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4 oranges sell for Nx, 1 orange will sell for \(\frac{Nx}{4}\) 24 oranges will sell for: \(\frac{Nx}{4} \times 24\) = n6x 3 mangoes sell for Ny, 1 mango will sell for \(\frac{Ny}{3}\) 12 mangoes will sell for \(\frac{Ny}{3} \times 12\) = 4Ny total money pay N6x + N4y = N(6x + 4y)

\(\frac{x^2 - y^2}{(x + y)^2} \div \frac{(x - y)^2}{(3x + 3y)}\) \(\frac{(x + y)(x - y)}{(x + y)(x + y)}\div \frac{(x - y)(x - y)}{3(x + y)}\) = \(\frac{3}{x - y}\)

\(\frac{2x - 5}{2} < \frac{(2 - x)}{1}\) 2x - 5 < 4 - 2x 2x + 2x < 4 + 5 4x < 9 x < \(\frac{9}{4}\) x < 2\(\frac{1}{4}\)

\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\) substitute x = 64 and y = 27 \(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\) = \(\frac{8 - 3}{27 - 16}\) = \(\frac{5}{11}\)

\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2) \(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2) x + 4y = 6......(iii) 3x - 4y = 2.....(iv) add (iii) and (iv) 4x = 8, x = \(\frac{8}{4}\) = 2 substitute x = 2 into equation (iii) x + 4y = 6 2 + 4y = 6 4y = 6 - 2 4y = 4 y = \(\frac{4}{4}\) = 1(x + y) 2 + 1 = 3

\(p^{\frac{1}{3}}\) = \(\frac{\sqrt[3]{q}}{r}\) cross multiply \(p^{\frac{1}{3}}\) = \(\frac{q^{1/3}}{r}\) r\(p^{\frac{1}{3}}\) = \(q^{\frac{1}{3}}\) take the cube of both sides \((rp^{\frac{1}{3}})^3\) = \((q^{\frac{1}{3}})^3\) \(r^3p^{\frac{3}{3}}\) = \(q^{\frac{3}{3}}\) \(r^3\)p = q ∴ q = p\(r^3\)

From \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ2 52 = 22 + MQ2 25 = 4 + MQ2 25 - 4 = MQ2 21 - MQ2 MQ2 = 21 MQ2 = \(\sqrt{21}\) Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm

Volume of a cube with side a cm = a\(^3\) a\(^3\) = 64 \(\implies\) a = 4cm Base area of a cube = a\(^2\) = 4\(^2\) = 16 cm\(^2\) \(\implies\) Base area of the cuboid = 16 cm\(^2\) Volume of cuboid = Base area x height 80 = 16 x h h = \(\frac{80}{16}\) = 5 cm

Sin x = \(\frac{5}{13}\) 0o \(\leq\) x \(\leq\) 90o, (cos x - tan x) AC2 = AB2 + BC2 132 = 52 + BC2 169 - 25 + BC2 169 - 25 = BC2 144 = BC2 Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\) BC = \(\sqrt{144}\) BC = 12 tan x = \(\frac{opp}{adj} = \frac{5}{12}\) BC = 12 cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\) \(\frac{144 - 65}{156} = \frac{79}{156}\)

Tan 50o = \(\frac{h}{6}\) h = 6 tan 50 = 6 x 1.1917 = 7.1505 = 7.15

From the diagram above, the bearing of X from Z = 180º + 60º = 240º (Note: alternate angles are equal and measurement here is clockwise)

The probability not scoring 85% is = 1 - pro(scoring 85%) = 1 - \(\frac{85}{100}\) = 1 - 0.85 = 0.15 not scoring would be less than 1. Hence option D is correct because the probability of an event cannot be more than 1.

2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2) \(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\) \(\log_x \frac{27}{8} = 3\) \(\frac{27}{8} = x^3\) \(x^3 = \frac{27}{8}\) x = \(\sqrt{\frac{27}{8}}\) = (\(\frac{27}{8}\))\(\frac{1}{3}\) = \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\) = \(\frac{3}{2}\)

p = (y : 2y \(\geq\) 6) 2y \(\geq\) 6 y \(\geq \frac{6}{2}\) y = \(\geq\) 3 and Q = (y : y -3 \(\leq\) 4) y - 3 \(\leq\) 4 y \(\leq\) 4 + 3 y \(\leq\) 7 therefore p = {3, 4, 5, 6, 7} and Q = {7, 6, 5, 4, 3....} P\(\cap\)Q = {3, 4, 5, 6, 7}

6k2 = 5k + 6 6k2 - 5k - 6 = 0 6k2 - 0k + 4k - 6 = 0 3k(2k - 3) + 2(2k - 3) = 0 (3k + 2)(2k - 3) = 0 3k + 2 = 0 or 2k - 3 = 0 3k = -2 or 2k = 3 k = \(\frac{-2}{3}\) or k = \(\frac{3}{2}\) k = (\(\frac{-2}{3}\), k = \(\frac{3}{2}\))