRS$ with $RS$ as the base such that $|PQ| = 7.8\text{ cm}$ , $|QR| = 5.6\text{ cm}$ and $\angle QRS = 120^\circ$ ; (ii) a rectangle $ABRS$ equal in area to the parallelogram $PQRS$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw the base $RS = 7.8$ cm. Construct $120^\circ$ at $R$ and mark $Q$ with $RQ = 5.6$ cm. Through $Q$ draw the line parallel to $RS$ and mark $P$ with $QP = 7.8$ cm. For the rectangle, draw perpendiculars to $RS$ at $R$ and $S$ to meet the line $QP$ (produced) at $B$ and $A$ . It has the same base and lies between the same parallels, so it has the same area. Measured: $|AP| = 2.8$ cm and $|AS| \approx 4.85$ cm.
(b)
Measure: (i) $|AP|$ ; (ii) $|AS|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Draw the base $RS = 7.8$ cm.
Construct $120^\circ$ at $R$ and mark $RQ = 5.6$ cm on the arm.
With centre $Q$ and radius 7.8 cm, and centre $S$ and radius 5.6 cm, draw arcs meeting at $P$ .
Join $PQ$ and $PS$ : $PQRS$ is the parallelogram.

(ii)

A rectangle on the same base between the same parallels has the same area.
Produce $QP$ and drop perpendiculars to it from $R$ and $S$ , meeting it at $B$ and $A$ .
$ABRS$ is the rectangle.

(b)

Measure: (i) $|AP| = 2.8$ cm.

(ii)

$|AS| \approx 4.9$ cm.
Check: $|AS|$ is the height, $5.6\sin 60^\circ \approx 4.85$ cm, and $|AP| = 5.6\cos 60^\circ = 2.8$ cm.

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Question 9

(a)
An object was thrown vertically upwards from the top of a cliff and its height, $y$ metres, above sea level after $t$ seconds is given by $y = -16t^2 + 64t + 5$ . Copy and complete the table of values for $0 \le t \le 4.0$ .

$t$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

$y$ 5 65 53

Model answer

$t$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

$y$ 5 33 53 65 69 65 53 33 5

For example $t = 0.5$ : $-4 + 32 + 5 = 33$ , and $t = 2.0$ : $-64 + 128 + 5 = 69$ . The values are symmetrical about $t = 2$ .
(b)
Using scales of 2 cm to 0.5 seconds on the $t$ -axis and 2 cm to 10 m on the $y$ -axis, draw the graph of $y = -16t^2 + 64t + 5$ for $0 \le t \le 4.0$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 0.5 s across, 2 cm to 10 m up. The path is symmetrical about $t = 2$ .

For (c): (i) at $t = 1.75$ s the height is about 68 m; (ii) the line $y = 50$ meets the curve at $t \approx 0.9$ s and $3.1$ s; (iii) the maximum height is 69 m.
(c)
Use the graph to find the: (i) height reached when $t = 1.75$ seconds; (ii) times the object was at a height of $50\text{ m}$ ; (iii) maximum height reached.
Separate values with commas, e.g. 3, −2

$t$	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
$y$	5	33	53	65	69	65	53	33	5

Try it on a graph

The height curve with the line y = 50.

Worked solution (try it first)

(a)

Substitute each $t$ into $y = -16t^2 + 64t + 5$ .
For example, $t = 0.5$ gives $-4 + 32 + 5 = 33$ and $t = 2$ gives $-64 + 128 + 5 = 69$ .
$t$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

$y$ 5 33 53 65 69 65 53 33 5

$t$	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
$y$	5	33	53	65	69	65	53	33	5

(b)

With 2 cm to 0.5 s on the $t$ -axis and 2 cm to 10 m on the $y$ -axis, plot the points and join them with a smooth curve shaped like an upside-down U.

(c)(i)

Go up from $t = 1.75$ to the curve and across: the height is about $68\text{ m}$ .

(ii)

Draw the line $y = 50$ .
It meets the curve at $t \approx 0.9\text{ s}$ and $t \approx 3.1\text{ s}$ .

(iii)

The highest point of the curve is at $t = 2$ : the maximum height is $69\text{ m}$ .
(Check: $\frac{dy}{dt} = -32t + 64 = 0$ at $t = 2$ .)

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Question 10

(a)
(i) Solve the inequality $\frac12x - \frac56(x + 2) \le 1 + x$ . (ii) Illustrate the solution on a number line.
Model answer
(i) $x \ge -2$ . (ii) A solid dot at $-2$ (it is included) with an arrow to the right.
(b)
From a point $P$ on level ground and directly west of a pole, the angle of elevation of the top of the pole is $45^\circ$ , and from a point $Q$ east of the pole, the angle of elevation of the top of the pole is $58^\circ$ . If $|PQ| = 10\text{ m}$ , calculate, correct to 2 significant figures, the: (i) distance from $P$ to the pole; (ii) height of the pole.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

Multiply every term by 6: $3x - 5(x + 2) \le 6 + 6x$ .
Expand: $3x - 5x - 10 \le 6 + 6x$ , so $-2x - 10 \le 6 + 6x$ .
Collect terms: $-8x \le 16$ .
Divide by $-8$ , which reverses the inequality sign: $x \ge -2$ .

(ii)

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

(b)

The pole stands between $P$ (to the west) and $Q$ (to the east).
Let the distance from $P$ to the foot of the pole be $x$ m.
Then $Q$ is $(10 - x)$ m from it.
The height $h$ from each side: $h = x\tan 45^\circ = x$ , and $h = (10 - x)\tan 58^\circ$ .
Set them equal: $x = (10 - x)(1.6003)$ .
So $x + 1.6003x = 16.003$ , $2.6003x = 16.003$ and $x \approx 6.154$ .

(i)

$P$ is about 6.2 m from the pole.

(ii)

The height is $h = x \approx 6.2$ m (2 significant figures).

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Question 11

(a)
The probabilities that Manful, John and Ernest will pass an examination are $\frac23$ , $\frac58$ and $\frac34$ respectively. Find the probability that all three will pass the examination.
(b)
The scores of students in a test were recorded as follows:

4 2 1 6 5 3 5 6 1 2 1 5 5 6 3 4 3 5 1 5

(i) Construct a frequency distribution table. (ii) Represent the information in a bar chart. (iii) Calculate the interquartile range.

Worked solution (try it first)

(a)

The three results are independent, so multiply: $\frac23 \times \frac58 \times \frac34 = \frac{30}{96}$
$= \frac{5}{16}$ .

(b)(i)

Tally each score:
Score 1 2 3 4 5 6

Frequency 4 2 3 2 6 3
The frequencies add up to 20, the number of scores.

Score	1	2	3	4	5	6
Frequency	4	2	3	2	6	3

(ii)

The scores are separate whole numbers, so draw a bar chart: one bar for each score, all the same width, with gaps between them, and heights 4, 2, 3, 2, 6, 3.
Label the axes "Score" and "Frequency".

(iii)

In order, the 20 scores are $1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6$ .
$Q_1$ is at position $\frac{20}{4} = 5$ : the 5th score is 2.
$Q_3$ is at position $\frac{3 \times 20}{4} = 15$ : the 15th score is 5.
Interquartile range $= Q_3 - Q_1 = 5 - 2 = 3$ .

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Question 12

(a)
In the diagram, $\overline{TS}$ is a tangent to the circle $PQRS$ at $S$ , and $O$ is the centre of the circle ( $QOS$ is a diameter). If $\angle TSP = 21^\circ$ and $\angle RQP = 100^\circ$ , find, with reasons: (i) $\angle SPR$ ; (ii) $\angle QSR$ .
Separate values with commas, e.g. 3, −2
(b)
Given the relation $T = \sqrt{\dfrac{U}{\frac1f + \frac1g}}$ : (i) make $g$ the subject of the relation; (ii) find $g$ when $T = 3$ , $f = 4$ and $U = 5$ .

Worked solution (try it first)

(a)(i)

$\angle PRS = \angle TSP = 21^\circ$ (angle in the alternate segment).
$PQRS$ is a cyclic quadrilateral, so $\angle RSP = 180^\circ - \angle RQP$
$= 180^\circ - 100^\circ$
$= 80^\circ$ (opposite angles).
In $\triangle PRS$ : $\angle SPR = 180^\circ - 80^\circ - 21^\circ$
$= 79^\circ$ (angles in a triangle).

(ii)

$\angle PQS = \angle PRS = 21^\circ$ (angles in the same segment).
$QS$ is a diameter, so $\angle QPS = 90^\circ$ (angle in a semicircle).
In $\triangle PQS$ : $\angle QSP = 180^\circ - 90^\circ - 21^\circ$
$= 69^\circ$ .
So $\angle QSR = \angle RSP - \angle QSP$
$= 80^\circ - 69^\circ$
$= 11^\circ$ .

(b)(i)

Square both sides: $T^2 = \frac{U}{\frac1f + \frac1g}$ .
Combine the fractions underneath: $\frac1f + \frac1g = \frac{g + f}{fg}$ , so $T^2 = \frac{Ufg}{g + f}$ .
Multiply both sides by $(g + f)$ : $T^2 g + T^2 f = Ufg$ .
Collect the $g$ terms on one side: $T^2 f = Ufg - T^2 g = g(Uf - T^2)$ .
So $g = \frac{fT^2}{Uf - T^2}$ .

(ii)

With $T = 3$ , $f = 4$ , $U = 5$ : $g = \frac{4 \times 9}{5 \times 4 - 9}$
$= \frac{36}{11}$
$= 3\frac{3}{11}$ .

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Question 13

(a)
If $\mathbf x = \begin{pmatrix} -2 \\ 4 \end{pmatrix}$ and $\mathbf y = \begin{pmatrix} -3 \\ 1 \end{pmatrix}$ , find, correct to 1 decimal place, $|\mathbf x + \mathbf y|$ .
(b)
$P(6, 4)$ , $Q(-2, -2)$ and $R(4, -6)$ are the vertices of triangle $PQR$ . (i) Determine the coordinates of $M$ and $S$ , the midpoints of $\overline{PQ}$ and $\overline{PR}$ respectively. (ii) Find $\overrightarrow{QR}$ and $\overrightarrow{MS}$ . (iii) State the relationship between $\overrightarrow{QR}$ and $\overrightarrow{MS}$ . (iv) Find the equation of $\overline{MS}$ .
Show the answer
(i) $M(2, 1)$ , $S(5, -1)$ ; (ii) $\overrightarrow{QR} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ , $\overrightarrow{MS} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ ; (iii) $\overrightarrow{QR} = 2\overrightarrow{MS}$ ; (iv) $2x + 3y = 7$

Worked solution (try it first)

(a)

Add the vectors component by component: $\mathbf x + \mathbf y = \begin{pmatrix} -2 + (-3) \\ 4 + 1 \end{pmatrix}$
$= \begin{pmatrix} -5 \\ 5 \end{pmatrix}$ .
Its length is $|\mathbf x + \mathbf y| = \sqrt{(-5)^2 + 5^2}$
$= \sqrt{50}$
$\approx 7.1$ .

(b)(i)

Midpoints average the coordinates: $M = \left(\frac{6 + (-2)}{2}, \frac{4 + (-2)}{2}\right)$
$= (2, 1)$ and $S = \left(\frac{6 + 4}{2}, \frac{4 + (-6)}{2}\right)$
$= (5, -1)$ .

(ii)

A vector from one point to another is "end minus start": $\overrightarrow{QR} = \begin{pmatrix} 4 - (-2) \\ -6 - (-2) \end{pmatrix}$
$= \begin{pmatrix} 6 \\ -4 \end{pmatrix}$ and $\overrightarrow{MS} = \begin{pmatrix} 5 - 2 \\ -1 - 1 \end{pmatrix}$
$= \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ .

(iii)

$\overrightarrow{QR} = 2\overrightarrow{MS}$ : $QR$ is parallel to $MS$ and twice as long.

(iv)

Gradient of $MS = \frac{-1 - 1}{5 - 2} = -\frac23$ .
Through $M(2, 1)$ : $y - 1 = -\frac23(x - 2)$ .
Multiply by 3: $3y - 3 = -2x + 4$ , so $2x + 3y = 7$ .

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