1. Problem 1: Ike and Seb's soft drink sharing. Let Ike originally have $x$ ml and Seb have $120 - x$ ml. Step 1: Dad pours one-third of Ike's drink into Seb's glass. Ike now has $x - \frac{x}{3} = \frac{2x}{3}$ ml. Seb now has $120 - x + \frac{x}{3} = 120 - \frac{2x}{3}$ ml. Step 2: Dad pours one-third of Seb's drink back into Ike's glass. One-third of Seb's current drink is $\frac{1}{3} \times \left(120 - \frac{2x}{3}\right) = 40 - \frac{2x}{9}$ ml. After pouring back: Ike has $\frac{2x}{3} + 40 - \frac{2x}{9} = \frac{6x}{9} + 40 - \frac{2x}{9} = \frac{4x}{9} + 40$ ml. Seb has $120 - \frac{2x}{3} - \left(40 - \frac{2x}{9}\right) = 120 - \frac{6x}{9} - 40 + \frac{2x}{9} = 80 - \frac{4x}{9}$ ml. Step 3: They have equal amounts now, so: $$\frac{4x}{9} + 40 = 80 - \frac{4x}{9}$$ Step 4: Solve for $x$: $$\frac{4x}{9} + \frac{4x}{9} = 80 - 40$$ $$\frac{8x}{9} = 40$$ $$x = 40 \times \frac{9}{8} = 45$$ So Ike originally had 45 ml, Seb had $120 - 45 = 75$ ml. Step 5: Find how much Ike's drink differs from Seb's: $75 - 45 = 30$ ml, so Ike had 30 ml less than Seb. Answer: (B) 少 30ml. 2. Problem 2: Hui's reading problem. Let the total number of pages be $P$. Step 1: First day, Hui reads $\frac{1}{5}P + 12$ pages. Pages left after day 1: $$P - \left(\frac{1}{5}P + 12\right) = \frac{4}{5}P - 12$$ Step 2: Second day, Hui reads $\frac{1}{4}$ of remaining pages plus 15 pages: $$\frac{1}{4} \left(\frac{4}{5}P - 12\right) + 15 = \frac{1}{5}P - 3 + 15 = \frac{1}{5}P + 12$$ Pages left after day 2: $$\left(\frac{4}{5}P - 12\right) - \left(\frac{1}{5}P + 12\right) = \frac{3}{5}P - 24$$ Step 3: Third day, Hui reads $\frac{1}{3}$ of remaining pages plus 18 pages: $$\frac{1}{3} \left(\frac{3}{5}P - 24\right) + 18 = \frac{1}{5}P - 8 + 18 = \frac{1}{5}P + 10$$ Pages left after day 3: $$\left(\frac{3}{5}P - 24\right) - \left(\frac{1}{5}P + 10\right) = \frac{2}{5}P - 34$$ Step 4: Given pages left for next day is 62: $$\frac{2}{5}P - 34 = 62$$ Step 5: Solve for $P$: $$\frac{2}{5}P = 96$$ $$P = 96 \times \frac{5}{2} = 240$$ Answer: (C) 240 pages.