WAEC · 2021 · Private · General Maths · Paper 2

1. Draw three overlapping circles M, P and C.
2. Nobody applied for both Mathematics and Chemistry, so the regions M and C only and all three are empty.
3. The 2 who applied for Mathematics and Physics are therefore M and P only.
4. Let $x$ be Physics and Chemistry only, $y$ Physics only and $a$ Chemistry only.
5. Physics: $2 + x + y = 5$, so $x + y = 3$.
6. Chemistry: $x + a = 3$.
7. All 10 applicants are in the diagram: $3 + 2 + x + y + a = 10$, so $x + y + a = 5$.
8. Since $x + y = 3$, $a = 2$.

(a)

1. From $x + a = 3$: $x = 1$.
2. One person applied for Physics and Chemistry only.

(b)

1. Chemistry only: $a = 2$.

(a)

1. $L$ is inside a square root, so square both sides: $S^2 = \frac{5d(L - d)}{7}$.
2. Multiply both sides by 7: $7S^2 = 5d(L - d) = 5dL - 5d^2$.
3. Add $5d^2$ to both sides: $7S^2 + 5d^2 = 5dL$.
4. Divide by $5d$: $L = \frac{7S^2 + 5d^2}{5d}$.

(b)

1. Substitute $d = 27$ and $S = 0.7$: $L = \frac{7 \times 0.49 + 5 \times 729}{5 \times 27}$
   $= \frac{3.43 + 3645}{135}$
   $= \frac{3648.43}{135}$
   $\approx 27.025$.
2. Correct to two decimal places: $L = 27.03$.

(a)

1. At the same time of day, heights and shadows are in the same ratio (similar triangles): $\frac{h}{30} = \frac{6}{9}$, so $h = 30 \times \frac23 = 20$ m.

(b)

1. Combine the right side: $\log(y + 1) + \log 4 = \log 4(y + 1)$.
2. So $\log(5y - 4) = \log(4y + 4)$, which gives $5y - 4 = 4y + 4$ and $y = 8$.
3. Check: $5y - 4 = 36$ and $y + 1 = 9$ are positive.

(a)

1. Area $= (x + 3)(x - 2) = 36$.
2. Expand: $x^2 + x - 6 = 36$, so $x^2 + x - 42 = 0$ and $(x + 7)(x - 6) = 0$.
3. $x = -7$ would make the sides $-4$ and $-9$, which is impossible, so $x = 6$.
4. The rectangle is 9 cm by 4 cm.

(b)

1. Perimeter $= 2(9 + 4) = 26$ cm.

1. Put the ages in order (the table in the question isn't) and find the mean first:

2. Age $x$	10	12	13	14	15	16	Total
   $f$	9	2	3	6	3	2	25
   $fx$	90	24	39	84	45	32	314

3. Mean $= \frac{314}{25} = 12.56$.

4. Age $x$	10	12	13	14	15	16	Total
   $\lvert x - 12.56\rvert$	2.56	0.56	0.44	1.44	2.44	3.44
   $f\lvert x - 12.56\rvert$	23.04	1.12	1.32	8.64	7.32	6.88	48.32

(a)

1. Mean deviation $= \frac{\sum f|x - \bar x|}{\sum f}$
   $= \frac{48.32}{25}$
   $= 1.9328$
   $\approx 1.9$.

(a)

1. A triangle on a 4 cm side has height 6 cm, so its sloping edge runs from a corner of the base to the apex: half the base (2 cm) and the height (6 cm) make a right-angled triangle.
2. Slant edge $= \sqrt{6^2 + 2^2}$
   $= \sqrt{40}$
   $\approx 6.32$ cm.

(b)

1. When the pyramid is folded up, the 6 cm height of that face runs from the middle of a short side to the apex.
2. The middle of a short side is 5 cm (half of 10) from the centre of the base.
3. Perpendicular height $= \sqrt{6^2 - 5^2}$
   $= \sqrt{11}$
   $\approx 3.32$ cm.

(c)

1. The height of the triangles on the 10 cm sides: from the middle of a long side (2 cm from the centre), $\sqrt{11 + 2^2} = \sqrt{15} \approx 3.873$ cm.
2. Total surface area $= \underbrace{10 \times 4}_{\text{base}} + 2 \times \frac12 \times 4 \times 6 + 2 \times \frac12 \times 10 \times \sqrt{15}$
   $= 40 + 24 + 38.73$
   $= 102.73\text{ cm}^2$.

(a)

1. Let the digits be $x$ (first), $y$ (middle) and $z$ (last).
2. The last digit is three times the first: $z = 3x$.
3. The middle digit is half the sum of the other two: $y = \frac{x + 3x}{2} = 2x$.
4. The digits add up to 12: $x + 2x + 3x = 12$, so $6x = 12$ and $x = 2$.
5. Then $y = 4$ and $z = 6$.
6. The number is 246.

(b)(i)

1. The $n$th term is $U_n = a + (n - 1)d$.
2. So $a + 6d = 15\frac14 = \frac{61}{4}$ and $a + 17d = 42\frac34 = \frac{171}{4}$.
3. Take the first from the second: $11d = \frac{110}{4}$, so $d = \frac{10}{4} = \frac52$.
4. Then $a = \frac{61}{4} - 6 \times \frac52$
   $= \frac{61}{4} - 15$
   $= \frac14$.
5. So $U_n = \frac14 + \frac52(n - 1)$
   $= \frac{1 + 10n - 10}{4}$
   $= \frac{10n - 9}{4}$.

(ii)

1. $U_6 = \frac{60 - 9}{4}$
   $= \frac{51}{4}$
   $= 12\frac34$.

(a)

1. "Partly constant and partly varies as three times the base area": $T = a + b(3A) = a + 3bA$, with $T$ in hours and $A$ in m².
2. $A = 24$, $T = 27$: $a + 72b = 27$.
3. $A = 48$, $T = 52$: $a + 144b = 52$.
4. Subtract: $72b = 25$, so $b = \frac{25}{72}$.
5. Then $a = 27 - 72 \times \frac{25}{72}$
   $= 27 - 25$
   $= 2$.
6. So $T = 2 + 3 \times \frac{25}{72}A$
   $= 2 + \frac{25}{24}A$.
7. For $A = 60$: $T = 2 + \frac{25}{24} \times 60$
   $= 2 + 62\frac12$
   $= 64\frac12$ hours.

(a)(i)

1. From 09:55 to 14:15 is 4 hours 20 minutes, which is $4\frac13$ hours.

(ii)

1. Speed $= \frac{2210}{4\frac13}$
   $= 2210 \times \frac{3}{13}$
   $= 510$ km/h.

(b)

1. Each year the value is multiplied by $1 - 0.085 = 0.915$.

(i)

1. After the third year: one more year from now, $102\,500 \times 0.915 = 93\,787.5$, which is ₦93,788 to the nearest naira.

(ii)

1. Two years ago it was worth $\frac{102\,500}{0.915^2} = \frac{102\,500}{0.837225}$
   $\approx ₦122,428$.

(a)

1. Running totals against the upper class boundaries:

2. Marks	1–10	11–20	21–30	31–40	41–50	51–60	61–70	71–80	81–90	91–100
   Upper boundary	10.5	20.5	30.5	40.5	50.5	60.5	70.5	80.5	90.5	100.5
   Cumulative frequency	2	8	25	55	88	139	168	177	180	180

(b)

1. Plot the cumulative frequencies at the upper boundaries, starting from $(0.5, 0)$, and draw a smooth S-shaped curve.

(c)(i)

1. There are 180 candidates, so the median is at 90.
2. Go across to the curve and down: about 51.
3. (Check: 90 is just above 88 at 50.5, and $50.5 + \frac{90 - 88}{51} \times 10 \approx 50.9$.)

(ii)

1. If $60\%$ passed, the lowest $40\%$ failed: $0.4 \times 180 = 72$ candidates are below the pass mark.
2. Go across from 72: about 45.5.
3. (Check: $40.5 + \frac{72 - 55}{33} \times 10 \approx 45.7$.)

(a)(i)

1. The angle at the centre is twice the angle at the circumference on the same arc: $\angle POS = 2 \times \angle PTS$
   $= 116^\circ$.

(ii)

1. $PRST$ is a cyclic quadrilateral, so opposite angles add up to $180^\circ$: $\angle PRS = 180^\circ - 58^\circ$
   $= 122^\circ$.

(b)

1. Total contributions: Ato $2 \times 2000 = 4000$, Bonsu 3000, Musah $3 \times 2500 = 7500$.
2. Ratio $4000 : 3000 : 7500 = 8 : 6 : 15$ (29 parts).
3. Ato's profit $= \frac{8}{29} \times 5220 =$ GH¢ 1,440.
4. As a percentage of his contribution: $\frac{1440}{4000} \times 100\% = 36\%$.

(a)

1. Draw the cliff with its midpoint $M$ halfway up, and the boat 120 m from the foot.
2. The angle of depression from $M$ equals the angle of elevation of $M$ from the boat, $35^\circ$.
3. So half the height is $120\tan 35^\circ \approx 84.02$ m, and the cliff is $2 \times 84.02 \approx 168.0$ m high.

(b)

1. With $a = 2$, $b = 3$, $c = -4$: $x = \frac{-3 \pm \sqrt{9 + 32}}{4}$
   $= \frac{-3 \pm \sqrt{41}}{4}$
   $= \frac{-3 \pm 6.403}{4}$.
2. So $x \approx 0.851$ or $x \approx -2.35$ (3 significant figures).

(a)

1. Taking every man to wear glasses or a watch (or both), let $x$ wear both.
2. Then $15 + 10 - x = 20$, so $x = 5$.
3. The probability that a man chosen at random wears both is $\frac{5}{20} = \frac14$.

(b)(i)

1. After a $10\%$ discount, Mr Adams paid $90\%$ of the marked price: $0.9 \times \text{marked price} = \text{\textdollar}36,000$, so the marked price is $\frac{36\,000}{0.9} = \text{\textdollar}40,000.00$.

(ii)

1. The company bought the car for $27,000 and sold it for $36,000, a profit of $9,000.
2. Percentage profit $= \frac{9000}{27\,000} \times 100$
   $= 33\frac13\%$.