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 crosses a river from port A to port B, which are just on the opposite side. The speed of the water is ${ V }{ W }$ and that of boat is ${ V }{ B }$ relative to still water. Assume ${ V }{ B }$ = 2${ V }{ W }$. What is the time taken by the boat, if it has to cross the river directly on the AB line
  1. $\frac { 2D }{ { V }_{ B }\sqrt { 3 } } $
  2. $\frac { \sqrt { 3 } D }{ 2{ V }_{ B } } $
  3. $\frac { D }{ { { V }_{ B } }\sqrt { 2 } } $
  4. $\frac { D\sqrt { 2 } }{ { { V }_{ B } } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To cross directly, the boat must head upstream at an angle. The velocity component perpendicular to the bank must be V_y = sqrt(V_B^2 - V_W^2). Since V_B = 2V_W, V_y = sqrt(4V_W^2 - V_W^2) = V_W * sqrt(3). Since V_W = V_B/2, V_y = (V_B/2) * sqrt(3). Time = D / V_y = 2D / (V_B * sqrt(3)).

Multiple choice

A boat moves relative to water with a velocity half of the river flow velocity if the angle from the direction of flow with which boat must move relative to stream directly to minimize drift is $\dfrac{2\times \pi }{n}$, then n is:

  1. 1

  2. 0

  3. 4

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

To minimize drift, the boat must head at an angle theta such that its velocity component perpendicular to the river flow cancels the river flow velocity. If v is boat speed and u is river speed, sin(theta) = v/u. Given v = u/2, sin(theta) = 1/2, so theta = 30 degrees or pi/6. The question asks for the angle from the direction of flow, which is 90 + 30 = 120 degrees or 2pi/3. Thus n=3.

Multiple choice

A boat crosses a river of width 200m in the shortest time and is found to experience a drift of 100m is reaching the opposite bank. The time taken now is 't'. If the same boat is to cross the river by shortest path, the time taken to cross will be:
  1. $2t$
  2. $\sqrt 2t$
  3. $3t$
  4. $\dfrac{2t}{\sqrt 3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let v_b be boat velocity and v_r be river velocity. Time t = width / v_b_y. Drift = v_r * t = 100. Shortest time path implies v_b is perpendicular to the bank. Shortest path (crossing straight) requires the boat to aim upstream such that the resultant velocity is perpendicular to the bank. Time = width / sqrt(v_b^2 - v_r^2). Given t = 200 / v_b, so v_b = 200/t. Drift = v_r * (200/v_b) = 100 => v_r = v_b/2. Time_shortest_path = 200 / sqrt(v_b^2 - (v_b/2)^2) = 200 / (v_b * sqrt(3)/2) = 400 / (v_b * sqrt(3)) = 2t / sqrt(3).

Multiple choice

Velocity of boat in still water is 13m/s. If water flowes in a river with a velocity of 5 m/s, what is the difference in times taken by him to cross the river in the shortest path and in the shortest time. If the width of the river is 156m.
  1. $1 sec$
  2. $\dfrac{13}{12}sec$
  3. $\dfrac{\sqrt3}{2}sec$
  4. $\sqrt3 sec$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shortest-time crossing takes 156/13 = 12 seconds. To follow the shortest path, the boat's effective crossing speed is sqrt(13^2 - 5^2) = 12 m/s, so the time is 13 seconds. The difference is 1 second.

Multiple choice

The speed of a boat is $5km/h$ in still water it crosses a river of width $1km$ along the shortest possible path in $15$minutes. The velocity of the river water is
  1. $1Km/h$
  2. $3Km/h$
  3. $4Km/h$
  4. $5Km/h$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Shortest path means the boat must head upstream to counteract the river current. Velocity of boat relative to ground = distance / time = 1 km / (15/60) hr = 4 km/h. Using the Pythagorean theorem where boat speed (5) is the hypotenuse and ground speed (4) is one leg, river speed v = sqrt(5^2 - 4^2) = sqrt(25 - 16) = 3 km/h.

Multiple choice

A man rowing a boat in river making an angle of $45^{o}$ with the straight course reaches the opposite point from the starting point. If velocity of water is $V$, then velocity of the boat w.r.t water is:
  1. $V/\sqrt{2}$
  2. $\sqrt{2}V$
  3. $2V$
  4. $V/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To reach the opposite point, the boat's velocity component perpendicular to the river must cancel the river's velocity. Let v be the boat's speed w.r.t water. The boat is angled at 45 degrees. The component of v along the river is v*cos(45) and perpendicular is v*sin(45). To cross straight, v*sin(45) = V. So v*(1/sqrt(2)) = V, which means v = sqrt(2)V.

Multiple choice

A boat having a speed of $5\ \text{km/h}$. in still water, crosses a river of width $1\ \text{km}$ along the shortest possible path in $15 \ :\text{minutes}$. The speed of the stream in $\text{km/h}$
  1. $1$
  2. $3$
  3. $4$
  4. $\sqrt {41}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Shortest path means the boat moves perpendicular to the stream. Speed of boat in still water (Vb) = 5 km/h. Distance = 1 km, Time = 15 min = 0.25 h. Resultant speed = 1 / 0.25 = 4 km/h. Since Vb^2 = Vs^2 + Resultant^2, 5^2 = Vs^2 + 4^2, so Vs^2 = 25 - 16 = 9, Vs = 3 km/h.

Multiple choice

A boat crosses a river at velocity $ \xrightarrow { v } $ (relative to water ) that is constant relative to water and perpendicular to the banks. the magnitude of the velocity $ \xrightarrow { V } $ is0.3 m/s . width of the river b=63 m. speed of the current varies according to the parabolic law. $ u\quad =u_{ 0 }-\dfrac { 4u_{ 0 } }{ b^{ 2 } } \left( x-\dfrac { b }{ 2 } \right) ^{ 2 } $ where x is distance from the banks and $ u_0 $ is constant equal to 5 //s. choose correct options.

  1. drift of the boat during the time it crosses the river is 1050 m.

  2. drift of the boat during the time it crosses the river is 700 m.

  3. time taken by the boat to crosses the river is 210 sec.

  4. time taken by the boat to crosses the river is 12.6 sec.
Reveal answer Fill a bubble to check yourself
A Correct answer