Two quadratic equations I and II are given below. Equation I: ax² - 33x + ab = 0 Equation II: cy² - 34y + 2cd = 0 Note: (i) Sum of the values of 'a' and 'c' is 5 (ii) One of the roots of the equation (

From (II) : One of the roots of equation (I) is 2nd smallest composite number. So, one of the roots of equation (I) = 6 Also, a = smallest odd prime number So, 'a' = 3 Equation I: 3x² − 33x + 3b = 0 3 X (6)² − 33 X (6) + 3b = 0 3 X 36 − 198 + 3b = 0 108 − 198 + 3b = 0 −90 + 3b = 0 3b = 90 'b' = 30 Now, 3x² − 33x + 90 = 0 x² − 11x + 30 = 0 x² − 5x − 6x + 30 = 0 x(x − 5) − 6(x − 5) = 0 (x − 5)(x − 6) = 0 'x' = 5 or 6 From (I) : a + c = 5 'c' = 5 - 3 = 2 Equation II: cy² - 34y + 2cd = 0 2 X (8)² − 34 X (8) + 4d = 0 2 X 64 − 272 + 4d = 0 128 − 272 + 4d = 0 −144 + 4d = 0 4d = 144 'd' = 36 Now, 2y² − 34y + 144 = 0 y² − 17y + 72 = 0 y² - 8y - 9y + 72 = 0 y(y - 8) - 9(y - 8) = 0 (y - 8)(y - 9) = 0 So, 'y' = 8 or 9 From I: x = 5 or 6 From II: y = 8 or 9 Therefore, x < y Hence, option b.