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?
please help me :(
If ‘v’ is the speed of the bout and ‘u’ is the speed of the stream then
48/(v+u) + 48/(v-u) = 14
4/(v+u) = 3/(v-u)
48 v =7(v+u) (v-u)
4v-4u = 3v+3u => v=7u
48•7u = 7(7u+u)(7u-u)
u=1 km/h
To find the rate of the stream, we'll first need to find the rate at which the boat moves in still water.
Let's assume that the speed of the boat in still water is 'b' km/h and the speed of the stream is 's' km/h.
Now, let's calculate the time taken by the boat to row 48 km upstream and downstream.
1. Time taken to row 48 km upstream:
The speed of the boat against the stream is (b - s) km/h.
Using the formula time = distance / speed, the time taken to row 48 km upstream is 48 / (b - s) hours.
2. Time taken to row 48 km downstream:
The speed of the boat with the stream is (b + s) km/h.
Similarly, using the formula time = distance / speed, the time taken to row 48 km downstream is 48 / (b + s) hours.
According to the given information, the total time taken for the round trip is 14 hours.
So, we can write the equation:
48 / (b - s) + 48 / (b + s) = 14
Now, let's simplify this equation to solve for 's'.
Multiply both sides of the equation by (b - s)(b + s) to eliminate the denominators:
48(b + s) + 48(b - s) = 14(b - s)(b + s)
Simplify further:
48b + 48s + 48b - 48s = 14(b^2 - s^2)
Combine like terms:
96b = 14b^2 - 14s^2
Rearrange the equation:
14s^2 = 14b^2 - 96b
Divide by 14:
s^2 = b^2 - 6b
Since we need to find the rate of the stream, 's', we still need another equation with 's' and 'b' to solve for 's'.
According to the given information, the boat can row 4 km with the stream in the same time as 3 km against the stream.
Let's set up another equation using the formula time = distance / speed:
4 / (b + s) = 3 / (b - s)
Cross-multiply:
4(b - s) = 3(b + s)
Expand and simplify:
4b - 4s = 3b + 3s
Combine like terms:
b - 4s = 3s
Add 4s to both sides:
b = 7s
Now we have two equations:
s^2 = b^2 - 6b
b = 7s
We can substitute the value of 'b' from the second equation into the first equation:
s^2 = (7s)^2 - 6(7s)
Simplify:
s^2 = 49s^2 - 42s
Subtract 49s^2 and 42s from both sides:
0 = 48s^2 - 42s
Factor out 's':
0 = 6s(8s - 7)
Set each factor equal to zero and solve for 's':
6s = 0 or 8s - 7 = 0
If 6s = 0, then s = 0, but a stream with zero velocity is not possible.
Therefore, we consider the second equation:
8s - 7 = 0
Add 7 to both sides:
8s = 7
Divide by 8:
s = 7/8
So, the rate of the stream is 7/8 km/h.