A ship leaves on a long voyage . When it is 180 km from the shore , a seaplane…

Concept: When two travellers start at different times but arrive at the same point together, equate the time each has actually been travelling using time = distance ÷ speed. If one traveller already has a head start, express both remaining distances in terms of one unknown and set the two times equal — this turns a two-body catch-up problem into a single linear equation.

Application:

1. Let the ship's speed be v, so the seaplane's speed is 10v (given ratio 10:1).

2. Let d = the distance from the shore where the seaplane catches up with the ship.

3. The seaplane is sent exactly when the ship is 180 km out, so from that moment the ship still has (d − 180) km left to cover, while the seaplane covers the full d km from the shore.

4. Both distances are covered in the same time interval, so (d − 180)/v = d/(10v).

5. Cancel v and cross-multiply: 10(d − 180) = d.

6. Expand and solve: 10d − 1800 = d → 9d = 1800 → d = 200.

Cross-check: in that same interval the ship travels 200 − 180 = 20 km while the seaplane travels the full 200 km — a ratio of 200:20 = 10:1, exactly the given speed ratio. So the seaplane catches up with the ship 200 km from the shore.