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Table of value/observation \(a_0\) = 2.00cm S/N x(cm) a(cm) v(cm) y=\(\frac{a}{a_{0}}\) p=\(\frac{1+y^{2}}{y}\) T=x+v(cm) 1 25.0 2.60 49.0 1.30 2.07 74.00 2 30.0 2.40 46.0 1.20 2.03 76.00 3 35.0 2.10 45.0 1.05 2.01 80.00 4 40.0 1.80 43.0 0.90 2.01 83.00 5 45.0 1.60 40.0 0.80 2.05 85.00 viii. viii. Slopes(s) = \(\frac{x_{2} - x_{1}}{y_{2} - y_{1}}\) = \(\frac{76 - 70}{2.073 - 2.040}\) = \(\frac{6}{0.033}\) = 181.8 x. Intercept (c) = 81 xi. Evaluation k = \(\frac{c}{2}\) = \(\frac{81}{2}\) = 40.5 xii. Precautions; (a) Conxial arrangement of optical instruments (b) Avoided parallax error in reading metre rule (c) Lens kept upright (d) Repeated readings shown on the table (e) Surface of lens cleaned (f) Avoided zero error on metre rule (b)i. The distance between the optical centre and the principal focus of the lens is 20cm or if the parallel beam of light is incident on the lens, it is brought to a focus at a point 20cm from the lens. ii. M = 5cm, F = 20c, x = u M = \(\frac{v}{u}\) 5 = \(\frac{v}{u}\) v = 5u Using the formula \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) \(\frac{1}{5u} + \frac{1}{u} = \frac{1}{20}\) \(\frac{1+5}{5u} = \frac{1}{20}\) 5u = 20(1 + 5) 5u = 20 x 6 5u = 120 u = \(\frac{120}{5}\) = 24 x = 24cm

Given: P = 4 atm , V = 10 dm\(^3\) , n = ? R = 0.082 (constant) T = 273 + 273 = 546K Using the ideal gas law, PV = nRT n = \(\frac{PV}{RT}\) n = \(\frac{{4}\times{10}}{{0.082}\times{546}}\) n = \(\frac{40}{44.772}\) n = 0.893 mole

Tap water is likely to contain less Ca\(^{2+}\) and Mg\(^{2+}\). The reason is because looking through all the given options, spring water, river water and sea water are natural water that has not undergone any treatment. Conversely, it is believed that before tap water is distributed for consumption, it must have been treated, thereby reducing the Ca\(^{2+}\) and Mg\(^{2+}\) content in the water.

Let the number of people that offer both Mathematics and Physics = y Then, \((32 - y) + y + (24 - y) + 4 = 40\) \(60 - y = 40 \implies y = 20\) \(\therefore\) 20 students offer both Mathematics and Physics.

\(\frac {x+2}{4}\) - \(\frac{2x - 3}{3} < 4\) \(\frac{3(x + 2) - 4(2x - 3)}{12} < 4\) \(3x + 6 - 8x + 12 < 48 \) \(18 - 5x < 48 \implies -5x < 30\) \(\therefore x > -6\)

Let the angle for school fees = x^\circ Then Housing = 2x^\circ 120^\circ + 90^\circ + x^\circ + 2x^\circ = 360^\circ 3x^\circ = 150^\circ \(\implies\) x^\circ = 50^\circ. Amount spent on housing = \(\frac{100}{360} \times 7200\) = N2000.

Profit (P) = 10\(_x\) − \(_x\)2 Maximum profit can be achieved when the differential of profit with respect to number of bags(x) is 0 i.e. \(\frac{dp}{dx}\) = 0 \(\frac{dp}{dx}\) = 10 - 2x = 0 10 = 2x Then x = \(\frac{10}{2}\) = 5 Answer is C

y = 23\(_{five}\) + 101\(_{three}\) 23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\) = 13\(_{ten}\) 101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\) = 10\(_{ten}\) y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\) = 23\(_{ten}\) = 10111\(_{two}\)

The diagram shows angles at a point, the total angle at a point is 360 x - 10 + 4x - 50 + 2x + 3x + 20 = 360 10x - 40 = 360 10x = 360 + 40 10x = 400 x = \(\frac{400}{10}\) x = 40

\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11 Multiply through by the LCM of 4 and 3 which is 12 12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12) 9t + 4(21 - t) = 132 9t + 84 - 4t = 132 5t + 84 = 132 5t = 132 - 84 = 48 t = \(\frac{48}{5}\) t = 9 \(\frac{3}{5}\) Answer is D

let the girls' initial pocket money be whole(1) \(\frac{1}{4}\) of 1 = books, \(\frac{1}{3}\) of 1 = dress fraction of her money left = 1 - \(\frac{1}{4}\) - \(\frac{1}{3}\) = \(\frac{5}{12}\)

No expanded working has been attached for this question yet. The verified answer is Option A.

Total surface area of a cupboard is given by the equation A = 2(lb + bh + lh) L = 12, b = 10, h = 8 A = 2((12 x 10) + (10 x 8) + (12 x 8)) A = 2(120 + 80 + 96) A = 2 x 296 A = 592cm\(^2\) Answer is B

0.04945 to 2s.f is 0.049

To find the probability of both John and James failing the examination: Probabilities of failing: \(P(\text{John fails}) = 1 - P(\text{John passes}) = 1 - \frac{3}{4} = \frac{1}{4}\) \(P(\text{James fails}) = 1 - P(\text{James passes}) = 1 - \frac{3}{5} = \frac{2}{5}\) Probability of both failing: \(P(\text{Both fail}) = P(\text{John fails}) \times P(\text{James fails}) = \frac{1}{4} \times \frac{2}{5} = \frac{2}{20} = \frac{1}{10}\) Thus, the probability of both boys failing the examination is: \(\frac{1}{10}\)