WAEC · 2016 · May/June · General Maths · Paper 2

Question 1

(a)
Without using mathematical tables or calculators, evaluate $\dfrac{0.09 \times 1.21}{3.3 \times 0.00025}$ , leaving the answer in standard form (scientific notation).
(b)
A principal of GH¢ 5,600.00 was deposited for 3 years at compound interest. If the interest earned was GH¢ 1,200.00, find, correct to 3 significant figures, the interest rate per annum.

Worked solution (try it first)

(a)

Write each decimal as a whole number times a power of 10: $\frac{9 \times 10^{-2} \times 121 \times 10^{-2}}{33 \times 10^{-1} \times 25 \times 10^{-5}} = \frac{9 \times 121}{33 \times 25} \times 10^{-4 - (-6)}$ .
$\frac{9 \times 121}{33 \times 25} = \frac{1089}{825}$
$= 1.32$ , and $10^{2} = 100$ , so the value is $132 = 1.32 \times 10^2$ .

(b)

The amount after 3 years is $5600 + 1200 = 6800$ .
Compound interest: $5600\left(1 + \frac{r}{100}\right)^3 = 6800$ , so $\left(1 + \frac{r}{100}\right)^3 = \frac{6800}{5600}$
$\approx 1.2143$ .
Take the cube root: $1 + \frac{r}{100} \approx 1.06686$ , so $r \approx 6.69\%$ per annum (3 significant figures).

Report a problem with this question

Question 2

(a)
Solve: $7(x + 4) - \frac23(x - 6) \le 2[x - 3(x + 5)]$ .
Show the answer
$x \le -6$
(b)
A transport company has a total of 20 vehicles made up of tricycles and taxicabs. Each tricycle carries 2 passengers while each taxicab carries 4 passengers. If the 20 vehicles carry a total of 66 passengers at a time, how many tricycles does the company have?

Worked solution (try it first)

(a)

Simplify the right side first: $2[x - 3(x + 5)] = 2[x - 3x - 15]$
$= 2(-2x - 15)$
$= -4x - 30$ .
Multiply every term by 3 to clear the fraction: $21(x + 4) - 2(x - 6) \le 3(-4x - 30)$ .
Expand: $21x + 84 - 2x + 12 \le -12x - 90$ .
So $19x + 96 \le -12x - 90$ .
Collect terms: $31x \le -186$ .
Divide by 31 (positive, so the sign stays the same): $x \le -6$ .

(b)

Let there be $t$ tricycles, so there are $20 - t$ taxicabs.
Passengers: $2t + 4(20 - t) = 66$ .
So $2t + 80 - 4t = 66$ , $-2t = -14$ and $t = 7$ .
The company has 7 tricycles.
Check: 7 tricycles carry 14 and 13 taxicabs carry 52, making 66 ✓.

Report a problem with this question

Question 3

(a)
In the diagram, $\angle RTS = 28^\circ$ , $\angle VRM = 46^\circ$ and $MQ$ is a tangent to the circle $VRSTU$ at the point $R$ . Find $\angle VUS$ .
(b)
A cylindrical tin, $7\text{ cm}$ high, is closed at one end. If its total surface area is $462\text{ cm}^2$ , calculate its radius. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

By the alternate segment theorem, the angle between the tangent $RQ$ and the chord $RS$ equals the angle in the alternate segment: $\angle SRQ = \angle RTS = 28^\circ$ .
Angles on the straight line $MQ$ at $R$ add up to $180^\circ$ : $\angle VRS = 180^\circ - 46^\circ - 28^\circ$
$= 106^\circ$ .
$VUSR$ is a cyclic quadrilateral, so its opposite angles add up to $180^\circ$ : $\angle VUS = 180^\circ - 106^\circ$
$= 74^\circ$ .

(b)

Closed at one end: the surface is the curved side plus one circle, $2\pi rh + \pi r^2 = 462$ .
With $h = 7$ : $\frac{22}{7}(14r + r^2) = 462$ , so $r^2 + 14r = 147$ .
$r^2 + 14r - 147 = 0$ factorises as $(r + 21)(r - 7) = 0$ .
The radius is positive, so $r = 7$ cm.

Report a problem with this question

Question 4

Probability

Score	1	2	3	4	5	6
Frequency	25	30	$x$	28	40	32

The table shows the outcome when a die is thrown a number of times. If the probability of obtaining a 3 is 0.225:

(a)
How many times was the die thrown?
(b)
Calculate the probability that a trial chosen at random gives a score of an even number or a prime number.

Worked solution (try it first)

(a)

The total number of throws is $25 + 30 + x + 28 + 40 + 32 = 155 + x$ .
So $P(3) = \frac{x}{155 + x} = 0.225$ .
Then $x = 0.225(155 + x) = 34.875 + 0.225x$ , so $0.775x = 34.875$ and $x = 45$ .
The die was thrown $155 + 45 = 200$ times.

(b)

Even scores: 2, 4, 6.
Prime scores: 2, 3, 5.
Even or prime: 2, 3, 4, 5, 6, with 2 counted only once.
(1 is neither even nor prime.)
These happened $30 + 45 + 28 + 40 + 32 = 175$ times, so the probability is $\frac{175}{200} = \frac78$ .

Report a problem with this question

Question 5

(a)
In the diagram, $PQST$ is a parallelogram, $PQMR$ is a straight line, $|TS| = 8\text{ cm}$ , $|SM| = 6\text{ cm}$ ( $SM \perp PR$ ) and the area of triangle $PSR = 36\text{ cm}^2$ . Find the value of $|QR|$ .
(b)
A tree and a flagpole are on the same horizontal ground. A bird on top of the tree observes the top and bottom of the flagpole below it at angles of depression of $45^\circ$ and $60^\circ$ respectively. If the tree is $10.65\text{ m}$ high, calculate, correct to 3 significant figures, the height of the flagpole.

Worked solution (try it first)

(a)

Opposite sides of a parallelogram are equal, so $|PQ| = |TS| = 8$ cm.
$SM$ is the height of triangle $PSR$ on the base $PR$ : $\frac12 \times |PR| \times 6 = 36$ , so $|PR| = 12$ cm.
So $|QR| = 12 - 8 = 4\text{ cm}$ .

(b)

Put the bird at the top $B$ of the tree, 10.65 m up.
The angle of depression of the bottom of the flagpole is $60^\circ$ , so the horizontal distance from the tree to the flagpole is $d = \frac{10.65}{\tan 60^\circ}$
$\approx 6.149$ m.
Looking at the top of the flagpole at $45^\circ$ , the bird looks down through $d\tan 45^\circ = 6.149$ m.
The flagpole is the rest of the tree's height: $10.65 - 6.149 \approx 4.50\text{ m}$ .

Report a problem with this question

Question 6

(a)
Find the sum of the Arithmetic Progression (A.P.) $1, 3, 5, \ldots, 101$ .
(b)
Out of 95 travellers interviewed, 7 travelled by bus and train only, 3 by train and car only and 8 travelled by all three means of transport. The number, $x$ , of travellers who travelled by bus only was equal to the number who travelled by bus and car only. If 47 people travelled by bus and 30 by train: (i) represent this information in a Venn diagram; (ii) calculate the: I. value of $x$ ; II. number who travelled by at least two means.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The first term is $a = 1$ and the common difference is $d = 2$ .
Find how many terms: $1 + 2(n - 1) = 101$ , so $n - 1 = 50$ and $n = 51$ .
$S_n = \frac{n}{2}(\text{first} + \text{last})$
$= \frac{51}{2}(1 + 101)$
$= 51 \times 51$
$= 2601$ .

(b)(i)

Draw three overlapping circles B (bus), T (train) and C (car) in a rectangle of 95.
Put 8 in the centre, 7 in B and T only, 3 in T and C only, and $x$ in both B only and B and C only.

(ii)

I.** The bus circle holds 47: $x + 7 + 8 + x = 47$ , so $2x + 15 = 47$ , $2x = 32$ and $x = 16$ .
II. At least two means: B and T only, T and C only, B and C only, and all three: $7 + 3 + 16 + 8 = 34$ .

Report a problem with this question

Question 7

(a)
Using the completing the square method, solve, correct to 2 decimal places, $\dfrac{x - 2}{4} = \dfrac{x + 2}{2x}$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $PQRST$ is a circle with centre $O$ . If $PS$ is a diameter, $RS \parallel QT$ , $|QR| = |RS|$ and $\angle QTS = 52^\circ$ , find: (i) $\angle SQT$ ; (ii) $\angle PQT$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Clear the fractions by multiplying both sides by $4 \times 2x$ : $2x(x - 2) = 4(x + 2)$ , so $2x^2 - 4x = 4x + 8$ .
Collect everything on one side and divide by 2: $x^2 - 4x - 4 = 0$ , so $x^2 - 4x = 4$ .
Complete the square by adding $\left(\frac{-4}{2}\right)^2 = 4$ to both sides: $x^2 - 4x + 4 = 8$ , so $(x - 2)^2 = 8$ .
Take square roots: $x - 2 = \pm\sqrt8 = \pm 2.828$ .
So $x = 4.83$ or $x = -0.83$ to 2 decimal places.

(b)(i)

$QRST$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$ : $\angle QRS = 180^\circ - 52^\circ$
$= 128^\circ$ .
$|QR| = |RS|$ , so triangle $QRS$ is isosceles and $\angle RSQ = \frac{180^\circ - 128^\circ}{2}$
$= 26^\circ$ .
$RS \parallel QT$ , so $\angle SQT = \angle RSQ = 26^\circ$ (alternate angles).

(ii)

$PS$ is a diameter, so $\angle PQS = 90^\circ$ (angle in a semicircle).
Then $\angle PQT = 90^\circ - 26^\circ$
$= 64^\circ$ .

Report a problem with this question

Question 8

(a)
In the diagram, $\angle KLM = x$ , $\angle LMK = y$ , $\angle KJH = r$ and $\angle KGF = 110^\circ$ . If $2x = r = y$ , find the value of $x$ .
(b)
Ten boys and twelve girls collected donations for a project. The total amount collected by the boys was ₦600.00 greater than that collected by the girls. If the average collection of the boys was ₦100.00 greater than the average collection of the girls, how much was collected by the two groups?

Worked solution (try it first)

(a)

$\angle JKM$ is an exterior angle of $\triangle KLM$ , so it equals the sum of the two opposite interior angles: $\angle JKM = x + y$ .
Then $\angle KGF = 110^\circ$ is an exterior angle of $\triangle JKG$ , so $(x + y) + r = 110^\circ$ .
Since $y = r = 2x$ : $x + 2x + 2x = 110^\circ$ , so $5x = 110^\circ$ and $x = 22^\circ$ .

(b)

Let the boys collect ₦ $B$ in total and the girls ₦ $G$ .
The boys collected ₦600 more: $B - G = 600$ (1).
The boys' average is $\frac{B}{10}$ and the girls' is $\frac{G}{12}$ .
The boys' average is ₦100 more: $\frac{B}{10} - \frac{G}{12} = 100$ .
Multiply by 60: $6B - 5G = 6000$ (2).
From (1), $B = G + 600$ .
Substitute into (2): $6(G + 600) - 5G = 6000$ , so $G + 3600 = 6000$ and $G = 2400$ .
Then $B = 3000$ .
Together they collected $3000 + 2400 =$ ₦5,400.00.

Report a problem with this question

Question 9

The weight (in kg) of 50 contestants at a competition is as follows:

65 66 67 66 64 66 65 63 65 68 64 62 66 64 67 65 64 66 65 67 65 67 66 64 65 64 66 65 64 65 66 65 64 65 63 63 67 65 63 64 66 64 68 65 63 65 64 67 66 64

(a)
Construct a frequency table for the discrete data.
Model answer

Weight (kg) 62 63 64 65 66 67 68

Frequency 1 5 12 14 10 6 2

One column for each weight (the data are discrete, so do not group them). Tally through the list, then check that the frequencies add up to 50.
(b)
Calculate, correct to 2 decimal places, the: (i) mean; (ii) standard deviation of the data.
Separate values with commas, e.g. 3, −2

Weight (kg)	62	63	64	65	66	67	68
Frequency	1	5	12	14	10	6	2

Worked solution (try it first)

(a)

The weights are separate whole numbers from 62 to 68, so make one row for each weight (don't group them).
Tally through the list:
Weight (kg) 62 63 64 65 66 67 68

Frequency 1 5 12 14 10 6 2
The frequencies add up to 50.

Weight (kg)	62	63	64	65	66	67	68
Frequency	1	5	12	14	10	6	2

(b)(i)

$\sum fx = 62 + 315 + 768 + 910 + 660 + 402 + 136$
$= 3253$ , so the mean is $\frac{3253}{50} = 65.06$ kg.

(ii)

Use an assumed mean of 65, with $d = x - 65 = -3, -2, -1, 0, 1, 2, 3$ .
Then $\sum fd = -3 - 10 - 12 + 0 + 10 + 12 + 6$
$= 3$ and $\sum fd^2 = 9 + 20 + 12 + 0 + 10 + 24 + 18$
$= 93$ .
Standard deviation $= \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2}$
$= \sqrt{\frac{93}{50} - \left(\frac{3}{50}\right)^2}$
$= \sqrt{1.86 - 0.0036}$
$= \sqrt{1.8564}$
$\approx 1.36$ kg.

Report a problem with this question

Question 10

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct: (i) $\triangle XYZ$ such that $|XY| = 10\text{ cm}$ , $\angle XYZ = 30^\circ$ and $\angle YXZ = 45^\circ$ ; (ii) the locus $l_1$ of points equidistant from $Y$ and $Z$ ; (iii) the locus $l_2$ of points on the line through $Z$ parallel to $XY$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $XY = 10$ cm. At $Y$ , construct $60^\circ$ and bisect it to get $30^\circ$ . At $X$ , construct $90^\circ$ and bisect it to get $45^\circ$ . The two arms meet at $Z$ . Then (ii) bisect $YZ$ perpendicularly to get $l_1$ , and (iii) draw the line through $Z$ parallel to $XY$ (copy an angle, or use two equal perpendiculars) to get $l_2$ .
(b)
Locate the point $M$ , the point of intersection of $l_1$ and $l_2$ .
Model answer
$M$ is where $l_1$ crosses $l_2$ . Because $MZ = MY$ and $\angle MZY = 30^\circ$ (alternate to $\angle ZYX$ ), triangle $ZMY$ is isosceles with base angles $30^\circ$ , so $\angle ZMY$ measures $120^\circ$ .
(c)
Measure $\angle ZMY$ .

Worked solution (try it first)

(a)(i)

Draw $XY = 10$ cm.
At $Y$ construct $60^\circ$ and bisect it to get $30^\circ$ .
At $X$ construct $90^\circ$ and bisect it to get $45^\circ$ .
The two arms meet at $Z$ .

(ii)

$l_1$ , equidistant from $Y$ and $Z$ : construct the perpendicular bisector of $YZ$ .

(iii)

$l_2$ : construct the line through $Z$ parallel to $XY$ .

(b)

$M$ is where $l_1$ and $l_2$ cross.

(c)

Measure $\angle ZMY = 120^\circ$ .
Check: $M$ is on $l_1$ , so $MZ = MY$ .
$l_2 \parallel XY$ , so $\angle MZY = \angle ZYX = 30^\circ$ (alternate angles).
Triangle $ZMY$ is isosceles with base angles $30^\circ$ , so $\angle ZMY = 180^\circ - 60^\circ$
$= 120^\circ$ .

Report a problem with this question

Question 11

(a)
If $\dfrac{3p + 4q}{3p - 4q} = 2$ , find $p : q$ .
Show the answer
$4 : 1$
(b)
The diagram shows the cross section $PQRSTU$ of a bridge with a semicircular hollow $TS$ in the middle. $|PU| = |QR| = 4\text{ m}$ and $|UT| = |SR| = 2\text{ m}$ . If the perimeter of the cross section is $34\text{ m}$ , calculate the: (i) length $PQ$ ; (ii) area of the cross section. $\left[\text{Take }\pi = \frac{22}{7}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Cross-multiply: $3p + 4q = 2(3p - 4q) = 6p - 8q$ .
So $12q = 3p$ , $p = 4q$ , and $p : q = 4 : 1$ .

(b)(i)

Let the semicircle have radius $r$ .
The width across the bottom is $2 + 2r + 2$ , so $|PQ| = 4 + 2r$ .
The perimeter goes round $PQ$ , down $QR$ (4), along $RS$ (2), round the semicircle ( $\frac{22}{7}r$ ), along $TU$ (2) and up $UP$ (4): $(4 + 2r) + 4 + 2 + \frac{22}{7}r + 2 + 4 = 34$ .
So $16 + \frac{36}{7}r = 34$ , $\frac{36}{7}r = 18$ and $r = 3.5$ m.
Then $|PQ| = 4 + 7 = 11$ m.

(ii)

Area $=$ rectangle $-$ semicircle $= 11 \times 4 - \frac12 \times \frac{22}{7} \times 3.5^2$
$= 44 - 19.25$
$= 24.75\text{ m}^2$ .

Report a problem with this question

Question 12

Trigonometric graphs

(a)

Copy and complete the table of values, correct to one decimal place, for the relation $y = 3\sin x + 2\cos x$ for $0^\circ \le x \le 360^\circ$ .

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$	$270^\circ$	$300^\circ$	$330^\circ$	$360^\circ$
$y$	$2.0$			$3.0$	$1.6$		$-2.0$		$-3.6$	$-3.0$			$2.0$

Model answer

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$	$210^\circ$	$240^\circ$	$270^\circ$	$300^\circ$	$330^\circ$	$360^\circ$
$y$	$2.0$	$3.2$	$3.6$	$3.0$	$1.6$	$-0.2$	$-2.0$	$-3.2$	$-3.6$	$-3.0$	$-1.6$	$0.2$	$2.0$

Work in degree mode, to one decimal place: for example $x = 30^\circ$ : $3(0.5) + 2(0.866) = 3.2$ , and $x = 150^\circ$ : $1.5 - 1.73 = -0.2$ .

(b)
Using scales of 2 cm to $30^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of the relation.
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The curve peaks at about $3.6$ near $56^\circ$ and dips to about $-3.6$ near $236^\circ$ .

For (c)(i): the roots of $3\sin x + 2\cos x = 0$ are where it crosses the $x$ -axis, $x \approx 146.3^\circ$ and $326.3^\circ$ . For (c)(ii): $2 + 2\cos x + 3\sin x = 0$ means $y = -2$ , which meets the curve at $x = 180.0^\circ$ and $292.6^\circ$ .
(c)(i)
Use the graph to solve $3\sin x + 2\cos x = 0$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Use the graph to solve $2 + 2\cos x + 3\sin x = 0$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

x in degrees. The x-axis gives (c)(i); the line y = −2 gives (c)(ii).

Worked solution (try it first)

(a)

In degree mode, to 1 decimal place.
For example, $x = 30^\circ$ : $3(0.5) + 2(0.866) = 1.5 + 1.73 = 3.2$ .
$x = 150^\circ$ : $1.5 - 1.73 = -0.2$ .
The full row is $2.0, 3.2, 3.6, 3.0, 1.6, -0.2, -2.0, -3.2, -3.6, -3.0, -1.6, 0.2, 2.0$ .

(b)

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

(c)(i)

$3\sin x + 2\cos x = 0$ where the curve crosses the $x$ -axis: $x \approx 146^\circ$ and $x \approx 326^\circ$ .

(ii)

Rearrange so one side is the relation you drew: $2 + 2\cos x + 3\sin x = 0$ is $3\sin x + 2\cos x = -2$ .
Draw the line $y = -2$ and read where it meets the curve: $x = 180^\circ$ and $x \approx 293^\circ$ .

Report a problem with this question

Question 13

(a)
Find the equation of a straight line which passes through the point $(2, -3)$ and is parallel to the line $2x + y = 6$ .
Show the answer
$2x + y - 1 = 0$
(b)
The operation $\Delta$ is defined on the set $T = \{2, 3, 5, 7\}$ by $x\,\Delta\,y = (x + y + xy) \bmod 8$ . (i) Construct the modulo 8 table for the operation $\Delta$ on the set $T$ . (ii) Use the table to find: I. $2\,\Delta\,(5\,\Delta\,7)$ ; II. $n$ if $2\,\Delta\,n = 5\,\Delta\,7$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$2x + y = 6$ is $y = -2x + 6$ , with gradient $-2$ .
A parallel line has the same gradient: $y = -2x + c$ .
Through $(2, -3)$ : $-3 = -4 + c$ , so $c = 1$ .
The line is $y = -2x + 1$ , or $2x + y - 1 = 0$ .

(b)(i)

Each entry is $x + y + xy$ reduced modulo 8.
For example $2\,\Delta\,3 = 2 + 3 + 6 = 11 \equiv 3$ and $5\,\Delta\,7 = 5 + 7 + 35 = 47 \equiv 7$ .
$\Delta$ 2 3 5 7

2 0 3 1 7

3 3 7 7 7

5 1 7 3 7

7 7 7 7 7

$\Delta$	2	3	5	7
2	0	3	1	7
3	3	7	7	7
5	1	7	3	7
7	7	7	7	7

(ii)

I.
$5\,\Delta\,7 = 7$ , so $2\,\Delta\,(5\,\Delta\,7) = 2\,\Delta\,7$
$= 7$ .
II. $2\,\Delta\,n = 5\,\Delta\,7 = 7$ : in the row of 2, the 7 is in the column of 7, so $n = 7$ .

Report a problem with this question