RS Aggarwal Summary: Boats and Streams
Introduction
Boats and Streams is a crucial topic that frequently appears in the quantitative aptitude section of competitive examinations. This chapter tests your understanding of relative speed, time-distance relationships, and problem-solving abilities under time constraints. Questions from this topic are designed to assess your analytical skills and ability to work with speed variations in different directions. Mastering this chapter will not only help you score easy marks but also strengthen your foundation in speed, time, and distance concepts that appear across various question types in competitive exams.
Important Facts and Formulae
1. Basic Terminology
- Downstream: The direction along the stream (with the current)
- Upstream: The direction against the stream (against the current)
2. Speed Calculations with Current
If the speed of a boat in still water is \(u\) km/hr and the speed of the stream is \(v\) km/hr, then:
\[\text{Speed downstream} = (u + v) \text{ km/hr}\]
\[\text{Speed upstream} = (u - v) \text{ km/hr}\]
Example 1:
The speed of a boat when travelling downstream is 32 km/hr, whereas when travelling upstream it is 28 km/hr. What is the speed of the boat in still water and the speed of the stream?
Solution:
\[\text{Speed of boat in still water} = \frac{1}{2}(32 + 28) = 30 \text{ km/hr}\]
\[\text{Speed of stream} = \frac{1}{2}(32 - 28) = 2 \text{ km/hr}\]
Example 2:
A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the river current in km/hr.
Solution:
\[\text{Rate downstream} = 15 \div \frac{15}{4} = 15 \times \frac{4}{15} = 4 \text{ km/hr}\]
\[\text{Rate upstream} = 5 \div \frac{5}{2} = 5 \times \frac{2}{5} = 2 \text{ km/hr}\]
\[\text{Speed of current} = \frac{1}{2}(4 - 2) = 1 \text{ km/hr}\]
3. Finding Speed from Upstream and Downstream Speeds
If the speed downstream is \(a\) km/hr and the speed upstream is \(b\) km/hr, then:
\[\text{Speed in still water} = \frac{1}{2}(a + b) \text{ km/hr}\]
\[\text{Rate of stream} = \frac{1}{2}(a - b) \text{ km/hr}\]
Example 3:
The speed of a motor boat is that of the current of water as 36 : 5. The boat goes along with the current in 5 hours 10 minutes. How much time will it take to come back?
Solution:
Let the speed of the motor boat and that of the current be \(36x\) km/hr and \(5x\) km/hr respectively.
\[\text{Speed downstream} = (36x + 5x) = 41x \text{ km/hr}\]
\[\text{Speed upstream} = (36x - 5x) = 31x \text{ km/hr}\]
Let the distance be \(d\) km. Then:
\[\frac{d}{41x} = 5\frac{10}{60} = \frac{31}{6} \text{ hours}\]
\[d = \frac{41x \times 31}{6}\]
\[\text{Time taken while coming back} = \frac{d}{31x} = \frac{41x \times 31}{6 \times 31x} = \frac{41}{6} \text{ hrs} = 6 \text{ hrs } 50 \text{ min}\]
Example 4:
A man can row 6 km/hr in still water. It takes him twice as long to row up as to row down the river. Find the rate of stream.
Solution:
Let man's rate upstream be \(x\) km/hr. Then, his rate downstream = \(2x\) km/hr.
\[\text{Rate in still water} = \frac{1}{2}(x + 2x) = \frac{3x}{2} \text{ km/hr}\]
So, \(\frac{3x}{2} = 6\) or \(x = 4\)
Rate upstream = 4 km/hr, Rate downstream = 8 km/hr
\[\text{Rate of stream} = \frac{1}{2}(8 - 4) = 2 \text{ km/hr}\]
4. Swimming Across a Stream
Suppose a man can swim in still water at the rate of \(u\) km/hr, the speed of current/stream is \(v\) km/hr, and the man wishes to cross the stream (of width \(x\) metres) straight along its width, then:
\[\text{Time taken to cross the river} = \text{Time taken to swim } x \text{ metres at } u \text{ km/hr}\]
Note: The stream sways the man such that both the distance and the effective velocity increase, and the time taken to cross the river remains unaffected.
5. Rate of Current when Crossing Times Differ
A man can swim directly across a stream of width \(x\) km in \(t\) hours when there is no current and in \(t'\) hours when there is a current. Then:
\[\text{Rate of current} = x\sqrt{\frac{1}{t^2} - \frac{1}{t'^2}} \text{ km/hr}\]
Concept: Average Speed in Round Trips
Example 5:
There is a road beside a river. Two friends started from a place A, moved to a temple situated at another place B and then returned to A again. One of them moves on a cycle at a speed of 12 km/hr, while the other sails on a boat at a speed of 10 km/hr. If the river flows at the speed of 4 km/hr, which of the two friends will return to place A first?
Solution:
Clearly, the cyclist moves both ways at a speed of 12 km/hr
\[\text{Average speed of the cyclist} = 12 \text{ km/hr}\]
The boat sailor moves downstream @ \((10 + 4) = 14\) km/hr and upstream @ \((10 - 4) = 6\) km/hr
\[\text{Average speed of the boat sailor} = \frac{2 \times 14 \times 6}{14 + 6} = \frac{168}{20} = 8.4 \text{ km/hr}\]
Since the average speed of the cyclist is greater, he will return to A first.
Problem-Solving Approach for Round Trips
Example 6:
A man can row 7½ kmph in still water. If in a river running at 1.5 km an hour, it takes him 50 minutes to row to a place and back, how far off is the place?
Solution:
\[\text{Speed downstream} = (7.5 + 1.5) = 9 \text{ kmph}\]
\[\text{Speed upstream} = (7.5 - 1.5) = 6 \text{ kmph}\]
Let the required distance be \(x\) km. Then:
\[\frac{x}{9} + \frac{x}{6} = \frac{50}{60}\]
\[\frac{2x}{18} + \frac{3x}{18} = \frac{5}{6}\]
\[\frac{5x}{18} = \frac{5}{6}\]
\[x = 3 \text{ km}\]
Hence, the required distance is 3 km.
Example 7:
A boat goes 8 km upstream and then returns. Total time taken is 4 hrs 16 minutes. If the velocity of current is 1 km/hr, find the actual velocity of the boat.
Solution:
Let the actual velocity of the boat be \(x\) km/hr. Then:
\[\text{Speed downstream} = (x + 1) \text{ km/hr}; \text{ Speed upstream} = (x - 1) \text{ km/hr}\]
\[\frac{8}{x + 1} + \frac{8}{x - 1} = 4 + \frac{16}{60} = \frac{64}{15}\]
\[\frac{8(x - 1) + 8(x + 1)}{(x + 1)(x - 1)} = \frac{64}{15}\]
\[\frac{16x}{x^2 - 1} = \frac{64}{15}\]
\[30x = 8(x^2 - 1)\]
\[4x^2 - 15x - 4 = 0\]
\[4x^2 - 16x + x - 4 = 0\]
\[4x(x - 4) + (x - 4) = 0\]
\[(x - 4)(4x + 1) = 0\]
\[x = 4\]
Hence, actual velocity of the boat = 4 km/hr
Solving Simultaneous Equations
Example 8:
A boatman rows to a place 45 km distant and back in 20 hours. He finds that he can row 12 km with the stream in the same time as 4 km against the stream. Find the speed of the stream.
Solution:
Suppose he moves 12 km downstream in \(x\) hours. Then:
\[\text{Speed downstream} = \frac{12}{x} \text{ km/hr}\]
\[\text{Speed upstream} = \frac{4}{x} \text{ km/hr}\]
\[\frac{45}{\frac{12}{x}} + \frac{45}{\frac{4}{x}} = 20\]
\[\frac{45x}{12} + \frac{45x}{4} = 20\]
\[\frac{45x + 135x}{12} = 20\]
\[\frac{180x}{12} = 20\]
\[x = \frac{4}{3}\]
\[\text{Speed downstream} = 12 \times \frac{3}{4} = 9 \text{ km/hr}\]
\[\text{Speed upstream} = 4 \times \frac{3}{4} = 3 \text{ km/hr}\]
\[\text{Speed of the stream} = \frac{1}{2}(9 - 3) = 3 \text{ km/hr}\]
Example 9:
A man can row 40 km upstream and 55 km downstream in 13 hours. Also, he can row 30 km upstream and 44 km downstream in 10 hours. Find the speed of the man in still water and the speed of the current.
Solution:
Let rate upstream = \(x\) km/hr and rate downstream = \(y\) km/hr. Then:
\[\frac{40}{x} + \frac{55}{y} = 13 \quad \text{...(i)}\]
\[\frac{30}{x} + \frac{44}{y} = 10 \quad \text{...(ii)}\]
Multiplying (ii) by 4 and (i) by 3 and subtracting, we get:
\[\frac{120}{x} + \frac{165}{y} = 39\]
\[\frac{120}{x} + \frac{176}{y} = 40\]
\[\frac{11}{y} = 1 \text{ or } y = 11\]
Substituting \(y = 11\) in (i), we get: \(x = 5\)
Therefore:
\[\text{Rate in still water} = \frac{1}{2}(11 + 5) = 8 \text{ kmph}\]
\[\text{Rate of current} = \frac{1}{2}(11 - 5) = 3 \text{ kmph}\]
Practice Questions (From Exercise)
Question 1
A boat goes 8 km in one hour along the stream and 2 km in one hour against the stream. The speed in km/hr of the stream is
(a) 2 (b) 3 (c) 4 (d) 5
Solution:
\[\text{Speed of stream} = \frac{1}{2}(8 - 2) = 3 \text{ km/hr}\]
Answer: (b)
Question 2
In one hour, a boat goes 11 km along the stream and 5 km against the stream. The speed of the boat in still water (in km/hr) is
(a) 3 (b) 5 (c) 8 (d) 9
Solution:
\[\text{Speed in still water} = \frac{1}{2}(11 + 5) = 8 \text{ km/hr}\]
Answer: (c)
Question 3
A man rows downstream 32 km and 14 km upstream. If he takes 6 hours to cover each distance, then the velocity (in kmph) of the current is
(a) ½ (b) 1 (c) 1½ (d) 2
Solution:
\[\text{Speed downstream} = \frac{32}{6} = \frac{16}{3} \text{ km/hr}\]
\[\text{Speed upstream} = \frac{14}{6} = \frac{7}{3} \text{ km/hr}\]
\[\text{Speed of current} = \frac{1}{2}\left[\frac{16}{3} - \frac{7}{3}\right] = \frac{1}{2} \times \frac{9}{3} = \frac{3}{2} = 1\frac{1}{2} \text{ km/hr}\]
Answer: (c)
Question 4
A boatman rows 1 km in 5 minutes, along the stream and 6 km in 1 hour against the stream. The speed of the stream is
(a) 3 kmph (b) 6 kmph (c) 10 kmph (d) 12 kmph
Solution:
\[\text{Speed downstream} = \frac{1}{\frac{5}{60}} = 12 \text{ km/hr}\]
\[\text{Speed upstream} = \frac{6}{1} = 6 \text{ km/hr}\]
\[\text{Speed of stream} = \frac{1}{2}(12 - 6) = 3 \text{ km/hr}\]
Answer: (a)
Question 5
A boat takes half time in moving a certain distance downstream than upstream. What is the ratio between the rate in still water and the rate of current?
(a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) 3 : 1
Solution:
Let distance = \(d\), speed in still water = \(u\), speed of current = \(v\)
\[\text{Time upstream} = \frac{d}{u - v}, \text{ Time downstream} = \frac{d}{u + v}\]
Given: \(\frac{d}{u + v} = \frac{1}{2} \times \frac{d}{u - v}\)
\[2(u - v) = (u + v)\]
\[2u - 2v = u + v\]
\[u = 3v\]
Ratio = 3 : 1
Answer: (d)
Question 6
If a man goes 18 km downstream in 4 hours and returns against the stream in 12 hours, then the speed of the stream in km/hr is
(a) 1 (b) 1.5 (c) 1.75 (d) 3
Solution:
\[\text{Speed downstream} = \frac{18}{4} = 4.5 \text{ km/hr}\]
\[\text{Speed upstream} = \frac{18}{12} = 1.5 \text{ km/hr}\]
\[\text{Speed of stream} = \frac{1}{2}(4.5 - 1.5) = 1.5 \text{ km/hr}\]
Answer: (b)
Question 7
A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?
Solution:
\[\text{Speed upstream} = \frac{2}{1} = 2 \text{ km/hr}\]
\[\text{Speed downstream} = \frac{1}{\frac{10}{60}} = 6 \text{ km/hr}\]
\[\text{Speed in still water} = \frac{1}{2}(2 + 6) = 4 \text{ km/hr}\]
\[\text{Time to go 5 km in stationary water} = \frac{5}{4} = 1.25 \text{ hours} = 1 \text{ hour } 15 \text{ minutes}\]
Question 8
Using the formula from fact III: If speed downstream is 15 km/hr and speed upstream is 9 km/hr, find the speed in still water and rate of stream.
Solution:
\[\text{Speed in still water} = \frac{1}{2}(15 + 9) = 12 \text{ km/hr}\]
\[\text{Rate of stream} = \frac{1}{2}(15 - 9) = 3 \text{ km/hr}\]
Question 9
A swimmer can swim 5 km/hr in still water. If the river flows at 2 km/hr, how long will it take to swim 12 km downstream and return?
Solution:
\[\text{Speed downstream} = 5 + 2 = 7 \text{ km/hr}\]
\[\text{Speed upstream} = 5 - 2 = 3 \text{ km/hr}\]
\[\text{Time} = \frac{12}{7} + \frac{12}{3} = \frac{12}{7} + \frac{28}{7} = \frac{40}{7} \text{ hours} \approx 5.71 \text{ hours}\]
Question 10
A boat's speed in still water is 8 km/hr. It can go 15 km upstream and return downstream to the starting point in 4 hours. Find the speed of the stream.
Solution:
Let speed of stream = \(v\) km/hr
\[\text{Speed downstream} = (8 + v) \text{ km/hr}\]
\[\text{Speed upstream} = (8 - v) \text{ km/hr}\]
\[\frac{15}{8 - v} + \frac{15}{8 + v} = 4\]
\[15(8 + v) + 15(8 - v) = 4(8 - v)(8 + v)\]
\[120 + 15v + 120 - 15v = 4(64 - v^2)\]
\[240 = 256 - 4v^2\]
\[4v^2 = 16\]
\[v^2 = 4\]
\[v = 2 \text{ km/hr}\]
Key Takeaways
- Always identify whether the motion is upstream or downstream
- Remember: Downstream speed is always greater than upstream speed
- Speed in still water is the average of upstream and downstream speeds
- Rate of current is half the difference between downstream and upstream speeds
- For round trip problems, calculate time for each direction separately
- Pay attention to unit conversions (hours to minutes and vice versa)
- Set up equations systematically for problems involving multiple conditions
FAQs on RS Aggarwal Summary: Boats and Streams
| 1. What is average speed in the context of round trips? |  |
| 2. How do you calculate average speed for a round trip when the speeds in both directions are different? | |
| 3. What is the significance of solving simultaneous equations in relation to boats and streams problems? | |
| 4. Can you explain the problem-solving approach for round trip questions in examinations? | |
| 5. What are some key takeaways regarding boats and streams for SSC CGL preparation? | |
