# Top 210 Most Asked Boat and Stream Questions [ 100% FREE ]

Navigating through competitive exams requires a clear understanding of various topics, one of which is boat and stream questions. The world of boats and streams is not just about boats moving on water; it’s about understanding the intricate dance between the boat and the water’s flow. And in the realm of competitive exams, this dance translates into boat and stream questions that challenge, intrigue, and test the understanding of aspirants.

One of the primary reasons behind the popularity of boat and stream questions in Government exams is their real-world applicability. Imagine a boat navigating the waters of a stream. The way it maneuvers against or along the stream can be mathematically modeled, and this forms the basis of boat and stream questions. Given their relevance, it’s no surprise that these questions often make an appearance in the quantitative aptitude section of various exams, carrying a weightage of 1-3 marks.

But before diving headlong into boat and stream questions, it’s vital to grasp the fundamental concepts associated with boats and streams. At the heart of these questions lies the boat and stream formula. This formula, crucial for solving boat and stream questions, breaks down the dynamics of boats in different water conditions. And for those eager to delve deep into the topic, our boat and stream formula pdf offers a detailed look into the mathematical intricacies of the concept.

Understanding boats and streams is about more than just formulas; it’s about visualizing various scenarios. Consider a boat moving against the flow of water. This is termed as ‘upstream’. The speed at which the boat moves in this scenario gives us the upstream speed. On the other hand, when the boat aligns with the water’s flow, moving with the current, it’s called ‘downstream’, and we measure its net speed as the downstream speed. And of course, there are times when waters are calm, and we’re dealing with ‘still water’ scenarios, where the water speed is zero.

For many aspirants, especially those who prefer reading in Hindi, we’ve curated boat and stream questions in Hindi. This ensures that everyone, regardless of their language preference, has equal access to this crucial topic. Moreover, our boat and stream questions pdf provides a handy compilation of these questions, making it easier for aspirants to practice and master the topic.

In conclusion, boat and stream questions are not just another topic; they are a doorway to understanding the beautiful interplay between motion and resistance, between man-made vessels and nature’s flow. Whether you’re seeking the boat and stream formula or practicing with our boats and streams problems, remember that each question brings you one step closer to mastering this vital topic for your exams.

## Top 210 Most Asked Boat and Stream Questions

21. A motorboat can cover a distance of 14.4 km downstream and 6.4 km upstream in 4 hours. It can also cover 3 km downstream and 1.2 km upstream in 48 minutes. What is the speed of the boat in still water(in km/hr)?

एक नाव 4 घंटे में 14.4 किमी धारा के अनुकूल और 6.4 किमी धारा के प्रतिकूल दूरी तय कर सकती है तथा वह 48 मिनट में 3 किमी धारा के अनुकूल और 1.2 किमी धारा प्रतिकुल तय कर सकती हैं। स्थिर पानी में नाव की चाल (किमी/घंटा) में कितनी है?

Option “C” is correct.

Given:

A motorboat covers a distance of 14.4 km downstream and 6.4 km upstream in 4 hours.

It also covers 3 km downstream and 1.2 km upstream in 48 minutes.

Concept:

The direction along the stream is called downstream.

The direction against the stream is called upstream.

Formula used:

Downstream speed of the boat is

⇒ Speed of boat + speed of stream

Speed of a boat in upstream

⇒ Speed of boat – speed of stream

Speed = distance/time

Calculation:

Let x = speed of motor boat in still water and a = speed of stream

Then the speed of motor boat in downstream be (x + a) and in upstream be (x – a)

Distance covered in downstream = 14.4 km

Distance covered in upstream = 6.4 km

Total time taken in whole journey= 4 hr

For total time,

⇒ [14.4/(x + a)] + [6.4/(x – a)] = 4

⇒ [18/(x + a)] + [8/(x – a)] = 5      —-(i)

After that it covers 3 km in downstream and 1.2 km in upstream in 48 minutes

Now for total time,

⇒ [3/(x + a)] + [1.2/(x – a)] = 48/60

⇒ [30/(x + a)] + [12/(x – a)] = 480/60

⇒ [15/(x + a)] + [6/(x – a)] = 4      —-(ii)

Now multiply equation (i) by 3 and equation (ii) by 4, and subtract equation (i) from (ii) we get

⇒ [60/(x + a)] + [24/(x – a)] – [54/(x + a)] – [24/(x – a)] = 16 – 15

⇒ 6/(x + a) = 1

⇒ x + a = 6      —-(iii)

Equation (iii) shows the downstream speed of motor boat.

Now put the value in equation (i) from equation (iii) we get

⇒ 18/6 + [8/(x – a)] = 5

⇒ 8/(x – a) = 5 – 3

⇒ x – a = 4      —-(vi)

On subtracting equation (vi) from (iii) we get

⇒ 2a = 2

⇒ a = 1 km/h

Now the speed of motor boat in still water

⇒ Total downstream speed of motor boat – speed of stream

⇒ 6 – 1 = 5 km/hr

∴ The speed of motor boat in still water is 5 km/hr.

22. A girl swims triple the distance in the direction of the current as compared to the distance she covers while swimming against the current. If her speed in still water is 3.5 km/hr and time taken by her in the direction of current and against the current is 36 minutes and 48 minutes respectively, then what will be the sum of speeds of swimming with and against the current?

एक लड़की धारा प्रवाह के खिलाफ तैरने के दौरान जितनी दूरी तय करती है, उसकी तुलना में वह धारा प्रवाह की दिशा में दूरी को तीन गुना कर देती है। यदि स्थिर पानी में उसकी गति 3.5 किमी/घंटा है और धारा प्रवाह की दिशा में और धारा प्रवाह खिलाफ 36 मिनट और 48 मिनट का समय लिया जाता है, तो वर्तमान के साथ और खिलाफ तैराकी की गति का योग क्या होगा?

Option “B” is correct.

SHORTCUT TRICK

Sum of downstream and upstream speeds = D + U

= B + S + B – S

= 2 × B

= 2 × 3.5

= 7 km/hr

IMPORTANT POINT

Sometimes extra data in the question is there to waste the time.

ALTERNATE METHOD

Given:

Speed of girl = 3.5 km/hr

Time taken by girl in downstream = 36 mins.

Time taken by girl in upstream = 48 mins.

Formula Used:

Distance = Speed × Time

D = B + S

U = B – S

B = (D + U)/2

S = (D – U)/2

where, D → Downstream speed, U → Upstream speed, B → Speed of boat in still water, S → Speed of stream.

Calculations:

Let the speed of current be x km/hr.

Let the downstream and upstream speed be D km/hr and U km/hr respectively.

Let the downstream and upstream distance be d1 km and d2 km respectively.

Distance = Speed × Time

For downstream, distance (d1) = D × 36

⇒ d1 = (3.5 + x) × 36

⇒ d1 = 36(3.5 + x)

For upstream, distance (d2) = U × 48

⇒ d= (3.5 – x) × 48

⇒ d= 48(3.5 – x)

∵ d= 3d2

⇒ 36(3.5 + x) = 3 × 48(3.5 – x)

⇒ (3.5 + x)/(3.5 – x) = 4

Using componendo – dividendo

⇒ 3.5/x = (4 + 1)/(4 – 1)

⇒ x = 2.1 km/hr

D = B + x

⇒ D = (3.5 + 2.1) km/hr

⇒ D = 5.6 km/hr

U = B – x

⇒ U = (3.5 – 2.1) km/hr

⇒ U = 1.4 km/hr

Sum of downstream and upstream speeds = D + U = (5.6 + 1.4) km/hr

⇒ D + U = 7 km/hr

∴ The sum of speeds of swimming with and against the current is 7 km/hr.

23. A man can row at a speed of 15/2 km/hr in still water. He goes to a certain point upstream and back to the starting point in a river which flows at 3/2 km/hr, then the average speed of the man for the total journey is:

एक आदमी स्थिर जल में 15/2 किमी/घंटे की गति से नाव चलाता है वह नदी में एक निश्चित बिंदु तक धारा के विपरीत जाता है और वापस अपने प्रारंभिक बिंदु पर आता है जिसमें धारा की गति 3/2 किमी./घंटा है, तो कुल यात्रा में आदमी की औसत गति ज्ञात कीजिये।

Option “B” is correct.

Let the total upstream and downstream distance be 2x km,

Average speed = Distance covered/Time taken

⇒ Average speed = 2x/ [{x/ (7.5 + 1.5} + {x/ (7.5 – 1.5}]

∴ Average speed = 7.2 km/hr

24. A boat can go 20 km downstream and 30 km upstream in 2 hours 20 minutes. Also, it can go 10 km downstream and 8 km upstream in 49 minutes. What is the speed of boat downstream in km/h?

एक नाव 20 किमी प्रवाह के अनुकूल दिशा में और 30 किमी प्रवाह के प्रतिकूल दिशा में 2 घंटे 20 मिनट में जा सकती है। इसके अतिरिक्त, यह 10 किमी प्रवाह के अनुकूल दिशा में और 8 किमी प्रवाह के प्रतिकूल दिशा में 49 मिनट में तय कर सकती है। नाव की प्रवाह के अनुकूल दिशा में गति किमी/घंटा में क्या है?

Option “D” is correct.

Time in hours = 2 + 20/60 = 7/3 hours

Time in hours = 49/60 hours

Let upstream speed and downstream speed be A kmph and B kmph respectively.

⇒ 30/A + 20/B = 7/3      —- (1)

⇒ 8/A + 10/B = 49/60

Solving,

⇒ 16/A + 20/B = 49/30      —- (2)

Subtracting,

⇒ 14/A = 7/10

⇒ A = 20

⇒ 20/B = 7/3 – 30/20

⇒ B = 24

Speed downstream = 24 kmph

25. Boat M can row 60 km downstream and 20 km upstream in the same time and Boat N can row 50 km downstream and 10 km upstream in the same time in the stream running at 10 km/h. Find the difference between the speeds of boats in still water.

नाव M 60 किमी अनुप्रवाह और 20 किमी उर्ध्वप्रवाह समान समय में चलती है और नाव N 50 किमी अनुप्रवाह और 10 किमी उर्ध्वप्रवाह समान समय में 10 किमी/घंटा वाली धारा में चलती है। स्थिर पानी में नाव की चालों का अंतर ज्ञात कीजिये

Option “A” is correct.

Let the speed of the boat M in still water be M km/h and speed of the boat N in still water be N km/h.

According to the question

⇒ 60/(M + 10) = 20/(M – 10)

⇒ 60M – 600 = 20M + 200

⇒ M = 20km/h

⇒ 50/(N + 10) = 10/(N – 10)

⇒ 10N + 100 = 50N – 500

⇒ N = 15 km/h

Required difference = 20 – 15 = 5 km/h

26. The total time taken by a boat to go 120 km upstream and came back to the starting point is 8 hours. If the speed of the stream is 25% of the speed of the boat in still water, then find the difference between the upstream speed and the downstream speed of the boat.

एक नाव द्वारा 120 किमी धारा के प्रतिकूल जाने में और प्रारंभिक बिंदु पर वापस आने में लिया गया कुल समय 8 घंटे है। यदि धारा की गति स्थिर पानी में नाव की गति की 25% है, तो धारा के प्रतिकूल और धारा के अनुकूल नाव की गति के बीच अंतर ज्ञात कीजिये।

Option “A” is correct.

Given:

Time to go 120 km upstream and come back to the starting point = 8 hours

The speed of the stream is 25% of the speed of the boat in still water

Formula Used:

Downstream Speed = Speed of boat + Speed of stream

Upstream Speed = Speed of boat – Speed of stream

Calculation:

Let the speed of the boat in still water be x km/h.

So, the speed of stream = x × 25/100 = (x/4) km/h

The upstream speed of boat = x – (x/4) = (3x/4) km/h

The downstream speed of boat = x + (x/4) = (5x/4) km/h

According to the question:

[120/ (3x/4)] + [120/ (5x/4)] = 8

⇒ (480/3x) + (480/5x) = 8

⇒ 480[(5 + 3) /15x] = 8

⇒ 15x = 480

⇒ x = 32

⇒ The upstream speed of the boat = 32 × 3/4 = 24 km/h

⇒ The downstream speed of the boat = 32 × 5/4 = 40 km/h

∴ The required difference = 40 – 24 = 16 km/h

27.A boat can go 3.6 km upstream and 5.4 km downstream in 54 minutes, while it can go 5.4 km upstream and 3.6 km downstream in 58.5 minutes. The time (in minutes) taken by the boat in going 10 km downstream is:

एक नाव 3.6 किमी धारा के प्रतिकूल और 5.4 किमी धारा के अनुकूल 54 मिनट में जा सकती है, जबकि यह 5.4 किमी के प्रतिकूल और 3.6 किमी  के अनुकूल  58.5 मिनट में जा सकती है। 10 किमी धारा के अनुकूल जाने में नाव द्वारा लिया गया समय (मिनट में) है:

Option “C” is correct.

Short Trick:

U = upstream and D = downstream

3.6(U) + 5.4(D) = 54 and 5.4(U) + 3.6(D) = 58.5

Divide by 18

2U + 3D = 30 and 3U + 2D = 32.5

Now, Multiply by 3 in (1) and by 2 in (2), and then subtract (2) from (1), then

5D = 25 min

∴10D = 50 min

Detailed Method:

Let the speed of boat be x km/hr and speed of current be y km/hr

Upstream speed = (x – y) km/hr

Downstream speed = (x + y) km/hr

According to the question

3.6/(x – y) + 5.4/(x + y) = 54/60

5.4/(x – y) + 3.6/(x + y) = 58.5/60

Let 1/(x – y) = a and 1/(x + y) = b, then

3.6a + 5.4b = 54/60   —(1)

5.4a + 3.6b = 58.5/60   —(2)

Add equation (1) and equation (2)

9 (a + b) = 112.5/60

⇒a + b = 12.5/60

⇒a + b = 5/24   —(3)

Subtract equation (2) from equation (1)

1.8 (b – a) = -4.5/60

⇒b – a = -2.5/60

⇒b – a = -1/24   —(4)

Add equation (3) and equation (4)

2b = 4/24

⇒b = 1/12

⇒(x + y) = 12 km/hr

Downstream speed = 12 km/hr

∴Time taken to cover 10 km distance with downstream = 10/12 hr or (10/12) × 60 = 50 min

28. In a stream running at 3 km/h, a motorboat goes 12 km upstream and back to the starting point in 60 min. Find the speed of the motorboat in still water. (in km/h)

3 किमी/घंटा की गति से बहने वाली एक नदी में मोटरबोट धारा के प्रतिकूल 12 किमी जाती है और 60 मिनट में वापस प्रारंभिक स्थान पर आती है। तो शांत जल में मोटरबोट की गति (किमी/घंटा में) ज्ञात कीजिए।

Option “B” is correct.

Given:

In a stream running at 3 km/h, a motorboat goes 12 km upstream and back to the starting point in 60 min

Formula used:

x = [-b ± √(b2 – 4ac)]/2a

Calculation:

Let the speed of boat in still water be x, then

⇒ 12/(x – 3) + 12/(x + 3) = 1

⇒ 12 [(x + 3 + x – 3)/(x2 – 9)] = 1

⇒ 12 × 2x = x2 – 9

⇒ x2 – 24x – 9 = 0

Compare on ax2 + bx + c = 0

a = 1, b = -24 and c = -9

As we know,

⇒ x = [-b ± √(b2 – 4ac)]/2a

⇒ x = [24 ± √(242 – 4 × 1 × (-9)]/2

⇒ x = [24 ± √(576 + 36)]/2

⇒ x = [24 ± √612]/2

⇒ x = (24 ± 6√17)/2

⇒ x = 12 ± 3√17

⇒ x = 3 (4 ± √17)

Speed of boat in still water is 3 (4 + √17)

29. Ram hires a motorboat to go to a place on the other side of a lake and come back to the starting point. He goes to his destination at a speed of 125 km/hr in still water. If the speed of stream is 25 km/hr and the distance between the 2 ends is 55 km. Find the average speed of Ram in the whole journey.

राम एक झील के दूसरी ओर एक स्थान पर जाने के लिए एक मोटरबोट को किराए पर लेता हैं और शुरुआती बिंदु पर वापस आता है। वह स्थिर पानी में 125 किमी/घंटा की गति से अपने गंतव्य तक जाता है। यदि धारा की गति 25 किमी/घंटा है और 2 छोरों के बीच की दूरी 55 किमी है। पूरी यात्रा में राम की औसत गति ज्ञात कीजिये।

Option “B” is correct.

Short Trick Method/Topper’s Approach:

Let the downstream and upstream speeds be D km/hr and U km/hr respectively.

Let the speed of motorboat in still water and speed of stream be B km/hr and S km/hr respectively.

Speed = Distance/Time

D = B + S

⇒ D = (125 + 25) km/hr

⇒ D = 150 km/hr

U = B – S

⇒ U = (125 – 25) km/hr

⇒ U = 100 km/hr

For equal distance, Average speed = 2v1v2/(v1 + v2)

⇒ Average speed = 2 × 100 × 150/(100 + 150)

⇒ Average speed = 120 km/hr

∴ The average speed of Ram in the whole journey is 120 km/hr.

ALTERNATE METHOD

Given:

Distance between the 2 ends = 55 km

Speed of Ram in still water = 125 km/hr

Speed of current = 25 km/hr

Concept Used:

Time = Distance/Speed

Average speed = Total distance/Total time

Formula Used:

D = B + S

U = B – S

where, D → Downstream speed, U → Upstream speed, B → Speed of boat in still water, S → Speed of stream.

Calculations:

Let the downstream and upstream speeds be D km/hr and U km/hr respectively.

Let the speed of motorboat in still water and speed of stream be B km/hr and S km/hr respectively.

Let the total time taken in downstream and upstream be t1 and t2 hours respectively.

Speed = Distance/Time

D = B + S

⇒ D = (125 + 25) km/hr

⇒ D = 150 km/hr

U = B – S

⇒ U = (125 – 25) km/hr

⇒ U = 100 km/hr

Time = Distance/Speed

t= 55/150 = 11/30 hrs

⇒ t= (11/30) × 60 = 22 mins.

t2 = 55/100 = 11/20 hrs

⇒ t2 = (11/20) × 60 = 33 mins.

Total time in the journey = (22 + 33) mins. = 55/60 hrs.

Average speed = Total distance/Total time

⇒ Average speed = (2 × 55)/(55/60)

⇒ Average speed = 120 km/hr

The average speed of Ram in the whole journey is 120 km/hr.

30. A boat goes 28 km downstream and while returning covered only 75% of distance covered in upstream. If boat takes 3 hour more to cover upstream than downstream, then find the speed of boat in still water (km/hr) if speed of current is 5/9 m/sec.

एक नाव धारा के अनुकूल 28 किमी जाता है जबकि धारा के प्रतिकूल वापस लौटते समय केवल 75% दूरी तय करता है। यदि नाव दूरी तय करने के लिए धारा के अनुकूल की तुलना में धारा के प्रतिकूल 3 घंटे अधिक समय लगाता हैं, यदि धारा की गति 5/9 मीटर/सेकंड है, तो स्थिर जल में नाव की गति (किमी/घंटा) ज्ञात करें।

Option “B” is correct.

Let the speed of the boat in still water be x km/hr

Speed of the stream = 5/9 m/s = 2 km/hr

Speed of downstream = (x + 2) km/hr

Speed of upstream = (x – 2) km/hr

(0.75 × 28)/(x – 2) – (28/(x + 2)) = 3

21/(x – 2) – 28/(x + 2) = 3

21x + 42 – 28x + 56 = 3x2 – 12

3x2 + 7x – 110 = 0

3x+ 22x – 15x – 110 = 0

3x (x – 5) + 22(x – 5) = 0

⇒ x = 5