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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

Pass mark = 50% No. of students that passed = f50 + f65 + f80 = 45 + 25 + 15 = 85

Percentage of students with marks ranging from 35 to 50 = \(\frac{f_{35} + f{40} + f{50}}{\sum f}\) = \(\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}\) x 100% = \(\frac{120}{180}\) x 100% = 66\(\frac{2}{3}\)%

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

In \(\bigtriangleup\) PQR, Q = S = 75o (Corresponding angle) R = Q = 75o (Base angles of an isosceles \(\bigtriangleup\)) But, y + 75o = 180o (Sum of angles in a straight line) y = 180 - 75 y = 105o

\(\bigtriangleup\) is similar to \(\bigtriangleup\) HIJ o Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\) \(\frac{LH}{JH} = \frac{KL}{JI}\)

From the diagram, OP bisects o 3x = 33o x = \(\frac{33^o}{3}\) = 11o

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

Using \(\frac{x}{a} + \frac{y}{b}\) < 1for the equation of the time where a = intercept on x-axis and b = intercept on y - axis \(\frac{x}{3} + \frac{y}{3} = 1\) = \(\frac{x + y}{3} = 1\) = x + y < 3

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