Given $ \frac{1}{x} + \frac{1}{y} = 1 $, we have $ \frac{x + y}{xy} = 1 $. Since $ xy = 12 $, it follows that $ x + y = 12 $. Check with roots of $ t^2 - (x+y)t + xy = 0 $: $ t^2 - 12t + 12 = 0 $. Discriminant = $ 144 - 48 = 96 $, roots are $ t = \frac{12 \pm \sqrt{96}}{2} $. However, directly, $ x + y = 12 $, but testing pairs (e.g., 3, 4) gives $ x + y = 7 $,

Solve $ 2x + 3 = 7x - 2 $. Rearrange: $ 3 + 2 = 7x - 2x $, so $ 5 = 5x $, and $ x = 1 $.

Add the equations: $ (2x + 3y) + (3x - 2y) = 12 + 5 $, giving $ 5x + y = 17 $. Subtract: $ (2x + 3y) - (3x - 2y) = 12 - 5 $, giving $ -x + 5y = 7 $. Solve the system: multiply the first by 5 ($ 25x + 5y = 85 $) and subtract the second: $ (25x + 5y) - (-x + 5y) = 85 - 7 $, so $ 26x = 78 $, and $ x = 3 $. Substitute into $ 5x + y = 17 $: $ 5(3) + y = 17 $, so $ y = 2 $. Thus, $ x + y = 3 + 2 = 4 $.

Let the greater number be $ x $, and the smaller number be $ y $. Given $ x - y = 30 $ and $ y = \frac{x}{2} + 13 $. Substitute $ y $ in the first equation: $ x - \left( \frac{x}{2} + 13 \right) = 30 $. Solving, $ \frac{x}{2} - 13 = 30 $, so $ \frac{x}{2} = 43 $, and $ x = 86 $.

Given $ x + y + z = 0 $, the expression is $ \frac{x^2}{yz} + \frac{y^2}{xz} + \frac{z^2}{xy} $. This simplifies to $ \frac{x^3 + y^3 + z^3}{xyz} $. Since $ x + y + z = 0 $, we use the identity $ x^3 + y^3 + z^3 = 3xyz $. Thus, $ \frac{3xyz}{xyz} = 3 $.

Using the alligation rule: $ (75 - 65) : (65 - 50) = 10 : 15 = 2 : 3 $. Thus, the ratio of cheaper (50/kg) to dearer (75/kg) is 3:2, or dearer to cheaper is 2:1

Let the original price be 100. After a 20% increase, price = $ 100 \times 1.2 = 120 $. After a 20% decrease, price = $ 120 \times 0.8 = 96 $. Net change = $ \frac{96 - 100}{100} \times 100 = -4\% $.

Total distance = 275 + 315 = 590 km. Time for the first part = $ \frac{275}{50} = 5.5 $ hours; second part = $ \frac{315}{70} \approx 4.5 $ hours. Total time = $ 5.5 + 4.5 = 10 $ hours. Average speed = $ \frac{590}{10} = 59 $ km/h (approximately 58.33 km/h for precision).