Practice MCQ Questions and Answer on Boats and Streams

1.

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 $$\eqalign{ & \frac{M}{4} = \frac{M}{8} + 4 \cr & \frac{M}{4} = \frac{{M + 32}}{8} \cr & M = 32 \cr & \therefore {\text{Distance = 32 }}km \cr} $$

2.

Speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is:

• (A) 16 hours
• (B) 18 hours
• (C) 20 hours
• (D) 24 hours

Show Answer

 Speed upstream = 7.5 kmph Speed downstream = 10.5 kmph ∴ Total time taken = $$\left( {\frac{{105}}{{7.5}} + \frac{{105}}{{10.5}}} \right)$$   hours = 24 hours

3.

A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:

• (A) 4 km/hr
• (B) 5 km/hr
• (C) 6 km/hr
• (D) 10 km/hr

Show Answer

 Let the speed of the stream be x km/hr Then, Speed downstream = (15 + x) km/hr Speed upstream = (15 - x) km/hr $$\eqalign{ & \therefore \frac{{30}}{{ {15 + x} }} + \frac{{30}}{{ {15 - x} }} = 4\frac{1}{2} \cr & \Rightarrow \frac{{900}}{{225 - {x^2}}} = \frac{9}{2} \cr & \Rightarrow 9{x^2} = 225 \cr & \Rightarrow {x^2} = 25 \cr & \Rightarrow x = 5\,km/hr \cr} $$

4.

On a river , Q is the mid - point between two points P and R on the same bank of the river. A boat can go from P to Q and back in 12 hours, and from P to R in 16 hours 40 minutes. How long would it take to go from R to P?

• (A) $${\text{3}}\frac{3}{7}$$ hr
• (B) 5 hr
• (C) $${\text{6}}\frac{2}{3}$$ hr
• (D) $${\text{7}}\frac{1}{3}$$ hr

Show Answer

 $$\eqalign{ & {\text{P - - - - - - - - - - }}\mathop {\text{ - }}\limits_{\text{Q}} {\text{ - - - - - - - - - - R}} \cr & {\text{ - - - - - - - 16}}{\text{.40 - - - - - - - > }} \cr & \because {\text{Time required travelling from P to R}} \cr & {\text{ = 16H 40M}} \cr & \because {\text{PQ = QR}}\left( {\because {\text{Q is mid point}}} \right){\text{ }} \cr & {\text{Required time to travel PQ}} \cr & {\text{ = }}\frac{1}{2}\left( {{\text{16H 40M}}} \right) = \,\,{\text{8H 20M}} \cr & {\text{total required time from}} \cr & {\text{P to Q + Q to P = 12H}} \cr & {\text{ = 8H 20M + Q to P = 12H}} \cr & {\text{Time Q to P}} \cr & {\text{ = 12H - 8H 20M}} \cr & {\text{ = 3H 40M}} \cr & \because \overrightarrow {{\text{QP}}} {\text{ = }}\frac{1}{2}\overrightarrow {{\text{RP}}} {\text{ }}\,\,\left( {{\text{Distance}}} \right) \cr & \therefore {\text{Required time to }}\overrightarrow {{\text{RP}}} {\text{ }} \cr & {\text{ = 2}} \times \left[ {\overrightarrow {{\text{QP}}} } \right] \cr & {\text{ = 2}}\left[ {3{\text{H 40M}}} \right]{\text{ }} \cr & {\text{ = 7 H 20M}} \cr & {\text{ = 7}}\frac{1}{3}{\text{hr}}{\text{ }} \cr} $$

5.

A boat can cover a distance of 56 km downstream in 3.5 hours. The ratio of the boat in still water and the speed of stream is 3 : 1. How much time (in hours) will the boat take to cover a distance of 41.6 km downstream?

• (A) 2.1
• (B) 1.5
• (C) 2.6
• (D) 1.8

Show Answer

 Speed of boat in downstream $$\eqalign{ & = \frac{{56}}{{3.5}} \cr & = \frac{{56}}{7} \times 2 \cr & = 16{\text{ km/hr}} \cr} $$ ∴ Required time $$ = \frac{{41.6}}{{16}} = 2.6{\text{ hrs}}$$

6.

A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water?

• (A) 40 minutes
• (B) 1 hour
• (C) 1 hr 15 min
• (D) 1 hr 30 min

Show Answer

 $$\eqalign{ & {\text{Rate}}\,{\text{downstream}} \cr & = \left( {\frac{1}{{10}} \times 60} \right)km/hr = 6\,km/hr \cr & {\text{Rate}}\,{\text{upstream}} = 2\,km/hr \cr & {\text{Speed}}\,{\text{in}}\,{\text{still}}\,{\text{water}} \cr & = \frac{1}{2}\left( {6 + 2} \right)km/hr = 4\,km/hr \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = {\frac{5}{4}} \,hrs \cr & = 1\frac{1}{4}hrs \cr & = 1\,hr\,15\,\min . \cr} $$

7.

A motorboat goes 24 km in 2 hour along the stream and 10 km in 1 hour against the stream. The speed of the motorboat in kilometres per hour, is:

• (A) 14
• (B) 10
• (C) 11
• (D) 12

Show Answer

 $$\eqalign{ & {\text{B}} + {\text{W}} = 12{\text{ km/h}} \cr & {\text{B}} - {\text{W}} = 10{\text{ km/h}} \cr & \overline {\,{\text{B}} = \frac{{22}}{2} = 11{\text{ km/h}}\,} \cr} $$

8.

The speed of a boat in still water in 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is:

• (A) 1.2 km
• (B) 1.8 km
• (C) 2.4 km
• (D) 3.6 km

Show Answer

 Speed downstream = (15 + 3) kmph = 18 kmph Distance travelled = $$\left( {18 \times \frac{{12}}{{60}}} \right)$$   km = 3.6 km

9.

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 $$\eqalign{ & = \frac{{60}}{{15 - 5}} + \frac{{60}}{{15 + 5}} \cr & = 6 + 3 \cr & = 9{\text{ hr}} \cr} $$

10.

A boat can go 40 km downstream and 25 km upstream in 7 hours 30 minutes. It can go 48 km downstream and 36 km upstream in 10 hours. What is the speed (in km/h) of the boat in still water?

• (A) 6
• (B) 12
• (C) 9
• (D) 15

Show Answer

 $$\eqalign{ & {\text{Speed of boat}} = x \cr & {\text{Speed of current}} = y \cr & {\text{According to the question,}} \cr & \frac{{40}}{{x + y}} + \frac{{25}}{{x - y}} = \frac{{15}}{2} \cr & {\text{Multiplying both sides by}}\frac{6}{5}{\text{ and}} \cr & \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = \frac{{18}}{2} \cr & \frac{{40}}{{x + y}} + \frac{{30}}{{x - y}} = 9\,.\,.\,.\,.\,.\,.\,.\,\left( 1 \right) \cr & \frac{{48}}{{x + y}} + \frac{{36}}{{x - y}} = 10\,.\,.\,.\,.\,.\,.\,.\,\left( 2 \right) \cr & \underline {\,\, - \,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{ - 6}}{{x - y}} = - 1 \cr & x - y = 6\,.\,.\,.\,.\,.\,.\,.\,\left( 3 \right) \cr & {\text{From equation }}\left( 2 \right), \cr & \frac{{48}}{{x + y}} + \frac{{36}}{6} = 10 \cr & \frac{{48}}{{x + y}} = 4 \cr & x + y = 12\,.\,.\,.\,.\,.\,.\,.\,\left( 4 \right) \cr & {\text{From equation }}\left( 3 \right){\text{ and }}\left( 4 \right) \cr & x - y = 6 \cr & x + y = 12 \cr & \overline {\,x = 9,\,y = 3\,} \cr & {\text{Speed of boat}} = 9 \cr} $$