WAEC · 2019 · Private, 2nd series · General Maths · Paper 2

(a)

1. Multiply and take the remainder on dividing by 8.
2. For example $5 \otimes 7 = 35 = 4 \times 8 + 3$, so the entry is 3.

3. $\otimes$	2	3	5	7
   2	4	6	2	6
   3	6	1	7	5
   5	2	7	1	3
   7	6	5	3	1

(b)(i)

1. In the row of 3, the entry 5 is in the column of 7: the solution set is $\{7\}$.

(ii)

1. On the diagonal, $n \otimes n = 1$ for $n = 3$, 5 and 7: the solution set is $\{3, 5, 7\}$.

(a)

1. Radius $= \frac{24}{2} = 12$ cm.
2. Curved surface $= \pi r l$
   $= \frac{22}{7} \times 12 \times 18.7$
   $\approx 705.26$, which is $705\text{ cm}^2$ to three significant figures.

(b)

1. Write every number as a power of 2: $128^x = 2^{7x}$, $16^{1 - x} = 2^{4 - 4x}$ and $8^{\frac23 x} = 2^{2x}$.
2. The left side is $\frac{2^{7x} \times 2^1}{2^{4 - 4x}} = 2^{7x + 1 - 4 + 4x}$
   $= 2^{11x - 3}$.
3. So $11x - 3 = 2x$, $9x = 3$ and $x = \frac13$.

(a)

1. Write each term as a multiple of $\sqrt6$.
2. $3\sqrt{\frac32} = \frac{3\sqrt3}{\sqrt2}$
   $= \frac{3\sqrt6}{2}$.
3. $4\sqrt{\frac23} = \frac{4\sqrt2}{\sqrt3}$
   $= \frac{4\sqrt6}{3}$.
4. $\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt6$.
5. Over the common denominator 6: $\frac{9\sqrt6}{6} - \frac{8\sqrt6}{6} - \frac{12\sqrt6}{6} = -\frac{11\sqrt6}{6}$.

(b)

1. $P(M \text{ fails}) = \frac13$ and $P(N \text{ fails}) = \frac15$.

(i)

1. Only one passes: $M$ passes and $N$ fails, $\frac23 \times \frac15 = \frac{2}{15}$.
2. Or $M$ fails and $N$ passes, $\frac13 \times \frac45 = \frac{4}{15}$.
3. Together: $\frac{6}{15} = \frac25$.

(ii)

1. At least one passes is everything except "both fail": $1 - \frac13 \times \frac15 = 1 - \frac{1}{15}$
   $= \frac{14}{15}$.

(a)

1. $\cos\theta = \frac{15}{17}$: the opposite side is $\sqrt{17^2 - 15^2} = \sqrt{64} = 8$, so $\tan\theta = \frac{8}{15}$.
2. Then $\frac{\tan\theta}{1 + 2\tan\theta} = \frac{\frac{8}{15}}{\frac{31}{15}}$
   $= \frac{8}{31}$.

(b)

1. Multiply both sides by $\log_{10} 64$: $\log_{10} y = \frac12\log_{10} 64$
   $= \log_{10} 64^{\frac12}$
   $= \log_{10} 8$.
2. So $y = 8$.

(a)

1. The reflex angle at $O$ is $196^\circ$, so the angle $MOP$ on the other side is $360^\circ - 196^\circ = 164^\circ$.
2. The angle at the circumference is half the angle at the centre: $\angle MNP = 82^\circ$.
3. $OM = OP$ (radii), so triangle $OMP$ is isosceles: $\angle OMP = \angle OPM$
   $= \frac{180^\circ - 164^\circ}{2}$
   $= 8^\circ$.
4. In triangle $MNP$: $\angle NMP = 52^\circ + 8^\circ$
   $= 60^\circ$ and $\angle NPM = m + 8^\circ$.
5. So $60^\circ + 82^\circ + m + 8^\circ = 180^\circ$, and $m = 30^\circ$.

(b)

1. Cassava uses $\frac25$, leaving $\frac35$.
2. Plantain uses $\frac13$ of the remainder: $\frac13 \times \frac35 = \frac15$.
3. Yam gets the rest: $1 - \frac25 - \frac15 = \frac25$ of the land.

(a)

1. In base 2: $2^m \times (2^{-3})^n = 2^7$, so $m - 3n = 7$.
2. And $4^m \div 2^{-4n} = 2^{2m} \times 2^{4n}$
   $= 2^{-4}$, so $2m + 4n = -4$, i.e. $m + 2n = -2$.
3. Subtract the second equation from the first: $-5n = 9$, so $n = -\frac95$.
4. Then $m = 7 + 3n = 7 - \frac{27}{5} = \frac85$.
5. $m - n = \frac85 + \frac95$
   $= \frac{17}{5}$
   $= 3\frac25$.

(b)

1. Gradient $= \frac{-\frac23 - \frac12}{1 - (-2)}$
   $= \frac{-\frac76}{3}$
   $= -\frac{7}{18}$.
2. Through $\left(-2, \frac12\right)$: $y - \frac12 = -\frac{7}{18}(x + 2)$.
3. Multiply by 18: $18y - 9 = -7x - 14$, so $18y + 7x + 5 = 0$.

(a)

1. Tally each score into its class:

2. Scores	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	5	12	6	3	4	5

3. The frequencies add up to 48.

(b)

1. Draw the histogram on the class boundaries $-0.5, 9.5, 19.5, \ldots, 99.5$, with touching bars of heights $1, 3, 3, 6, 5, 12, 6, 3, 4, 5$.

(c)(i)

1. The tallest bar is 49.5–59.5.
2. Draw the two crossed lines from its top corners to the neighbouring bars' top corners, and read down: about 55.
3. (By formula: $49.5 + \frac{12 - 5}{(12 - 5) + (12 - 6)} \times 10 \approx 54.9$.)

(ii)

1. Promoted means a score of 30 or more.
2. Below 30 there are $1 + 3 + 3 = 7$ students, so $48 - 7 = 41$ are promoted: $\frac{41}{48} \times 100 \approx 85.4\%$.

(a)

1. $T_5 = a + 4d = 11$ and $T_8 = a + 7d = 20$.
2. Subtract: $3d = 9$, so $d = 3$ and $a = 11 - 12 = -1$.

(i)

1. $T_{12} = a + 11d = -1 + 33 = 32$.

(ii)

1. $S_{12} = \frac{12}{2}(a + T_{12})$
   $= 6(-1 + 32)$
   $= 186$.

(b)

1. After a $15\%$ discount he collected $85\%$ of the marked price: marked price $= \frac{36\,000}{0.85} \approx ₦42,352.94$.
2. The discount was $42\,352.94 - 36\,000 \approx ₦6,352.94$.

(a)(i)

1. Draw three overlapping circles P, B and M in a rectangle of 110 students.
2. Put 5 in the centre.
3. Take it away from each pair to get the "two subjects only" regions: P and M only $10 - 5 = 5$, B and M only $8 - 5 = 3$, P and B only $6 - 5 = 1$.
4. Then the "one subject only" regions: Physics only $= 25 - 5 - 1 - 5 = 14$, Biology only $= 45 - 1 - 3 - 5 = 36$, Mathematics only $= 48 - 5 - 3 - 5 = 35$.

(ii)

1. Biology but neither Physics nor Mathematics is Biology only: 36.

(iii)

1. Inside the circles: $14 + 36 + 35 + 5 + 3 + 1 + 5 = 99$.
2. So $110 - 99 = 11$ were not selected for any subject.

(b)

1. The five numbers add up to $5 \times 15 = 75$: $22 + 18 + (2x + 1) + 10 + 20 = 75$, so $71 + 2x = 75$ and $x = 2$.
2. The numbers are $22, 18, 5, 10, 20$.
3. In order: $5, 10, 18, 20, 22$.
4. The median is 18.

1. He paid $20\%$ himself, $0.2 \times 350\,000 = 70\,000$ dollars, and borrowed the other $280,000.

(a)

1. Interest $= \frac{280\,000 \times 7 \times 8}{100}$
   $= 156\,800$ dollars.
2. Total cost $= 350\,000 + 156\,800 = 506\,800$, that is $506,800.

(b)

1. The increase is the interest: $\frac{156\,800}{350\,000} \times 100\% = 44.8\%$.

(c)

1. Total spent $= 506\,800 + 10\,000 = 516\,800$ dollars.
2. He sold for $460,000, a loss of $516\,800 - 460\,000 = 56\,800$ dollars.
3. Percentage loss $= \frac{56\,800}{516\,800} \times 100\%$
   $\approx 10.99\%$.

(a)

1. Put each $x$ into $y = x^2 + \frac12$.
2. For $x = \pm4$: $16.5$.
3. For $x = \pm3$: $9.5$.
4. For $x = \pm2$: $4.5$.
5. For $x = \pm1$: $1.5$.
6. For $x = 0$: $0.5$.
7. So the row is $16.5, 9.5, 4.5, 1.5, 0.5, 1.5, 4.5, 9.5, 16.5$.

(b)

1. With 2 cm to 1 unit across and 2 cm to 2 units up, plot the nine points and join them with a smooth U-shaped curve, symmetrical about the $y$-axis.

(c)(i)

1. Add $\frac12$ to both sides of $x^2 = 5$: $x^2 + \frac12 = 5.5$, that is $y = 5.5$.
2. Draw the line $y = 5.5$.
3. It meets the curve at $x \approx -2.2$ and $x \approx 2.2$.
4. (Exactly, $\pm\sqrt5 \approx \pm 2.24$.)

(ii)

1. Moving from left to right, the curve goes down until its lowest point at $x = 0$.
2. So $y$ decreases as $x$ increases for $-4 \le x < 0$.

1. Draw north at $K$.
2. $J$ is 20 km from $K$ on $065^\circ$ and $T$ is 8 km from $K$ on $155^\circ$.
3. The angle between them is $\angle JKT = 155^\circ - 65^\circ$
   $= 90^\circ$, so triangle $JKT$ is right-angled at $K$.

(a)(i)

1. $|TJ| = \sqrt{20^2 + 8^2}$
   $= \sqrt{464}$
   $\approx 21.54$ km, which is 22 km to the nearest kilometre.

(ii)

1. At $J$: $\tan\angle KJT = \frac{8}{20} = 0.4$, so $\angle KJT \approx 21.8^\circ$.
2. At $J$, the direction back to $K$ is $065^\circ + 180^\circ = 245^\circ$, and $T$ is $21.8^\circ$ further round anticlockwise.
3. Bearing of $T$ from $J$ $= 245^\circ - 21.8^\circ$
   $= 223.2^\circ$
   $\approx 223^\circ$.

(a)(i)

1. $OA = OD$ (radii), so triangle $OAD$ is isosceles: $\angle OAD = \angle ODA$
   $= \frac{180^\circ - 130^\circ}{2}$
   $= 25^\circ$.
2. The angle at the centre is twice the angle at the circumference on the same arc, so $\angle ABD = \frac12 \times 130^\circ$
   $= 65^\circ$.
3. In triangle $ABD$: $\angle BAD = 26^\circ + 25^\circ$
   $= 51^\circ$, so $\angle ADB = 180^\circ - 65^\circ - 51^\circ$
   $= 64^\circ$.
4. Then $\angle ODB = \angle ADB - \angle ODA$
   $= 64^\circ - 25^\circ$
   $= 39^\circ$.

(ii)

1. $OB = OD$ (radii), so triangle $OBD$ is isosceles with $\angle OBD = \angle ODB = 39^\circ$: $\angle BOD = 180^\circ - 2 \times 39^\circ$
   $= 102^\circ$.
2. (Check: it is twice $\angle BAD = 51^\circ$.)

(b)

1. There are $3 \times 3 = 9$ equally likely pairs.
2. Their sums: with 1: 4, 5, 6.
3. With 2: 5, 6, 7.
4. With 6: 9, 10, 11.
5. Greater than seven: 9, 10 and 11, so $P = \frac39 = \frac13$.