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 :
1.75 km
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:
We are given the following information about Arun's swim:
Using the given speeds, we can calculate Arun's effective speed in each direction:
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.
Let \(t_{down}\) be the time taken to swim downstream and \(t_{up}\) be the time taken to swim upstream.
Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \):
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 \(t_{down}\):
\( \frac{d}{3} - \frac{d}{7} = \frac{1}{3} \)
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.
Let's quickly verify the answer:
\( \frac{1}{3} \) hours is equal to \( \frac{1}{3} \times 60 = 20 \) minutes. This matches the condition given in the problem.
| 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 |
Based on the calculations using the given speeds and time difference, the distance between the two points is 1.75 km.
| 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 |
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:
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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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) \( x^3 \) |
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