In each of the following questions two equations are given. Solve these equations and give answer : (i) (3x^2 – x – 70 = 0) (ii) (y^2 – 31y + 240 = 0)

For equation (i): 3x^2 - x - 70 = 0. Using quadratic formula: x = [1 ± √(1 + 840)]/6 = [1 ± √841]/6 = [1 ± 29]/6. Roots are x = 30/6 = 5 and x = -28/6 = -14/3 ≈ -4.67. For equation (ii): y^2 - 31y + 240 = 0. Roots are y = [31 ± √(961 - 960)]/2 = [31 ± 1]/2. Roots are y = 32/2 = 16 and y = 30/2 = 15. So x can be 5 or -4.67, while y can be 15 or 16. In all cases, x < y (5 < 15, 5 < 16, -4.67 < 15, -4.67 < 16). Option A is correct.