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The volume of the solid = vol. of cone + vol. of hemisphere volume of cone = \(\frac{1}{3} \pi r^2 h\) = \(\frac{1 \pi}{3} \times (3)^2 x 6 = 18 \pi cm^2\) vol. of hemisphere = \(\frac{4 \pi r^3}{6} = \frac{2 \pi r^3}{3}\) = \(\frac{2 \pi}{3} \times (3)^3 = 18\pi cm^3\) vol. of solid = 18\(\pi\) + 18\(\pi\) = 36\(\pi\)cm3

(i) (ii) Average production in the period = \(\frac{4100 + 2500 + 1500 + 1800 + 9200}{5}\) = \(\frac{19100}{5}\) = \(3820\)

(a) Let \(< SQR = \theta ; < QRS = 90^\circ - \theta ; < QPS = \theta\) In \(\Delta\) PQS, \(\tan \theta = \frac{x}{9} ... (1)\) In \(\Delta\) QRS, \(\tan \theta = \frac{16}{x} ... (2)\) From (1) and (2), \(\frac{x}{9} = \frac{16}{x}\) \(x^{2} = 16 \times 9 \implies x = \sqrt{16 \times 9} = 12 cm\) (b) (i) From \(\Delta\) QRS, \(\tan \theta = \frac{16}{12} = 1.333\) \(\theta = \tan^{-1} (1.333) = 53.1^\circ\) \(\therefore < QRS = 90^\circ - 53.1^\circ = 36.9^\circ\) \(\approxeq 37^\circ\) (to the nearest degree) (ii) In \(\Delta\) PQS, \(|PQ|^{2} = 12^{2} + 9^{2}\) = \(144 + 81 = 225\) \(|PQ| = \sqrt{225} = 15 cm\)

(a) Length of the arc XY = Circumference of the base. \(\frac{\theta}{360^\circ} \times 2\pi r\) where \(\theta\) = 270^\circ ; r = 7cm. Length of arc XY = \(\frac{270^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = 33cm\) Let the radius of the base of the cone = x \(2 \pi r = 33cm\) \(2 \times \frac{22}{7} \times x = 33\) \(\frac{44}{7} \times x = 33 \implies x = \frac{33 \times 7}{44}\) \(x = 5.25cm\) (b) Area of the base of the cone = \(\pi r^{2}\) = \(\frac{22}{7} \times \frac{21}{4} \times \frac{21}{4}\) = \(\frac{693}{8} cm^{2} = 86.625 cm^{2}\) \(\approxeq 87 cm^{2}\)

(a) A = {m, a, p, e}; B = {a, e, i, o, u} ; C = {l, m, n, o, p, q, r, s, t, u} (i) \(A \cup B = {a, e, i, m, o, p, u}\) (ii) \(A \cup C = {a, e, l, m, n, o, p, q, r, s, t, u}\) (iii) \(A \cup (B \cap C)\) \(B \cap C = {o, u}\) \(A \cup (B \cap C) = {m, a, p, e, o, u}\) Biology students = n(B) = 190 Chemistry students = n(A) = 300 Let \(n(A \cap B) = a\) (i) From the Venn diagram, \((300 - a) + a + (190 - a) + 70 = 400\) \(560 - a = 400 \implies a = 160\) \(\therefore\) 160 students offer both Biology and Chemistry. (ii) Students offering at least one of Biology and Chemistry = \(n(A) + n(B) + n(A \cap B)\) = \((300 - 160) + (190 - 160) + 160\) = \(140 + 30 + 160\) = 330 students 330 students offer at least one of Biology and Chemistry.

(a)(i) Marks (%) Freq Cum Freq 0 - 9 10 10 10 - 19 40 50 20 - 29 80 130 30 - 39 140 270 40 - 49 170 440 50 - 59 130 570 60 - 69 100 670 70 - 79 70 740 80 - 89 40 780 90 - 99 20 800 (ii) (iii) From the Ogive, the 50th percentile is 70 marks. (b) Those who scored below 25% (24% and below) = 85 students Those who scored 75% and below = 710 \(\therefore\) Those who scored above 75% = 800 - 710 = 90 students Those to be retained without award = 800 - (85 + 90) = 800 - 175 = 625 students.

\(\frac{x^{2} - y^{2}}{2x^{2} + xy - y^{2}}\) \(2x^{2} + xy - y^{2} = 2x^{2} - xy + 2xy - y^{2}\) = \(x(2x - y) + y(2x - y) \) = \((x + y)(2x - y)\) \(\frac{x^{2} - y^{2}}{2x^{2} + xy - y^{2}} = \frac{(x + y)(x - y)}{(x + y)(2x - y)}\) = \(\frac{x - y}{2x - y}\)

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o 50 + 20 + 1 = 7110 Convert 7110 to base 8 \(\begin{array}{c|c} 8 & 71 \\ 8 & 8 R 7\\8 & 1 R 0\\8 & 0 R 1\end{array}\) = 1078

The least length is \(\frac{40}{48}\) = \(\frac{5}{8}\) for the rod to be cut in exactly equal trips Ratio \(\frac{5}{6}\) : \(\frac{48}{40}\) \(\frac{\frac{5}{6}}{\frac{40}{48}}\) = 1 \(\frac{5}{6}\) x \(\frac{48}{40}\) = \(\frac{240}{240}\) = 1 The least length = 240cm

\(\frac{0.021741 \times 1.2047}{0.023789}\) = \(\frac{0.022 \times 1.2}{0.024}\) (to 216) = \(\frac{0.0264}{0.024}\) = 1.1

Let the length of the square be \( L \). The area of the square is: \(A_s = L^2\) After modifications, the new length \( L_r \) and width \( W_r \) of the rectangle are: \(L_r = 1.2L, \quad W_r = 0.8L\) The area of the rectangle \( A_r \) is: \(A_r = L_r \times W_r = (1.2L)(0.8L) = 0.96L^2\) The ratio of the areas is: \(\text{Ratio} = \frac{A_r}{A_s} = \frac{0.96L^2}{L^2} = 0.96\) Thus, the ratio of the area of the rectangle to the area of the square is \( 0.96 \). Or \(\frac{96}{100}\) = 24:25

Total amount invested by A and B = N5,000 farm yield was sold for N15,000.00 profit = 15,000.00 - 5,000.00 = N10,000.00 Profit was shared in ratio 2 : 3 2 + 3 = 5 A received \(\frac{2}{5}\) of profit = \(\frac{2}{5}\) x 10,000 = N4,000.00 A receive \(\frac{3}{5}\) of profit = \(\frac{3}{5}\) x 10,000 = N6,000.00 Difference in profit received = N6,0000 - N4,000.00 = N2,000.00

I = \(\frac{PRT}{100}\) 100(A - P) = prt P = \(\frac{100A}{100 + RT}\) = \(\frac{100 \times 7000}{100 + (4 \times 3)}\) p = \(\frac{700,000}{112}\) = N6,250.00

profit 8% of N1.35 = \(\frac{8}{100}\) x N1.35 = N0.08 Cost price = N1.35 - N0.10 = N1.25 If he sells the 20 oranges for N1.10 now %loss = \(\frac{\text{actual loss}}{\text{Cost price}}\) x 100 \(\frac{1.25 - 1.10}{1.25}\) x 100 = \(\frac{0.15 \times 100}{1.25}\) = \(\frac{15}{1.25}\) = 12%

Let x represents number of boys that can work at \(\frac{5}{4}\) the rate at which the 10 girls work For 1hr. x boys will work for \(\frac{\frac{1}{5}}{4}\) x 10 x = \(\frac{4}{5}\) x 10 = 8 boys 8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs For 3 hrs, \(\frac{12 \times 5}{3}\) boys will be needed = 20 boys