This post is a collection of Boat and Stream quantitative aptitude questions. All the questions have been previously asked in competition exams like GRE, GMAT, Math Olympiad, CAT, CMAT,NMAT, SNAP, SSC, IBPS, SBI, AFCAT, NDA etc

There are total of 10 questions, all have been fully solved step by step for your understanding. In order to solve these questions on your own, you should have conceptual understanding of the chapter Speed, Time and Distance.

**Boat and Stream Aptitude Questions**

Q1. The speed of motor boat is that of the current of water as 36:5. The boat goes along with the current in 5 hours 10 minutes. It will come back in

(a) 5 hrs 50 min (b) 6 hrs

(c) 6 hrs 50 min (d) 12 hrs 10 min

Read SolutionSolution

Let the speed of motorboat = 36 km/hr

Speed of current = 5 km/hr

Downstream Speed (V1) = 36 + 5 = 41

Upstream Speed (V2) = 36 -5 = 31

Time (T1) = 5 hrs 10 mins = 31/6 hrs

Time (T2) = x

Now,

\propto \frac { 1 }{ T } \\\ \\ \frac { { V }<em>{ 1 } }{ { V }</em>{ 2 } } =\frac { { T }<em>{ 2 } }{ { T }</em>{ 1 } } \\\ \\ \frac { 41 }{ 31 } =\frac { x }{ { 31 }/{ 6 } } \\\ \\ \ x=\frac { 41 }{ 6 } =\quad 6\quad hrs\quad 50\quad min

**option (c) is the right answer**

Q2. In a fixed time, a boy swims double the distance along the current that he swims against the current. If the speed of the current is 3 km/hr, the speed of the boy in still water is:

(a) 6 km/hr (b) 9km/hr

(c) 10 km/hr (d) 12 km/hr

Read SolutionSolution

Let the downstream distance (D1) = 2x

and upstream distance (D2) = x

Speed of Stream = 3 km/hr

**Speed of Boy / Speed of Stream= (D1 + D2) / (D1 – D2)**

Speed of Boy / 3 = (2x + x) / (2x – x)

Speed of Boy = 9 km/hr

**option (b) is the right answer**

Q3. A man goes downstream with a boat to some destination and returns upstream to his original place in 5 hours. If the speed of the boat in still water and the stream are 10 km/hr and 4 km/hr respectively, the distance of the destination from the starting place is:

(a) 16 km (b) 18 km

(c) 21 km (d) 25 km

Read Solution**Solution**

Speed of boat (x) = 10 km/hr

Speed of Stream = 4 km/hr

Time (T) = 5 hrs

D=\frac { T({ x }^{ 2 }-{ y }^{ 2 }) }{ 2x } \\\ \\ D=\frac { 5({ 10 }^{ 2 }-{ 4 }^{ 2 }) }{ 2\times 10 } \\\ \\ \ D=\frac { 84 }{ 4 } =21\quad km

**option (c) is the right answer**

Q4. Two boats A and B start towards each other from two places, 108 km apart. Speed of the boat A and B in still water are 12 km/hr and 15km/hr respectively. If A proceeds down and B up the stream, the will meet after.

(a) 4.5 hrs (b) 4 hrs

(c) 5.4 hrs (d) 6 hrs

Read SolutionSolution

Let the speed of the stream be x km/hr

Both Boat meet after **t hours**

Distance between boat A & B = 108 km/hr

Speed of Boat A = 12 km/hr

Speed of Boat B = 15 km/hr

Now,

(12+x)t+(15-x)t=108\\\ \\ 12t+xt+15t-xt=108\\\ \\ 27t=108 \\\ \\ t=\frac { 108 }{ 27 } =4hrs

**Hence both boat meet after 4 hours**

option (b) is the right answer

Q5. A boat covers 24 km upstream and 36 km downstream in 6 hours, while it covers 36 km upstream and 24 km downstream in 6 hours. The speed of the current is:

(a) 1 km/hr (b) 2 km/hr

(c) 1.5 km/hr (d) 2.5 km/hr

Read SolutionSolution

Let the speed of the boat be x km/hr

and speed of the current be y km/hr

**Case 01:**

Upstream Speed = (x – y) km/hr, upstream distance = 24 km

Downstream Speed = ( x + y ) km/hr, downstream distance = 36 km

now

Upstream Distance/Speed + Downstream Distance/speed = 6 hours

24/(x – y) + 36/ (x + y) = 6

On solving we get

60x-12y=6({ x }^{ 2 }-{ y }^{ 2 })\quad \quad …(a)

**Case 02:**

Upstream Speed = (x – y) km/hr, upstream distance = 36 km

Downstream Speed = ( x + y ) km/hr, downstream distance = 24 km

Upstream Distance/Speed + Downstream Distance/speed = 6.5 hours

36/(x – y) + 24/(x+y) = 6.5

On solving we get

60x+12y=\frac { 13 }{ 2 } ({ x }^{ 2 }-{ y }^{ 2 })\quad \quad …(b)
Divide\quad eq\quad 'b'\quad from\quad eq\quad 'a'\\\ \\ \\\ \\ \frac { 13 }{ 2\times 6 } =\frac { 60x+12y }{ 60x-12y } \\\ \\ \ 65x-13y=60x+12y\\\ \\ 5x=25y\\\ \\ x=5y \\\ \\ \\\ \\
Put\quad the\quad value\quad of\quad x\quad in\quad eq\quad 'a'\\\ \\ 60(5y)-12y=6[{ (5y) }^{ 2 }-{ y }^{ 2 }]\\\ \\ \ y=\frac { 288y }{ 144y } \\\ \\ y=2\quad kmph

**option (b) is the right answer**

(Q6). A man can swim at the rate of 4 km/hr in still water. If the speed of water is 2 km/hr, then the time taken by him to swim 10 km upstream is:

(a) 2 hrs (b) 3 hrs

(c) 5 hrs (d) 4 hrs

Read SolutionSolution

Speed of Man = 4 km/hr

Speed of Water = 2 km/hr

Upstream speed = 4 – 2 = 2 km/hr

Distance = 10 Km

Upstream Time = Distance /Upstream Speed = 10/2 = 5 hours**Option (c) is the right answer**

(Q7). The speed of a boat along the stream is 12 km/hr and against the stream is 8 km/hr. The time taken by the boat to sail 24 km in still water is:

(a) 2 hrs (b) 3 hrs

(c) 2.4 hrs (d) 1.2 hrs

Read SolutionSolution

Downstream Speed = 12 km/hr

Upstream Speed = 8 km/hr**Speed of Boat = (Upstream Speed + Downstream Speed) / 2**

Speed of Boat = (12 + 8)/2 => 10 Km/hr

Time taken to cover 24 Km = Distance/Speed => 24/10 => 2.4 hours

**Option (c) is the right answer**

(Q8). A swimmer swims from a point A against a current for 5 minutes and then swims backwards in favor of the current for next 5 minutes and comes to point B. If AB is 100 meters, the speed of current (in km per hour) is:

(a) 0.4 (b) 0.2

(c) 1 (d) 0.6

Read Solution**Solution**

Let A, B and C are the points lie on a line

Upstream Distance = AC = d

Speed of swimmer = x m/min

Speed of current = y m/min

AB = 100 (given)

BC = 100 + d

Upstream Speed = ( x – y) m/min

Downstream Speed = ( x + y ) m/min

Upstream Time = Upstream Distance/ Upstream Speed

5 = d / (x – y)

d = 5 (x – y) —-(a)

Again

Time Taken = BC / Downstream Speed

5=\frac { 100+d }{ x+y } \\\ \\ 5=\frac { 100+5(x-y) }{ (x+y) } \quad \quad (from\quad 'a')\\\ \\ 5(x+y)=100+5x-5y\\\ \\ 5x+5y-5x+5y=100\\\ \\ 10y=100\ y=\frac { 100 }{ 10 } =10\quad m/min\\\ \\ \\\ \\ y=\frac { 10 }{ 1000 } \times 60=0.6\quad kmph

**option (d) is the right answer**

(Q9). A person can row a distance of one km upstream in ten minutes and downstream in four minutes. What is the speed of the stream?

(a) 4.5 kmph (b) 4 kmph

(c) 9 kmph (d) 5.6 kmph

Read SolutionSolution

Speed in still water = x km/hr

Speed of current = y km/hr

Upstream Time = 10 min

Downstream Time = 4 min

Upstream speed = Distance/Time

x – y = \frac { 1 }{ { 10 }/{ 60 } } = 6 km/hr

Downstream Speed = Distance / Time

x + y = \frac { 1 }{ { 4 }/{ 60 } } = 15 km/hr

Speed of Current = (Downstream speed – Upstream Speed) / 2

y = (15 – 6) / 2 = 4.5 km/hr

**Option (a) is the right answer**

(Q10). A man rows down a river 15 km in 3 hrs with the stream and returns in 15/2 hrs. The rate at which he rows in still water is:

(a) 2.5 kmph (b) 1.5 kmph

(c) 3.5 kmph (d) 4.5 kmph

Read Solution**Solution**

Downstream Speed = x

Upstream Speed = y

Upstream Time = 15/2 hours

Downstream Time = 3 hours

Distance = 15 Km

x= Distance / Downstream time => 15 / 3 => 5 km/hr

y = Distance / Upstream time => \frac { 15 }{ { 15 }/{ 2 } } = 2 km/hr

**option (c) is right answer**