# SSC CGL Boat and Stream Questions PDF

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## SSC CGL Boat and Stream Questions PDF:

Download SSC CGL Boat and Stream questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Boat and Stream objective questions for SSC exams.

Question 1: A boat covers 12 km upstream in 4 hours and can cover the same distance downstream in 3 hours. What is the speed (in km/hr) of the boat in still water?

a) 3.5
b) 3
c) 2.5
d) 2

Question 2: Two boat are travelling with speed of 36 km/hr and 54 km/hr respectively towards each other. What is the distance (in metres) between the two boats one second before they collide?

a) 10
b) 15
c) 25
d) 5

Question 3: A boat goes 15 km upstream and 22 km downstream is 5 hours. It goes 20 km upstream and $\ \frac{55}{2}\$km downstream in $\ \frac{13}{2}\$hours. What is the speed (in km/hr) of stream ?

a) 3
b) 5
c) 8
d) 11

Question 4: A boat travels 60 kilometers downstream and 20 kilometers upstream in 4 hours. The same boat travels 40 kilometers downstream and 40 kilometers upstream in 6 hours. What is the speed (in km/hr) of the stream?

a) 24
b) 16
c) 18
d) 20

Question 5: A man rows 750 m in 675 seconds against the stream and returns in $\ 7\frac{1}{2}\$minutes. Its rowing speed in still water is (in km/hr).

a) 5.5
b) 5.75
c) 5
d) 5.25

Question 6: If a boat goes a certain distance at 20 km/hr and comes back the same distance at 30 km/hr. What is the average speed (in km/hr) for the total journey?

a) 12
b) 24
c) 26
d) 25

Question 7: A boat took 4 hours to travel 20km upstream. Find the speed of the boat in still water if the speed of the stream is 2 kmph ?

a) 5 kmph
b) 6 kmph
c) 7 kmph
d) 8 kmph

Question 8: A boat goes 8 km upstream and 12 km downstream in 7hours. It goes 9 km upstream and 18 km downstream in 9 hours. What is the speed (in km/h) of the boat in still water?

a) 5
b) 4
c) 2
d) 3

Question 9: A boat covers 143 km upstream in 13 hours and the same distance downstream in 11 hours. What is the speed (in km/hr) of the boat in still water?

a) 10
b) 12
c) 14
d) 8

Question 10: A boat covers a distance of 14 km upstream and 16 km downstream in 9 hours. It covers a distance of 12 km upstream and 40 km downstream in 11 hours. What is the speed (in km/hr) of the boat in still water?

a) 5
b) 2
c) 3
d) 4

Question 11: Speed of a boat along and against the current are 14 kms/hr and 8 kms/hr respectively. The speed of the current is

a) 11 kms/hr
b) 6 kms/hr
c) 5.5 kms/hr
d) 3 kms/hr

Question 12: Speed of a boat is 8 km/hr in still water and the speed of the stream is 2 km/hr. If the boat takes 8 hours to go to a place and come back, then what is the distance (in km) of the place?

a) 24
b) 30
c) 45
d) 42

Question 13: Speeds of a boat along the current and against the current are 14 km/hr and 7 km/hr respectively. What is the speed of boat (in km/hr) in still water?

a) 3.5
b) 7.5
c) 10.5
d) 9.5

Question 14: Speeds of a boat along the current and against the current are 16 km/hr and 12 km/hr respectively. What is the speed (in km/hr) of the current?

a) 1
b) 2
c) 3
d) 4

Question 15: A man can row 32 km upstream in 8 hours and the same distance downstream can be covered in 4 hours. If the speed of the stream and boat is constant throughout then what is the speed of stream?

a) 1 km/hr
b) 1.5 km/hr
c) 2 km/hr
d) 2.5 km/hr

Question 16: A boat going upstream can cover a certain distance in 6 hours. The speed of the stream is half the speed of boat in still water. How much time will the boat take to cover the same distance downstream?

a) 3 hours
b) 2 hours
c) 1 hour
d) Cannot be determined

Question 17: A boat goes 4 km upstream and 4 km downstream in 1 hour. The same boat goes 5 km downstream and 3 km upstream in 55 minutes. What is the speed (in km/hr) of boat in still water?

a) 6.5
b) 7.75
c) 9
d) 10.5

Question 18: A boat is sailing towards a lighthouse of height 20√3 m at a certain speed. The angle of elevation of the top of the lighthouse changes from 30° to 60° in 10 seconds. What is the time taken (in seconds) by the boat to reach the lighthouse from its initial position?

a) 10
b) 15
c) 20
d) 60

Question 19: If a boat goes a certain distance at 30 km/hr and comes back the same distance at 60 km/hr. What is the average speed (in km/hr) for the total journey?

a) 45
b) 5
c) 40
d) 35

Question 20: The average weight of 15 oarsmen in a boat is increased by 1.6 kg when one of the crew, who weighs 42 kg is replaced by a new man. Find the weight of the new man (in kg).

a) 67
b) 65
c) 66
d) 43

Question 21: If the speed of a boat in still water is 20 km/hr and the speed of the current is 5 km/hr, then the time taken by the boat to travel 100 km with the current is

a) 2 hrs
b) 3 hrs
c) 4 hrs
d) 7 hrs

Question 22: A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream in 6 hours and 30 minutes. The speed of the boat in still water is

a) 8 km/hr
b) 9 km/hr
c) 12 km/hr
d) 10 km/hr

Question 23: A boat can travel with a speed of 13 km/hr in still water. If the speed of stream is 4 km/hr in the same direction, time taken by boat to go 63 km in opposite direction is

a) 9 hrs
b) 4 hrs
c) 7 hrs
d) $3\frac{9}{17}$ hrs

Question 24: 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 km/h
b) 4 km/h
c) 9 km/h
d) 5.6 km/h

Question 25: Ravi rows a boat upstream in a river from point A to B in 4 hours. Had the river been still, he could have completed the journey in 3 hours. How much time will it take for Ravi to row the boat from point B to A?

a) 2 hour 24 minutes
b) 2 hours
c) 2 hour 48 minutes
d) 2 hour 12 minutes.

Let speed of boat in still water = $x$ km/hr and speed of current = $y$ km/hr
Speed upstream = $(x-y)$ km/hr
Using, speed = distance/time
=> $x-y=\frac{12}{4}=3$ —————-(i)
Similarly, downstream speed = $x+y=\frac{12}{3}=4$ —————(ii)
Adding equations (i) and (ii), we get : $2x=3+4=7$
=> $x=\frac{7}{2}=3.5$ km/hr
=> Ans – (A)

Relative speed of boats (since they are travelling towards each other = $36+54=90$ km/hr
= $(90\times\frac{5}{18})$ m/s = $25$ m/s
Distance between them one second before they collide = Distance covered in 1 second = 25 m
=> Ans – (C)

Let speed of boat = $x$ km/hr and speed of stream = $y$ km/hr
=> Downstream speed = $(x+y)$ km/hr and Upstream speed = $(x-y)$ km/hr
According to ques,
=> $\frac{15}{x-y}+\frac{22}{x+y}=5$
and $\frac{20}{x-y}+\frac{27.5}{x+y}=6.5$
Let $\frac{1}{x-y}=m$ and $\frac{1}{x+y}=n$
=> $15m+22n=5$ and $20m+27.5n=6.5$
Solving above equations, we get : $m=\frac{1}{5}$ and $n=\frac{1}{11}$
Thus, $x-y=5$ and $x+y=11$
Subtracting both equation, => $2y=11-5=6$
=> $y=\frac{6}{2}=3$
$\therefore$ Speed of stream = 3 km/hr
=> Ans – (A)

Let speed of boat = $x$ km/hr and speed of stream = $y$ km/hr
Thus, downstream speed = $(x+y)$ km/hr and upstream speed = $(x-y)$ km/hr
Using, time = distance/speed
=> $(\frac{60}{x+y})+(\frac{20}{x-y})=4$
=> $\frac{15}{x+y}+\frac{5}{x-y}=1$ —————(i)
Similarly, $(\frac{40}{x+y})+(\frac{40}{x-y})=6$
=> $\frac{1}{x+y}+\frac{1}{x-y}=\frac{3}{20}$ ————(ii)
Solving equations (i) and (ii), we get : $x=24$ and $y=16$
$\therefore$ Speed of stream = 16 km/hr
=> Ans – (B)

Boat’s upstream speed($S_{u}$) $= \frac{750}{675} = \frac{10}{9}$ m/sec
Boat’s downstream speed($S_{d}$) $= \frac{750}{450} = \frac{5}{3}$ m/sec
Boat’s speed in still water $= \frac{1}{2}\times(S_{u}+(S_{d})$
$= \frac{1}{2}\times(\frac{10}{9}+\frac{5}{3})$
$= \frac{1}{2}\times(\frac{25}{9})$
$= \frac{25}{18}$ m/sec
Converting it into km/hr
$= \frac{25}{18}\times\frac{18}{5} = 5$ km/hr

average speed = $\frac{2ab}{a+b}=\frac{2(20)(30)}{20+30}=24$
So the answer is option B.

Let x be the speed of the boat in still water
During upstream, speed = x-2 kmph
Speed = Distance/time
x-2 = 20/4
x-2 = 5
x = 7 kmph
So the answer is option C.

Let speed of boat in still water = $x$ km/hr and speed of current = $y$ km/hr
According to ques,
=> $\frac{12}{x+y}+\frac{8}{x-y}=7$ —————-(i)
and $\frac{18}{x+y}+\frac{9}{x-y}=9$ —————-(ii)
Applying the operation : $3\times(i)-2\times(ii)$
=> $\frac{24}{x-y}-\frac{18}{x-y}=21-18$
=> $\frac{6}{x-y}=3$
=> $x-y=\frac{6}{3}=2$ ————–(iii)
Substituting it in equation (i), => $\frac{12}{x+y}+\frac{8}{2}=7$
=> $\frac{12}{x+y}=7-4=3$
=> $x+y=\frac{12}{3}=4$ ————(iv)
Now, adding equations (iii) and (iv), we get :
=> $2x=2+4=6$
=> $x=\frac{6}{2}=3$
=> Ans – (D)

Let speed of the boat in still water = $x$ km/hr and speed of current = $y$ km/hr
According to ques,
=> $x-y=\frac{143}{13}=11$ ————-(i)
and $x+y=\frac{143}{11}=13$ ———–(ii)
Adding both equations, we get : $2x=11+13=24$
=> $x=\frac{24}{2}=12$ km/hr
=> Ans – (B)

Let speed of boat in still water = $x$ km/hr and speed of current = $y$ km/hr
According to ques,
=> $\frac{16}{x+y}+\frac{14}{x-y}=9$ —————-(i)
and $\frac{40}{x+y}+\frac{12}{x-y}=11$ —————-(ii)
Applying the operation : $5\times(i)-2\times(ii)$
=> $\frac{70}{x-y}-\frac{24}{x-y}=45-22$
=> $\frac{46}{x-y}=23$
=> $x-y=\frac{46}{3}=2$ ————–(iii)
Substituting it in equation (i), => $\frac{16}{x+y}+\frac{14}{2}=9$
=> $\frac{16}{x+y}=9-7=2$
=> $x+y=\frac{16}{2}=8$ ————(iv)
Now, adding equations (iii) and (iv), we get :
=> $2x=2+8=10$
=> $x=\frac{10}{2}=5$
=> Ans – (A)

If Upstream speed = $x$ km/hr and downstream speed = $y$ km/hr
The, speed of boat = $\frac{x+y}{2}$ km/hr and speed of current = $\frac{y-x}{2}$ km/hr
According to ques, upstream speed (x) = 8 km/hr
Downstream speed (y) = 14 km/hr
=> Speed of current = $\frac{14-8}{2}$
= $\frac{6}{2}=3$ km/hr
=> Ans – (D)

Speed of boat in still water = 8 km/hr and speed of stream = 2 km/hr
Let distance covered = $d$ km
According to ques,
=> $\frac{d}{(8-2)}+\frac{d}{(8+2)}=8$
=> $\frac{d}{6}+\frac{d}{10}=8$
=> $\frac{10d+6d}{60}=8$
=> $\frac{4d}{15}=8$
=> $d=\frac{15\times8}{4}=30$ km
=> Ans – (B)

Downstream speed of boat = 14 km/hr
Upstream speed of boat = 7 km/hr
=> Speed of boat (in km/hr) in still water = $\frac{1}{2}\times(14+7)$
= $\frac{21}{2}=10.5$ km/hr
=> Ans – (C)

Downstream speed of boat = 16 km/hr
Upstream speed of boat = 12 km/hr
=> Speed of current (in km/hr) = $\frac{1}{2}\times(16-12)$
= $\frac{4}{2}=2$ km/hr
=> Ans – (B)

We have been given that speed while travelling upstream is 4 km/hr and while rowing downstream it is 8 km/hr.
So if x is the speed of the boat and y is the speed of the stream then we have
x + y = 8
x – y = 4
Hence, 2x = 12 => x = 6
=> y = 2

Let the speed of the boat be 2x. Hence, the speed of the stream will be ‘x’. Thus, net speed while going upstream = x and net speed while going downstream = 3x. Hence, the time taken while travelling downstream will be one third of the time taken while going upstream. Thus, the correct answer will 6/3 = 2 hours

Let speed of boat in still water = $x$ km/hr and speed of current = $y$ km/hr
The boat goes 4 km upstream and 4 km downstream in 1 hour
Using, time = distance/speed
=> $\frac{4}{x+y}+\frac{4}{x-y}=1$
Similarly, $\frac{5}{x+y}+\frac{3}{x-y}=\frac{55}{60}$
Let $\frac{1}{x+y}=w$ and $\frac{1}{x-y}=z$
=> $4w+4z=1$ and $5w+3z=\frac{55}{60}$
Solving above equations, we get : $w=\frac{1}{12}$ and $z=\frac{1}{6}$
=> $x+y=12$ ———(i)
and $x-y=6$ ———–(ii)
Adding equations (i) and (ii), => $2x=12+6=18$
=> $x=\frac{18}{2}=9$ km/hr
=> Ans – (C)

Given : CD = $20\sqrt3$ m and time taken to reach B from A = 10 seconds
To find : Time taken to reach D from A
Solution : In $\triangle$ BCD,
=> $tan(60^\circ)=\frac{CD}{BD}$
=> $\sqrt3=\frac{20\sqrt3}{BD}$
=> $BD=20$ m
Similarly, in $\triangle$ ACD,
=> $tan(30^\circ)=\frac{CD}{AD}$
=> $\frac{1}{\sqrt3}=\frac{20\sqrt3}{20+AB}$
=> $AB+20=60$
=> $AB=60-20=40$ m
=> Speed of boat (while travelling from A to B) = distance/time
= $\frac{40}{10}=4$ m/s
=> AD = BD + AB = 20 + 40 = 60 m
$\therefore$ Time taken to reach D from A = $\frac{60}{4}=15$ seconds
=> Ans – (B)

average speed = $\frac{2ab}{a+b}=\frac{2(30)(60)}{30+60}=40$
So the answer is option C.

Let the average weight of the original 15 oarsmen = $A$
=> Total weight of 15 oarsmen = $15A$
Let the weight of new man be $x$
After he replaces the man with weight 42 kg, the average weight increases by 1.6 kg
=> $\frac{15A + x – 42}{15} = A + 1.6$
=> $15A + x – 42 = 15A + 24$
=> $x = 24+42 = 66$

Speed of boat = 20 km/hr and speed of current = 5 km/hr
Distance to travel = 100 km
Speed downstream (with the current) = 20 + 5 = 25 km/hr
=> Time taken = distance / speed
= $\frac{100}{25}=4$ hours
=> Ans – (C)

Let speed of boat in still water = $x$ km/h
and speed of stream = $y$ km/h
=> Upstream speed of boat = $(x-y)$ km/h
Downstream speed = $(x+y)$ km/h
Acc to ques :
=> $\frac{24}{x-y} + \frac{28}{x+y} = 6$
and $\frac{30}{x-y} + \frac{21}{x+y} = 6\frac{1}{2}$
Solving above equations, we get :
$x$ = 10 km/h and $y$ = 4 km/h

Speed of boat = 13 km/hr and speed of stream = 4 km/hr
When going in opposite direction i.e. upstream, relative speed = 13-4 = 9 km/hr
=> Time taken to go 63 km upstream = $\frac{distance}{speed}$
= $\frac{63}{9}$ = 7 hours

Distance(D) is given as 1 km
Let the speed of boat and river be B km/hr and R km/hr
in case of upstream , the speed of boat = (B-R) km/hr
in case of downstream, the speed of boat = (B+R) km/hr
in upstream it takes 10 minutes to move 1 km and takes 4 minutes in downstream for the
same distance
So , using distance = speed x time
$\frac{1}{B-R}$ = $\frac{10}{60}$
$\frac{1}{B+R}$ = $\frac{4}{60}$
B-R = 6 …..(1)
B+R = 15 ………(2)
Solving equations 1 and 2
B = 10.5 km/hr and R = 4.5 Km/hr

Let the distance between the 2 points A and B be D, the speed of Ravi be r and the speed of the stream be s.

D/(r-s) = 4
D = 4r – 4s ——-(1)

D/r = 3
D = 3r ——–(2).

Substituting (2) in (1), we get,

3r = 4r – 4s
r = 4s

While moving downstream, Ravi will travel at r + s = 4s + s = 5s
Distance to be covered = 3r = 3*4s = 12s
Time taken = 12s/5s = 2.4 hours = 2 hours 24 minutes.
Therefore, option A is the right answer.