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(a) \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = \sqrt{3 \times 3} + \sqrt{3 \times 2} - 5\sqrt{2 \times 3} - 5\sqrt{2 \times 2}\) = \(3 + \sqrt{6} - 5\sqrt{6} - 5(2)\) = \(-7 - 4\sqrt{6}\) \(\therefore\) a = -7 and b = -4. (b) \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\) \(\frac{2^{1 - y + y - 1}}{2^{y + 2}} = (2^{3})^{2 - 3y}\) \(2^{0 - (y + 2)} = 2^{6 - 9y} \implies 2^{- y - 2} = 2^{6 - 9y}\) \(- y - 2 = 6 - 9y \implies - y + 9y = 6 + 2\) \(8y = 8 \implies y = 1\).

(a) Let B represent those with degree. Let D represent those with diploma. Let x represent those who have both degree and diploma. (17 - x) + x + (15 - x) + 4 = 30 36 - x = 30 x = 36 - 30 = 6. (b) \(< PNM = 75^\circ\) (corresponding angle) \(\therefore NQR = 125^\circ - 75^\circ \) = \(50^\circ\) \(< NPM = \frac{180^\circ - 125^\circ}{2}\) = \(\frac{55^\circ}{2}\) = \(27.5^\circ\). (Base angles of an isosceles triangle).

(a) \(\log (\frac{35 \times 49}{40 \div 56})\) Given \(\log 5 = 0.6990 , \log 7 = 0.8451 , \log 8 = 0.9031\) \(\log (\frac{35 \times 49}{40 \div 56} = \log (\frac{(7 \times 5) \times 7^{2}}{(8 \times 5) \div (8 \times 7)}\) = \(\log (\frac{7^{3} \times 5}{5 \div 7}\) = \(\log 7^{3} + \log 5 - (\log 5 - \log 7)\) = \(3 \log 7 + \log 5 - \log 5 + \log 7\) = \(4 \log 7\) = \(4 \times 0.8471\) = \(3.3804\) (b) (i) Since there are 60 more adults than children, then the number of adults = x + 60 \(\therefore D 5(x + 60) + D x(2) = D 1280\) \(5x + 300 + 2x = 1280\) \(7x + 300 = 1280 \implies 7x = 1280 - 300 = 980\) \(x = \frac{980}{7} = 140\) There were 140 children. (ii) Ratio of children to adults = \(x : x + 60\) = \(140 : (140 + 60)\) = \(140 : 200\) = \(7 : 10\) (iii) Total number of persons at the show : 140 + 200 = 340 persons. Total amount gotten = D 1280 Average paid per person = \( D \frac{1280}{340}\) = \(D \frac{64}{17}\) = D 3.765 (iv) Percentage profit Profit : D (1280 - 720) = D 560 % profit : \(\frac{560}{720} \times 100% = \frac{700}{9} %\) = \(77.78%\)

(a) Marks frequency mid-point \(fx\) 21 - 30 2 25.5 51 31 - 40 10 35.5 355 41 - 50 12 45.5 546 51 - 60 15 55.5 832.5 61 - 70 8 65.5 524 71 - 80 3 75.5 226.5 \(\sum\) 50 2535 (b) Mean (x) = \(\frac{\sum fx}{\sum f}\) = \(\frac{2535}{50}\) = \(50.7\) (c) Number of students that scored above 60% : 8 + 3 = 11 %age : \(\frac{11}{50} \times 100% = 22%\)

1101\(_2\) + 10111\(_2\) + 111\(_2\) = 101011\(_2\)

125\(_x\) = 20\(_{10}\) 1 x X\(^2\) + 2 x X\(^1\) + 5x X \(^0\) = 20 X2 + 2X + 5 = 20 X\(^2\) + 2X - 15 = 0 (X + 5)( X - 3) = 0 X + 5 implies X = -5 X - 3 implies X = 3 But X cannot be negative ∴X = 3

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

NB: This is a correction to the solution in the video lesson. 123456 \(\approxeq\) 123000.

\(I = \frac{P \times T \times R}{100}\) = \(\frac{7500 \times 8 \times 5}{100}\) = N3000

N85 - N60 = N25 increase ∴percentage increase = \(\frac{25}{60}\times \frac{100}{1}\) = \(\frac{125}{3}\) = 41.67% = 42%

\(16^{\frac{-1}{2}}\times 4^{\frac{-1}{2}} \times 27^{\frac{1}{3}}\) =\(\frac{1}{16^{\frac{1}{2}}}\times \frac{1}{4^{\frac{1}{2}}}\times 27^{\frac{1}{3}}\) =\(\frac{1}{\sqrt{16}} \times \frac{1}{\sqrt{4}} \times \sqrt[3]{27}\) =\(\frac{1}{4} \times \frac{1}{2} \times \frac{3}{1}\) = \(\frac{3}{8}\)

If log\(_{x^{\frac{1}{2}}}\)64 = 3 (X\(^{\frac{1}{2}})^3\) = 64 (X\(^{\frac{1}{2}})^3\) = 4\(^3\) X\(^{\frac{1}{2}}\) = 4 X = 4\(^2\) X = 16

\(\frac{1+\sqrt{2}}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}}\) = \(\frac{1+(1+\sqrt{2})+\sqrt{2}(1+\sqrt{2})}{1^2 - (\sqrt{2})^2}\) = \(\frac{(1+\sqrt{2}+\sqrt{2}+2)}{1-2}\) = \(\frac{3+2\sqrt{2}}{-1}\) = - 3 - 2\(\sqrt{2}\) ∴ X and Y = -3 and -2

X = {1,5,10} Y = {3,4,6} X∩Y = ∅

n(M) only = 30-10 = 20 n(E) only = 20-10 = 10 n(M∩E) = 10 ∴M∪E = 20+10+10 = 40