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No. of times = \(\frac{\text{Total distance}}{\text{Circumference of circle}}\) = \(\frac{\text{Total distance}}{\pi d}\) = \(\frac{1000m}{\frac{22}{7} \times 100m}\) = \(\frac{1000 \times 7}{2200} = 3.187\) = 3(approx.) nearest whole no.

Pr. (winning 100m race) = \(\frac{1}{8}\) Pr. (losing 100m race) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\) Pr. (winning high jump) = \(\frac{1}{4}\) Pr. (losing high jump ) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump) = (\(\frac{1}{8} \times \frac{3}{4}\)) + (\(\frac{7}{8} \times \frac{1}{4}\)) = \(\frac{3}{32} + \frac{7}{32}\) = \(\frac{10}{32}\) = \(\frac{5}{16}\)

Joining SQ. In \(\bigtriangleup\) SPQ, (22o + a) + 55o + (41o + b) = 180o 121o + a + b = 180o a + b = 180 - 121 a + b = 59o.....(1) In \(\bigtriangleup\) SRPQ; R + a + b = 180o R + 59o = 180o (in (1), a + b = 59o) R = 180 - 59 R = 121o

4m - 15o = m + 75o (Vertically opposite angles are equal) 4m - m = 75 + 15 3m = 90 m = \(\frac{90}{3}\) m = 30o

Join SR o(Angle in a semicircle) o (Angle between a chord and a tangent = Angle in the alternate segment) o 90v + 55o + o o - 145o = 35o

3x = 2 x 60 = 2y (Angle at centre = 2 x angle at circumference) 3x = 2 x 60 x = \(\frac{2 \times 60}{3}\) = 40o 2 x 60 = 2y y = 60o x + y = 40 + 60 = 100o

F = 54o (Alternate triangle angles) o(Angle between a chord and a tangent = angles in the alternate segment) Now, o(Angles on a straight line) 54o + o = 180o o - 108o = 72o

Total surface area = sum of the area of the \(\bigtriangleup\) S + Area of the rectangle = 2 x Area of \(\bigtriangleup\) PTQ + 2 x Area of \(\bigtriangleup\) QUR + Area of rectangle PQRS 2 x \(\frac{1}{2}\)(8 x 4) + 2 x \(\frac{1}{2}\)(5 x 4) + 8 x 5 = 32 + 20 + 40 = 92cm2

Area of remaining = Area of the rectangle - Area of the semi-circle 22 x 8 - \(\frac{1}{2}\)x \pir\(^2\) Here r = \(\frac{14}{2}\)cm = 7cm Area of remaining = 176 - \(\frac{1}{2}\) x \(\frac{22}{7}\) x 7 x 7 = 176 - 77 = 99cm\(^2\)

POQ in a straight line Hence, o 56o + o o - 56o = 124o Now, in \(\bigtriangleup\) QOR OR = OQ = Radius o 4x + 124 = 180 4x = 180 - 124 4x = 56 x = \(\frac{56}{4}\) x = 14o

The bearing of x from y = 270o + \(\theta\) where \(\theta\) + 50o = y in \(\bigtriangleup\) OXY O + X + Y = 180o Where O = 40o + 30o = 70o 70o + 30o + y = 180o y + 100o = 180o y = 180o - 100o = 30o \(\theta\) + 50o = 80o 80o - 50o = 30o The bearing of x from y = 270o + 30o = 300o

(a) \(\therefore r = 36^\circ + 55^\circ = 91^\circ\) (b) \(< BAS = 60^\circ\) (angles in alternate segments) \(< BAE = 180^\circ - 5x^\circ\) (adjacent angles on a transversal) \(\therefore < BAS + < BAE + < EAT = 180^\circ\) \(\therefore 60^\circ + 180^\circ - 5x + x = 180^\circ\) \(\implies 240^\circ - 180^\circ = 4x\) \(60^\circ = 4x \implies x = \frac{60}{4} = 15^\circ\)

(a) M = {(4, 6), (5, 5), (6,4)} N = {(1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)} (b) Pr(M or N) = Pr(M) + Pr(N) Total sample = 36. \(Pr(M) = \frac{3}{36} = \frac{1}{12}\) \(Pr(N) = \frac{6}{36} = \frac{1}{6}\) \(Pr(\text{M or N}) = \frac{1}{12} + \frac{1}{6}\) = \(\frac{2 + 1}{12}\) = \(\frac{3}{12} = \frac{1}{4}\) (c) Yes, they are mutually exclusive since \(M \cap N = \emptyset\).

(a) 1km = 100,000 cm. Area = \(\frac{85 \times 100,000 \times 100,000}{20,000 \times 20,000}\) = \(85 \times 5 \times 5\) = \(2125 cm^{2}\) (b) Perimeter of path = \(2(length + width)\) = \(2(30 + (12 + a)) = 2(42 + a)\) = \(84 + 2a\) Area of field = \(length \times width\) = \(18 \times a\) = 18a \(\therefore 18a = 84 + 2a\) \(18a - 2a = 84 \implies 16a = 84\) \(a = \frac{84}{16}\) = \(5.25 cm\) (c) Area of major sector = \(\frac{360 - x}{360} \times \pi r^{2}\) \(27.5 = \frac{360 - x}{360} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) \(27.5 = \frac{360 - x}{360} \times \frac{77}{2}\) \(360 - x = \frac{27.5 \times 360 \times 2}{77}\) \(360 - x = 257.14\) \(x = 360^\circ - 257.14^\circ = 102.86^\circ\) \(x \approxeq 103^\circ\) (to the nearest degree).

(a) (b) /XT/ = 3.4 cm.