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Error in length = 60cm - 50cm = 10cm Error in breadth = 40cm - 35cm = 5cm (i) % error in length = \(\frac{10}{50} \times 100% = 20%\) % error in breadth = \(\frac{5}{40} \times 100% = 12.5%\) (ii) Area of the actual table = \(50cm \times 40cm = 2000cm^{2}\) Area of the table in error = \(60cm \times 35cm = 2100cm^{2}\) Error in area = \(2100cm^{2} - 2000cm^{2} = 100cm^{2}\) % error in area = \(\frac{100}{2000} \times 100% = 5%\) (b) Product of the error in length and in breadth = \(20 \times 12.5 = 250%\)

(i) From the figure, \(\sin 30^\circ = \frac{4}{RT} \implies RT = \frac{4}{\sin 30^\circ}\) = \(\frac{4}{0.5} = 8m\) (ii) |RQ| = |XY| \(\therefore |XY|^{2} = 8^{2} - 4^{2} = 64 - 16 = 48\) \(|XY| = \sqrt{48} = 4\sqrt{3} m = 6.928m\) But |PY| = |PX| + |XY| = 20 + 6.928 = 26.928m \(\approxeq\) 26.9m. (iii) Let the angle of elevation = \(\theta\) \(\tan \theta = \frac{TY}{PY} = \frac{7}{26.928}\) \(\tan \theta = 0.26 \implies \theta = \tan^{-1} (0.26)\) \(\theta = 14.57^\circ \approxeq 14.6^\circ\) (to 1 decimal place)

(a) Formula : Volume = \(\frac{1}{3} \pi r^{2} h\) \(\pi = 3.142; r = 4cm; h = 6cm\) \(Volume = \frac{1}{3} \times 3.142 \times 4^{2} \times 6 = 100.544 cm^{3}\) (b) Volume of hemisphere = \(\frac{1}{2} \times \frac{4}{3} \pi r^{3}\) = \(\frac{2}{3} \times \frac{22}{7} \times 5^{3}\) = \(\frac{5500}{21} cm^{3} = 261.905 cm^{3}\) Volume of water discharged per second = \(\pi r^{2} h\) = \(\frac{22}{7} \times 0.2m \times 0.2m \times 2m = \frac{1.76}{7}\) = \(0.2514 m^{3} / sec\) \(\therefore \text{The time to fill the tank = } \frac{261.905}{0.2514 \times 60}\) = \(\frac{261.905}{15.084} = 17.363 mins\) \(\approxeq 17.4 mins\)

(a) Class Interval Tally Freq 45 - 49 |||| | 6 50 - 54 |||| |||| 9 55 - 59 |||| |||| 10 60 - 64 |||| |||| || 12 65 - 69 |||| || 7 70 - 74 |||| | 6 (b) Class Interval Mid-value (x) x - 62 \((x - 62)^{2}\) \(f\) \(f(x - 62)\) \(f(x - 62)^{2}\) 45 - 49 47 -15 225 6 -90 1350 50 - 54 52 -10 100 9 -90 900 55 - 59 57 -5 25 10 -50 250 60 - 64 62 0 0 12 0 0 65 - 69 67 5 25 7 35 175 70 - 74 72 10 100 6 60 600 \(\sum\) 50 -135 3275 \(Mean (\bar{x}) = A + \frac{\sum f(x - A)}{\sum f}\) = \(62 + \frac{-135}{50}\) = \(62 - 2.7 = 59.3\) Standard deviation = \(\sqrt{\frac{\sum f(x - A)}{\sum f}}\) = \(\sqrt{\frac{3275}{50}}\) = \(\sqrt{65.5}\) = \(8.093 \approxeq 8.1\) (to 1 decimal place)

(b) |AC| = 7cm ; |QB| = 2.9cm.

(a) Let S = sample space = integers from 5 to 25 inclusive. F = multiple of 5 ; T = multiple of 3. S = {5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25} F = {5, 10, 15, 20, 25}; T = {6, 9, 12, 15, 18, 21, 24} P(F or T) = \(\frac{5}{21} + \frac{7}{21} = \frac{12}{21} = \frac{4}{7}\) (b) Let B = blue balls and R = red balls n(B) = 4 ; n(R) = 6. (i) P(both are red) = \(\frac{6}{10} \times \frac{6}{10} = \frac{36}{100}\) = \(\frac{9}{25}\) (ii) P(both are same colour) = P(both are red) or P(both are blue) = \(\frac{9}{25} + (\frac{2}{5} \times \frac{2}{5})\) = \(\frac{9}{25} + \frac{4}{25}\) = \(\frac{13}{25}\)

(a) \(y = 2x^{2} - 5x + 1\) x -3 -2 -1 0 1 2 3 4 5 y 34 19 8 1 -2 -1 4 13 26 Table of value for \(y = x + 6\) x -2 0 2 y 4 6 8 (b) (d) (i) Least value of y = -2 occurs at x = 1 (ii) Solution of \(2x^{2} - 5x + 1 = x + 6\) are x = -0.7 and x = 3.7.

(a) The number that speak Hausa = n(H) = 25 The number that speak Igbo = n(I) = 16 The number that speak Yoruba = n(Y) = 21 Total no of students = U = 40 \(n(H \cap I) = 8 ; n(H \cap Y) = 11 ; n(Y \cap I) = 6\) (b) \(U = n(H) + n(I) + n(Y) - n(H \cap Y) - n(H \cap I) - n(I \cap Y) + n(H \cap Y \cap I)\) \(40 = 6 + x + 2 + x + 4 + x + 8 - x + 11 - x + 6 - x + x\) \(40 = 37 + x \implies x = 40 - 37 = 3\)

(a) Distance along arc PQ = \(\frac{\theta}{360^\circ} \times 2 \pi r\) where \(r = R \cos \theta\) \(|PQ| = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 6400 \times \cos 40^\circ\) \(|PQ| = \frac{1}{6} \times 2 \times \frac{22}{7} \times 6400 \times 0.7660\) = \(5134.74 km \approxeq 5135km\) (to the nearest whole number) (b) Difference in latitude between Q and T = 40^\circ - 28^\circ = 12^\circ \(\therefore |QT| \text{along the line of longitude} = \frac{12}{360} \times 2 \times \frac{22}{7} \times 6400\) \(\frac{1}{30} \times 2 \times \frac{22}{7} \times 6400 = 1340.95km \approxeq 1341km\) (c) Average speed = \(\frac{\text{Total distance covered}}{\text{Total time taken}}\) = \(\frac{5134.74 + 1340.95}{12} = 539.64km/hr\) \(\approxeq 540km/hr\) (to 3 significant figures)

Degree for food items = 360^\circ - (120^\circ + 80^\circ + 43^\circ + 30^\circ + 18^\circ) = 360^\circ - 291^\circ = 69^\circ ∴∴ Amount spent on food items = \(\frac{69}{360}\)\times12,000.00 = N2300.00

Area of a sector \(\frac{\theta}{360^\circ} * \pir^2\) = \(\frac{135}{360} * 3.142 * 4 * 4\) = \(18\frac{6}{7}cm^2\)

This histogram is transferred into this frequency table \(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\) Students who scored more than 40 = 6 + 6 + 2 = 14 i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)%

loga3 + loga5 - loga6 log(a3 * a5/a6) = loga64 a2 = 64 a = 8

ar = 8 .....(1) ar3 = 32 .....(2) divide (2) by (1) r2 = 4 r = 2 ∴ a = 4 S = a(rn - 1)/r - 1 S = 4(24 - 1)/2 - 1 = 60

Area of \(\Delta\)QST = 52cm\(^2\); height = 8cm b=base length; but A = 1/2 b x h =52=1/2 x b x 8cm = 104 =8b b=104/8 = 13cm; ST= 13cm; PS = 6cm TP = 13 - 6 = 7cm; area of \(\Delta\)QST = 1/2 b x h = 1/2 x 7 x 8/1 = 28cm\(^2\)