Solving the Motorboat and River Flow Problem

This problem involves a motorboat traveling downstream and upstream in a river. When the boat travels downstream, its speed is increased by the speed of the river flow. When it travels upstream, its speed is decreased by the speed of the river flow.

Understanding Downstream and Upstream Speed

• Downstream Speed: Speed of boat in still water + Speed of river flow.
• Upstream Speed: Speed of boat in still water - Speed of river flow.

Given Information

• Speed of motorboat in still water ($v_m$) = 20 km/h
• Distance traveled downstream = 150 km
• Distance traveled upstream = 150 km (returns to the starting point)
• Total time for the round trip = 16 hours

Finding the Speed of the River Flow

Let the speed of the river flow be $v_r$ km/h.

The downstream speed will be $(v_m + v_r) = (20 + v_r)$ km/h.

The upstream speed will be $(v_m - v_r) = (20 - v_r)$ km/h.

The formula relating time, distance, and speed is:

$$ \text{Time} = \frac{\text{Distance}}{\text{Speed}} $$

Time taken to travel downstream ($t_{downstream}$):

$$ t_{downstream} = \frac{150}{20 + v_r} \text{ hours} $$

Time taken to travel upstream ($t_{upstream}$):

$$ t_{upstream} = \frac{150}{20 - v_r} \text{ hours} $$

The total time for the round trip is given as 16 hours. So, the sum of the downstream time and upstream time must equal 16 hours.

$$ t_{downstream} + t_{upstream} = 16 $$

Substitute the expressions for $t_{downstream}$ and $t_{upstream}$:

$$ \frac{150}{20 + v_r} + \frac{150}{20 - v_r} = 16 $$

Solving the Equation for $v_r$

We need to solve this equation for $v_r$.

Factor out 150 from the left side:

$$ 150 \left( \frac{1}{20 + v_r} + \frac{1}{20 - v_r} \right) = 16 $$

Combine the fractions inside the parenthesis:

$$ 150 \left( \frac{(20 - v_r) + (20 + v_r)}{(20 + v_r)(20 - v_r)} \right) = 16 $$

Simplify the numerator and the denominator (using the difference of squares formula, $(a+b)(a-b) = a^2 - b^2$):

$$ 150 \left( \frac{20 - v_r + 20 + v_r}{20^2 - v_r^2} \right) = 16 $$

$$ 150 \left( \frac{40}{400 - v_r^2} \right) = 16 $$

Multiply 150 by 40:

$$ \frac{6000}{400 - v_r^2} = 16 $$

Rearrange the equation to solve for $400 - v_r^2$:

$$ 400 - v_r^2 = \frac{6000}{16} $$

Calculate the division:

$$ 400 - v_r^2 = 375 $$

Rearrange to solve for $v_r^2$:

$$ v_r^2 = 400 - 375 $$

$$ v_r^2 = 25 $$

Take the square root of both sides to find $v_r$. Since speed must be a positive value:

$$ v_r = \sqrt{25} $$

$$ v_r = 5 $$

The speed of the river flow is 5 km/h.

Verification

Let's check if this value works:

• Downstream speed = $20 + 5 = 25$ km/h. Time downstream = $150 / 25 = 6$ hours.
• Upstream speed = $20 - 5 = 15$ km/h. Time upstream = $150 / 15 = 10$ hours.
• Total time = $6 + 10 = 16$ hours. This matches the given total time.

Therefore, the speed of the flow of the river is 5 km/h.

Revision Table: Motorboat and Stream Concepts

Additional Information: Boat and Stream Problems

Boat and stream problems are common in quantitative aptitude. They are based on the relative speeds of the boat and the stream. It's important to remember that the stream's speed helps the boat when going downstream and opposes it when going upstream.

• The boat's speed in still water is its actual capability without any external influence from the water current.
• The stream's speed is the speed of the water current itself.
• For upstream travel to be possible, the speed of the boat in still water must be greater than the speed of the stream ($v_m > v_r$). If $v_m \le v_r$, the boat cannot move against the current.
• Problems often involve finding one of the speeds or the distance or time given the others, using the formulas derived from the concepts of downstream and upstream speeds.