WAEC · 2019 · May/June · General Maths · Paper 2

Question 1

(a)
Given that $110_x = 40_{\text{five}}$ , find the value of $x$ .
(b)
Simplify $\dfrac{15}{\sqrt{75}} + \sqrt{108} + \sqrt{432}$ , leaving the answer in the form $a\sqrt b$ , where $a$ and $b$ are positive integers.

Worked solution (try it first)

(a)

Change both to base ten: $110_x = x^2 + x$ and $40_{\text{five}} = 4 \times 5 = 20$ .
So $x^2 + x - 20 = 0$ , $(x + 5)(x - 4) = 0$ , and the base is $x = 4$ (a base can't be negative).

(b)

Rationalise and simplify each term: $\frac{15}{\sqrt{75}} = \frac{15}{5\sqrt3}$
$= \frac{3}{\sqrt3}$
$= \sqrt3$ .
$\sqrt{108} = 6\sqrt3$ .
$\sqrt{432} = \sqrt{144 \times 3}$
$= 12\sqrt3$ .
Total: $\sqrt3 + 6\sqrt3 + 12\sqrt3 = 19\sqrt3$ .

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Question 2

(a)
Find the equation of the line which passes through the points $A(-2, 7)$ and $B(2, -3)$ .
Show the answer
$2y + 5x - 4 = 0$
(b)
Given that $\dfrac{5b - a}{8b + 3a} = \dfrac15$ , find, correct to two decimal places, the value of $\dfrac ab$ .

Worked solution (try it first)

(a)

Gradient $= \frac{y_2 - y_1}{x_2 - x_1}$
$= \frac{-3 - 7}{2 - (-2)}$
$= \frac{-10}{4}$
$= -\frac52$ .
Using the point $A(-2, 7)$ : $y - 7 = -\frac52(x + 2)$ .
Multiply by 2: $2y - 14 = -5x - 10$ .
So $2y + 5x - 4 = 0$ .

(b)

One fraction equals another, so cross-multiply: $5(5b - a) = 1 \times (8b + 3a)$ .
Expand: $25b - 5a = 8b + 3a$ .
Collect the $a$ terms on one side and the $b$ terms on the other: $17b = 8a$ .
Divide both sides by $8b$ : $\frac ab = \frac{17}{8} = 2.125$ .
Correct to two decimal places: $2.13$ .

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Question 3

(a)
Ali, Musah and Yusif shared ₦420,000.00 in the ratio $3 : 5 : 8$ respectively. Find the sum of Ali's and Yusif's shares.
(b)
Solve: $2\left(\frac18\right)^x = 32^{x - 1}$ .

Worked solution (try it first)

(a)

The ratio has $3 + 5 + 8 = 16$ parts, so one part is $\frac{420\,000}{16} = 26\,250$ .
Ali gets $3 \times 26\,250 = ₦78,750$ and Yusif $8 \times 26\,250 = ₦210,000$ .
Together: ₦288,750.

(b)

Write both sides as powers of 2: $2 \times (2^{-3})^x = 2^{1 - 3x}$ and $32^{x - 1} = 2^{5(x - 1)} = 2^{5x - 5}$ .
So $1 - 3x = 5x - 5$ , $8x = 6$ and $x = \frac34$ .

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Question 4

(a)
In the diagram, $PQRS$ is a quadrilateral, $\angle PQR = \angle PRS = 90^\circ$ , $|PQ| = 3\text{ cm}$ , $|QR| = 4\text{ cm}$ and $|PS| = 13\text{ cm}$ . Find the area of the quadrilateral.

Worked solution (try it first)

(a)

Split the quadrilateral along the diagonal $PR$ into two right-angled triangles.
In triangle $PQR$ (right angle at $Q$ ): $|PR| = \sqrt{3^2 + 4^2} = 5$ cm.
In triangle $PRS$ (right angle at $R$ ): $|RS| = \sqrt{13^2 - 5^2}$
$= \sqrt{144}$
$= 12$ cm.
Area $= \frac12 \times 3 \times 4 + \frac12 \times 5 \times 12$
$= 6 + 30$
$= 36\text{ cm}^2$ .

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Question 5

Probability

Three red balls, five green balls and a number of blue balls are put together in a sack. One ball is picked at random from the sack. If the probability of picking a red ball is $\frac16$ , find:

(a)
the number of blue balls in the sack;
(b)
the probability of picking a green ball.

Worked solution (try it first)

(a)

Let there be $b$ blue balls.
Then there are $3 + 5 + b = 8 + b$ balls in the sack, and $P(\text{red}) = \frac{3}{8 + b} = \frac16$ .
Cross-multiply: $8 + b = 18$ , so $b = 10$ blue balls.

(b)

There are 18 balls, 5 of them green: $P(\text{green}) = \frac{5}{18}$ .

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Question 6

(a)
The force of attraction, $F$ , between two bodies varies directly as the product of their masses, $m_1$ and $m_2$ , and inversely as the square of the distance, $d$ , between them. Given that $F = 20\text{ N}$ when $m_1 = 25\text{ kg}$ , $m_2 = 10\text{ kg}$ and $d = 5\text{ m}$ , find: (i) an expression for $F$ in terms of $m_1$ , $m_2$ and $d$ ; (ii) the distance $d$ when $F = 30\text{ N}$ , $m_1 = 7.5\text{ kg}$ and $m_2 = 4\text{ kg}$ .
(b)
The diagram is a pentagon with interior angles $x$ , $(x + 20^\circ)$ , $(x + 40^\circ)$ , $(x + 80^\circ)$ and $(x + 60^\circ)$ . Find the value of $x$ .

Worked solution (try it first)

(a)(i)

$F$ varies directly as the product $m_1m_2$ (on top) and inversely as $d^2$ (underneath): $F = \dfrac{km_1m_2}{d^2}$ .
Put in $F = 20$ , $m_1 = 25$ , $m_2 = 10$ , $d = 5$ : $20 = \dfrac{k \times 25 \times 10}{25}$
$= 10k$ , so $k = 2$ .
The expression is $F = \dfrac{2m_1m_2}{d^2}$ .

(ii)

Put in $F = 30$ , $m_1 = 7.5$ , $m_2 = 4$ : $30 = \dfrac{2 \times 7.5 \times 4}{d^2}$
$= \dfrac{60}{d^2}$ .
So $d^2 = \dfrac{60}{30} = 2$ and $d = \sqrt2 \approx 1.41$ m.

(b)

The interior angles of a pentagon add up to $(5 - 2) \times 180^\circ = 540^\circ$ .
So $x + (x + 20) + (x + 40) + (x + 80) + (x + 60) = 540$ , which gives $5x + 200 = 540$ , $5x = 340$ and $x = 68^\circ$ .

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Question 7

The data show the marks obtained by students in a Biology test.

50 56 25 56 68 73 66 64 56 48 20 39 9 50 46 54 54 40 50 96 36 44 18 97 65 21 60 44 54 32 92 49 37 94 72 88 89 35 59 34 15 88 53 16 84 52 72 46 60 42

(a)
Construct a frequency distribution table using the class intervals $0$ – $9$ , $10$ – $19$ , $20$ – $29$ , …
Model answer

Class interval Frequency

0–9 1

10–19 3

20–29 3

30–39 6

40–49 8

50–59 12

60–69 6

70–79 3

80–89 4

90–99 4

Total 50

Tally each score into its class, then count; the frequencies must add up to 50.
(b)
Draw a cumulative frequency curve for the distribution.
Model answer
Plot each cumulative frequency against the upper class boundary of its class, starting from $(-0.5, 0)$ where the cumulative frequency is 0 and ending at $(99.5, 50)$ . Join the points with a smooth rising S-shaped curve (an ogive), not straight lines. Label both axes. Readings from a hand-drawn curve differ a little from person to person; examiners accept a small range, usually about ±1.

For (c): the median is the 25th mark. Read across from 25: about 52.9. Read up from 65.5: about 37 students scored below 66, so about 13 of the 50 (26%) scored at least 66.
(c)
Use the graph to estimate the: (i) median; (ii) percentage of students who scored at least 66 marks, correct to the nearest whole number.
Separate values with commas, e.g. 3, −2

Class interval	Frequency
0–9	1
10–19	3
20–29	3
30–39	6
40–49	8
50–59	12
60–69	6
70–79	3
80–89	4
90–99	4
Total	50

Try it on a graph

The ogive: cumulative frequency at each upper class boundary.

Worked solution (try it first)

(a)

Tally each mark into its class:
Marks 0–9 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 90–99

Frequency 1 3 3 6 8 12 6 3 4 4
The frequencies add up to 50.

Marks	0–9	10–19	20–29	30–39	40–49	50–59	60–69	70–79	80–89	90–99
Frequency	1	3	3	6	8	12	6	3	4	4

(b)

The cumulative frequencies are $1, 4, 7, 13, 21, 33, 39, 42, 46, 50$ .
Plot them at the upper class boundaries $9.5, 19.5, \ldots, 99.5$ , start at $(-0.5, 0)$ , and draw a smooth curve.

(c)(i)

The median is at $\frac{50}{2} = 25$ .
Go across to the curve and down: about 53.
(Check: 25 lies between 21 at 49.5 and 33 at 59.5, and $49.5 + \frac{4}{12} \times 10 \approx 52.8$ .)

(ii)

"At least 66" means 66 or more.
Go up from 65.5 and across: about 36.6, so about 37 students scored less than 66 and about $50 - 37 = 13$ scored at least 66.
As a percentage: $\frac{13}{50} \times 100 = 26\%$ .

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Question 8

(a)
Solve the inequality $\frac13x - \frac14(x + 2) \ge 3x - 1\frac13$ .
Show the answer
$x \le \frac27$
(b)
In the diagram, $ABC$ is a right-angled triangle on a horizontal ground and $|AD|$ is a vertical tower. $\angle BAC = 90^\circ$ , $\angle ACB = 35^\circ$ , $\angle ABD = 52^\circ$ and $|BC| = 66\text{ m}$ . Find, correct to two decimal places, the: (i) height of the tower; (ii) angle of elevation of the top of the tower from $C$ .
A 3-D sketch: triangle ABC lies on the ground and AD is vertical.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Multiply every term by 12: $4x - 3(x + 2) \ge 36x - 16$ .
So $x - 6 \ge 36x - 16$ , and $-35x \ge -10$ .
Divide by $-35$ , which reverses the sign: $x \le \frac{10}{35} = \frac27$ .

(b)(i)

In triangle $ABC$ the right angle is at $A$ , so $BC = 66$ m is the hypotenuse.
$AB$ is opposite the $35^\circ$ angle at $C$ : $|AB| = 66\sin 35^\circ \approx 37.86$ m.
The tower $AD$ stands at $A$ , and from $B$ its top is at $52^\circ$ : $|AD| = |AB|\tan 52^\circ$
$\approx 37.86 \times 1.2799$
$\approx 48.45$ m.

(ii)

$|AC| = 66\cos 35^\circ \approx 54.06$ m.
From $C$ : $\tan\theta = \frac{48.45}{54.06}$
$\approx 0.8962$ , so $\theta \approx 41.87^\circ$ .

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Question 9

Trigonometric graphs

(a)
Copy and complete the table of values for $y = 2\cos x + 3\sin x$ for $0^\circ \le x \le 360^\circ$ .

$x$ $0^\circ$ $60^\circ$ $120^\circ$ $180^\circ$ $240^\circ$ $300^\circ$ $360^\circ$

$y$ $2.0$ $-3.6$

Model answer

$x$ 0° 60° 120° 180° 240° 300° 360°

$y$ 2.0 3.6 1.6 −2.0 −3.6 −1.6 2.0

For example, at $x = 60^\circ$ : $y = 2(0.5) + 3(0.866) = 3.6$ (1 d.p.).
(b)
Using a scale of 2 cm to $60^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of $y = 2\cos x + 3\sin x$ for $0^\circ \le x \le 360^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). With points only every $60^\circ$ , draw a smooth wave: it peaks at about $3.6$ near $56^\circ$ and dips to about $-3.6$ near $236^\circ$ .

For (c): (i) draw $y = -1$ : $x \approx 162.4^\circ$ and $310.2^\circ$ . (ii) At $x = 342^\circ$ , $y \approx 1.0$ .
(c)(i)
Using the graph, solve $2\cos x + 3\sin x = -1$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Using the graph, find, correct to one decimal place, the value of $y$ when $x = 342^\circ$ .

$x$	0°	60°	120°	180°	240°	300°	360°
$y$	2.0	3.6	1.6	−2.0	−3.6	−1.6	2.0

Try it on a graph

x in degrees. The line y = −1 gives the solutions of (c)(i).

Worked solution (try it first)

(a)

In degree mode, to 1 decimal place.
For example, $x = 60^\circ$ : $2(0.5) + 3(0.866) = 1 + 2.598 = 3.6$ .
$x = 300^\circ$ : $1 - 2.598 = -1.6$ .
The full row is $2.0, 3.6, 1.6, -2.0, -3.6, -1.6, 2.0$ .

(b)

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

(c)(i)

Draw the line $y = -1$ and read down from where it meets the curve: $x \approx 162^\circ$ and $x \approx 310^\circ$ .

(ii)

Read up from $x = 342^\circ$ to the curve and across: $y \approx 1.0$ .
(By calculation, $2\cos 342^\circ + 3\sin 342^\circ \approx 1.90 - 0.93$
$= 0.97$ .)

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Question 10

Commercial arithmetic

(a)
A woman bought 130 kg of tomatoes for ₦52,000.00. She sold half of the tomatoes at a profit of $30\%$ . The rest of the tomatoes began to go bad; she then reduced the selling price per kg by $12\%$ . Calculate the new selling price per kg.
(b)
Calculate the percentage profit on the entire sales if she threw away 5 kg of bad tomatoes.

Worked solution (try it first)

(a)

Cost per kg $= \frac{52\,000}{130} = ₦400$ .
At $30\%$ profit the selling price was $1.3 \times 400 = ₦520$ per kg.
Reduced by $12\%$ : $0.88 \times 520 = ₦457.60$ per kg.

(b)

She sold 65 kg at ₦520: ₦33,800.
Of the other 65 kg she threw away 5 kg and sold 60 kg at ₦457.60: ₦27,456.
Total sales $= ₦61,256$ .
Profit $= 61\,256 - 52\,000 = ₦9256$ .
Percentage profit $= \frac{9256}{52\,000} \times 100\%$
$\approx 17.8\%$ .

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Question 11

(a)
The third and sixth terms of a Geometric Progression (G.P.) are $\frac14$ and $\frac{1}{32}$ respectively. Find the: (i) first term and the common ratio; (ii) seventh term.
Separate values with commas, e.g. 3, −2
(b)
Given that 2 and $-3$ are the roots of the equation $ax^2 + bx + c = 0$ , find the values of $a$ , $b$ and $c$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

The $n$ th term of a G.P. is $ar^{n - 1}$ .
So $T_3 = ar^2 = \frac14$ and $T_6 = ar^5 = \frac{1}{32}$ .
Divide the sixth term by the third, so that $a$ cancels: $\frac{ar^5}{ar^2} = r^3$
$= \frac{1}{32} \div \frac14$
$= \frac{4}{32}$
$= \frac18$ .
So $r = \frac12$ .
Put $r = \frac12$ into $ar^2 = \frac14$ : $a \times \frac14 = \frac14$ , so $a = 1$ .
The first term is 1 and the common ratio is $\frac12$ .

(ii)

$T_7 = ar^6$
$= 1 \times \left(\frac12\right)^6$
$= \frac{1}{64}$ .

(b)

A root of 2 gives the factor $(x - 2)$ , and a root of $-3$ gives $(x + 3)$ .
So the equation is $(x - 2)(x + 3) = 0$ .
Expand: $x^2 + 3x - 2x - 6 = x^2 + x - 6 = 0$ .
Compare with $ax^2 + bx + c = 0$ : $a = 1$ , $b = 1$ and $c = -6$ .

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Question 12

(a)
Given that $\sin y = \frac{8}{17}$ , find the value of $\dfrac{\tan y}{1 + 2\tan y}$ .
(b)
An amount of ₦300,000.00 was shared among Otobo, Ada and Adeola. Otobo received ₦60,000.00, Ada received $\frac{5}{12}$ of the remainder, while the rest went to Adeola. In what ratio was the money shared?
Show the answer
$3 : 5 : 7$

Worked solution (try it first)

(a)

$\sin y = \frac{8}{17}$ : the adjacent side is $\sqrt{17^2 - 8^2} = \sqrt{225} = 15$ , so $\tan y = \frac{8}{15}$ .
Then $\frac{\tan y}{1 + 2\tan y} = \frac{\frac{8}{15}}{1 + \frac{16}{15}}$
$= \frac{\frac{8}{15}}{\frac{31}{15}}$
$= \frac{8}{31}$ .

(b)

After Otobo's ₦60,000, the remainder is ₦240,000.
Ada received $\frac{5}{12} \times 240\,000 =$ ₦100,000, and Adeola the rest, ₦140,000.
The ratio is $60\,000 : 100\,000 : 140\,000 = 3 : 5 : 7$ .

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Question 13

(a)
In the diagram, $\overline{RS}$ and $\overline{RT}$ are tangents to the circle with centre $O$ . $\angle TUS = 68^\circ$ , $\angle SRT = x$ and $\angle UTO = y$ . Find the value of $x$ .
(b)
Two tanks $A$ and $B$ are filled to capacity with diesel. Tank $A$ holds 600 litres of diesel more than tank $B$ . If 100 litres of diesel were pumped out of each tank, tank $A$ would then contain 3 times as much as tank $B$ . Find the capacity of each tank.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Join $O$ to $S$ and $T$ .
$\angle SOT = 2\angle SUT$
$= 2 \times 68^\circ$
$= 136^\circ$ (angle at the centre is twice the angle at the circumference).
A tangent meets the radius at $90^\circ$ , so $\angle OSR = \angle OTR = 90^\circ$ .
The angles of quadrilateral $OSRT$ add up to $360^\circ$ : $x = 360^\circ - 90^\circ - 90^\circ - 136^\circ$
$= 44^\circ$ .

(b)

Let tank $B$ hold $b$ litres.
Tank $A$ holds 600 litres more, $b + 600$ .
After 100 litres are pumped out of each, they hold $b + 500$ and $b - 100$ , and $A$ then holds three times as much as $B$ : $b + 500 = 3(b - 100)$ .
Expand: $b + 500 = 3b - 300$ .
So $2b = 800$ and $b = 400$ .
Tank $A$ holds 1000 litres and tank $B$ holds 400 litres.

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