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

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: \(\frac{1}{10}\)

Length of the arc = \(\frac{\theta}{360} \times 2\pi r\) = \(\frac{300}{360} \times 2 \times \frac{22}{7} \times 14\) = \(\frac{220}{3}\) = 73.33 cm Perimeter of the sector = Length of the arc + 2r = 73.33 + 2(14) = 101.33 cm

\(\frac {25^{\frac{2}{3}} \div 25^{\frac{1}{6}}} {(\frac{1}{5})^{\frac{7}{6}} \div (\frac{1}{5})^{\frac{1}{6}}}\) = \(\frac{25^{\frac{2}{3} - \frac{1}{6}}}{(\frac{1}{5})^{\frac{7}{6} - \frac{1}{6}}}\) = \(\frac{25^{\frac{1}{2}}}{(\frac{1}{5})}\) = \(5 \div \frac{1}{5}\) = 25

P = (\(\frac{Q( R - T )}{ 15})^\frac{1}{3}\) take cube of both sides \(P^3 =\frac{Q( R - T)}{ 15}\) cross multiply \(15P^3 = Q( R - T)\) \(\frac{15P^3}{Q}\) = R - T T = R - \(\frac{15P^3}{Q}\)

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