The speed of the boat on each leg of the trip is the composite of two speeds. One of those is the speed of the current. The other of those is the speed of the boat relative to the water (the speed in still water).

When the boat travels with the current, its speed is the sum of (the speed of the water relative to the shore) and (the speed of the boat relative to the water). When the boat travels against the current, the rate at which it makes distance will be (the speed of the boat relative to the water) minus (the speed of the water relative to the shore).

To solve the problem, we need to know the upstream speed (given as 5 km/hr) and the downstream speed (can be computed from given information).

The total trip distance is (5 km/hr)*(5 hr) = 25 km. The return speed is (25 km)/(10 hr) = 2.5 km/hr.

If we let b represent the speed of the boat relative to the water (the speed in still water) and c represent the speed of the current relative to the shore, then we have two equations for effective travel speed.

B + c = 5 km/hr

b - c = 2.5 km/hr

These two linear equations can be solved in any of the usual ways. I like to add them to eliminate c, and subtract them to eliminate b.

(b + c) + (b - c) = (5 km/hr) + (2.5 km/hr)

2b = 7.5 km/hr (collect terms)

b = 3.75 km/hr (divide by 2)

This answer is all we need to know to answer the question.

When we subtract the second equation from the first, we get

(b + c) - (b - c) = (5 km/hr) - (2.5 km/hr)

2c = 2.5 km/hr (collect terms)

c = 1.25 km/hr (divide by 2)

The speed of the current is 1.25 km/hr. (We don't need to know this to answer the question, but it is easy to figure. You may want to find it in some other problem, so I showed how you can do it.)

When the boat travels with the current, its speed is the sum of (the speed of the water relative to the shore) and (the speed of the boat relative to the water). When the boat travels against the current, the rate at which it makes distance will be (the speed of the boat relative to the water) minus (the speed of the water relative to the shore).

To solve the problem, we need to know the upstream speed (given as 5 km/hr) and the downstream speed (can be computed from given information).

The total trip distance is (5 km/hr)*(5 hr) = 25 km. The return speed is (25 km)/(10 hr) = 2.5 km/hr.

If we let b represent the speed of the boat relative to the water (the speed in still water) and c represent the speed of the current relative to the shore, then we have two equations for effective travel speed.

B + c = 5 km/hr

b - c = 2.5 km/hr

These two linear equations can be solved in any of the usual ways. I like to add them to eliminate c, and subtract them to eliminate b.

(b + c) + (b - c) = (5 km/hr) + (2.5 km/hr)

2b = 7.5 km/hr (collect terms)

b = 3.75 km/hr (divide by 2)

This answer is all we need to know to answer the question.

**The speed of the boat in still water is 3.75 km/hr.**When we subtract the second equation from the first, we get

(b + c) - (b - c) = (5 km/hr) - (2.5 km/hr)

2c = 2.5 km/hr (collect terms)

c = 1.25 km/hr (divide by 2)

The speed of the current is 1.25 km/hr. (We don't need to know this to answer the question, but it is easy to figure. You may want to find it in some other problem, so I showed how you can do it.)