A motor boat covers 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find its speed in still water and the speed of the stream.
Boat in still water = 10 km/h; Stream = 4 km/h.
Step 1: Understand the problem.
- When a boat goes upstream (against the flow of water), its speed becomes slower: \(u = b - s\).
- When a boat goes downstream (with the flow of water), its speed becomes faster: \(d = b + s\).
Here, \(b\) is the speed of the boat in still water, and \(s\) is the speed of the stream.
Step 2: Use the first condition (7 hours).
The boat goes 30 km upstream and 28 km downstream in 7 hours. Time = Distance ÷ Speed, so:
\(\dfrac{30}{u} + \dfrac{28}{d} = 7\) … (Equation 1)
Step 3: Use the second condition (5 hours).
The boat goes 21 km upstream and 21 km downstream in 5 hours. So:
\(\dfrac{21}{u} + \dfrac{21}{d} = 5\)
Simplify: \(21 \left( \dfrac{1}{u} + \dfrac{1}{d} \right) = 5\)
So: \(\dfrac{1}{u} + \dfrac{1}{d} = \dfrac{5}{21}\) … (Equation 2)
Step 4: Make substitution.
Let \(A = \dfrac{1}{u}\) and \(B = \dfrac{1}{d}\).
From Equation 1: \(30A + 28B = 7\).
From Equation 2: \(A + B = \dfrac{5}{21}\).
Step 5: Solve the equations.
Multiply Equation 2 by 28: \(28A + 28B = \dfrac{140}{21} = \dfrac{20}{3}\).
Now subtract this from Equation 1: \((30A + 28B) - (28A + 28B) = 7 - \dfrac{20}{3}\).
\(2A = \dfrac{21 - 20}{3} = \dfrac{1}{3}\).
So, \(A = \dfrac{1}{6}\). That means \(u = 6\).
From Equation 2: \(\dfrac{1}{6} + B = \dfrac{5}{21}\).
So, \(B = \dfrac{5}{21} - \dfrac{1}{6} = \dfrac{1}{14}\). That means \(d = 14\).
Step 6: Find boat and stream speed.
\(b = \dfrac{u + d}{2} = \dfrac{6 + 14}{2} = 10\).
\(s = \dfrac{d - u}{2} = \dfrac{14 - 6}{2} = 4\).
Final Answer: Boat in still water = 10 km/h, Stream = 4 km/h.