WAEC · 2025 · May/June · General Maths · Paper 2

Question 1 ✱✱

(a)
Factorize $px - 2qx - 4qy + 2py$ .
(b)
Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ , $P = \{1, 2, 4, 6, 10\}$ and $Q = \{2, 3, 6, 9\}$ , show that $(P \cup Q)' = P' \cap Q'$ .
Show the answer
Both sides equal $\{5, 7, 8\}$ .

Worked solution (try it first)

(a)

Group the terms in $x$ and the terms in $y$ : $(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q)$ .
$(p - 2q)$ is a common factor: $px - 2qx - 4qy + 2py = (p - 2q)(x + 2y)$ .

(b)

Left side: $P \cup Q = \{1, 2, 3, 4, 6, 9, 10\}$ , so $(P \cup Q)' = \{5, 7, 8\}$ .
Right side: $P' = \{3, 5, 7, 8, 9\}$ and $Q' = \{1, 4, 5, 7, 8, 10\}$ .
The elements in both: $P' \cap Q' = \{5, 7, 8\}$ .
Both sides equal $\{5, 7, 8\}$ , so $(P \cup Q)' = P' \cap Q'$ .

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Question 2 ✱✱

Variation

$y$ is partly constant and partly varies inversely as the square of $x$ .

(a)
Write down the relationship between $x$ and $y$ .
Show the answer
$y = a + \dfrac{b}{x^2}$ , where $a$ and $b$ are constants
(b)
When $x = 1$ , $y = 11$ , and when $x = 2$ , $y = 5$ . Find $y$ when $x = 4$ .

Worked solution (try it first)

(a)

Partly constant ( $a$ ) and partly varies inversely as the square of $x$ ( $\frac{b}{x^2}$ ): $y = a + \dfrac{b}{x^2}$ , where $a$ and $b$ are constants.

(b)

$x = 1$ , $y = 11$ : $a + b = 11$ .
$x = 2$ , $y = 5$ : $a + \frac{b}{4} = 5$ .
Subtract: $b - \frac{b}{4} = 6$ , so $\frac{3b}{4} = 6$ and $b = 8$ .
Then $a = 11 - 8 = 3$ , and $y = 3 + \dfrac{8}{x^2}$ .
When $x = 4$ : $y = 3 + \dfrac{8}{16} = 3\frac12$ .

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Question 3 ✱✱

(a)
In the diagram, $PQSR$ and $SRYZ$ are parallelograms and $PQYZ$ is a straight line. If $|QY| = 2\text{ cm}$ and $|RS| = 3\text{ cm}$ , find $|PZ|$ .
(b)
Towns $P$ and $Q$ lie on latitude $56^\circ\text{N}$ , with longitudes $25^\circ\text{E}$ and $95^\circ\text{E}$ . Find the distance $PQ$ along the parallel of latitude, correct to the nearest km. $\left[R = 6400\text{ km}, \pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

Opposite sides of a parallelogram are equal.
In $PQSR$ , $PQ = RS = 3$ cm.
In $SRYZ$ , $YZ = RS = 3$ cm.
Along the straight line: $PZ = PQ + QY + YZ = 3 + 2 + 3 = 8$ cm.

(b)

Both towns are on latitude $56^\circ$ N, and the difference in longitude is $95^\circ - 25^\circ = 70^\circ$ .
Along the parallel: $PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times \cos 56^\circ$
$\approx 4374$ km.

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Question 4 ✱✱

(a)
A pack of 52 playing cards is shuffled and one card is drawn. Find the probability that it is either a five or a red nine.
(b)
The bearing of $Q$ from $P$ is $150^\circ$ and the bearing of $R$ from $Q$ is $060^\circ$ . If $|PQ| = 5\text{ m}$ and $|QR| = 3\text{ m}$ , find the bearing of $R$ from $P$ , correct to the nearest degree.

Worked solution (try it first)

(a)

There are four fives and two red nines (hearts and diamonds), with no card counted twice: 6 cards.
Probability $= \frac{6}{52} = \frac{3}{26}$ .

(b)

Break each leg into east and north parts.
$P$ to $Q$ on $150^\circ$ , 5 m: east $5\sin 150^\circ = 2.5$ , north $5\cos 150^\circ \approx -4.330$ (so 4.330 south).
$Q$ to $R$ on $060^\circ$ , 3 m: east $3\sin 60^\circ \approx 2.598$ , north $3\cos 60^\circ = 1.5$ .
Altogether, $R$ is $5.098$ m east and $2.830$ m south of $P$ .
That direction is south-east of $P$ , with angle $\tan^{-1}\frac{2.830}{5.098} \approx 29.0^\circ$ below due east.
Bearing of $R$ from $P$ $= 090^\circ + 29.0^\circ$
$\approx 119^\circ$ .

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Question 5 ✱✱

Statistics: data & averages

The table shows the marks scored by 32 students.

Mark	1	2	3	4	5	6	7	8	9	10
Frequency	2	3	4	4	4	4	5	3	2	1

(a)
Find the (i) mean, correct to two decimal places; (ii) median; (iii) mode.
Separate values with commas, e.g. 3, −2
(b)
Find the percentage of students who scored at least 8 marks.

Worked solution (try it first)

(a)(i)

$\sum fx = 1(2) + 2(3) + 3(4) + 4(4) + 5(4) + 6(4) + 7(5) + 8(3) + 9(2) + 10(1)$
$= 2 + 6 + 12 + 16 + 20 + 24 + 35 + 24 + 18 + 10$
$= 167$ , and $\sum f = 32$ .
Mean $= \frac{167}{32} \approx 5.22$ .

(ii)

With 32 marks, the median is halfway between the 16th and 17th.
Running totals: $2, 5, 9, 13, 17, \ldots$ .
The 14th to 17th marks are all 5, so the median is 5.

(iii)

The highest frequency is 5, for the mark 7, so the mode is 7.

(b)

"At least 8" means 8, 9 or 10: $3 + 2 + 1 = 6$ students.
As a percentage: $\frac{6}{32} \times 100 = 18.75\%$ .

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Question 6 ✱✱

(a)
Using mathematical tables, find: (i) $2\sin63.35^\circ$ ; (ii) $\log(\cos44.74^\circ)$ ; (iii) $k$ , given that $\log k - \log(k - 2) = \log 5$ .
Separate values with commas, e.g. 3, −2
(b)
Use logarithm tables to evaluate $\dfrac{3.68^2 \times 6.705}{\sqrt{0.3581}}$ .

Worked solution (try it first)

(a)(i)

From the tables, $\sin 63.35^\circ \approx 0.8938$ , so $2\sin 63.35^\circ \approx 1.788$ .

(ii)

$\cos 44.74^\circ \approx 0.7103$ , and $\log 0.7103 = \bar{1}.8514$ (which is $-0.1486$ ).

(iii)

$\log k - \log(k - 2) = \log\frac{k}{k - 2}$ , so $\frac{k}{k - 2} = 5$ .
Then $k = 5k - 10$ , $4k = 10$ and $k = 2\frac12$ .

(b)

Add and subtract logarithms: $\log(3.68^2) = 2 \times 0.5658 = 1.1316$ .
$\log 6.705 = 0.8264$ .
$\log\sqrt{0.3581} = \frac12(\bar{1}.5540)$
$= \bar{1}.7770$ .
So the log of the answer is $1.1316 + 0.8264 - \bar{1}.7770 = 1.9580 - \bar{1}.7770$
$= 2.1810$ .
The antilog of $2.1810$ is about $151.7$ .

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Question 7 ✱✱

(a)
Given $p = x + ym^3$ , find $m$ in terms of $p$ , $x$ and $y$ .
Show the answer
$m = \sqrt[3]{\dfrac{p - x}{y}}$
(b)
By completing the square, find the roots of $x^2 - 6x + 7 = 0$ , correct to one decimal place.
Separate values with commas, e.g. 3, −2
(c)
The product of two consecutive positive odd numbers is 195. Form a quadratic equation and solve it to find the numbers.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Take $x$ from both sides: $ym^3 = p - x$ .
Divide by $y$ : $m^3 = \frac{p - x}{y}$ .
Take the cube root: $m = \sqrt[3]{\frac{p - x}{y}}$ .

(b)

Move the number across: $x^2 - 6x = -7$ .
Half of $-6$ is $-3$ , and $(-3)^2 = 9$ : add 9 to both sides.
$x^2 - 6x + 9 = 2$ , so $(x - 3)^2 = 2$ .
Take the square root, remembering $\pm$ : $x - 3 = \pm\sqrt2$ .
So $x = 3 + \sqrt2 \approx 4.4$ or $x = 3 - \sqrt2 \approx 1.6$ (to one decimal place).

(c)

Let the smaller odd number be $n$ .
The next odd number is $n + 2$ .
Their product is 195: $n(n + 2) = 195$ , so $n^2 + 2n - 195 = 0$ .
Two numbers that multiply to $-195$ and add to 2 are 15 and $-13$ : $(n + 15)(n - 13) = 0$ .
So $n = 13$ or $n = -15$ .
The numbers are positive, so $n = 13$ : the numbers are 13 and 15.

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Question 8 ✱✱

Trigonometric graphs

(a)
Complete the table of values for $y = 2\cos2x - 1$ for $x = 0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 180^\circ$ .
Separate values with commas, e.g. 3, −2
(b)
Draw the graph of $y = 2\cos2x - 1$ for $0^\circ \le x \le 180^\circ$ . On the same axes, draw $y = \frac{1}{180}(x - 360)$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). The curve starts at 1, falls to $-3$ at $90^\circ$ and rises back to 1 at $180^\circ$ . The line $y = \frac{1}{180}(x - 360)$ runs from $(0^\circ, -2)$ to $(180^\circ, -1)$ .

For (c): $2\cos 2x + \frac12 = 0$ is $2\cos 2x - 1 = -1.5$ , so draw $y = -1.5$ (not the sloping line): $x \approx 52.2^\circ$ and $127.8^\circ$ . The sloping line would solve a different equation; it meets the curve at $x \approx 55.1^\circ$ and $131.1^\circ$ .
(c)
Use the graph to find the values of $x$ for which $2\cos2x + \frac12 = 0$ .
Separate values with commas, e.g. 3, −2

Try it on a graph

x is in degrees. For (c), 2cos 2x + ½ = 0 is where the curve meets y = −1.5.

Worked solution (try it first)

(a)

Double $x$ first, then take the cosine.
$x = 0^\circ$ : $2\cos 0^\circ - 1 = 1$ .
$x = 30^\circ$ : $2\cos 60^\circ - 1 = 0$ .
$x = 60^\circ$ : $2\cos 120^\circ - 1 = -2$ .
$x = 90^\circ$ : $2\cos 180^\circ - 1 = -3$ .
By symmetry, $x = 120^\circ, 150^\circ, 180^\circ$ give $-2, 0, 1$ .

(b)

Plot the seven points and join them with a smooth curve.
The line $y = \frac{1}{180}(x - 360)$ passes through $(0, -2)$ and $(180, -1)$ : draw it with a ruler.

(c)

Rearrange to match the curve: $2\cos 2x + \frac12 = 0$ is $2\cos 2x - 1 = -1\frac12$ .
Draw $y = -1.5$ and read where it meets the curve: $x \approx 52^\circ$ and $x \approx 128^\circ$ .

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Question 9 ✱✱

Construction & loci

(a)
Using a ruler and a pair of compasses only, construct $\triangle ABC$ with $|AB| = 7.5\text{ cm}$ , $|BC| = 8.1\text{ cm}$ and $\angle ABC = 105^\circ$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $BC = 8.1$ cm. $105^\circ = 90^\circ + 15^\circ$ : construct $90^\circ$ at $B$ , then bisect the $30^\circ$ between the $90^\circ$ and $120^\circ$ lines to add $15^\circ$ . Mark $A$ with $BA = 7.5$ cm and join $AC$ .
(b)
Locate $D$ on $BC$ such that $|BD| : |DC| = 3 : 2$ , and through $D$ construct the line perpendicular to $BC$ .
Model answer
To divide $BC$ in the ratio $3 : 2$ , draw a line from $B$ at any angle and step off 5 equal lengths with the compasses. Join the fifth mark to $C$ , and draw a parallel through the third mark to cut $BC$ at $D$ . Then $BD = \frac35 \times 8.1 = 4.86$ cm. At $D$ , construct the perpendicular to $BC$ . It meets $AC$ at $P$ , and $|BP|$ measures about 5.4 cm.
(c)
If the perpendicular meets $AC$ at $P$ , measure $|BP|$ (cm).

Worked solution (try it first)

(a)

Draw $BC = 8.1$ cm.
At $B$ construct $105^\circ$ : construct $90^\circ$ and $120^\circ$ , then bisect the angle between them.
Mark $BA = 7.5$ cm on the arm and join $AC$ .

(b)

$|BD| : |DC| = 3 : 2$ is 5 equal parts.
Draw a line from $B$ and step off 5 equal lengths.
Join the 5th mark to $C$ , and through the 3rd mark draw a line parallel to it, cutting $BC$ at $D$ ( $|BD| = \frac35 \times 8.1 = 4.86$ cm).
At $D$ construct the perpendicular to $BC$ : mark equal distances either side of $D$ on $BC$ and bisect the line between them.

(c)

The perpendicular meets $AC$ at $P$ .
Measure $|BP| \approx 5.4$ cm.
(Check: with $B$ at the origin, $A \approx (-1.94, 7.24)$ and $C = (8.1, 0)$ .
The perpendicular $x = 4.86$ meets $AC$ at height about 2.34, so $|BP| = \sqrt{4.86^2 + 2.34^2} \approx 5.4$ cm.)

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Question 10 ✱✱

(a)
A man travels from $X$ on a bearing of $060^\circ$ to $Y$ , $20\text{ km}$ away. From $Y$ he travels to $Z$ on a bearing of $195^\circ$ . If $Z$ is directly east of $X$ , find, correct to three significant figures: (i) $|YZ|$ ; (ii) $|ZX|$ .
Separate values with commas, e.g. 3, −2
(b)
An aircraft flies due south from latitude $36^\circ\text{N}$ to latitude $36^\circ\text{S}$ along the same longitude. (i) Find the distance travelled, correct to three significant figures. (ii) If its speed is $800\text{ km/h}$ , find the time taken, to the nearest hour. $\left[\pi = \frac{22}{7}, R = 6400\text{ km}\right]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Break each leg into north and east parts.
$X$ to $Y$ on $060^\circ$ , 20 km: north $20\cos 60^\circ = 10$ km, east $20\sin 60^\circ \approx 17.32$ km.
$Y$ to $Z$ on $195^\circ$ is $15^\circ$ west of due south.
$Z$ is due east of $X$ , so $Z$ is level with $X$ : the leg $YZ$ must come 10 km back south.

(i)

$|YZ|\cos 15^\circ = 10$ , so $|YZ| = \frac{10}{0.9659} \approx 10.4$ km.

(ii)

Going from $Y$ to $Z$ also moves $10.35\sin 15^\circ \approx 2.68$ km west.
So $|ZX| = 17.32 - 2.68 \approx 14.6$ km.

(b)(i)

Both places are on the same line of longitude, so the aircraft flies along a great circle through an angle of $36^\circ + 36^\circ = 72^\circ$ .
Distance $= \frac{72}{360} \times 2 \times \frac{22}{7} \times 6400$
$\approx 8045.7$ km, which is 8050 km to three significant figures.

(ii)

Time $= \frac{8045.7}{800} \approx 10.06$ , which is 10 hours to the nearest hour.

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Question 11 ✱✱

Dispersion & cumulative frequency

The table shows the scores of 2000 candidates in an examination.

Marks (%)	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90
Frequency	68	184	294	402	480	310	164	98

(a)
Prepare a cumulative frequency table.
Separate values with commas, e.g. 3, −2
(b)
Draw the cumulative frequency curve.
Model answer
Plot each cumulative frequency against the upper class boundary of its class, starting from $(10.5, 0)$ where the cumulative frequency is 0 and ending at $(90.5, 2000)$ . 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. Use a scale that fits 2000 on the vertical axis, e.g. 2 cm to 250 candidates.

Worked solution (try it first)

(a)

Add the frequencies one class at a time, and write the upper class boundary beside each running total:
Marks 11–20 21–30 31–40 41–50 51–60 61–70 71–80 81–90

Upper boundary 20.5 30.5 40.5 50.5 60.5 70.5 80.5 90.5

Cumulative frequency 68 252 546 948 1428 1738 1902 2000
The last cumulative frequency is 2000, the number of candidates, which checks the additions.

Marks	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90
Upper boundary	20.5	30.5	40.5	50.5	60.5	70.5	80.5	90.5
Cumulative frequency	68	252	546	948	1428	1738	1902	2000

(b)

Plot each cumulative frequency against its upper class boundary, starting from $(10.5, 0)$ , the lower boundary of the first class.
Join the points with one smooth S-shaped curve, drawn freehand.

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