**Problem Analysis**

Let's assume the speed of the boat in still water is 'b' km/h and the speed of the stream is 's' km/h.

When the boat is moving upstream, i.e., against the current, the effective speed is reduced by the speed of the stream. So, the speed of the boat while going upstream is (b - s) km/h.

When the boat is moving downstream, i.e., with the current, the effective speed is increased by the speed of the stream. So, the speed of the boat while going downstream is (b + s) km/h.

We are given two sets of data:

1. Boat goes 30 km upstream and 44 km downstream in 10 hours.
2. Boat goes 40 km upstream and 55 km downstream in 13 hours.

We can use these two sets of data to form two equations and solve them simultaneously to find the values of 'b' and 's'.

**Forming Equations**

Let's form the equations based on the given information:

Equation 1:
30/(b - s) + 44/(b + s) = 10

Equation 2:
40/(b - s) + 55/(b + s) = 13

**Solving the Equations**

To solve the equations, we can use the method of substitution or elimination. Here, we will use the method of substitution.

From Equation 1, we can express 30/(b - s) as (10 - 44/(b + s)) and substitute it into Equation 2:

(10 - 44/(b + s)) + 55/(b + s) = 13

Simplifying the equation:

10 - 44/(b + s) + 55/(b + s) = 13

10 + 11/(b + s) = 13

11/(b + s) = 3

Cross multiplying:

11 = 3(b + s)

11 = 3b + 3s

3b + 3s = 11 ------ Equation 3

From Equation 3, we can express 3s as (11 - 3b) and substitute it into Equation 1:

30/(b - s) + 44/(b + s) = 10

30/(b - s) + 44/((b + s)) = 10

30/(b - s) + 44/((b + (11 - 3b)/3)) = 10

Simplifying the equation:

30/(b - s) + 44/((4b + 11)/3) = 10

30/(b - s) + 132/(4b + 11) = 10

Multiplying the entire equation by (b - s)(4b + 11) to eliminate the denominators:

30(4b + 11) + 132(b - s) = 10(b - s)(4b + 11)

120b + 330 + 132b - 132s = 40b^2 - 110bs + 440b - 110s

Simplifying the equation:

160b + 330 - 132s = 40b^2 - 110bs + 440b - 110s

Rearranging the terms:

40b^