A BOATMAN ROWS TO A PLACE 48KM DISTANT AND BACK IN 14 HOURS; HE FINDS HE CAN ROW 4KM WITH THE STREAM IN THE SAME TIME AS 3KM AGAINST THE STREAM. FIND THE RATE OF THE STREAM?

If his rowing speed is r, and the current speed is c, then

(r+c)/4 = (r-c)/3
48/(r+c) + 48/(r-c) = 14

c = 1

Suppose he move 4 km downstream in x hours. Then,

Speed downstream = 4 km/hr.
x
Speed upstream = 3 km/hr.
x
48 + 48 = 14 or x = 1 .
(4/x) (3/x) 2
So, Speed downstream = 8 km/hr, Speed upstream = 6 km/hr.

Rate of the stream = 1 (8 - 6) km/hr = 1 km/hr.
2

Let's represent the speed of the boat as B and the speed of the stream as S.

When the boat is rowing with the stream, its effective speed is increased by the speed of the stream. So, the effective speed is B + S.

Similarly, when the boat is rowing against the stream, its effective speed is decreased by the speed of the stream. So, the effective speed is B - S.

According to the problem, the boat can row 4 km with the stream in the same time it takes to row 3 km against the stream. This can be represented as:

4/(B + S) = 3/(B - S)

To simplify this equation, we can cross-multiply:

4(B - S) = 3(B + S)

Expanding both sides of the equation:

4B - 4S = 3B + 3S

Simplifying:

B = 7S

Now, we know that the boatman rows 48 km in a total of 14 hours. We can use this information to find another equation.

The time taken to row 48 km with the stream is 48/(B + S), and the time taken to row 48 km against the stream is 48/(B - S).

According to the problem, the total time is 14 hours:

48/(B + S) + 48/(B - S) = 14

Substituting the value of B from the previous equation:

48/(7S + S) + 48/(7S - S) = 14

Simplifying:

48/8S + 48/6S = 14

Multiplying both sides by the common denominator 24S:

3(6) + 4(8) = 14(24S)

18 + 32 = 336S

50 = 336S

Dividing both sides by 336:

S = 50/336

Simplifying:

S ≈ 0.149

Therefore, the rate of the stream is approximately 0.149 km/h.

To solve this problem, we can use the concept of relative speed. Let's call the speed of the boat in still water as 'b' km/h and the speed of the stream as 's' km/h.

When the boat is rowing downstream, the speed of the boat is increased by the speed of the stream, so its effective speed becomes (b + s) km/h. In this case, the boat can row 4 km in the same time it takes to row 3 km against the stream.

We know that the time taken to row a given distance is equal to the distance divided by the speed. So, the time taken to row 4 km downstream is 4/(b + s) hours, and the time taken to row 3 km upstream is 3/(b - s) hours.

According to the given information, the boatman rows to a place 48 km distant and back in 14 hours. This can be represented by the equation:

4/(b + s) + 3/(b - s) = 14

Now we need to solve the equation to find the values of 'b' and 's' in order to determine the rate of the stream.

To simplify the equation, we can multiply through by (b + s)(b - s) to eliminate the denominators:

4(b - s) + 3(b + s) = 14(b + s)(b - s)

Expanding and simplifying the equation gives us:

4b - 4s + 3b + 3s = 14(b^2 - s^2)

Combining like terms:

7b - s = 14(b^2 - s^2)

We also know that the total distance rowed is 48 km. We can express this as:

2(4/(b + s)) + 2(3/(b - s)) = 48

Dividing through by 2 to simplify:

4/(b + s) + 3/(b - s) = 24

Now we have two equations:

7b - s = 14(b^2 - s^2)
4/(b + s) + 3/(b - s) = 24

We can further simplify the second equation by cross multiplying:

4(b - s) + 3(b + s) = 24(b - s)(b + s)

Expanding and simplifying:

4b - 4s + 3b + 3s = 24(b^2 - s^2)

Combining like terms:

7b - s = 24(b^2 - s^2)

Now we have two equations:

7b - s = 14(b^2 - s^2)
7b - s = 24(b^2 - s^2)

Since the left sides of both equations are the same, we can set the right sides equal to each other:

14(b^2 - s^2) = 24(b^2 - s^2)

Dividing through by (b^2 - s^2) (assuming b ≠ s) gives:

14 = 24

However, this is contradictory and impossible. Therefore, there must be an error in the problem statement or calculations. Please double-check the given information and try again.