Question

David can row a boat in still water at the rate of 5 km/hr. He rowed in a river downstream to meet his friend. After returning back, he observed that the duration of the upstream journey was three times that of the downstream journey. The speed of the stream was:

• A. 2 km/hr
• B. 2.5 km/hr
• C. 3 km/hr
• D. 3.5 km/hr

Answer 1

The Correct Option is B

To solve the problem of determining the speed of the stream, we begin by denoting the speed of the stream as \( x \) km/hr.

Given:

• Speed of David in still water = 5 km/hr.
• Let the distance of the journey downstream (and also upstream) be \( d \) km.
• Time taken downstream = \( \frac{d}{5+x} \) hours.
• Time taken upstream = \( \frac{d}{5-x} \) hours.
• Time for upstream journey = 3 times time for downstream journey.

Therefore, we can set up the equation:

\[\frac{d}{5-x} = 3 \times \frac{d}{5+x}\]

By canceling \( d \) from both sides, we have:

\[\frac{1}{5-x} = \frac{3}{5+x}\]

Cross-multiplying gives:

\[5+x = 3(5-x)\]

Expanding and simplifying:

\[5+x = 15-3x\]

\[4x = 10\]

\[x = 2.5\]

Thus, the speed of the stream is 2.5 km/hr.

Answer 2

Let the speed of the stream be \( x \). The speed of the boat downstream is \( 5 + x \), and the speed upstream is \( 5 - x \).

The time for the downstream journey is:

\[ t_d = \frac{D}{5 + x}. \]

The time for the upstream journey is:

\[ t_u = \frac{D}{5 - x}. \]

It is given that the time for the upstream journey is three times that of the downstream journey:

\[ t_u = 3t_d \implies \frac{D}{5 - x} = 3 \cdot \frac{D}{5 + x}. \]

Cancel \( D \) (since \( D > 0 \)):

\[ \frac{1}{5 - x} = \frac{3}{5 + x}. \]

Cross-multiply:

\[ 5 + x = 3(5 - x) \implies 5 + x = 15 - 3x \implies 4x = 10 \implies x = 2.5. \]

Thus, the speed of the stream is 2.5 km/hr.