Solving Boat Speed Problems: Upstream and Downstream Travel

This problem involves the concept of relative speed in the context of a boat traveling in a stream. The speed of the boat is affected by the speed of the stream depending on whether the boat is moving upstream (against the current) or downstream (with the current).

Understanding Key Concepts: Boat and Stream Speed

• Speed of the boat in still water ($V_b$): This is the boat's own speed without any influence from a stream.
• Speed of the stream ($V_s$): This is the speed of the water flow.
• Speed downstream ($V_d$): When the boat travels with the stream, the speeds add up. ${V_d = V_b + V_s}$.
• Speed upstream ($V_u$): When the boat travels against the stream, the stream's speed is subtracted from the boat's speed. ${V_u = V_b - V_s}$. For upstream travel to be possible, the boat's speed in still water must be greater than the stream's speed (${V_b > V_s}$).
• Time, Distance, and Speed Relationship: ${ \text{Time} = \frac{\text{Distance}}{\text{Speed}} }$.

Step-by-Step Solution: Finding Stream Speed

We are given the initial boat speed in still water, ${V_b = 20}$ km/h, and a distance of ${D = 24}$ km. We know that the time taken to go 24 km upstream is 30 minutes (which is 0.5 hours) more than the time taken to cover the same distance downstream.

Let ${V_s}$ be the speed of the stream in km/h.

• Downstream speed: ${V_d = V_b + V_s = 20 + V_s}$ km/h.
• Upstream speed: ${V_u = V_b - V_s = 20 - V_s}$ km/h.
• Time taken to go 24 km downstream: ${T_d = \frac{24}{20 + V_s}}$ hours.
• Time taken to go 24 km upstream: ${T_u = \frac{24}{20 - V_s}}$ hours.

According to the problem statement, ${T_u - T_d = 0.5}$ hours.

So, we have the equation:

${ \frac{24}{20 - V_s} - \frac{24}{20 + V_s} = 0.5 }$

To solve for ${V_s}$, we can find a common denominator:

${ 24 \left( \frac{(20 + V_s) - (20 - V_s)}{(20 - V_s)(20 + V_s)} \right) = 0.5 }$

${ 24 \left( \frac{20 + V_s - 20 + V_s}{400 - V_s^2} \right) = 0.5 }$

${ 24 \left( \frac{2V_s}{400 - V_s^2} \right) = 0.5 }$

${ \frac{48V_s}{400 - V_s^2} = 0.5 }$

${ 48V_s = 0.5 (400 - V_s^2) }$

${ 48V_s = 200 - 0.5V_s^2 }$

Multiply by 2 to remove the decimal:

${ 96V_s = 400 - V_s^2 }$

Rearrange into a quadratic equation:

${ V_s^2 + 96V_s - 400 = 0 }$

We can solve this quadratic equation using the quadratic formula: ${V_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$, where ${a=1}$, ${b=96}$, and ${c=-400}$.

${ V_s = \frac{-96 \pm \sqrt{96^2 - 4(1)(-400)}}{2(1)} }$

${ V_s = \frac{-96 \pm \sqrt{9216 + 1600}}{2} }$

${ V_s = \frac{-96 \pm \sqrt{10816}}{2} }$

The square root of 10816 is 104.

${ V_s = \frac{-96 \pm 104}{2} }$

We get two possible values for ${V_s}$: ${V_s = \frac{-96 + 104}{2} = \frac{8}{2} = 4}$ or ${V_s = \frac{-96 - 104}{2} = \frac{-200}{2} = -100}$. Since speed cannot be negative, we take the positive value.

So, the speed of the stream is ${V_s = 4}$ km/h.

Calculating Time with Increased Boat Speed

The problem states that the speed of the boat in still water is increased by 2 km/h. The new boat speed in still water is ${V_b' = 20 + 2 = 22}$ km/h.

The speed of the stream remains the same, ${V_s = 4}$ km/h.

• New downstream speed: ${V_d' = V_b' + V_s = 22 + 4 = 26}$ km/h.
• New upstream speed: ${V_u' = V_b' - V_s = 22 - 4 = 18}$ km/h.

We need to find the total time taken to go 39 km downstream and 30 km upstream with these new speeds.

• Time taken to go 39 km downstream: ${T_d' = \frac{39 \text{ km}}{26 \text{ km/h}} = \frac{3}{2} = 1.5}$ hours.
• Time taken to go 30 km upstream: ${T_u' = \frac{30 \text{ km}}{18 \text{ km/h}} = \frac{5}{3}}$ hours.

The total time taken is the sum of these two times:

${ \text{Total Time} = T_d' + T_u' = 1.5 + \frac{5}{3} }$ hours.

To add these, convert 1.5 to a fraction: ${1.5 = \frac{3}{2}}$.

${ \text{Total Time} = \frac{3}{2} + \frac{5}{3} = \frac{3 \times 3}{2 \times 3} + \frac{5 \times 2}{3 \times 2} = \frac{9}{6} + \frac{10}{6} = \frac{19}{6} }$ hours.

Convert the total time from hours to hours and minutes.

${ \frac{19}{6} \text{ hours} = \frac{18}{6} \text{ hours} + \frac{1}{6} \text{ hours} = 3 \text{ hours} + \frac{1}{6} \text{ hours} }$.

To convert the fraction of an hour to minutes, multiply by 60:

${ \frac{1}{6} \text{ hours} \times 60 \text{ minutes/hour} = 10 \text{ minutes} }$.

So, the total time is 3 hours and 10 minutes.

Summary of Calculations

The total time taken to go 39 km downstream and 30 km upstream with the increased boat speed is 3 hours and 10 minutes.

Revision Table: Key Concepts Review

Additional Information: Related Quantitative Aptitude Concepts

Boat and stream problems are a common type in quantitative aptitude tests. They are a specific application of the concept of relative speed. Other related topics include:

• Relative Speed: How the speed of one object appears from the perspective of another moving object. When objects move in the same direction, relative speed is the difference; when they move in opposite directions, relative speed is the sum.
• Speed, Distance, and Time: The fundamental relationship between these three quantities forms the basis of these problems.
• Solving Quadratic Equations: As seen in this problem, solving for an unknown speed (like the stream speed) can sometimes lead to a quadratic equation.
• Converting Units: Be careful to convert between hours and minutes consistently. 30 minutes = 0.5 hours.

Understanding how the stream affects the boat's effective speed is crucial. Always clearly define the speeds involved (boat in still water, stream, upstream, downstream) before setting up equations based on the given distances and times.