WAEC · 2022 · May/June · General Maths · Paper 2

(a)

1. In a G.P., each term divided by the one before gives the same common ratio: $\frac{9}{7 - 2x} = \frac{5x + 17}{9}$.
2. Cross-multiply: $81 = (7 - 2x)(5x + 17)$.
3. Expand: $81 = 35x + 119 - 10x^2 - 34x$
   $= -10x^2 + x + 119$.
4. Rearrange: $10x^2 - x - 38 = 0$.
5. Factorise: $(x - 2)(10x + 19) = 0$.
6. So $x = 2$ or $x = -\frac{19}{10}$.

(b)

1. The numbers are in the ratio $3 : 4$, so let them be $3k$ and $4k$.
2. Thrice the first plus twice the second is 68: $9k + 8k = 68$.
3. So $17k = 68$ and $k = 4$.
4. The numbers are 12 and 16.
5. The smaller is 12.

(a)

1. The bracket is raised to the power $-\frac32$.
2. Undo it by raising both sides to the power $-\frac23$: $y^{-\frac23} = \frac{pr}{m} - p^2 r$.
3. Multiply every term by $m$: $m\,y^{-\frac23} = pr - p^2 m r$.
4. Take out $r$: $m\,y^{-\frac23} = r(p - p^2 m)$.
5. So $r = \frac{m\,y^{-\frac23}}{p - p^2 m}$.

(b)

1. $(-8)^{-\frac23} = \frac{1}{(-8)^{\frac23}}$
   $= \frac{1}{\left(\sqrt[3]{-8}\right)^2}$
   $= \frac{1}{(-2)^2}$
   $= \frac14$.
2. With $m = 1$ and $p = 3$: $r = \frac{1 \times \frac14}{3 - 9 \times 1}$
   $= \frac{\frac14}{-6}$
   $= -\frac{1}{24}$.

1. The perimeter of a segment is its chord plus its arc.
2. The perpendicular from the centre bisects the chord and the $72^\circ$ angle, so half the chord is $24.5\sin 36^\circ \approx 14.40$ m and the chord is about $28.80$ m.
3. Arc $= \frac{72}{360} \times 2 \times \frac{22}{7} \times 24.5$
   $= 30.8$ m.
4. Perimeter $\approx 28.80 + 30.8 = 59.6$ m.

(a)

1. $BCDE$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$: $(2x + 40) + (2y + 10) = 180$, which gives $x + y = 65$.
2. The angle at the centre is twice the angle at the circumference on the same arc $BD$: $5x - 35 = 2(2y + 10)$, which gives $5x - 4y = 55$.
3. From the first equation, $x = 65 - y$.
4. Substitute: $5(65 - y) - 4y = 55$, so $325 - 9y = 55$, $9y = 270$ and $y = 30$.
5. Then $x = 35$.

(b)

1. With $x = 35$: $\angle BCD = 110^\circ$ and $\angle BAD = 140^\circ$.
2. The angles of quadrilateral $ABCD$ add up to $360^\circ$.
3. So $\angle ABC = 360^\circ - 140^\circ - 110^\circ - 40^\circ$
   $= 70^\circ$.

(a)

1. $m = \tan 30^\circ = \frac{1}{\sqrt3}$ and $n = \tan 45^\circ = 1$.
2. So $\frac{m - n}{mn} = \frac{\frac{1}{\sqrt3} - 1}{\frac{1}{\sqrt3}}$.
3. Multiply the top and bottom by $\sqrt3$: $\frac{1 - \sqrt3}{1} = 1 - \sqrt3$.
4. This is in the form $p + \sqrt q$ with $p = 1$ and the surd term $-\sqrt3$.

(b)

1. Let $x$ women wear both.
2. Every woman wears at least one (glasses or watch), so $15 + 10 - x = 20$ and $x = 5$.
3. Probability $= \frac{5}{20} = \frac14$.

(a)

1. On the $x$-axis, 2 cm represents 2 units.
2. On the $y$-axis, 2 cm represents 10 units.

(b)

1. The curve crosses the $x$-axis at $x = -2$ and $x = 4$, so $(x + 2)$ and $(x - 4)$ are factors.
2. It has a highest point, so the $x^2$ term is negative: $y = -(x + 2)(x - 4) = -x^2 + 2x + 8$.
3. So $m = -1$, $n = 2$ and $r = 8$.
4. (Check: the curve crosses the $y$-axis at $r = 8$ ✓.)

(c)

1. From the graph, $P(-5, -27)$ and $Q(3, 5)$.
2. The rise is $5 - (-27) = 32$ and the run is $3 - (-5) = 8$, so the gradient is $\frac{32}{8} = 4$.

(d)

1. $y > 0$ where the curve is above the $x$-axis, between the roots: $-2 < x < 4$.

(a)

1. Cost $= 180 \times 250 = ₦45,000$.
2. A discount of 5 kobo in the naira is $5\%$, so the rest sold at $0.95 \times 250 = ₦237.50$ each.
3. Sales $= 300y + 237.5(180 - y) = 62.5y + 42\,750$.
4. Profit: $62.5y + 42\,750 - 45\,000 = 7125$, so $62.5y = 9375$ and $y = 150$.

(b)(i)

1. Profit $= s - c$
   $= 33x - \frac{x^2}{20} - (24x + 103)$
   $= 9x - \frac{x^2}{20} - 103$.

(ii)

1. At $x = 20$: cost $= 24(20) + 103 = 583$, sales $= 660 - 20 = 640$, profit $= 57$.
2. Percentage profit $= \frac{57}{583} \times 100\%$
   $\approx 9.78\%$.

(a)

1. The percentages must add up to 100.
2. The others total $35 + 7.5 + 10 + 15 + 17.5 = 85$, so savings are $100 - 85 = 15\%$.

(b)

1. $100\%$ is the whole circle, $360^\circ$, so $1\% = 3.6^\circ$.
2. Food and drinks $35 \times 3.6^\circ = 126^\circ$, fuel $27^\circ$, rent $36^\circ$, building project $54^\circ$, education $63^\circ$, savings $54^\circ$.
3. Check: they add up to $360^\circ$.
4. Draw the sectors with a protractor and label each with the item and its angle.

(c)(i)

1. Annual tax $= 12\%$ of $28,800 $= 0.12 \times 28\,800$
   $= \text{\textdollar}3,456$.
2. Monthly tax $= \frac{3456}{12}$
   $= \text{\textdollar}288.00$.

(ii)

1. After tax he has $28\,800 - 3456 = \text{\textdollar}25,344$ a year, which is $\frac{25\,344}{12} = \text{\textdollar}2,112$ a month.
2. He saves $15\%$ of it: $0.15 \times 2112 = \text{\textdollar}316.80$ a month.

(a)

1. In degree mode, to 1 decimal place.
2. For example, $x = 20^\circ$: $3(0.342) + 7(0.940) = 1.03 + 6.58$
   $\approx 7.6$.
3. $x = 160^\circ$: $3(0.342) + 7(-0.940) \approx -5.6$.
4. The full row is $7.0, 7.6, 7.3, 6.1, 4.2, 1.7, -0.9, -3.4, -5.6, -7.0$.

(b)

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

(c)(i)

1. Read up from $x = 150^\circ$ to the curve and across: $y \approx -4.6$.

(ii)

1. $y > 0$ where the curve is above the $x$-axis.
2. It crosses the axis at about $113^\circ$, so $y > 0$ for $0^\circ \le x < 113^\circ$.

(a)

1. $\sum f = 2 + 6 + 5 + x + 6 + 9 + 8 + 5$
   $= 41 + x$ and $\sum fx = 6 + 24 + 25 + 6x + 42 + 72 + 72 + 50$
   $= 291 + 6x$.
2. The mean is 7: $\frac{291 + 6x}{41 + x} = 7$.
3. So $291 + 6x = 287 + 7x$, which gives $x = 4$.
4. Check: $\sum f = 45$, $\sum fx = 315$, and $\frac{315}{45} = 7$.

(b)

1. With the mean 7:

2. Age $x$	3	4	5	6	7	8	9	10	Total
   $f$	2	6	5	4	6	9	8	5	45
   $(x - 7)^2$	16	9	4	1	0	1	4	9
   $f(x - 7)^2$	32	54	20	4	0	9	32	45	196

3. Standard deviation $= \sqrt{\frac{196}{45}}$
   $= \sqrt{4.3556}$
   $\approx 2.087$ years.

(a)

1. The exterior angles of any polygon add up to $360^\circ$: $42 + 38 + 57 + x + (x + y) + (2x - 15) + (3x - y) = 360$.
2. The $y$ terms cancel: $122 + 7x = 360$, so $7x = 238$ and $x = 34$.
3. $x$ is $7^\circ$ less than $y$, so $y = 34 + 7 = 41$.

(b)

1. $\angle XZY$ stands on the arc $XY$, and $\angle XOY = 146^\circ$ is the angle at the centre on the same arc: $\angle XZY = \frac12 \times 146^\circ$
   $= 73^\circ$.
2. $OX = OZ$ (radii), so triangle $OXZ$ is isosceles and $\angle OZX = \angle OXZ = 34^\circ$.
3. So $\angle OZY = \angle XZY - \angle OZX$
   $= 73^\circ - 34^\circ$
   $= 39^\circ$.
4. $OY = OZ$ (radii), so triangle $OYZ$ is isosceles and $\angle OYZ = \angle OZY = 39^\circ$.

(a)

1. Read "will not win any of the races" as: the probability of losing each race is $\frac14$, so the probability of winning each race is $\frac34$, and the races are independent.

(i)

1. Only the second race: lose, win, lose: $\frac14 \times \frac34 \times \frac14 = \frac{3}{64}$.

(ii)

1. All three: $\frac34 \times \frac34 \times \frac34 = \frac{27}{64}$.

(iii)

1. Only two: win-win-lose, win-lose-win or lose-win-win.
2. Each has probability $\frac34 \times \frac34 \times \frac14 = \frac{9}{64}$, so together $3 \times \frac{9}{64} = \frac{27}{64}$.

(b)

1. $V = \frac13\pi r^2 h$.
2. With the same radius, the volume is proportional to the height: $V = 1200 \times \frac{84}{24}$
   $= 1200 \times 3.5$
   $= 4200\text{ cm}^3$.

(a)

1. A closed cylinder has two circular ends and a curved side: total surface area $= 2\pi r^2 + 2\pi rh$.
2. With $r = 3.5$: $2 \times \frac{22}{7} \times 3.5^2 + 2 \times \frac{22}{7} \times 3.5 \times h = 209$.
3. So $77 + 22h = 209$, $22h = 132$ and $h = 6$ cm.

(b)(i)

1. Draw the tree upright with $X$ and $Y$ on the same side of it, $Y$ 19 m further away.
2. The angle of elevation is $43^\circ$ from $X$ (nearer) and $38^\circ$ from $Y$ (farther).

(ii)

1. Let $X$ be $y$ m from the foot.
2. Then $h = y\tan 43^\circ$ and $h = (y + 19)\tan 38^\circ$.
3. Set them equal: $y(\tan 43^\circ - \tan 38^\circ) = 19\tan 38^\circ$, so $y = \frac{19 \times 0.7813}{0.9325 - 0.7813}$
   $\approx 98.2$ m.
4. So $h = 98.2 \times 0.9325 \approx 91.5$ m.
5. (Rounding the tangents early can move the last digit.)