WAEC · 2024 · May/June · General Maths · Paper 2

(a)

1. $t$ varies directly as $V$ (on top) and inversely as $P$ (underneath): $t = \dfrac{kV}{P}$.
2. With 5 pumps, 20 vehicles took 10 minutes: $10 = \dfrac{20k}{5} = 4k$, so $k = \frac52$.
3. The relationship is $t = \dfrac{5V}{2P}$.

(b)

1. $V = 50$, $P = 2$: $t = \dfrac{5 \times 50}{2 \times 2}$
   $= \dfrac{250}{4}$
   $= 62.5$ minutes.

(c)

1. $t = 20$, $V = 40$: $20 = \dfrac{5 \times 40}{2P}$
   $= \dfrac{100}{P}$, so $P = \dfrac{100}{20} = 5$ pumps.

1. Time $= \frac{\text{distance}}{\text{speed}}$ for each part, and 90 minutes $= 1.5$ hours: $\frac{2x + 13}{67.5} + \frac{5x - 20}{72} = 1.5$.
2. Multiply every term by 1080 (which both 67.5 and 72 divide into: $1080 \div 67.5 = 16$ and $1080 \div 72 = 15$): $16(2x + 13) + 15(5x - 20) = 1620$.
3. Expand: $32x + 208 + 75x - 300 = 1620$.
4. So $107x - 92 = 1620$, $107x = 1712$ and $x = 16$.
5. Check: the distances are 45 km and 60 km, taking $\frac{45}{67.5} = \frac23$ h and $\frac{60}{72} = \frac56$ h, a total of $1\frac12$ h ✓.

(a)

1. The perimeter of the floor is its circumference: $2 \times \frac{22}{7} \times r = 66$, so $r = 10.5$ m.
2. Area of the floor $= \frac{22}{7} \times 10.5^2$
   $= 346.5\text{ m}^2$.
3. Each tile is $40\text{ cm} = 0.4$ m square, so its area is $0.4 \times 0.4 = 0.16\text{ m}^2$.
4. Number of tiles $= \frac{346.5}{0.16} = 2165.6$, which is 2166 tiles to the nearest whole number.

(a)

1. $OBCD$ is a rhombus, so $OB = BC = CD = DO$.
2. Also $OB = OC = OD$ (radii).
3. So triangles $OBC$ and $OCD$ are equilateral, each with $60^\circ$ angles.
4. So $\angle BOD = 60^\circ + 60^\circ$
   $= 120^\circ$.
5. The angle at the circumference is half the angle at the centre: $t = \angle BAD$
   $= \frac12 \times 120^\circ$
   $= 60^\circ$.

(b)

1. $OA = OB$ (radii), so $\angle OAB = \angle OBA = y$.
2. Likewise $OA = OD$ gives $\angle OAD = \angle ODA = y$.
3. So $t = \angle OAB + \angle OAD = 2y$, and $2y = 60^\circ$ gives $y = 30^\circ$.

(c)

1. $\angle ADC = \angle ADO + \angle ODC$
   $= 30^\circ + 60^\circ$
   $= 90^\circ$.
2. (This fits: $A$, $O$ and $C$ are on one line, so $AC$ is a diameter and $\angle ADC$ is an angle in a semicircle.)

1. Let there be $s$ silver marbles, so $s + 7$ marbles in all.
2. Without replacement, the first is silver with probability $\frac{s}{s + 7}$, and the second with probability $\frac{s - 1}{s + 6}$ (one silver marble and one marble in total have gone).
3. So $\frac{s}{s + 7} \times \frac{s - 1}{s + 6} = \frac{1}{15}$.
4. Cross-multiply: $15s(s - 1) = (s + 7)(s + 6)$, so $15s^2 - 15s = s^2 + 13s + 42$.
5. Collect terms: $14s^2 - 28s - 42 = 0$.
6. Divide by 14: $s^2 - 2s - 3 = 0$, which factorises as $(s - 3)(s + 1) = 0$.
7. So $s = 3$ or $s = -1$.
8. A number of marbles can't be negative, so there are 3 silver marbles.
9. Check: $\frac{3}{10} \times \frac{2}{9} = \frac{6}{90}$
   $= \frac{1}{15}$.

(a)

1. For consecutive terms of a G.P., the ratios are equal: $\frac{4x + 3}{x + 2} = \frac{7x + 24}{4x + 3}$.
2. Cross-multiply: $(4x + 3)^2 = (x + 2)(7x + 24)$.
3. Expand both sides: $16x^2 + 24x + 9 = 7x^2 + 38x + 48$.
4. Collect everything on one side: $9x^2 - 14x - 39 = 0$.
5. Factorise: $(x - 3)(9x + 13) = 0$, so $x = 3$ or $x = -\frac{13}{9}$.

(b)

1. When $x = 3$ the terms are $5, 15, 45$, so the common ratio is $\frac{15}{5} = 3$.
2. When $x = -\frac{13}{9}$: $x + 2 = \frac59$, $4x + 3 = -\frac{52}{9} + \frac{27}{9}$
   $= -\frac{25}{9}$ and $7x + 24 = -\frac{91}{9} + \frac{216}{9}$
   $= \frac{125}{9}$.
3. The common ratio is $-\frac{25}{9} \div \frac59 = -5$.

(a)

1. There are 20 children: $2 + (2x - 1) + (y + 2) + 4 + 2 + (y - 1) = 20$, so $8 + 2x + 2y = 20$ and $x + y = 6$.
2. $x : y = 1 : 2$ means $y = 2x$.
3. Then $x + 2x = 6$, so $x = 2$ and $y = 4$.

(b)

1. The frequencies are $2, 3, 6, 4, 2, 3$ (total 20).
2. $\sum fx = 5(2) + 6(3) + 7(6) + 8(4) + 9(2) + 10(3)$
   $= 10 + 18 + 42 + 32 + 18 + 30$
   $= 150$.
3. Mean age $= \frac{150}{20} = 7.5$ years.

(a)

1. Draw the pole $FB$ upright with the bird at the top $B$.
2. Badu $D$ is nearer the pole (the larger angle, $48^\circ$) and Abu $A$ is 46 m further back on the same line (the smaller angle, $40^\circ$).

(b)

1. Let Badu be $x$ m from the foot.
2. From each observer: $h = x\tan 48^\circ$ and $h = (x + 46)\tan 40^\circ$.
3. Set them equal: $x(\tan 48^\circ - \tan 40^\circ) = 46\tan 40^\circ$.
4. So $x = \frac{46 \times 0.8391}{1.1106 - 0.8391}$
   $= \frac{38.60}{0.2715}$
   $\approx 142.2$ m.
5. Then $h = 142.2 \times 1.1106 \approx 157.9$ m.

1. The radius is $\frac{26}{2} = 13$ cm.
2. Let $M$ be the midpoint of $PQ$.
3. $OM \perp PQ$ and $OM = 5$ cm.

(a)

1. In the right-angled triangle $OMP$: $\cos\angle MOP = \frac{5}{13}$, so $\angle MOP \approx 67.38^\circ$ and $\angle POQ = 2 \times 67.38^\circ$
   $\approx 134.76^\circ$, which is $135^\circ$ to the nearest degree.
2. Also $PM = \sqrt{13^2 - 5^2} = 12$ cm, so $PQ = 24$ cm.

(b)

1. Sector $= \frac{134.76}{360} \times \frac{22}{7} \times 13^2$
   $\approx 198.8\text{ cm}^2$.
2. Triangle $OPQ = \frac12 \times 24 \times 5$
   $= 60\text{ cm}^2$.
3. Segment $\approx 198.8 - 60 = 138.8$, which is $139\text{ cm}^2$ to the nearest whole number.

(a)

1. Let the total be $T$ acres.
2. The wife gets $\frac25T$, leaving $\frac35T$.
3. The family gets $\frac13$ of that, $\frac15T$.
4. The children get the rest: $T - \frac25T - \frac15T = \frac25T$.
5. The children share in the ratio $3 : 5 : 2$ (10 parts).
6. The smallest share (2 parts) is 8 acres, so one part is 4 acres and the children have $10 \times 4 = 40$ acres.

(i)

1. $\frac25T = 40$, so $T = 100$ acres.

(ii)

1. The wife received $\frac25 \times 100 = 40$ acres.

(b)

1. Balance after the deposit: $1600 - 400 = 1200$ dollars.
2. Interest for 12 months (1 year) at $25\%$: $0.25 \times 1200 = 300$ dollars.
3. Total cost $= 400 + 1200 + 300 = 1900$, that is $1,900.

(a)

1. Find where the lines cross.
2. Take $x - y = 4$ from $x + 2y = 7$: $3y = 3$, so $y = 1$ and $x = 4 + 1 = 5$.
3. The point is $(5, 1)$.
4. The line passes through $(0, 0)$ and $(5, 1)$: gradient $= \frac{1 - 0}{5 - 0} = \frac15$.
5. Through the origin, $y = \frac15 x$, that is $x - 5y = 0$.

(b)(i)

1. Interior + exterior $= 180^\circ$, in the ratio $4 : 1$ (5 parts).
2. The exterior angle is one part: $\frac{180^\circ}{5} = 36^\circ$.

(ii)

1. Number of sides $= \frac{360^\circ}{36^\circ} = 10$.

(iii)

1. Sum of the interior angles $= (10 - 2) \times 180^\circ = 1440^\circ$.

(a)(i)

1. $T$ lies on $PO$, so $\angle TOQ = 56^\circ$.
2. $OT = OQ$ (radii), so $\angle OQT = \frac{180^\circ - 56^\circ}{2}$
   $= 62^\circ$.
3. A tangent is perpendicular to the radius at the point of contact, so $\angle OQP = 90^\circ$.
4. Then $\angle TQP = 90^\circ - 62^\circ$
   $= 28^\circ$.

(ii)

1. The angle at the circumference is half the angle at the centre on the same arc $TQ$: $\angle TSQ = \frac12 \times 56^\circ$
   $= 28^\circ$.
2. In triangle $STV$: $\angle SVT = 109^\circ$ and $\angle TSV = 28^\circ$, so $\angle STV = 180^\circ - 109^\circ - 28^\circ$
   $= 43^\circ$.
3. $V$ lies on $TO$, so $\angle VTQ = \angle OTQ$, and $\angle OTQ = \angle OQT = 62^\circ$ (isosceles triangle $OTQ$).
4. So $\angle QTS = \angle STV + \angle VTQ$
   $= 43^\circ + 62^\circ$
   $= 105^\circ$.

(b)

1. Factorise each part: $2n^2 - 3n - 2 = (2n + 1)(n - 2)$, $2n^2 + 3n + 1 = (2n + 1)(n + 1)$, $n^2 - 1 = (n - 1)(n + 1)$ and $n^2 - 4 = (n - 2)(n + 2)$.
2. Cancel the common factors $(2n + 1)$, $(n + 1)$ and $(n - 2)$.
3. What's left is $\dfrac{n - 1}{n + 2}$.

(a)

1. By the alternate segment theorem, the angle between the tangent $TQ$ and the chord $TB$ equals the angle in the alternate segment: $\angle BTQ = \angle BAT = 44^\circ$.
2. $TC$ bisects it, so $\angle BTC = 22^\circ$.
3. Also by the alternate segment theorem, $\angle ABT = \angle PTA = 60^\circ$.
4. $ABC$ is a straight line, so $\angle ABT$ is an exterior angle of triangle $BTC$, and it equals the sum of the two opposite interior angles: $60^\circ = 22^\circ + \angle ACT$.
5. So $\angle ACT = 38^\circ$.

(b)(i)

1. $2 \times \frac{22}{7} \times r = 11$, so $r = 1.75$ m.

(ii)

1. $h = 6 \times 1.75 + 3 = 13.5$ m.

(iii)

1. Volume $= \frac{22}{7} \times 1.75^2 \times 13.5$
   $\approx 129.9\text{ m}^3$.