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o(o o (sum of interior o + 55o + o

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

student that like swimming = x + 2 where 2 is the number of students who like reading, dancing and swimming. To find x from the venn diagram of swimming 6 +3 + 2 + x = 16 11 + x = 16 x = 16 - 11 = 5 no. of students that like dancing and swimming x + 2 = 5 + 2 = 7

(a) \(\frac{\frac{1}{2} of \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}}\) Numerator - \(\frac{1}{2} \times \frac{1}{4} \div \frac{1}{3} = (\frac{1}{2} \times \frac{1}{4}) \div \frac{1}{3}\) \(\frac{1}{8} \times 3 = \frac{3}{8}\) Denominator - \(\frac{1}{6} - \frac{3}{4} + \frac{1}{2} = \frac{1}{6} - (\frac{3}{4} - \frac{1}{2})\) \(\frac{1}{6} - (\frac{3 - 2}{4}) = \frac{1}{6} - \frac{1}{4}\) \(\frac{2 - 3}{12} = \frac{-1}{12}\) \(\therefore \frac{\frac{1}{2} \times \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}} = \frac{3}{8} \div \frac{-1}{12}\) \(\frac{3}{8} \times -12 = \frac{-9}{2}\) = \(-4.5\) (b) \(\sqrt{x} = 10^{\bar{1}.6741}}\) Squaring both sides, we have \(x = (10^{\bar{1}.6741}})^{2}\) \(x = 10^{\bar{1}.3482}\) (checking antilogarithm) \(x = 0.2229\)

(a) \(t = \sqrt{\frac{pq}{r} - r^{2}q}\) Squaring both sides, \(t^{2} = \frac{pq}{r} - r^{2}q\) \(t^{2} = \frac{pq - r^{3}q}{r}\) \(t^{2}r = pq - r^{3}q \implies t^{2}r = q(p - r^{3})\) \(q = \frac{t^{2}r}{p - r^{3}}\). (b) \(9^{(1 - x)} = 27^{y}\) \(\implies 3^{2(1 - x)} = 3^{3y}\) \(2 - 2x = 3y ..... (1)\) \(x - y = -1\frac{1}{2} ....... (2)\) From (2), \(x = -1\frac{1}{2} + y\) \(\therefore 2 - 2(-1\frac{1}{2} + y) = 3y\) \(2 + 3 - 2y = 3y \implies 5 = 3y + 2y = 5y\) \(\implies y = 1\) \(x = -1\frac{1}{2} + 1 = -\frac{1}{2}\) \(\therefore x + y = -\frac{1}{2} + 1 = \frac{1}{2}\)

(a) Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\) \(280 = \frac{\theta}{360} \times \frac{22}{7} \times 21 \times 21\) \(280 = \frac{1386 \theta}{360}\) \(\theta = \frac{280 \times 360}{1386}\) \(\theta = 72.72^\circ\) Perimeter of sector = \(2r + \frac{\theta}{360} \times 2\pi r\) = \(2(21) + \frac{72.72}{360} \times 2 \times \frac{22}{7} \times 21\) = \(42 + (0.202 \times 2 \times 66)\) = \(42 + 26.667\) = \(68.667 cm\) \(\approxeq 68.7 cm\) (b) When the sector is bent to form a cone, its radius becomes the slant height of the cone. The radius of the base of the cone is obtained from the relation \(r = \frac{R \theta}{360}\), where r = radius of the base of the cone, R = radius of the sector, \theta = angle of the sector. Therefore, r = \(\frac{21 \times 800}{11 \times 360}\) = \(\frac{140}{33}\) If y is the vertical angle of the cone, then \(sin \frac{y}{2} = \frac{r}{l}\) = \(\frac{140}{33 \times 21}\) = 0.2020 Hence, required angle = y = 2 x sin\(^{-1}\) (0.2020) = 23\(^o\).

Income = $1,300 grant Grant on different aspects: Magazine = \(\frac{10}{100} \times 1300 = $130\) New books = \(\frac{35}{100} \times 1300 = $455\) Damaged books = \(\frac{15}{100} \times 1300 = $195\) New furniture = \(\frac{30}{100} \times 1300 = $390\) Library staff = \(\frac{10}{100} \times 1300 = $130\) For pie chart, Magazine = \(\frac{130}{1300} \times 360^\circ = 36^\circ\) New books = \(\frac{455}{1300} \times 360^\circ = 126^\circ\) Damaged books = \(\frac{195}{1300} \times 360^\circ = 54^\circ\) New furniture = \(\frac{390}{1300} \times 360^\circ = 108^\circ\) Library staff = \(\frac{130}{1300} \times 360^\circ = 36^\circ\) (b) Percentage increase = \(\frac{455 - 390}{390} \times 100% =16.67%\)

(a) \(n(U) = 40\) \(n(M) = 18\) \(n(E) = 16\) \(n(A) = 19\) \(n(M \cap E) = 5\) \(n(A \cap E) = 2\) \(n(M) only = 6\) \(n(A) only = 9\) Accounts: x + 5 + 2 + 9 = 19 x + 16 = 19 x = 19 - 16 = 3. Mathematics : y + 6 + x + 5 = 18 y + 6 + 3 + 5 = 18 y + 14 = 18 y = 18 - 14 = 4. Economics : z + y + x + 2 = 16 z + 4 + 3 + 2 = 16 z + 9 = 16 z = 16 - 9 = 7. Let the number of students that failed be r. 40 - r = 6 + 5 + 9 + 2 + 3 + 4 + 7 40 - r = 36 r = 40 - 36 = 4. (b) Percentage that failed at least one of Economics and Mathematics = 100% - (% of people that passed at least one of Economics and Mathemics) Number that passed at least one of Economics and Mathematics = n(E) + n(M) + n(E and M) = 6 + 7 + 4 = 17 students passed at least one of Economics and Mathematics % passed = \(\frac{17}{40} \times 100% = 42.5%\) Percentage that failed at least one of Economics and Mathematics = 100% - 42.5% = 57.5% (c) Probability of failed in Accounts = 1 = Probability of passed in Accounts. P(passed Accounts) = \(\frac{n(A)}{n(U)}\) = \(\frac{19}{40}\) P(failed accounts) = \(1 - \frac{19}{40}\) = \(\frac{21}{40} = 0.525\)

(a) Total surface area of a sphere = \(4\pi r^{2}\) Let the TSA of the smaller sphere be \(S_{1}\) with radius r and the bigger sphere be \(S_{2}\) with radius R. \(\frac{S_{1}}{S_{2}} = \frac{9}{49}\) \(\frac{9}{49} = \frac{4\pi r^{2}}{4\pi R^{2}}\) \(\implies \frac{9}{49} = \frac{12^{2}}{R^{2}}\) \(\frac{49 \times 144}{9} = R^{2}\) \(R = \sqrt{49 \times 16} = 28 cm\) Volume of bigger sphere = \(\frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \times \frac{22}{7} \times 28 \times 28 \times 28\) = \(\frac{275,968}{3}\) = \(91989.33 cm^{3}\) \(\approxeq 91989 cm^{3}\) (nearest whole number). (b) \(|ZX|^{2} = |YZ|^{2} + |YX|^{2} - 2|YZ||YX| \cos Y\) = \(5^{2} + 3^{2} - 2(5)(3) \cos 135\) = \(25 + 9 - 30(-0.7071)\) = \(34 + 21.213\) \(|ZX|^{2} = 55.213\) \(|ZX| = 7.43 km \approxeq 7 km\) (ii) Using sine rule, \(\frac{\sin \theta}{5} = \frac{\sin 135}{7.43}\) \(\sin \theta = \frac{5 \times \sin 135}{7.43}\) = \(\frac{3.5355}{7.43}\) \(\sin \theta = 0.4758\) \(\theta = \sin^{-1} (0.4758)\) = \(28.41^\circ \approxeq 28^\circ\) (to the nearest degree). Bearing of Z from X = 270^\circ + 28.41^\circ = 298.41^\circ \(\approxeq\) 298^\circ.

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

TUVW is a cyclic quad 3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary) 3χ + 108 = 180 3χ = 180 - 108 3χ = 72 χ = 72/3 χ = 24o

\(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}\) = \(\frac{1}{\left(\frac{81}{16}\right)^{\frac{1}{4}}}\times \frac{1}{2}\) = \(\left(\frac{16}{81}\right)^{\frac{1}{4}}\times \frac{1}{2}\) = \(\left(\frac{2}{3}\right)^{4\times \frac{1}{4}}\) =\(\frac{2}{3}\times \frac{1}{2}\) = \(\frac{1}{3}\)

\(\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}}\) = \((\frac{2\sqrt{3} + \sqrt{5}}{\sqrt{5} - \sqrt{3}})(\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}})\) = \(\frac{2\sqrt{15} + 6 + 5 + \sqrt{15}}{5 - 3}\) = \(\frac{3\sqrt{15} + 11}{2}\)

Let C n G = X n C only = 40 - x n G only = 30 - x 40 - X + 30 - X + X = 50 70 - X = 50 X = 70 - 50 X = 20

\(\frac{x^{3} + 3x^{2} - 10x}{2x^{2} - 8} = \frac{x(x^{2} + 3x - 10)}{2(x^{2} - 4)}\) = \(\frac{x(x^{2} - 2x + 5x - 10)}{2(x - 2)(x + 2)}\) = \(\frac{x(x - 2)(x + 5)}{2(x - 2)(x + 2)}\) = \(\frac{x(x + 5)}{2(x + 2)}\)