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Question

Arun's speed of swimming in still water is 5 km/hr. He swims between two points in a river and returns back to the same starting point. He took 20 minutes more to cover the distance upstream than downstream. If the speed of the stream is 2 km/hr, then the distance between the two points is :

The correct answer is

1.75 km

Understanding River Speed and Swimming Concepts

This problem involves understanding how the speed of a river affects a swimmer's speed when moving upstream (against the current) and downstream (with the current).

Here's a breakdown of the key concepts:

  • Speed in Still Water: This is the swimmer's own speed without any influence from the current. Let's call it \(u\).
  • Speed of the Stream/Current: This is the speed at which the river water flows. Let's call it \(v\).
  • Speed Downstream: When the swimmer goes with the current, the stream adds to their speed. Effective speed = Speed in still water + Speed of stream = \(u + v\).
  • Speed Upstream: When the swimmer goes against the current, the stream opposes their movement. Effective speed = Speed in still water - Speed of stream = \(u - v\). Note that the speed in still water must be greater than the speed of the stream for the swimmer to make progress upstream.
  • Time, Distance, and Speed Relationship: The fundamental formula connecting these is \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

Analyzing the Given Information

We are given the following information about Arun's swim:

  • Speed in still water (\(u\)) = 5 km/hr
  • Speed of the stream (\(v\)) = 2 km/hr
  • Let the distance between the two points be \(d\) km.
  • Time taken upstream is 20 minutes more than time taken downstream.

Calculating Upstream and Downstream Speeds

Using the given speeds, we can calculate Arun's effective speed in each direction:

  • Speed downstream = \(u + v = 5 + 2 = 7\) km/hr
  • Speed upstream = \(u - v = 5 - 2 = 3\) km/hr

Notice that Arun's speed in still water (5 km/hr) is indeed greater than the stream's speed (2 km/hr), so he can swim upstream.

Setting Up the Time Equation

Let \(t_{down}\) be the time taken to swim downstream and \(t_{up}\) be the time taken to swim upstream.

Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):

  • \(t_{down} = \frac{d}{\text{Speed downstream}} = \frac{d}{7}\) hours
  • \(t_{up} = \frac{d}{\text{Speed upstream}} = \frac{d}{3}\) hours

We are told that the time taken upstream is 20 minutes more than downstream. First, convert the time difference to hours:

\(20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hours}\)

So, the relationship between the times is \(t_{up} - t_{down} = \frac{1}{3}\).

Substitute the expressions for \(t_{up}\) and \(t_{down}\):

\( \frac{d}{3} - \frac{d}{7} = \frac{1}{3} \)

Solving for the Distance

Now we need to solve the equation for \(d\). To combine the terms on the left side, find a common denominator, which is the least common multiple of 3 and 7, which is 21.

\( \frac{7d}{21} - \frac{3d}{21} = \frac{1}{3} \)

\( \frac{7d - 3d}{21} = \frac{1}{3} \)

\( \frac{4d}{21} = \frac{1}{3} \)

To isolate \(d\), multiply both sides of the equation by 21:

\( 4d = \frac{1}{3} \times 21 \)

\( 4d = 7 \)

Now, divide both sides by 4:

\( d = \frac{7}{4} \)

Converting the fraction to a decimal:

\( d = 1.75 \)

So, the distance between the two points is 1.75 km.

Verification

Let's quickly verify the answer:

  • Distance \(d = 1.75\) km
  • Speed downstream = 7 km/hr
  • Speed upstream = 3 km/hr
  • Time downstream = \( \frac{1.75}{7} = \frac{7/4}{7} = \frac{1}{4} \) hours
  • Time upstream = \( \frac{1.75}{3} = \frac{7/4}{3} = \frac{7}{12} \) hours
  • Time difference = \( \frac{7}{12} - \frac{1}{4} = \frac{7}{12} - \frac{3}{12} = \frac{4}{12} = \frac{1}{3} \) hours

\( \frac{1}{3} \) hours is equal to \( \frac{1}{3} \times 60 = 20 \) minutes. This matches the condition given in the problem.

Summary of Calculations

Variable Value Unit
Speed in still water (\(u\)) 5 km/hr
Speed of stream (\(v\)) 2 km/hr
Speed downstream \(u+v=7\) km/hr
Speed upstream \(u-v=3\) km/hr
Time difference 20 minutes (\(1/3\) hr) hr
Distance (\(d\)) 1.75 km

Conclusion on Distance Calculation

Based on the calculations using the given speeds and time difference, the distance between the two points is 1.75 km.

Revision Table: Speed, Distance, Time Concepts

Concept Formula Notes
Speed Downstream \(u+v\) Swimmer's speed + Stream's speed
Speed Upstream \(u-v\) Swimmer's speed - Stream's speed
Time \(\frac{\text{Distance}}{\text{Speed}}\) Key relationship
Distance \(\text{Speed} \times \text{Time}\) Rearrangement of the formula
Speed \(\frac{\text{Distance}}{\text{Time}}\) Rearrangement of the formula

Additional Information: Solving River Speed Problems

Problems involving boats or swimmers in rivers are common in quantitative aptitude. They test your understanding of relative speed. Here are a few additional points to keep in mind:

  • Always identify the speed in still water and the speed of the stream clearly.
  • Remember that the stream helps downstream motion and hinders upstream motion.
  • If you are given the upstream and downstream speeds, you can find the speed in still water and the speed of the stream:
    • Speed in still water (\(u\)) = \( \frac{\text{Speed Downstream} + \text{Speed Upstream}}{2} \)
    • Speed of stream (\(v\)) = \( \frac{\text{Speed Downstream} - \text{Speed Upstream}}{2} \)
  • Pay close attention to units (km/hr, m/s, minutes, hours) and convert them to be consistent before performing calculations.

Mastering these basic formulas and concepts will help you solve a wide variety of river-based speed, distance, and time problems.

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Important Questions from Differential Equations

  1. Anubhav spent 14% of his income on electricity bills, 28% on rent and 18% on shopping. If 4/5 of the remaining amount is ₹ 5120, how much did he spend on electricity bills?

  2.  If \( f(x) = 2 \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \), then \( f(x) \) is:

  3. Match List-I with List-II:

    List-I List-II
    (A) Integrating factor of \( xdy - (y + 2x^2)dx = 0 \) (I) \( \frac{1}{x} \)
    (B) Integrating factor of \( (2x^2 - 3y)dx = xdy \) (II) \( x \)
    (C) Integrating factor of \( (2y + 3x^2)dx + xdy = 0 \) (III) \( x^2 \)
    (D) Integrating factor of \( 2xdy + (3x^3 + 2y)dx = 0 \) (IV) \( x^3 \)

    Choose the correct answer from the options given below:

  4. If t = e2x and y = loge(t2), then d2y/dx2  is :

  5. Degree of the differential equation \( \frac{d^2y}{dx^2} + 3 \left( \frac{dy}{dx} \right)^{\frac{1}{2}} = y^2 + e^x \) is:

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