### (a)

**Step 1 — Assume the variable.**
Let the usual speed of the aircraft be $x$ km/h.

**Step 2 — Write the two times.**
$$\text{Scheduled time} = \frac{600}{x} \text{ h}, \qquad \text{Actual time} = \frac{600}{x - 200} \text{ h}$$

**Step 3 — Form the equation ($30$ min $= \frac12$ h).**
$$\frac{600}{x-200} - \frac{600}{x} = \frac12$$

**Step 4 — Simplify.**
$$600\left[\frac{x - (x-200)}{x(x-200)}\right] = \frac12 \ \Rightarrow\ \frac{600 \times 200}{x(x-200)} = \frac12$$
$$120000 \times 2 = x^2 - 200x$$

**Step 5 — Form the quadratic equation.**
$$x^2 - 200x - 240000 = 0$$

**Step 6 — Factorise.**
$$x^2 - 600x + 400x - 240000 = 0$$
$$x(x - 600) + 400(x - 600) = 0 \ \Rightarrow\ (x - 600)(x + 400) = 0$$

**Step 7 — Solve.**
$$x = 600 \quad\text{or}\quad x = -400 \ (\text{rejected, speed cannot be negative})$$

**Step 8 — Find the scheduled duration.**
$$\text{Scheduled duration} = \frac{600}{600} = 1 \text{ hour}$$

$$\boxed{\text{Scheduled duration of the flight} = 1 \text{ hour}}$$

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### (b)

**Step 1 — Assume the variables.**
Let the larger pipe fill the pool in $x$ hours.
Then the smaller pipe takes $(x + 10)$ hours.

**Step 2 — Write the work done in 1 hour.**
$$\text{Larger pipe} = \frac{1}{x}, \qquad \text{Smaller pipe} = \frac{1}{x+10}$$

**Step 3 — Form the equation.**
$$\frac{4}{x} + \frac{9}{x+10} = \frac12$$

**Step 4 — Take LCM and cross-multiply.**
$$\frac{4(x+10) + 9x}{x(x+10)} = \frac12$$
$$2\left[4x + 40 + 9x\right] = x^2 + 10x$$
$$2(13x + 40) = x^2 + 10x$$

**Step 5 — Form the quadratic equation.**
$$26x + 80 = x^2 + 10x \ \Rightarrow\ x^2 - 16x - 80 = 0$$

**Step 6 — Factorise.**
$$x^2 - 20x + 4x - 80 = 0$$
$$x(x - 20) + 4(x - 20) = 0 \ \Rightarrow\ (x - 20)(x + 4) = 0$$

**Step 7 — Solve.**
$$x = 20 \quad\text{or}\quad x = -4 \ (\text{rejected})$$

**Step 8 — Final answer.**
$$\boxed{\text{Larger pipe} = 20 \text{ hours}, \qquad \text{Smaller pipe} = 30 \text{ hours}}$$