A can rows a certain distance in downstream in 10 hours and returns from the place in 20 hours. If the speed of current is 5 km/h then find the speed of A in still water.
15 km/h
Boat Speed Calculation: Downstream and Upstream
This problem involves calculating the speed of a boat in still water when given information about its journey downstream and upstream, along with the speed of the current.
Understanding the Concepts
When a boat travels downstream (with the current), its effective speed is the sum of its speed in still water and the speed of the current. When it travels upstream (against the current), its effective speed is the difference between its speed in still water and the speed of the current.
- Let $s_a$ be the speed of A in still water (in km/h).
- Let $s_c$ be the speed of the current (given as 5 km/h).
- Speed downstream = $s_a + s_c$
- Speed upstream = $s_a - s_c$
- Distance = Speed $\times$ Time
Step-by-Step Solution
-
Define Variables:
- Speed of A in still water = $s_a$ km/h
- Speed of current = $s_c = 5$ km/h
- Time taken for downstream journey = $t_{down} = 10$ hours
- Time taken for upstream journey = $t_{up} = 20$ hours
-
Express Speeds:
- Speed downstream = $s_a + 5$ km/h
- Speed upstream = $s_a - 5$ km/h
-
Express Distance:
The distance covered in both directions is the same. Let this distance be $d$.
- Distance downstream: $d = (\text{Speed downstream}) \times t_{down} = (s_a + 5) \times 10$
- Distance upstream: $d = (\text{Speed upstream}) \times t_{up} = (s_a - 5) \times 20$
-
Equate Distances and Solve for $s_a$:
Since the distance is the same for both journeys:
$$ (s_a + 5) \times 10 = (s_a - 5) \times 20 $$
Now, we solve this equation for $s_a$:
$$ 10s_a + 50 = 20s_a - 100 $$
Rearrange the terms to group $s_a$ terms on one side and constants on the other:
$$ 50 + 100 = 20s_a - 10s_a $$
$$ 150 = 10s_a $$
Divide by 10 to find $s_a$:
$$ s_a = \frac{150}{10} $$
$$ s_a = 15 $$
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Conclusion:
The speed of A in still water is 15 km/h.
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