WAEC · 2023 · Private · General Maths · Paper 2

(a)

1. Work year by year: add $10\%$ interest (× 1.1), then the chief's ₦100,000.
2. End of year 1: $400\,000 \times 1.1 + 100\,000 = 540\,000$.
3. Year 2: $540\,000 \times 1.1 + 100\,000 = 694\,000$.
4. Year 3: $694\,000 \times 1.1 + 100\,000 = 863\,400$.
5. Year 4: $863\,400 \times 1.1 + 100\,000 = 1\,049\,740$.
6. The estimate of ₦1,049,740 is reached after 4 years.

(a)

1. If 3 and $-\frac27$ are the roots, the factors are $(x - 3)$ and $\left(x + \frac27\right)$, or $(7x + 2)$ to avoid fractions.
2. $(x - 3)(7x + 2) = 7x^2 + 2x - 21x - 6$
   $= 7x^2 - 19x - 6$.
3. So $7x^2 - 19x - 6 = 0$: $p = 7$, $q = -19$, $r = -6$.

(b)

1. Let the longer parallel side be $x$ cm.
2. The shorter is $\frac{x}{3}$.
3. Area: $\frac12\left(x + \frac{x}{3}\right) \times 5 = 120$, so $\frac{4x}{3} \times \frac52 = 120$, $\frac{10x}{3} = 120$ and $x = 36$.
4. The parallel sides are 36 cm and 12 cm.

(a)

1. Triangle $PQR$ is isosceles with $|PQ| = |QR|$, so the line from $Q$ to the midpoint $M$ of $PR$ is perpendicular to $PR$.
2. $|PM| = \frac12 \times 16 = 8$ cm.
3. In the right-angled triangle $QMP$: $|QM| = \sqrt{17^2 - 8^2}$
   $= \sqrt{289 - 64}$
   $= \sqrt{225}$
   $= 15$ cm.

(b)

1. The centre $O$ of the circle is on the perpendicular bisector of the chord $PR$, which is the line $QM$.
2. Let the radius be $r$: $|OQ| = |OP| = r$ and $|OM| = 15 - r$.
3. In the right-angled triangle $OMP$: $r^2 = (15 - r)^2 + 8^2$
   $= 225 - 30r + r^2 + 64$.
4. The $r^2$ terms cancel: $30r = 289$, so $r = \frac{289}{30} \approx 9.63$.
5. Correct to the nearest whole number, the radius is 10 cm.

(a)

1. Gradient $= \frac{5 - n}{n - (-2)}$
   $= \frac{5 - n}{n + 2}$
   $= \frac43$.
2. Cross-multiply: $3(5 - n) = 4(n + 2)$.
3. So $15 - 3n = 4n + 8$, $7n = 7$ and $n = 1$.

(b)

1. $\tan x = \frac{15}{8}$: the hypotenuse is $\sqrt{15^2 + 8^2} = 17$, so $\sin x = \frac{15}{17}$ and $\cos x = \frac{8}{17}$.
2. Then $5\sin x - 6\cos x = \frac{75}{17} - \frac{48}{17}$
   $= \frac{27}{17}$.

1. Put the heights in order: $110, 120, 120, 125, 128, 130, 135, 140, 142, 145$.

(a)

1. Range $= 145 - 110 = 35$ cm.

(b)

1. $Q_1$ is at position $\frac{10}{4} = 2.5$: halfway between the 2nd and 3rd heights, $\frac{120 + 120}{2} = 120$.
2. $Q_3$ is at position $\frac{3 \times 10}{4} = 7.5$: halfway between the 7th and 8th heights, $\frac{135 + 140}{2} = 137.5$.
3. Interquartile range $= 137.5 - 120 = 17.5$ cm.

(c)

1. The heights add up to 1295, so the mean is $\frac{1295}{10} = 129.5$ cm.

1. List the heights from shortest to tallest.
2. They form an A.P. with first term $a$ (the shortest, unknown), $d = 0.05$, last term $l = 1.90$ and $n$ terms (the number of students).
3. From the last term: $a + 0.05(n - 1) = 1.90$, so $a = 1.95 - 0.05n$.
4. From the sum, $S_n = \frac{n}{2}(a + l)$: $\frac{n}{2}(a + 1.90) = 23.25$.
5. Put in $a$: $n(1.95 - 0.05n + 1.90) = 46.5$, that is $3.85n - 0.05n^2 = 46.5$.

(a)

1. Multiply by 20 and rearrange: $n^2 - 77n + 930 = 0$.
2. Factorise: $(n - 15)(n - 62) = 0$, so $n = 15$ or $n = 62$.
3. $n = 62$ gives $a = 1.95 - 3.10 = -1.15$ m, a negative height, so reject it.
4. There are 15 students.

(b)

1. $a = 1.95 - 0.05 \times 15$
   $= 1.95 - 0.75$
   $= 1.20$.
2. The shortest student is $1.20$ m tall.
3. (Check: $\frac{15}{2}(1.20 + 1.90) = 23.25$ ✓.)

(a)

1. Rent and food take $\frac13 + \frac16 = \frac12$ of his income, leaving $\frac12$.
2. He saves $\frac34$ of that: $\frac34 \times \frac12 = \frac38$.
3. What's left is $\frac12 - \frac38 = \frac18$ of his income.

(i)

1. $\frac18$ of his income is ₦125,000, so his income is $8 \times 125\,000 = ₦1,000,000$.

(ii)

1. Saved: $\frac38 \times 1\,000\,000 = ₦375,000$.

(b)

1. Multiply every term by 15 (the LCM of 5 and 3): $3(2 + 3x) - 5(1 - 7x) < 15(2x + 9)$, so $6 + 9x - 5 + 35x < 30x + 135$.
2. Then $44x + 1 < 30x + 135$, so $14x < 134$ and $x < \frac{67}{7}$ (that is, $x < 9\frac47$).

(a)(i)

1. Draw the building $B$, the tree $T$ 8 km due south of it, and Musa $M$ 8 km due west of the tree, so the angle at $T$ is a right angle.

(ii)

1. By Pythagoras, $|BM| = \sqrt{8^2 + 8^2}$
   $= \sqrt{128}$
   $\approx 11.31\text{ km}$.

(iii)

1. Triangle $BTM$ is right-angled and isosceles, so $\angle TBM = 45^\circ$.
2. From $B$, Musa is $45^\circ$ west of due south.
3. Bearings are measured clockwise from north: $180^\circ + 45^\circ = 225^\circ$.

(b)

1. Let the distance be $D$ km.
2. At 5 km/h the ride takes $\frac D5$ hours.
3. At 6 km/h it takes $\frac D6$ hours.
4. The slower ride takes 45 minutes $= \frac34$ hour longer: $\frac D5 - \frac D6 = \frac34$.
5. Multiply by 60: $12D - 10D = 45$.
6. So $2D = 45$ and $D = 22.5$.
7. The distance is 22.5 km.

(a)

1. The tankers take $3 \times 6500 = 19\,500$ litres.
2. $1\text{ m}^3 = 1000$ litres, so that is $19.5\text{ m}^3$.
3. The tank's radius is 1.75 m, so its base is $\frac{22}{7} \times 1.75^2 = 9.625\text{ m}^2$.
4. The water removed is a cylinder of that base: $9.625 \times h = 19.5$, so $h \approx 2.03$ m.

(b)

1. Gradient $= \frac{-6 - (-5)}{-4 - 2}$
   $= \frac{-1}{-6}$
   $= \frac16$.
2. Through $(2, -5)$: $y + 5 = \frac16(x - 2)$.
3. Multiply by 6: $6y + 30 = x - 2$, so $6y - x + 32 = 0$.

(a)

1. Draw north at Uju $U$ and put Adamu $A$ 15 m away on $305^\circ$.
2. Draw north at $A$ and put Bola $B$ 25 m from $A$ on $065^\circ$.
3. Join $U$ and $B$.

(b)

1. At $A$, the direction back to $U$ is $305^\circ - 180^\circ = 125^\circ$ and the direction to $B$ is $065^\circ$, so $\angle UAB = 125^\circ - 65^\circ$
   $= 60^\circ$.
2. Cosine rule: $|UB|^2 = 15^2 + 25^2 - 2(15)(25)\cos 60^\circ$
   $= 225 + 625 - 375$
   $= 475$.
3. So $|UB| \approx 21.8$ m.

(c)

1. Cosine rule for the angle at $U$: $\cos\angle AUB = \frac{15^2 + 21.79^2 - 25^2}{2 \times 15 \times 21.79}$
   $\approx \frac{74.8}{653.7}$
   $\approx 0.1144$, so $\angle AUB \approx 83.4^\circ$.
2. At $U$, Adamu is on $305^\circ$ and Bola is $83.4^\circ$ further round clockwise: $305^\circ + 83.4^\circ = 388.4^\circ$, which is $028^\circ$ after taking away $360^\circ$.

(a)

1. The class marks are $3, 8, 13, 18, 23, 28$.

2. Ages	1–5	6–10	11–15	16–20	21–25	26–30	Total
   $f$	7	12	4	6	11	10	50
   $x$	3	8	13	18	23	28
   $fx$	21	96	52	108	253	280	810

3. Mean $= \frac{810}{50} = 16.20$ years.

(b)

1. The distances of the class marks from 16.2 are $|x - 16.2| = 13.2, 8.2, 3.2, 1.8, 6.8, 11.8$.
2. Multiply by the frequencies: $92.4, 98.4, 12.8, 10.8, 74.8, 118$, which add up to 407.2.
3. Mean deviation $= \frac{\sum f|x - \bar x|}{\sum f}$
   $= \frac{407.2}{50}$
   $= 8.144$
   $\approx 8.14$ years.

(a)(i)

1. $PQ \parallel MN$, and $QM$ crosses both, so the alternate angles are equal: $\angle PQR = \angle QMN = 53^\circ$.
2. $|PR| = |QR|$, so triangle $PQR$ is isosceles and $\angle QPR = \angle PQR = 53^\circ$.
3. Then $\angle PRQ = 180^\circ - 53^\circ - 53^\circ$
   $= 74^\circ$.

(ii)

1. $PN$ also crosses the parallel lines, so $\angle MNP = \angle QPN = 53^\circ$ (alternate angles).
2. So $\frac32 y - 13 = 53$, $\frac32 y = 66$ and $y = 44$.

(b)(i)

1. Yellow balls $= \frac25 \times 15 = 6$.

(ii)

1. Green balls $= 15 - 6 = 9$.

(a)(i)

1. Triangle $ABE$ is right-angled at $E$: $\tan\angle BAE = \frac{18}{7}$
   $\approx 2.571$, so $\angle BAE \approx 68.75^\circ$.

(ii)

1. In right-angled triangle $BED$: $|ED| = \frac{18}{\tan 55^\circ}$
   $\approx 12.604$ cm, so $|AD| = 7 + 12.604 = 19.604$ cm.
2. The parallel sides are $BC = 9$ and $AD = 19.604$, and the height is $BE = 18$: area $= \frac12(9 + 19.604) \times 18$
   $\approx 257.44\text{ cm}^2$.

(b)

1. A $25\%$ profit means the selling price is $125\%$ of the cost: $y = 1.25x = \frac54x$.
2. So $4y = 5x$, or $x = \frac45y$.