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2 + 2 + 5 + 8 + 12 + H + 7 + 3 + 2 = 52

Modal class ≈ 40.5 That is, the class having the highest frequency.

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

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

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

o = 180o (o - 110o = 70o In \(\bigtriangleup\) XTU, o i.e. o = 180 o = 70o Also o = 180 o in \(\bigtriangleup\) SYT + o(Sum of interior o - 60 2(o TYW = \(\frac{300}{2}\) = 150o < SYT

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

o (Corresponding angles) 245 + o (o

(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) \(\frac{x^{2} - 4}{x^{2} + x} \div \frac{x^{2} - 4x + 4}{x + 1}\) = \(\frac{x^{2} - 4}{x^{2} + x} \times \frac{x + 1}{x^{2} - 4x + 4}\) = \(\frac{(x - 2)(x + 2)}{x(x + 1)} \times \frac{x + 1}{x^{2} - 2x - 2x + 4}\) (Using difference of two squares) = \(\frac{(x - 2)(x + 2)}{x} \times \frac{1}{(x - 2)^{2}}\) = \(\frac{(x + 2)}{x(x - 2)}. (b) Roots of \(ax^{2} +bx + c = mx + k\) For graph \(y = ax^{2} + bx + c\), \(x = -1\) and \(x = 2\) \(x + 1 = 0\) and \(x - 2 = 0\) \((x + 1)(x - 2) = 0\) \(x^{2} - 2x + x - 2 = 0\) \(x^{2} - x - 2 = 0 ..... (i)\) \(-x^{2} + x + 2 = 0 ...... (ii)\) For graph \(y = mx + k\) \(m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{1 - (-1)}{1.3 - (-1.3)}\) \(m = \frac{2}{2.6} = 0.769\) \(k = 0.25\) \(x^{2} - x - 2 = 0.769x + 0.25\) \(x^{2} - x - 0.769x - 2 - 0.25 = 0\) \(x^{2} - 1.769x - 2.25 = 0\) Using formula method, \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-1.769) \pm \sqrt{(-1.769)^{2} - 4(1)(-2.25)}}{2(1)}\) \(x = \frac{1.769 \pm \sqrt{3.1294 + 9}}{2}\) \(x = \frac{1.769 \pm 3.483}{2}\) \(x = \frac{1.769 + 3.483}{2}\) or \(x = \frac{1.769 - 3.483}{2}\) \(x = \frac{5.252}{2} \approxeq 2.626\) or \(x = \frac{-1.714}{2} \approxeq -0.857\) (ii) The coordinates of points L, M and N are (-1, 0), (0, 2) and (2, 0) respectively. Using coordinate for L: \(0 = a(-1^{2}) + b(-1) + c \implies 0 = a - b + c ..... (1)\) Using M : \(2 = a(0^{2}) + b(0) + c \implies c = 2 ...... (2)\) Using N : \(0 = a(2^{2}) + b(2) + c \implies 0 = 4a + 2b + c .... (3)\) From (2), c = 2 \(\therefore (1) : a - b + 2 = 0 \implies a - b = -2 .... (1a)\) \(\therefore (3) : 4a + 2b + 2 = 0 \implies 4a + 2b = -2 ..... (3a)\) Solving equations (1a) and (3a) simultaneously, we get \(a = -1 , b = 1 , c = 2\) \(\therefore\) Equation of the curve : \(-x^{2} + x + 2 = 0\) (iii) Line of symmetry = \(x = \frac{-b}{2a}\) \(y = -x^{2} + x + 2 \) Line of symmetry = \(\frac{-1}{2(-1)}\) \(x = \frac{1}{2}\)

Given \(\sin x = 0.6 = \frac{3}{5}\) \(\therefore Opp = 3; Hyp = 5\) Using Pythagoras theorem, \(5^{2} = 3^{2} + Adj^{2}\) \(Adj = \sqrt{5^{2} - 3^{2}}\) \(Adj = \sqrt{25 - 9} = \sqrt{16} = 4\) \(\therefore \cos x = \frac{4}{5} = 0.8\) \(2\cos x + 3\sin x = 2(\frac{4}{5}) + 3(\frac{3}{5})\) = \(\frac{8}{5} + \frac{9}{5}\) = \(\frac{17}{5}\). (b) Join O to X and O to Y. Considering either \(\Delta OXC\) or \(\Delta OYC\), \(\sin 65 = \frac{r}{30}\) \(r = 30 \sin 65\) \(r = 30 \times 0.9063\) \(r = 27.189 cm \approxeq 27.2 cm\) (to one decimal place) (b) Area of shaded portion = Area of triangle ABC - Area of semi-circle WXYZ. = \(\frac{1}{2} bh - \frac{1}{2} \pi r^{2}\) = \((\frac{1}{2} \times 2c \times 30) - (\frac{1}{2} \times \frac{22}{7} \times (27.189)^{2})\) From \(\Delta COA, \tan 65 = \frac{c}{30}\) \(\implies c = 30 \tan 65 \approxeq 64.34 cm\) \(\therefore\) Area of shaded portion = \(30(64.34) - \frac{8131.659}{7}\) = \(1930.056 - 1161.666\) = \(768.39 cm^{2}\)