Question:
A boy can swim with a speed of \(18\) kmph in still water. If the speed of the water in a river is \(10.8\) kmph, then the average speed of the boy in travelling downstream for a distance of \(120\) m and upstream for a distance of \(90\) m over the river is
A boy can swim with a speed of \(18\) kmph in still water. If the speed of the water in a river is \(10.8\) kmph, then the average speed of the boy in travelling downstream for a distance of \(120\) m and upstream for a distance of \(90\) m over the river is
Show Hint
Average speed is always
\[
\boxed{\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}.}
\]
Compute the time for each part of the journey separately before finding the average speed.
Updated On: Jul 18, 2026
- \(3\ \mathrm{ms^{-1}}\)
- \(3.5\ \mathrm{ms^{-1}}\)
- \(4\ \mathrm{ms^{-1}}\)
- \(4.5\ \mathrm{ms^{-1}}\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the downstream and upstream speeds.
Speed of the boy in still water
\[
18\text{ kmph}=5\text{ ms}^{-1}.
\]
Speed of the river
\[
10.8\text{ kmph}=3\text{ ms}^{-1}.
\]
Hence,
Downstream speed
\[
5+3=8\text{ ms}^{-1}.
\]
Upstream speed
\[
5-3=2\text{ ms}^{-1}.
\]
Step 2: Calculate the total time. Time taken downstream: \[ t_1=\frac{120}{8}=15\text{ s}. \] Time taken upstream: \[ t_2=\frac{90}{2}=45\text{ s}. \] Thus, \[ \text{Total time} = 15+45 = 60\text{ s}. \]
Step 3: Find the average speed. Total distance travelled: \[ 120+90=210\text{ m}. \] Therefore, \[ \text{Average speed} = \frac{210}{60} = 3.5\text{ ms}^{-1}. \] Hence, \[ \boxed{3.5\text{ ms}^{-1}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.
Step 2: Calculate the total time. Time taken downstream: \[ t_1=\frac{120}{8}=15\text{ s}. \] Time taken upstream: \[ t_2=\frac{90}{2}=45\text{ s}. \] Thus, \[ \text{Total time} = 15+45 = 60\text{ s}. \]
Step 3: Find the average speed. Total distance travelled: \[ 120+90=210\text{ m}. \] Therefore, \[ \text{Average speed} = \frac{210}{60} = 3.5\text{ ms}^{-1}. \] Hence, \[ \boxed{3.5\text{ ms}^{-1}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.
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