Practice MCQ Questions and Answer on Boats and Streams

31.

In one hour, a boat goes 11 km/hr along the stream and 5 km/hr against the stream. The speed of the boat in still water (in km/hr) is:

• (A) 3 kmph
• (B) 5 kmph
• (C) 8 kmph
• (D) 9 kmphr

Show Answer

 Speed in still water = $$\frac{1}{2}$$(11 + 5) kmph = 8 kmph

32.

A man can row 6 km/hr in still water. If the speed of the current is 2 km/hr he takes 4 hours more in upstream than the downstream. The distance is?

• (A) 30 km
• (B) 24 km
• (C) 20 km
• (D) 32 km

Show Answer

 Speed of man in still water, x = 6 km/h Speed of current, y = 2 km/h Let distance = M According to question Upstream time = Downstream time + 4 $$\eqalign{ & \frac{M}{4} = \frac{M}{8} + 4 \cr & \frac{M}{4} = \frac{{M + 32}}{8} \cr & M = 32 \cr & \therefore {\text{Distance = 32 }}km \cr} $$

33.

The speed of a boat in still water is 15 km/h, and the speed of the current is 5 km/h. In how much time (in hours) will the boat travel a distance of 60 km upstream and the same distance downstream?

• (A) 10 hr
• (B) 9 hr
• (C) 20 hr
• (D) 12 hr

Show Answer

 Speed of boat in still water = 15 km/hr Speed of stream = 5 km/hr Total time of upstream and downstream $$\eqalign{ & = \frac{{60}}{{15 - 5}} + \frac{{60}}{{15 + 5}} \cr & = 6 + 3 \cr & = 9{\text{ hr}} \cr} $$

34.

Speed of a along and against the current are 14 kms/hr and 8 kms/hr respectively. The speed of the current is?

• (A) 11 km/hr
• (B) 6 km/hr
• (C) 5.5 km/hr
• (D) 3 km/hr

Show Answer

 S + W = 14 . . . . . . (i) S - W = 8 . . . . . . . .(ii) _________________ from equation (i) & (ii) S = 11 km/hr, W = 3 km/hr Speed of current = 3 km/hr

35.

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

• (A) 2 km/hr
• (B) 2.5 km/hr
• (C) 3 km/hr
• (D) 4 km/hr

Show Answer

 Let the speed of the stream be x km/hr Then speed downstream = (10 + x) km/hr Speed upstream $$\eqalign{ & {\text{ = }}\left( {10 - x} \right)km/hr \cr & \therefore \frac{{26}}{{\left( {10 + x} \right)}} = \frac{{14}}{{\left( {10 - x} \right)}} \cr & \Rightarrow 260 - 26x = 140 + 14x \cr & \Rightarrow 40x = 120 \cr & \Rightarrow x = 3\,km/hr \cr} $$

36.

The speed of a boat downstream is 15 km/hr and the speed of current is 3 km/hr. Find the total time taken by the boat to cover 15 km upstream and 15 km downstream.

• (A) 2 hours 40 minutes
• (B) 2 hours 42 minutes
• (C) 2 hours 10 minutes
• (D) 2 hours 30 minutes

Show Answer

 Given, Speed of boat in downstream = 15 km/hr Speed of current = 3 km/hr Speed of boat in still water = 12 km/h Time taken at upstream $$\eqalign{ & = \frac{{15}}{{12 - 3}} \cr & = \frac{{15}}{9}{\text{hr}} \cr & = 1\,{\text{hour}}\,40\,{\text{minutes}} \cr} $$ Time taken at downstream $$\eqalign{ & = \frac{{15}}{{12 + 3}} \cr & = \frac{{15}}{{15}}{\text{hr}} \cr & = 1\,{\text{hour}}\, \cr} $$ ∴ Total time = 2 hours 40 minutes

37.

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?

• (A) 4 kmph
• (B) 6 kmph
• (C) 8 kmph
• (D) Data inadequate

Show Answer

 Rate downstream = $$\frac{{16}}{2}$$ kmph = 8 kmph Rate upstream = $$\frac{{16}}{4}$$ kmph = 4 kmph ∴ Speed in still water = $$\frac{1}{2}$$(8 + 4) kmph = 6 kmph

38.

A boat can go 5 km upstream and $$7\frac{1}{2}$$ km downstream in 45 minutes. If can also go 5 km downstream and 2.5 km upstream in 25 minutes. How much time (in minutes) will it take to go 6 km upstream?

• (A) 30
• (B) 36
• (C) 24
• (D) 32

Show Answer

 $$\eqalign{ & \frac{5}{{x - y}} + \frac{{15}}{{2\left( {x + y} \right)}} = \frac{3}{4}{\text{hr }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & \frac{{2.5 \times 2}}{{x - y}} + \frac{{5 \times 2}}{{x + y}} = \frac{{5 \times 2}}{{12}}{\text{hr }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{ii}}} \right) \cr & \frac{5}{{2\left( {x + y} \right)}} = \frac{1}{{12}} \cr & x + y = 30 \cr & {\text{Put }}x + y = 30{\text{ in equation }}\left( {\text{i}} \right) \cr & \frac{5}{{x - y}} + \frac{{15}}{{2 \times 30}} = \frac{3}{4} \cr & \frac{5}{{x - y}} = \frac{1}{2} \cr & x - y = 10 \cr & {\text{Now,}} \cr & \,x + y = 30 \cr & \underline {\,x - y = 10\,} \cr & x = 20 \cr & y = 10 \cr & \frac{{6\,{\text{km}}}}{{x - y}} = \frac{6}{{10}} \times 60 = 36\min \cr} $$

39.

A man can row at a speed of $$4\frac{1}{2}$$ km/hr in still water. If he takes 2 times as long to row a distance upstream as to row the same distance downstream, then the speed of stream (in km/hr) is-

• (A) 1
• (B) 1.5
• (C) 2
• (D) 2.5

Show Answer

 $$\eqalign{ & {\text{Speed of man in still water , }} \cr & x{\text{ }} = {\text{ }}\frac{9}{2}km/hr{\text{ }} \cr & {\text{let speed of stream = }}y{\text{ }}km/h \cr & {\text{Downstream speed = }} {\frac{9}{2} + y} \cr & {\text{Upstream speed = }} {\frac{9}{2} - y} \cr & {\text{Accroding to questions,}} \cr & {\text{2}} \times {\frac{{{\text{Distance}}}}{{ {\frac{9}{2} + y} }}} = \frac{{{\text{Distance}}}}{{ {\frac{9}{2} - y} }} \cr & \Rightarrow \frac{2}{{\frac{9}{2} + y}} = \frac{1}{{\frac{9}{2} - y}} \cr & \Rightarrow \frac{{2 \times 2}}{{9 + 2y}} = \frac{2}{{9 - 2y}} \cr & \Rightarrow \frac{2}{{9 + 2y}} = \frac{1}{{9 - 2y}} \cr & \Rightarrow 18 - 4y = 9 + 2y \cr & \Rightarrow 6y = 9 \cr & \Rightarrow y = \frac{9}{6} = \frac{3}{2} = 1.5\,km/h \cr} $$

40.

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. The speed of the stream is:

• (A) 1.5 km/h
• (B) 3.5 km/h
• (C) 1.8 km/h
• (D) 1 km/h

Show Answer

 $$\eqalign{ & {\text{Let, speed of boat}} = x \cr & {\text{Speed of stream}} = y \cr & \Rightarrow \frac{{48}}{{x + y}} + \frac{{48}}{{x - y}} = 14\,.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr & {\text{and, }}\frac{4}{{x + y}} = \frac{3}{{x - y}} \cr & \frac{{x - y}}{{x + y}} = \frac{3}{4} = \frac{6}{8} \cr & {\text{From equation }}\left( {\text{i}} \right), \cr & \frac{{48}}{8} + \frac{{48}}{6} = 14 \cr & 14 = 14\,\,\,\left( {{\text{Satisfied}}} \right) \cr & {\text{So, }}x - y = 6 \cr & \,\,\,\,\,\,\,\,\,x + y = 8 \cr & \,\,\,\,\,\overline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 1\,\,} {\text{km/hr}} \cr} $$