Quantitative Aptitude

Boats and Streams

369 Questions
All Recover still-water and stream speeds 136 Calculate one-way and round-trip time 61 Ratio of boat to stream speed 54 Use time differences and equal-distance ratios 39 Perpendicular river crossing 36 Round-trip average speed 14 Angles of drift 12 Handle drifting objects 8 Solve meeting and separation problems 6 Fixed-time distance ratios 3

Boats and Streams Questions

Multiple choice

A boat can travel 200 km upstream in 20 hours. Another boat can travel 150 km downstream in 10 hours. The speed of the stream is same in both the cases. If the speed of first boat is 50% higher than the speed of the second boat, find the speed (in kmph) of the stream?

  1. 12 km/hr

  2. 14 km/hr

  3. 5 km/hr

  4. 8 km/hr

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

First boat: upstream speed = 200/20 = 10 km/h. x - y = 10. Second boat: downstream speed = 150/10 = 15 km/h. z + y = 15. First boat 50% faster than second: x = 1.5z. Solving: x = 10, y = 5. Stream speed = 5 km/h. Option C correct.

Multiple choice

In a river, the ratio of the speed of the stream and the speed of a boat in still water is 3 : 7. Again, the ratio of the speed of the stream to the speed of another boat in still water is 6 : 8. What is the ratio of the speed of the first boat to that of the second boat in still water?

  1. 3 : 4

  2. 7 : 4

  3. 4 : 7

  4. 14 : 15

  5. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the first boat, stream:boat = 3:7, so if stream speed is 3k, boat speed is 7k. For the second boat, stream:boat = 6:8 = 3:4, so if stream speed is 3m, boat speed is 4m. Since the stream is the same, 3k = 3m, meaning k = m. The ratio of boat speeds in still water is 7k:4k = 7:4.

Multiple choice

Anil looked up at the top of a lighthouse from his boat and found the angle of elevation to be 30 degrees. After sailing against the direction of stream at the speed of 25m/s for 2.5seconds towards the lighthouse, he found that the angle of elevation changed to 45 degrees. Find the height of the lighthouse, if speed of stream is 5m/s.

  1. 25(√3 +1) m

  2. 50(√3 – 1) m

  3. 25(√3 – 1) m

  4. 50(√3 +1) m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sailing at 25 m/s for 2.5 seconds covers 62.5 m against a 5 m/s stream, so effective horizontal distance = (25-5) × 2.5 = 50 m. At 30° elevation: tan30° = h/d₁, so d₁ = h/√3. At 45° elevation: tan45° = h/d₂, so d₂ = h. Since d₁ - d₂ = 50 m, we get h/√3 - h = -50, giving h = 50√3/(√3-1) = 25(√3+1) m.

Multiple choice

A boat is moving 2 km against the current of the stream in 1 hour and moves 1 km in the direction of the current in 10 minutes. How long will it take the boat to go 5 km in stationary water?

  1. 1 hr 20 minutes

  2. 1 hr 30 minutes

  3. 1 hr 15 minutes

  4. 30 minutes

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let boat speed be b km/hr and stream speed be s km/hr. Against current: b - s = 2 km/hr (2 km in 1 hour). With current: b + s = 6 km/hr (1 km in 10 minutes). Solving: adding equations gives 2b = 8, so b = 4 km/hr. Time for 5 km in still water = 5/4 = 1.25 hours = 1 hr 15 min.

Multiple choice

A man can row a boat in still water at a speed of 5 m/s. He covers a stretch of 200 m in a river downstream during high and low tides in 10 sec and 25 sec respectively. What is the ratio of the speed (in m/s) of the water flowing in the river during high and low tides?

  1. 5 : 3

  2. 5 : 1

  3. 3 : 2

  4. 4 : 3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Downstream speed = boat speed + current speed. In high tide: 200m in 10s = 20 m/s. Current speed = 20-5 = 15 m/s. In low tide: 200m in 25s = 8 m/s. Current speed = 8-5 = 3 m/s. Ratio of current speeds = 15:3 = 5:1.

Multiple choice

A boat can go across a lake and return in time $T_{0}$ at a speed $v$. On a rough day there is a uniform current at speed $v_{1}$ to help the onward journey and impede the return journey. If the time taken to go across and return on the same day be $T$, then $T/T_{0}$ will be
  1. $\dfrac {1}{(1 - v_{1}^{2}/ v^{2})}$
  2. $\dfrac {1}{(1 + v_{1}^{2}/ v^{2})}$
  3. $(1 - v_{1}^{2} / v^{2})$
  4. $\left (1 + \dfrac {v_{1}^{2}}{v_{2}}\right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

T0 = 2d/v. With current v1, time T = d/(v+v1) + d/(v-v1) = 2dv / (v^2 - v1^2). Then T/T0 = (2dv / (v^2 - v1^2)) / (2d/v) = v^2 / (v^2 - v1^2) = 1 / (1 - v1^2/v^2).

Multiple choice

A boat is sent across a river with a velocity of $8$ km/hr. If the resultant velocity of boat is $10$ km/hr, then velocity of the river is
  1. $10 km/hr$
  2. $8km/hr$
  3. $6km/hr$
  4. $4km/hr$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The resultant velocity of the boat is the vector sum of the boat's velocity relative to the water and the river's velocity. Using the Pythagorean theorem, 10^2 = 8^2 + v_river^2, which gives v_river = sqrt(100 - 64) = sqrt(36) = 6 km/hr.

Multiple choice

A boat moves with a speed of $5km/h$ relative to water in a river flowing with a speed of $3 km/h$ and a width of $1km$. The minimum time taken around a round trio is
  1. $5$ minutes
  2. $20$ minutes
  3. $30$ minutes
  4. $60$ minutes
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the fastest crossing, the boat must aim upstream so its downstream drift is canceled. Its effective speed across the river is sqrt(5^2 - 3^2) = 4 km/h, so the round-trip time is 2/4 hour = 30 minutes.