WAEC · 2017 · May/June · General Maths · Paper 2

Question 1

(a)
If $(y - 1)\log_{10} 4 = y\log_{10} 16$ , without using mathematical tables or a calculator, find the value of $y$ .
(b)
When I walk from my house at $4\text{ km/h}$ , I get to the office 30 minutes later than when I walk at $5\text{ km/h}$ . Calculate the distance between my house and the office.

Worked solution (try it first)

(a)

Write 16 as a power of 4: $16 = 4^2$ , so $\log_{10} 16 = 2\log_{10} 4$ .
The equation becomes $(y - 1)\log_{10} 4 = 2y\log_{10} 4$ .
Divide both sides by $\log_{10} 4$ : $y - 1 = 2y$ .
So $y = -1$ .

(b)

Let the distance be $x$ km.
Time $= \frac{\text{distance}}{\text{speed}}$ , so walking at 4 km/h takes $\frac x4$ hours and at 5 km/h takes $\frac x5$ hours.
The slower walk takes 30 minutes $= \frac12$ hour longer: $\frac x4 - \frac x5 = \frac12$ .
Multiply by 20: $5x - 4x = 10$ , so $x = 10$ .
The office is 10 km from the house.

Report a problem with this question

Question 2

(a)
Solve the equation $\frac23(3x - 5) - \frac35(2x - 3) = 3$ .
(b)
In the diagram, $P$ , $U$ , $T$ , $S$ lie on a straight line and $P$ , $Q$ , $R$ on another. $\angle STQ = m$ , $\angle TUQ = 80^\circ$ , $\angle UPQ = r$ , $\angle PQU = n$ and $\angle RQT = 88^\circ$ . Find the value of $(m + n)$ .
The values of r, m and n are not given; the drawing uses r = 20°.

Worked solution (try it first)

(a)

Multiply every term by 15, the LCM of 3 and 5: $10(3x - 5) - 9(2x - 3) = 45$ .
Expand: $30x - 50 - 18x + 27 = 45$ .
So $12x - 23 = 45$ , $12x = 68$ and $x = \frac{68}{12} = 5\frac23$ .

(b)

$\angle TUQ = 80^\circ$ is an exterior angle of $\triangle PUQ$ , so it equals the two opposite interior angles: $r + n = 80^\circ$ (1).
At $T$ , the angle inside $\triangle PQT$ is $180^\circ - m$ (angles on a straight line).
$\angle RQT = 88^\circ$ is an exterior angle of $\triangle PQT$ : $r + (180^\circ - m) = 88^\circ$ , so $m - r = 92^\circ$ (2).
Add (1) and (2): the $r$ terms cancel, giving $m + n = 172^\circ$ .

Report a problem with this question

Question 3

(a)
The angle of depression of a point $P$ on the ground from the top $T$ of a building is $23.6^\circ$ . If the distance from $P$ to the foot of the building is $50\text{ m}$ , calculate, correct to the nearest metre, the height of the building.
(b)
In the diagram, $PQT$ and $SRU$ are parallel lines, $QS \parallel TR$ , $|SR| = 6\text{ cm}$ and $|RU| = 10\text{ cm}$ . If the area of $\triangle TRU = 45\text{ cm}^2$ , calculate the area of the trapezium $QTUS$ .

Worked solution (try it first)

(a)

Draw the building $FT$ and the point $P$ on the ground 50 m from $F$ .
The angle of depression from $T$ equals the angle of elevation from $P$ (alternate angles), so $\angle FPT = 23.6^\circ$ .
Then $\tan 23.6^\circ = \frac{h}{50}$ , so $h = 50 \times 0.4369 \approx 21.8$ m, which is 22 m to the nearest metre.

(b)

The height of $\triangle TRU$ on the base $RU$ is the distance between the parallel lines: $\frac12 \times 10 \times h = 45$ , so $h = 9$ cm.
$QTRS$ has both pairs of opposite sides parallel, so it is a parallelogram and $|QT| = |SR| = 6$ cm.
The parallel sides of the trapezium are $QT = 6$ cm and $SU = 6 + 10 = 16$ cm.
Area $= \frac12(6 + 16) \times 9$
$= 99\text{ cm}^2$ .

Report a problem with this question

Question 4

Sequences & series (AP, GP)

(a)
If the sixth term of an Arithmetic Progression (A.P.) is 37 and the sum of the first six terms is 147, find the: (i) first term; (ii) sum of the first fifteen terms.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(i)

The sixth term is the last of the first six terms, so use $S_n = \frac{n}{2}(a + l)$ with $l = 37$ : $S_6 = \frac62(a + 37) = 3(a + 37) = 147$ .
So $a + 37 = 49$ , and the first term is $a = 12$ .

(ii)

Find $d$ from the sixth term: $T_6 = a + 5d = 37$ , so $12 + 5d = 37$ , $5d = 25$ and $d = 5$ .
$S_{15} = \frac{15}{2}[2a + 14d]$
$= \frac{15}{2}[24 + 70]$
$= \frac{15}{2} \times 94$
$= 705$ .

Report a problem with this question

Question 5

Out of 120 customers in a shop, 45 bought bags and shoes. All the customers bought either bags or shoes, and 11 more customers bought shoes than bags.

(a)
Illustrate this information in a diagram.
Model answer
Two overlapping circles, $B$ (bags) and $S$ (shoes), in a rectangle for all 120 customers: 45 in the overlap, $x$ in bags only and $x + 11$ in shoes only, and 0 outside (everyone bought one or the other). Then $x + 45 + x + 11 = 120$ gives $x = 32$ , so the regions are 32, 45 and 43.
(b)
Find the number of customers who bought shoes.
(c)
Calculate the probability that a customer selected at random bought bags.

Worked solution (try it first)

(a)

Draw two overlapping circles, B (bags) and S (shoes), inside a rectangle of 120 customers.
Put 45 in the overlap.
Let $x$ customers buy bags only.
Then "11 more bought shoes than bags" makes shoes only $x + 11$ .
Nobody is outside both circles.
Everyone is in the diagram, so $x + 45 + (x + 11) = 120$ .
That gives $2x + 56 = 120$ , $2x = 64$ and $x = 32$ .

(b)

Shoes: $45 + (32 + 11) = 88$ customers.

(c)

Bags: $32 + 45 = 77$ customers, so $P(\text{bought bags}) = \frac{77}{120}$
$\approx 0.642$ .

Report a problem with this question

Question 6

(a)
A manufacturing company requires 3 hours of direct labour to process every ₦87.00 worth of raw materials. If the company uses ₦30,450.00 worth of raw materials, what amount should it budget for direct labour at ₦18.25 per hour?
(b)
An investor invested ₦ $x$ in bank $M$ at the rate of $6\%$ simple interest per annum and ₦ $y$ in bank $N$ at the rate of $8\%$ simple interest per annum. If a total of ₦8,000,000.00 was invested in the two banks and the investor received a total of ₦2,320,000.00 as interest from the two banks after 4 years, calculate the: (i) values of $x$ and $y$ ; (ii) interest paid by the second bank.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Every ₦87.00 of raw materials needs 3 hours.
₦30,450.00 is $\frac{30\,450}{87} = 350$ lots of ₦87.00.
So the labour needed is $350 \times 3 = 1050$ hours.
At ₦18.25 an hour, the budget is $1050 \times 18.25 =$ ₦19,162.50.

(b)(i)

Simple interest $= \frac{\text{principal} \times \text{rate} \times \text{time}}{100}$ .
In 4 years, bank $M$ pays $\frac{x \times 6 \times 4}{100} = 0.24x$ and bank $N$ pays $\frac{y \times 8 \times 4}{100} = 0.32y$ .
Two facts give two equations: $x + y = 8\,000\,000$ (1) and $0.24x + 0.32y = 2\,320\,000$ (2).
Multiply (1) by 0.24: $0.24x + 0.24y = 1\,920\,000$ (3).
Take (3) from (2): $0.08y = 400\,000$ , so $y = 5\,000\,000$ .
Then $x = 8\,000\,000 - 5\,000\,000$
$= 3\,000\,000$ .
So $x =$ ₦3,000,000.00 and $y =$ ₦5,000,000.00.

(ii)

Bank $N$ paid $0.32 \times 5\,000\,000 =$ ₦1,600,000.00 in interest.

Report a problem with this question

Question 7

(a)
Copy and complete the table of values for $y = 2x^2 - 7x - 9$ for $-3 \le x \le 6$ .

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

$y$ $13$ $-9$ $-14$ $-12$ $6$

Model answer

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

$y$ $30$ $13$ $0$ $-9$ $-14$ $-15$ $-12$ $-5$ $6$ $21$

For example $x = -3$ : $18 + 21 - 9 = 30$ , and $x = 2$ : $8 - 14 - 9 = -15$ .
(b)
Using scales of 2 cm to 1 unit on the $x$ -axis and 2 cm to 4 units on the $y$ -axis, draw the graph of $y = 2x^2 - 7x - 9$ for $-3 \le x \le 6$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 1 unit across, 2 cm to 4 units up.

For (c): (i) $2x^2 - 7x = 26$ is $2x^2 - 7x - 9 = 17$ , so draw $y = 17$ : $x \approx −2.3 and 5.8$ . (ii) The minimum point is about $(1.75, -15.1)$ . (iii) $2x^2 - 7x < 9$ means $y < 0$ : the curve is below the $x$ -axis for $-1 < x < 4.5$ .
(c)
Use the graph to estimate the: (i) roots of the equation $2x^2 - 7x = 26$ ; (ii) coordinates of the minimum point of $y$ ; (iii) range of values for which $2x^2 - 7x < 9$ .
Separate values with commas, e.g. 3, −2

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$		$13$		$-9$	$-14$		$-12$		$6$

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$y$	$30$	$13$	$0$	$-9$	$-14$	$-15$	$-12$	$-5$	$6$	$21$

Try it on a graph

The curve with the line y = 17.

Worked solution (try it first)

(a)

Put each $x$ into $y = 2x^2 - 7x - 9$ .
For example $x = -3$ : $18 + 21 - 9 = 30$ .
$x = 2$ : $8 - 14 - 9 = -15$ .
The row is $30, 13, 0, -9, -14, -15, -12, -5, 6, 21$ .

(b)

Plot the ten points with the given scales and join them with a smooth U-shaped curve.

(c)(i)

Make the equation match the curve: $2x^2 - 7x = 26$ is $2x^2 - 7x - 9 = 17$ , that is $y = 17$ .
Draw the line $y = 17$ and read where it meets the curve: $x \approx -2.3$ and $x \approx 5.8$ .

(ii)

The lowest point of the curve is about $(1.75, -15.1)$ .

(iii)

$2x^2 - 7x < 9$ is $2x^2 - 7x - 9 < 0$ , that is $y < 0$ : where the curve is below the $x$ -axis.
It crosses the axis at $x = -1$ and $x = 4.5$ , so $-1 < x < 4.5$ .

Report a problem with this question

Question 8

Marks	1	2	3	4	5
Number of students	$m + 2$	$m - 1$	$2m - 3$	$m + 5$	$3m - 4$

The table shows the distribution of marks scored by some students in a test.

(a)
If the mean mark is $3\frac{6}{23}$ , find the value of $m$ .
(b)
Find the: (i) interquartile range; (ii) probability of selecting a student who scored at least 4 marks in the test.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$\sum f = (m + 2) + (m - 1) + (2m - 3) + (m + 5) + (3m - 4)$
$= 8m - 1$ and $\sum fx = (m + 2) + 2(m - 1) + 3(2m - 3) + 4(m + 5) + 5(3m - 4)$
$= 28m - 9$ .
The mean is $3\frac{6}{23} = \frac{75}{23}$ , so $\frac{28m - 9}{8m - 1} = \frac{75}{23}$ .
Cross-multiply: $23(28m - 9) = 75(8m - 1)$ , so $644m - 207 = 600m - 75$ , $44m = 132$ and $m = 3$ .

(b)(i)

With $m = 3$ the frequencies are $5, 2, 3, 8, 5$ , a total of 23.
Running totals: $5, 7, 10, 18, 23$ .
$Q_1$ is at position $\frac{23}{4} = 5.75$ , which rounds up to the 6th mark: 2.
$Q_3$ is at position $\frac{3 \times 23}{4} = 17.25$ , which rounds up to the 18th mark: 4.
Interquartile range $= 4 - 2 = 2$ .

(ii)

"At least 4 marks" means 4 or 5: $8 + 5 = 13$ students.
The probability is $\frac{13}{23}$ .

Report a problem with this question

Question 9

Circle geometry

(a)
$PQ$ is a tangent to a circle $RST$ at the point $S$ . $PRT$ is a straight line, $\angle TPS = 34^\circ$ and $\angle TSQ = 65^\circ$ . (i) Illustrate the information in a diagram. (ii) Find the value of: (I) $\angle RTS$ ; (II) $\angle SRP$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $XVY$ and $XWZ$ are straight lines cutting the circle $VYZW$ . $|VZ| = |YZ|$ , $\angle YXZ = 20^\circ$ and $\angle ZVY = 52^\circ$ . Calculate the size of $\angle WYZ$ .

Worked solution (try it first)

(a)(i)

Draw the circle through $R$ , $S$ and $T$ .
Draw the tangent $PQ$ touching it at $S$ , and the straight line from $P$ through $R$ to $T$ .
Mark $\angle TPS = 34^\circ$ and $\angle TSQ = 65^\circ$ .

(ii)

(I)** $\angle TSQ$ is an exterior angle of triangle $PST$ , so it equals the sum of the two opposite interior angles: $65^\circ = 34^\circ + \angle PTS$ .
So $\angle RTS = 31^\circ$ .
(II) By the alternate segment theorem, the angle between the tangent $SQ$ and the chord $ST$ equals the angle in the alternate segment: $\angle SRT = 65^\circ$ .
$P$ , $R$ , $T$ are on a straight line, so $\angle SRP = 180^\circ - 65^\circ$
$= 115^\circ$ .

(b)

$X$ , $V$ , $Y$ are on a straight line, so $\angle XVZ = 180^\circ - 52^\circ$
$= 128^\circ$ .
In triangle $XVZ$ : $\angle XZV = 180^\circ - 20^\circ - 128^\circ$
$= 32^\circ$ .
$|VZ| = |YZ|$ , so $\angle VYZ = \angle ZVY = 52^\circ$ .
$\angle VYW$ and $\angle VZW$ stand on the same arc $VW$ , so $\angle VYW = 32^\circ$ (angles in the same segment).
So $\angle WYZ = 52^\circ - 32^\circ$
$= 20^\circ$ .

Report a problem with this question

Question 10

(a)
Given that $\sin x = \frac{5}{13}$ , $0^\circ < x < 90^\circ$ , find $\dfrac{\cos x - 2\sin x}{2\tan x}$ .
(b)
A ladder $LA$ leans against a vertical pole at a point $L$ which is $9.6$ metres above the ground. Another ladder $LB$ , 12 metres long, leans on the opposite side of the pole at the same point $L$ . If $A$ and $B$ are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to two significant figures, the: (i) length of ladder $LA$ ; (ii) angle which $LA$ makes with the ground.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

$\sin x = \frac{5}{13}$ gives a right-angled triangle with sides 5, 12 and 13, so $\cos x = \frac{12}{13}$ and $\tan x = \frac{5}{12}$ .
Then $\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - \frac{10}{13}}{\frac{10}{12}}$
$= \frac{2}{13} \times \frac{12}{10}$
$= \frac{12}{65}$ .

(b)

Let the foot of the pole be $F$ .

(i)

For ladder $LB$ : $|BF| = \sqrt{12^2 - 9.6^2}$
$= \sqrt{51.84}$
$= 7.2$ m.
$A$ and $B$ are on opposite sides of the pole and 10 m apart, so $|AF| = 10 - 7.2 = 2.8$ m.
Then $|LA| = \sqrt{9.6^2 + 2.8^2}$
$= \sqrt{100}$
$= 10\text{ m}$ .

(ii)

In triangle $LFA$ : $\tan\angle LAF = \frac{9.6}{2.8}$
$\approx 3.429$ , so the angle is about $73.7^\circ$ , which is $74^\circ$ to two significant figures.

Report a problem with this question

Question 11

(a)
It takes 8 students two-thirds of an hour to fill 12 tanks with water. How many tanks of water will 4 students fill in one-third of an hour at the same rate?
(b)
A chord, $20\text{ cm}$ long, is $12\text{ cm}$ from the centre of the circle. Calculate, correct to one decimal place, the: (i) angle subtended by the chord at the centre of the circle; (ii) perimeter of the minor segment cut off by the chord. $\left[\text{Take }\pi = \frac{22}{7}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The number of tanks is proportional to the number of students and to the time.
Half the students ( $\frac48$ ) for half the time ( $\frac{1/3}{2/3}$ ) fill $12 \times \frac12 \times \frac12 = 3$ tanks.

(b)(i)

The perpendicular from the centre to the chord bisects it, making a right-angled triangle with sides 12 cm (the distance) and 10 cm (half the chord).
Half the angle: $\tan\alpha = \frac{10}{12}$ , so $\alpha \approx 39.8^\circ$ and the angle at the centre is $79.6^\circ$ .

(ii)

The radius is the hypotenuse: $r = \sqrt{12^2 + 10^2}$
$= \sqrt{244}$
$\approx 15.62$ cm.
Arc $= \frac{79.6}{360} \times 2 \times \frac{22}{7} \times 15.62$
$\approx 21.7$ cm.
Perimeter of the segment $=$ arc $+$ chord $= 21.7 + 20 = 41.7$ cm.

Report a problem with this question

Question 12

(a)
Using the method of completing the square, solve, correct to 2 decimal places, the equation $3y^2 - 5y + 2 = 0$ .
Separate values with commas, e.g. 3, −2
(b)
Given that $M = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}$ , $N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}$ and $MN = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ , find the matrix $N$ .
Show the answer
$N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

Worked solution (try it first)

(a)

Divide by 3 so that $y^2$ stands alone, and move the number across: $y^2 - \frac53y = -\frac23$ .
Half of $-\frac53$ is $-\frac56$ .
Add its square, $\frac{25}{36}$ , to both sides: $\left(y - \frac56\right)^2 = \frac{25}{36} - \frac{24}{36}$
$= \frac{1}{36}$ .
Take square roots: $y - \frac56 = \pm\frac16$ , so $y = \frac56 + \frac16 = 1$ or $y = \frac56 - \frac16 = \frac23$ .
To 2 decimal places, $y = 1.00$ or $y = 0.67$ .

(b)

Multiply row by column: $MN = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix}$ .
Match each entry with $\begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ .
The first column gives $m + 2n = 2$ and $4m + 3n = 3$ .
Take 4 times the first from the second: $-5n = -5$ , so $n = 1$ and then $m = 0$ .
The second column gives $x + 2y = 1$ and $4x + 3y = 4$ .
In the same way, $-5y = 0$ , so $y = 0$ and then $x = 1$ .
So $N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ .
Check: $\begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$ ✓.

Report a problem with this question

Question 13

(a)
The operation $(*)$ is defined on the set of real numbers, $\mathbb R$ , by $x * y = \dfrac{x + y}{2}$ , $x, y \in \mathbb R$ . (i) Evaluate $3 * \frac25$ . (ii) If $8 * y = 8\frac14$ , find the value of $y$ .
Separate values with commas, e.g. 3, −2
(b)
In $\triangle ABC$ , $\overrightarrow{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}$ . If $P$ is the midpoint of $\overline{AB}$ , express $\overrightarrow{CP}$ as a column vector.
Show the answer
$\begin{pmatrix} -5 \\ 11 \end{pmatrix}$

Worked solution (try it first)

(a)(i)

Put $x = 3$ and $y = \frac25 = 0.4$ into the rule: $3 * \frac25 = \frac{3 + 0.4}{2}$
$= \frac{3.4}{2}$
$= 1.7$ .

(ii)

$8 * y = \frac{8 + y}{2}$ , and $8\frac14 = \frac{33}{4}$ .
So $\frac{8 + y}{2} = \frac{33}{4}$ .
Multiply both sides by 2: $8 + y = \frac{33}{2} = 16\frac12$ .
So $y = 16\frac12 - 8 = 8\frac12$ .

(b)

$P$ is the midpoint of $AB$ , so $\overrightarrow{AP} = \frac12\overrightarrow{AB}$
$= \frac12\begin{pmatrix} -4 \\ 6 \end{pmatrix}$
$= \begin{pmatrix} -2 \\ 3 \end{pmatrix}$ .
Go from $C$ to $A$ , then from $A$ to $P$ : $\overrightarrow{CP} = \overrightarrow{CA} + \overrightarrow{AP}$ , where $\overrightarrow{CA} = -\overrightarrow{AC}$
$= \begin{pmatrix} -3 \\ 8 \end{pmatrix}$ .
So $\overrightarrow{CP} = \begin{pmatrix} -3 + (-2) \\ 8 + 3 \end{pmatrix}$
$= \begin{pmatrix} -5 \\ 11 \end{pmatrix}$ .

Report a problem with this question