Solving Boat and River Speed Ratio Problems

This problem involves calculating the ratio of the speed of a boat in still water to the speed of the river stream, given information about the time taken for a round trip under different conditions. Let's break down the problem step by step.

Defining Variables

• Let \(V_b\) be the speed of the boat in still water.
• Let \(V_s\) be the speed of the river stream.
• Let \(D\) be the distance between points A and B.

When the boat travels downstream (with the current), its effective speed is the sum of its speed in still water and the speed of the stream: \(V_{downstream} = V_b + V_s\).

When the boat travels upstream (against the current), its effective speed is the difference between its speed in still water and the speed of the stream: \(V_{upstream} = V_b - V_s\). For the boat to be able to travel upstream, we must have \(V_b > V_s\).

Case 1: Original Round Trip Time

The time taken for the round trip journey is the sum of the time taken to travel downstream from A to B and the time taken to travel upstream from B to A (or vice versa). The distance for both legs of the journey is \(D\).

Time Downstream \(t_{down} = \frac{D}{V_b + V_s}\)

Time Upstream \(t_{up} = \frac{D}{V_b - V_s}\)

The total time for the round trip is given as \(T\).

$$T = t_{down} + t_{up} = \frac{D}{V_b + V_s} + \frac{D}{V_b - V_s}$$

Combining the terms, we get:

$$T = D \left( \frac{1}{V_b + V_s} + \frac{1}{V_b - V_s} \right)$$

$$T = D \left( \frac{(V_b - V_s) + (V_b + V_s)}{(V_b + V_s)(V_b - V_s)} \right)$$

$$T = D \left( \frac{2V_b}{V_b^2 - V_s^2} \right) \quad \text{(Equation 1)}$$

Case 2: Modified Round Trip Time

In this case, the speed of the boat in still water becomes 2 times the original speed, i.e., the new boat speed is \(2V_b\). The speed of the river stream remains \(V_s\).

New Downstream Speed \(V'_{downstream} = 2V_b + V_s\)

New Upstream Speed \(V'_{upstream} = 2V_b - V_s\)

The new time taken for the same round trip journey is given as \(T' = \frac{80}{161} T\).

The expression for \(T'\) is:

$$T' = \frac{D}{2V_b + V_s} + \frac{D}{2V_b - V_s}$$

Combining the terms, we get:

$$T' = D \left( \frac{1}{2V_b + V_s} + \frac{1}{2V_b - V_s} \right)$$

$$T' = D \left( \frac{(2V_b - V_s) + (2V_b + V_s)}{(2V_b + V_s)(2V_b - V_s)} \right)$$

$$T' = D \left( \frac{4V_b}{(2V_b)^2 - V_s^2} \right)$$

$$T' = D \left( \frac{4V_b}{4V_b^2 - V_s^2} \right) \quad \text{(Equation 2)}$$

Relating the Two Cases and Solving for the Ratio

We are given that \(T' = \frac{80}{161} T\). Substitute the expressions for \(T'\) and \(T\) from Equation 2 and Equation 1:

$$D \left( \frac{4V_b}{4V_b^2 - V_s^2} \right) = \frac{80}{161} \left( D \left( \frac{2V_b}{V_b^2 - V_s^2} \right) \right)$$

We can cancel out \(D\) from both sides (assuming \(D > 0\)):

$$\frac{4V_b}{4V_b^2 - V_s^2} = \frac{80}{161} \frac{2V_b}{V_b^2 - V_s^2}$$

We can also cancel out \(V_b\) from both sides (assuming \(V_b > 0\)):

$$\frac{4}{4V_b^2 - V_s^2} = \frac{80}{161} \frac{2}{V_b^2 - V_s^2}$$

$$\frac{4}{4V_b^2 - V_s^2} = \frac{160}{161(V_b^2 - V_s^2)}$$

Cross-multiply:

$$4 \times 161(V_b^2 - V_s^2) = 160(4V_b^2 - V_s^2)$$

$$644(V_b^2 - V_s^2) = 160(4V_b^2 - V_s^2)$$

Divide both sides by 4:

$$161(V_b^2 - V_s^2) = 40(4V_b^2 - V_s^2)$$

Expand both sides:

$$161V_b^2 - 161V_s^2 = 160V_b^2 - 40V_s^2$$

Rearrange the terms to group \(V_b^2\) and \(V_s^2\):

$$161V_b^2 - 160V_b^2 = 161V_s^2 - 40V_s^2$$

$$V_b^2 = 121V_s^2$$

Take the square root of both sides (since speeds must be positive):

$$\sqrt{V_b^2} = \sqrt{121V_s^2}$$

$$V_b = 11V_s$$

The problem asks for the ratio of the speed in still water to the speed of the river, which is \(\frac{V_b}{V_s}\).

$$\frac{V_b}{V_s} = \frac{11V_s}{V_s} = 11$$

The ratio is 11 : 1.

Let's verify the ratio against the options provided.

Our calculated ratio is 11 : 1, which matches Option 4.

Revision Table: Boat and Stream Concepts

Additional Information: Solving Boat and Stream Problems

Boat and stream problems are a common type of question in quantitative aptitude tests. They rely on understanding how the speed of the water affects the effective speed of the boat. Key principles include:

• The stream speed adds to the boat's speed when going downstream.
• The stream speed subtracts from the boat's speed when going upstream.
• If the time for the round trip is given, you sum the time taken for each leg (downstream and upstream).
• If the speeds are given and you need to find time or distance, use the basic formula: Time = Distance / Speed.
• If the times for downstream and upstream journeys are given over the same distance, you can often find the ratio of speeds or individual speeds.
• When speeds or times change, set up equations for the different scenarios and solve them simultaneously or by substitution, as demonstrated in this solution.

Mastering the formulas for downstream and upstream speeds and applying the time-distance-speed relationship correctly are crucial for solving boat and stream problems effectively.