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A man can row \(5 \mathrm{~km} / \mathrm{h}\) in still water. If the rate of current is \(1 \mathrm{~km} / \mathrm{h}\), it takes \(\frac{5}{4}\) hours to row to a place and back How far is the place? (b) \(2.5 \mathrm{~km}\) (a) \(2 \mathrm{~km}\) (d) \(4 \mathrm{~km}\) (c) \(3 \mathrm{~km}\)

Short Answer

Expert verified
Answer: 3 km

Step by step solution

01

Determine the effective speeds downstream and upstream

To find the effective speed of the man during downstream and upstream, we'll add or subtract the speed of the current, respectively, from his speed in still water. Downstream, the current helps the man, so the effective speed is the sum of the man's speed and the current's speed, whereas upstream, the current hinders the man, so the effective speed is the difference between the man's speed and the current's speed. Effective speed downstream = speed in still water + rate of current = 5 + 1 = 6 km/h Effective speed upstream = speed in still water - rate of current = 5 - 1 = 4 km/h
02

Calculate the distances for downstream and upstream

We know the time taken for the entire journey (to the place and back) is 5/4 hours. We also know that time = distance / speed. Since the total time is the sum of the downstream time and the upstream time, we can create an equation: \(\frac{5}{4} = \frac{D}{6} + \frac{D}{4}\) where D is the distance to the place.
03

Solve for the distance

Now, we can solve for D: Multiply both sides of the equation by 12 to get rid of the denominators: \(12 \times \frac{5}{4} = 12 \times \left( \frac{D}{6} + \frac{D}{4} \right)\) \(15 = 2D + 3D\) Combine the terms on the right side: \(15 = 5D\) Now, divide both sides by 5: \(D = 3\) The distance to the place is 3 km, which corresponds to option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effective Speed Calculation
Understanding 'effective speed' is crucial for solving problems involving different rates in opposite directions or with varying conditions. Imagine you are walking with the wind at your back; naturally, you move faster than if you walk into the wind. This concept works similarly with 'effective speed' in navigation.

To calculate effective speed, especially when dealing with downstream and upstream scenarios, you essentially account for the assistance or resistance on the object moving. If you are moving downstream, the current aids your motion, and thus, your effective speed becomes the sum of your speed in still water and the current's speed. Conversely, when going upstream, the current opposes you, so your effective speed is your speed in still water minus the current's speed.
Downstream and Upstream
The terms 'downstream' and 'upstream' are derived from river navigation and have become staple concepts in problems dealing with boats, swimmers, or any object moving in a flowing medium. 'Downstream' refers to moving along with the direction of the current, making the motion easier and increasing the effective speed. On the other hand, 'upstream' denotes movement against the current, resulting in a reduced effective speed.

When solving problems, you must adjust the speed of the moving object by either adding the current's speed (downstream) or subtracting it (upstream). This simple adjustment helps in finding the actual speed with which the distance will be covered in each direction.
Equation Solving
Equation solving is the process of finding the value(s) of variable(s) that satisfy a given mathematical statement. For speed, distance, and time problems, you often set up an equation that relates these three variables. It is vital to keep the equation balanced, meaning whatever operation you do to one side, you must do to the other.

In our example, we equated the sum of the times taken to travel downstream and upstream to the total time taken for the round trip. By appropriately converting the time into fractions and finding a common denominator, we could simplify the equation and solve for the unknown distance, 'D', effectively.
Quantitative Aptitude
Quantitative aptitude involves the ability to understand and solve mathematical problems. It is a measure of one's numerical ability and accuracy in mathematical calculations. The fundamental concepts such as ratio, proportion, and the ability to deal with abstract concepts such as speed, distance, and time are essential components of quantitative aptitude.

Developing quantitative aptitude is about practicing problem-solving strategies and understanding the basic principles that underpin each type of question, like the speed, distance, and time problems we're discussing. Taking a systematic approach to problems by breaking them down into simpler parts, as we did in determining speed, calculating distance, and then solving the equation step by step, helps in enhancing one's quantitative skills.

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Most popular questions from this chapter

Pankaj walked at \(5 \mathrm{~km} / \mathrm{h}\) for certain part of the journey and then he took an auto for the remaining part of the jourme: travelling at \(25 \mathrm{~km} / \mathrm{h}\). If he took 10 hours for the entire journey. What part of journey did he travelled by auto if the average speced of the entire joumey be \(17 \mathrm{~km} / \mathrm{h}\) : (a) \(750 \mathrm{~km}\) (b) \(100 \mathrm{~km}\) (c) \(150 \mathrm{~km}\) (d) \(200 \mathrm{~km}\)

Two horses start trotting towards each other, one from \(A\) to \(B\) and another from \(B\) to \(A\). They cross each other after one hour and the first horse reaches \(B, 5 / 6\) hour before the second horse reaches \(A\). If the distance between \(A\) and \(B\) is \(50 \mathrm{~km}\). What is the speed of the slower horse? (a) \(30 \mathrm{~km} / \mathrm{h}\) (b) \(15 \mathrm{~km} / \mathrm{h}\) (c) \(25 \mathrm{~km} / \mathrm{h}\) (d) \(20 \mathrm{~km} / \mathrm{h}\)

The difference between downstream speed and upstream speed is \(3 \mathrm{~km} / \mathrm{h}\) and the total time taken during upstreast and downstream is 3 hours. What is the downstream speed, if the downstream and upstream distance are \(3 \mathrm{~km}\) each? (a) \(2.5 \mathrm{~km} / \mathrm{h}\) (b) \(4.33 \mathrm{~km} / \mathrm{h}\) (c) \(4 \mathrm{~km} / \mathrm{h}\) (d) \(3.3 \mathrm{~km} / \mathrm{h}\)

Abhinav leaves Mumbai at 6 am and reaches Bangalore at \(10 \mathrm{am}\) Praveen leaves Bangalore at 8 am and reaches Mumbai at \(11: 30\) am. At what time do they cross each other? (a) \(10 \mathrm{am}\) (b) \(8: 32 \mathrm{am}\) (c) \(8: 56 \mathrm{am}\) (d) \(9: 20 \mathrm{am}\)

If a train \(225 \mathrm{~m}\) long passes a telegraphic pole in 9 seconds, then the time taken (in seconds) by it to cross a tunnel \(450 \mathrm{~m}\) long is: (a) \(8 \mathrm{~s}\) (b) \(10 \mathrm{~s}\) (c) \(27 \mathrm{~s}\) (d) none of these

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