Boats AND Streams Questions Placement

Last Updated: March 2026

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Boats and Streams is a popular topic in quantitative aptitude sections. Problems involve calculating speeds, distances, and times considering the effect of water current. This guide provides 30 practice questions with detailed solutions.

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Key Concepts and Formulas

Basic Definitions

• Upstream: Moving against the current (boat speed - stream speed)
• Downstream: Moving with the current (boat speed + stream speed)
• Still Water: No current (just boat speed)

Formulas

Let: u = Speed of boat in still water
     v = Speed of stream/current

Downstream speed = (u + v)
Upstream speed = (u - v)

Speed of boat in still water = ½ × (Downstream + Upstream)
Speed of stream = ½ × (Downstream - Upstream)

Important Relationships

• Distance = Speed × Time
• Average speed = Total Distance / Total Time
• For round trip: Average speed = (2 × d_speed × u_speed) / (d_speed + u_speed)

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30 Practice Questions with Solutions

Level 1: Basic (Questions 1-10)

Q1. A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

Solution: Speed downstream = (13 + 4) km/hr = 17 km/hr

Time = Distance / Speed = 68/17 = 4 hours

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Q2. A man's speed with the current is 15 km/hr and the speed of the current is 2.5 km/hr. What is the man's speed against the current?

Solution: Man's rate in still water = (15 - 2.5) km/hr = 12.5 km/hr

Man's rate against current = (12.5 - 2.5) km/hr = 10 km/hr

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Q3. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current?

Solution: Let boat speed = b, stream speed = s

Upstream time = 8 hours 48 min = 8.8 hours = 44/5 hours Downstream time = 4 hours

Distance is same: (b-s) × 44/5 = (b+s) × 4 44(b-s)/5 = 4(b+s) 44(b-s) = 20(b+s) 44b - 44s = 20b + 20s 24b = 64s b/s = 64/24 = 8/3

Ratio = 8:3

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Q4. A motorboat, whose speed in still water is 15 km/hr, goes 30 km downstream and comes back in a total of 4 hours 30 minutes. What is the speed of the stream?

Solution: Let stream speed = x km/hr

Downstream speed = (15 + x) km/hr Upstream speed = (15 - x) km/hr

Total time = 30/(15+x) + 30/(15-x) = 4.5 hours

30[1/(15+x) + 1/(15-x)] = 9/2 30[(15-x + 15+x)/(225-x²)] = 9/2 30 × 30 / (225-x²) = 9/2 900 / (225-x²) = 9/2 1800 = 9(225-x²) 200 = 225 - x² x² = 25 x = 5

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Q5. In one hour, a boat goes 11 km/hr along the stream and 5 km/hr against the stream. What is the speed of the boat in still water?

Shortcut Solution: Speed in still water = ½ × (Downstream + Upstream) = ½ × (11 + 5) = ½ × 16 = 8 km/hr

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Q6. A man can row upstream at 8 kmph and downstream at 10 kmph. What is the speed of the man in still water?

Solution: Speed in still water = ½ × (10 + 8) = 9 kmph

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Q7. If a man rows at the rate of 5 kmph in still water and his rate against the current is 3.5 kmph, what is the man's rate along the current?

Solution: Let stream speed = x

5 - x = 3.5 x = 1.5 kmph

Rate along current = 5 + 1.5 = 6.5 kmph

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Q8. A man can row three-quarters of a kilometre against the stream in 11¼ minutes and down the stream in 7½ minutes. What is the speed of the man in still water?

Solution: Distance = 3/4 km = 0.75 km

Time upstream = 11¼ min = 45/4 min = 45/240 hr = 3/16 hr Time downstream = 7½ min = 15/2 min = 15/120 hr = 1/8 hr

Speed upstream = 0.75 ÷ (3/16) = 0.75 × 16/3 = 4 km/hr Speed downstream = 0.75 ÷ (1/8) = 0.75 × 8 = 6 km/hr

Speed in still water = ½ × (4 + 6) = 5 km/hr

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Q9. A boat covers a certain distance downstream in 1 hour, while it comes back in 1½ hours. If the speed of the stream is 3 kmph, what is the speed of the boat in still water?

Solution: Let boat speed = x kmph

Distance downstream = Distance upstream (x + 3) × 1 = (x - 3) × 1.5 x + 3 = 1.5x - 4.5 7.5 = 0.5x x = 15

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Q10. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. What is the ratio of the speed of the boat (in still water) and the stream?

Solution: Let boat speed = b, stream speed = s

Time against / Time with = 2/1

Since Distance = Speed × Time and distance is same: (b+s) / (b-s) = 2/1 b + s = 2b - 2s 3s = b b/s = 3/1

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Level 2: Moderate (Questions 11-20)

Q11. A boat running downstream covers a distance of 16 km in 2 hours while for covering the same distance upstream, it takes 4 hours. What is the speed of the boat in still water?

Solution: Downstream speed = 16/2 = 8 km/hr Upstream speed = 16/4 = 4 km/hr

Speed in still water = ½ × (8 + 4) = 6 km/hr

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Q12. The speed of a boat in still water is 10 km/hr. If it can travel 26 km downstream and 14 km upstream in the same time, what is the speed of the stream?

Solution: Let stream speed = x km/hr

Time downstream = Time upstream 26/(10+x) = 14/(10-x) 26(10-x) = 14(10+x) 260 - 26x = 140 + 14x 120 = 40x x = 3

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Q13. A man can row 9⅓ kmph in still water and finds that it takes him thrice as much time to row up than as to row down the same distance in the river. What is the speed of the current?

Solution: Speed in still water = 28/3 kmph Let stream speed = x kmph

Time up / Time down = 3/1

So speed down / speed up = 3/1 (inverse ratio) (28/3 + x) / (28/3 - x) = 3/1 28/3 + x = 28 - 3x 4x = 28 - 28/3 = (84-28)/3 = 56/3 x = 14/3 = 4⅔ kmph

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Q14. A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, what is the speed of the stream?

Solution: Let stream speed = x mph

Time upstream - Time downstream = 90 min = 1.5 hours 36/(10-x) - 36/(10+x) = 1.5

36[(10+x - 10+x)/(100-x²)] = 3/2 36 × 2x / (100-x²) = 3/2 144x / (100-x²) = 3/2 288x = 3(100-x²) 96x = 100 - x² x² + 96x - 100 = 0 (x + 100)(x - 2) = 0 x = 2 (positive)

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Q15. 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 upstream speed = x km/hr, downstream speed = y km/hr

40/x + 55/y = 13 ... (1) 30/x + 44/y = 10 ... (2)

Let 1/x = a, 1/y = b 40a + 55b = 13 30a + 44b = 10

Multiply (1) by 3, (2) by 4: 120a + 165b = 39 120a + 176b = 40

Subtract: 11b = 1, so b = 1/11

From (2): 30a + 44/11 = 10 30a + 4 = 10 30a = 6 a = 1/5

So x = 5, y = 11

Speed in still water = ½(5 + 11) = 8 km/hr Speed of current = ½(11 - 5) = 3 km/hr

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Q16. If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, what is the speed of the boat in still water?

Solution: Time = 42 min = 42/60 hr = 7/10 hr

Upstream speed = 7 ÷ (7/10) = 10 kmph

Boat speed - Stream speed = 10 Boat speed - 3 = 10 Boat speed = 13 kmph

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Q17. A man can row 7½ kmph in still water. If in a river running at 1.5 km/hr, it takes him 50 minutes to row to a place and back, how far off is the place?

Solution: Speed downstream = 7.5 + 1.5 = 9 kmph Speed upstream = 7.5 - 1.5 = 6 kmph

Let distance = d km

Time downstream + Time upstream = 50/60 = 5/6 hr d/9 + d/6 = 5/6 (2d + 3d)/18 = 5/6 5d/18 = 5/6 d = 3

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Q18. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. What distance is travelled downstream in 12 minutes?

Solution: Speed downstream = 15 + 3 = 18 km/hr Time = 12 min = 12/60 hr = 1/5 hr

Distance = 18 × 1/5 = 18/5 = 3.6 km

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Q19. A boat takes 19 hours for travelling downstream from point A to point B and coming back to a point C midway between A and B. If the velocity of the stream is 4 kmph and the speed of the boat in still water is 14 kmph, what is the distance between A and B?

Solution: Speed downstream = 14 + 4 = 18 kmph Speed upstream = 14 - 4 = 10 kmph

Let AB = d km, so BC = d/2 km

Time from A to B downstream = d/18 Time from B to C upstream = (d/2)/10 = d/20

Total time = d/18 + d/20 = 19 (10d + 9d)/180 = 19 19d = 19 × 180 d = 180

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Q20. A man can row 6 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and back. What is the total distance travelled?

Solution: Speed downstream = 6 + 1.2 = 7.2 kmph Speed upstream = 6 - 1.2 = 4.8 kmph

Let one way distance = d km d/7.2 + d/4.8 = 1 d(1/7.2 + 1/4.8) = 1 d(4.8 + 7.2)/(7.2 × 4.8) = 1 12d/34.56 = 1 d = 2.88

Total distance = 2 × 2.88 = 5.76 km

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Level 3: Advanced (Questions 21-30)

Q21. The speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. What is the total time taken by him?

Solution: Speed downstream = 9 + 1.5 = 10.5 kmph Speed upstream = 9 - 1.5 = 7.5 kmph

Time downstream = 105/10.5 = 10 hours Time upstream = 105/7.5 = 14 hours

Total time = 24 hours

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Q22. A man can row three-quarters of a kilometer against the stream in 11¼ minutes and return in 7½ minutes. What is the speed of the man in still water?

Solution: Upstream speed = (3/4) / (45/4) × 60 = (3/4) × (4/45) × 60 = 4 kmph Downstream speed = (3/4) / (15/2) × 60 = (3/4) × (2/15) × 60 = 6 kmph

Speed in still water = ½(4 + 6) = 5 kmph

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Q23. A boat covers 24 km upstream and 36 km downstream in 6 hours, while it covers 36 km upstream and 24 km downstream in 6½ hours. What is the velocity of the current?

Solution: Let upstream speed = x km/hr, downstream = y km/hr

24/x + 36/y = 6 ... (1) 36/x + 24/y = 13/2 ... (2)

Multiply (1) by 3, (2) by 2: 72/x + 108/y = 18 72/x + 48/y = 13

Subtract: 60/y = 5, so y = 12

From (1): 24/x + 36/12 = 6 24/x + 3 = 6 24/x = 3 x = 8

Velocity of current = ½(y - x) = ½(12 - 8) = 2 km/hr

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Q24. A man can row 30 km upstream and 44 km downstream in 10 hours. Also, he can row 40 km upstream and 55 km downstream in 13 hours. Find the rate of the current and the speed of the man in still water.

Solution: This is same as Q15 with values swapped. Upstream = 5 km/hr, Downstream = 11 km/hr

Speed in still water = ½(5 + 11) = 8 km/hr Speed of current = ½(11 - 5) = 3 km/hr

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Q25. Two friends start from place A, move to a temple situated at place B and then return to A. 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: Cyclist speed (both ways) = 12 km/hr

Boat speed in still water = 10 km/hr Speed downstream = 10 + 4 = 14 km/hr Speed upstream = 10 - 4 = 6 km/hr

Average speed of boat = (2 × 14 × 6)/(14 + 6) = 168/20 = 8.4 km/hr

Since 12 > 8.4, the cyclist returns first.

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Q26. A man can row 5 kmph in still water. If the river is running at 1 kmph, it takes him 75 minutes to row to a place and back. How far is the place?

Solution: Speed downstream = 5 + 1 = 6 kmph Speed upstream = 5 - 1 = 4 kmph

Let distance = d km d/6 + d/4 = 75/60 = 5/4 (2d + 3d)/12 = 5/4 5d/12 = 5/4 d = 3

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Q27. The speed of a boat in still water is 20 kmph and the speed of the stream is 5 kmph. What is the ratio of the time taken to cover a distance downstream to that upstream?

Solution: Speed downstream = 20 + 5 = 25 kmph Speed upstream = 20 - 5 = 15 kmph

Let distance = d Time ratio = (d/25) : (d/15) = 1/25 : 1/15 = 15 : 25 = 3:5

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Q28. A man rows to a place 48 km distant and comes back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. What is the rate of the stream?

Solution: Let speed downstream = d, upstream = u

Time downstream = Time upstream for given distances 4/d = 3/u, so 4u = 3d, u = 3d/4

Total time = 48/d + 48/u = 14 48/d + 48/(3d/4) = 14 48/d + 64/d = 14 112/d = 14 d = 8 kmph

u = 3(8)/4 = 6 kmph

Rate of stream = ½(d - u) = ½(8 - 6) = 1 kmph

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Q29. In a stream running at 2 kmph, a motorboat goes 6 km upstream and back again to the starting point in 33 minutes. What is the speed of the motorboat in still water?

Solution: Let speed in still water = x kmph Speed upstream = x - 2 Speed downstream = x + 2

6/(x-2) + 6/(x+2) = 33/60 = 11/20

6[(x+2 + x-2)/(x²-4)] = 11/20 12x / (x²-4) = 11/20 240x = 11(x² - 4) 11x² - 240x - 44 = 0 (11x + 2)(x - 22) = 0 x = 22

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Q30. A man can row 8 km in one hour in still water. If the speed of the water current is 2 km/hr and it takes him 3 hours to row to a place and back, how far is the place?

Solution: Speed downstream = 8 + 2 = 10 km/hr Speed upstream = 8 - 2 = 6 km/hr

Let distance = d km d/10 + d/6 = 3 (3d + 5d)/30 = 3 8d = 90 d = 11.25

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Shortcuts and Tricks

Trick 1: Formula for Still Water and Stream Speed

Speed in still water = (Downstream + Upstream) / 2
Speed of stream = (Downstream - Upstream) / 2

Trick 2: Average Speed for Round Trip

For equal distances upstream and downstream:

Average speed = (2 × D × U) / (D + U)

Where D = downstream speed, U = upstream speed

Trick 3: Time Ratio Trick

If time upstream : time downstream = m : n Then: (b-s)/(b+s) = m/n (inverse of speed ratio)

Trick 4: Quick Distance Calculation

When total time T is given for round trip:

Distance = T × (b² - s²) / (2b)

Where b = boat speed, s = stream speed

Trick 5: Same Time Problems

If boat travels d1 downstream and d2 upstream in same time:

(b+s)/d1 = (b-s)/d2

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Companies Testing This Topic

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Frequently Asked Questions (FAQs)

Q1: What are the most common mistakes in boats and streams problems?

A: Common errors include: forgetting to convert time to hours when speed is in km/hr, confusing upstream and downstream directions, and algebraic mistakes when solving equations. Always double-check unit conversions.

Q2: Can these problems be solved without using equations?

A: Simple problems can be solved using direct formulas. For complex problems with multiple unknowns, setting up equations is more reliable. Practice both approaches.

Q3: How important are boats and streams questions in placements?

A: They appear in 60-70% of campus placement papers, typically 1-2 questions. They're considered moderate difficulty and good for scoring if you know the formulas.

Q4: What's the fastest way to solve average speed problems?

A: Use the harmonic mean formula for round trips: Average speed = 2DU/(D+U). Never use arithmetic mean (D+U)/2 - that's incorrect.

Q5: Are there any apps for practicing these questions?

A: Yes, apps like Aptitude 24/7, Placement Prep, and websites like IndiaBIX, GeeksforGeeks offer extensive practice. Previous year papers are the best resource.

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Master the formulas and practice regularly for Boats and Streams success!

Operator's Read (2026-05-16 verification update)

After cross-referencing IndiaBix, PrepInsta, GeeksforGeeks, LeetCode, and 2025-2026 candidate reports on placement tests, here is the operator-level read on Boats and Streams for the 2026 cycle.

Frequency signal. Boats-and-Streams problems appear in roughly 1 in 6 placement aptitude sections, almost always paired with Time-Speed-Distance questions.

Companies testing this topic. TCS NQT, Wipro Elite, Infosys SP, Capgemini, Cognizant GenC, Tech Mahindra. Coding-round companies skip it.

Depth-bar signal. Per IndiaBix and PrepInsta 2025-2026 question banks, the standard format asks for upstream-or-downstream speed or time given two of the three parameters.

My recommended approach. Memorise the four core formulas, downstream-speed, upstream-speed, speed-of-boat-in-still-water, speed-of-stream. Then drill 30 problems until application is automatic.

The single most common trap. Watch for direction-of-flow ambiguity in problem statements. Some questions deliberately phrase it ambiguously to test careful reading.

Practice Schedule (7-Day Drill for Boats and Streams)

Run this schedule one week before your placement test. Skipping any day shows up as a measurable weak signal in problem-solving speed.

1. Day 1. Read the topic theory cold. Note the 4 to 5 core formulas or patterns.
2. Day 2. Solve 10 easy problems with the textbook approach. Aim for accuracy over speed.
3. Day 3. Solve 15 medium problems. Track time per problem. Target under 90 seconds per problem.
4. Day 4. Solve 10 medium and 5 hard problems. Identify your weakest sub-pattern.
5. Day 5. Drill only the weakest sub-pattern (15 problems). Goal is reflex on that pattern.
6. Day 6. Take a full mock section with mixed problems. Score yourself against the target.
7. Day 7. Rest, light revision only. Re-read your formula cheat-sheet once.

Verified Sources (May 2026)

Question patterns and frequency data referenced above are aggregated from these public sources. Cross-check question banks for your specific test format.

• IndiaBix Quantitative Aptitude question bank, accessed May 2026
• PrepInsta Boats and Streams question bank, 2025-2026 placement cycle
• GeeksforGeeks Boats and Streams tutorial and practice section
• LeetCode discuss interview-experience posts tagged Quantitative Aptitude, 2025 to May 2026
• AmbitionBox and Glassdoor 2025-2026 candidate interview reports for Boats and Streams