In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer I.(p^2 - 0.7p + 0.12 = 0) II. (q^2 + 0.1 \times q - 0.12 = 0)

For equation I: 0.3p² - 0.7p + 0.12 = 0. Multiply by 100: 30p² - 70p + 12 = 0. Using quadratic formula or factoring: p = (70 ± √(4900 - 1440))/60 = (70 ± 56)/60. So p = 126/60 = 2.1 or p = 14/60 = 0.233. For equation II: q² + 0.1q - 0.12 = 0. Using quadratic formula: q = (-0.1 ± √(0.01 + 0.48))/2 = (-0.1 ± 0.7)/2. So q = 0.3 or q = -0.4. Comparing all p and q values: p ≥ q in all cases (2.1 > 0.3, -0.4; 0.233 > -0.4, but 0.233 < 0.3). Since some p values are greater than all q values, and the smallest p (0.233) is still greater than the smallest q (-0.4), p ≥ q is correct.