# 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

 201. The ratio of speed of stream and boat is 1: 7. If a boat covers 120 km upstream and downstream in 3.5 hours, then find the sum of speed of boat and stream.धारा और नाव की गति का अनुपात 1: 7 है। यदि एक नाव धारा के प्रतिकूल और धारा के अनुकूल 120 किमी 3.5 घंटे में तय करती है, तो नाव और धारा की गति का योग ज्ञात कीजिए।A. 70 kmphB. 50 kmphC. 140 kmphD. 80 kmphE. 120 kmph Option “D” is correct.The ratio of speed of stream and boat = 1: 7 (x, 7x)[120/(7x – x)] + [120/(7x + x)] = 3.535/x = 3.5x = 10 Required sum = 8 * 10 = 80 kmph 202. Ratio of the speed of the boat and stream is 7:3 respectively. If the boat covers 180 km upstream and the same distance in downstream in 10.5 hours, then find the time taken by boat to cover 210 km in still water?नाव और धारा की गति का अनुपात क्रमशः 7:3 है। यदि नाव धारा के प्रतिकूल 180 किमी और धारा के अनुकूल समान दूरी 10.5 घंटे में तय करती है, तो स्थिर जल में नाव द्वारा 210 किमी की दूरी तय करने में कितना समय लगेगा?A. 6 hoursB. 5 hoursC. 3 hoursD. 7 hoursE. None of theseOption “B” is correct. Speed of the boat=7xSpeed of the stream=3x180/(7x+3x)+180/(7x-3x)=10.518/x+45/x=10.563/10.5=xx=6Speed of the boat=7*6=42 kmphRequired time=210/42=5 hours 203. Time taken by a boat to cover ‘D’ km upstream is 6 hours and the same boat covers 1.25D km downstream in 3 hours. The boat takes 13 hours to cover 75% of D + 4 km downstream and 200% of D – 8 km upstream. Find the distance covered by the boat in 3 hours in downstream?एक नाव द्वारा धारा के प्रतिकूल ‘D’ किमी की दूरी तय करने में 6 घंटे का समय लगता है और वही नाव धारा के अनुकूल 1.25D किमी को 3 घंटे में तय करती है। नाव को धारा के अनुकूल D का 75% + 4 किमी और धारा के प्रतिकूल D का 200% – 8 किमी को तय करने में 13 घंटे लगते हैं। धारा के अनुकूल नाव द्वारा 3 घंटे में तय की गई दूरी ज्ञात कीजिए?A. 60 kmB. 54 kmC. 48 kmD. 57 kmE. None of theseOption “A” is correct.Downstream speed = 1.25D/3 km/hrUpstream speed = D/6 km/hrATQ,(2D – 8) * (6/D) + (0.75D + 4) * (3/1.25D) = 1317.25D – 48 = 16.25DD = 48Downstream speed = (1.25 x 48)/3 =20 km/hrRequired downstream distance = 20 *3 = 60 km 204. A boat covers 540 km downstream in t minutes. If the speed of the boat is increased by 20%, then the time taken by the same boat to cover the same downstream distance is (t – 24) minutes. If the ratio of the speed of the boat in still water to current is 5:4, then find the upstream speed of the boat?एक नाव t मिनट में अनुप्रवाह 540 किमी की दूरी तय करती है। यदि नाव की गति में 20% की वृद्धि की जाती है, तो समान नाव द्वारा समान अनुप्रवाह दूरी को तय करने में लिया गया समय (t – 24) मिनट है। यदि शांत जल में नाव की गति से धारा का अनुपात 5:4 है, तो नाव की उर्ध्वप्रवाह गति ज्ञात कीजिए?A. 12 kmphB. 15 kmphC. 18 kmphD. 20 kmphE. None of theseOption “B” is correct.Speed of boat = 5xSpeed of current = 4x540/(9x) – 540/(5x * 120/100 + 4x) = 24/60x = 15Speed of upstream = 5x-4x =15 kmph 205. A yacht can row at the speed of 20 meters per second in still water. If the speed of stream is 1/9 times the still water speed of the yacht then, by what percent should the speed of yacht be increased so that it can cover 1362 km downstream in 15 hours?एक नाव शांत पानी में 20 मीटर प्रति सेकंड की गति से तैर सकता है। यदि धारा की गति नाव की स्थिर पानी की गति का 1/9 गुना है, तो नाव की गति को कितने प्रतिशत तक बढ़ाया जाना चाहिए ताकि वह 15 घंटे में 1362 किमी अनुप्रवाह तय कर सके?A. 20%B. 10%C. 15%D. 25%E. None of the aboveOption “C” is correct.After increasing speed, downstream speed of the yacht = 1362/15 = 90.8 km per hourIn still water, the speed of the yacht = 20*18/5 = 72 km per hourThe speed of stream = 72/9 = 8 km per hourLet it increased its still water speed by a% then (100 + a)% of 72 + 8 = 90.8(100 + a)% of 72 = 82.8(18 (a + 100))/25 = 82.8a + 100 = 115Therefore, a = 15% 206. A boat covers 450 km along the stream in 15 hours and the same boat covers 140 km against the stream in 7 hours. What is the ratio of the speed of the stream and the speed of the boat in still water?एक नाव धारा के साथ 450 किमी की दूरी 15 घंटे में तय करती है और वही नाव 7 घंटे में धारा के विपरीत 140 किमी की दूरी तय करती है। शांत जल में धारा की गति और नाव की गति का अनुपात कितना है?A. 2:1B. 3:1 C. 5:4D. 5:3E. None of theseOption “E” is correct.Speed of the boat along the stream = 450/15 = 30 kmphSpeed of the boat against the stream = 140/7 = 20 kmphSpeed of the boat in still water = (30 + 20)/2 = 25 kmphSpeed of the stream = (30 – 20)/2 = 5 kmphRequired ratio = 5:25 = 1:5 207. The upstream speed of the boat is 60% of the downstream speed of the boat. If the boat covers 128 km in still water in 4 hours, then find the speed of the stream.नाव की धारा के प्रतिकूल गति नाव की धारा के अनुकूल गति का 60% है। यदि नाव स्थिर जल में 128 किमी की दूरी 4 घंटे में तय करती है, तो धारा की गति ज्ञात कीजिए।A. 8 km/hrB. 10 km/hrC. 4 km/hrD. 12 km/hrE. None of theseOption “A” is correct.The ratio of the downstream to the upstream speed of the boat = 100:60 = 5:3The speed of the boat in still water = 128/4 = 32 km/hr(5x + 3x)/2 = 328x = 32 * 2x = 8The downstream speed of the boat = 5 * 8 = 40 km/hrThe upstream speed of the boat = 3 * 8 = 24 km/hrThe speed of the stream = (40 – 24)/2 = 16/2 = 8 km/hr 208. The time taken by the boat to row 120 km along with stream is 2.5 hours and the time taken by same boat to cover the same distance against stream is 5 hours, then find the speed of the boat in still water.धारा के अनुकूल 120 किमी नाव की सैर करने में लिया गया समय 2.5 घंटे है और धारा के प्रतिकूल उसी दूरी को तय करने में उसी नाव के द्वारा लिया गया समय 5 घंटे है, फिर ठहरे हुए पानी में नाव की गति को ज्ञात कीजिए।A. 36 kmphB. 45 kmphC. 54 kmphD. 60 kmphE. 28 kmphOption “A” is correct.Speed of the boat = xSpeed of the stream = y120/x + y = 2.5120/(x – y) = 52.5 * (x + y) = 5 * (x – y)x + y = 2x – 2yx = 3y120/(3y + y) = 2.54y = 48y = 12 kmphx = 3 * 12 = 36 kmph 209. The time taken by the boat to cover x km downstream is equal to the time taken by the boat to cover (x-80) km upstream. The ratio of the speed of the boat in still water to stream is 4:1. If the difference of the downstream and upstream speed of the boat is 20 km/hr, then find the value of x?नाव द्वारा x किमी धारा के अनुकूल तय करने में लिया गया समय नाव द्वारा धारा के प्रतिकूल (x-80) किमी की दूरी तय करने में लगने वाले समय के बराबर है। स्थिर जल में नाव की गति से धारा का अनुपात 4:1 है। यदि नाव की धारा के अनुकूल और धारा के प्रतिकूल गति का अंतर 20 किमी/घंटा है, तो x का मान ज्ञात कीजिए?A. 240B. 160C. 200D. 120E. None of theseOption “C” is correct.The speed of the stream = (Downstream speed – upstream speed)/2 = 20/2 = 10 km/hrThe speed of the boat in still water = 10 * 4/1 = 40 km/hrx/(40+10) = (x-80)/(40-10)x/50 = (x-80)/30x – 80 = 3x/52x = 400x = 200 km 210. The total time taken to travel a distance of x Km against the stream and to travel a distance of (x+100) km along the stream is 7 hours. The stream’s speed is 14.28% of the downstream speed. If the speed of the stream is 4 kmph, then find the value of x.धारा के विपरीत x किमी की दूरी तय करने और धारा के अनुकूल (x+100) किमी की दूरी तय करने में कुल 7 घंटे लगते हैं। धारा की गति धारा के अनुकूल गति की 14.28% है। यदि धारा की गति 4 किमी प्रति घंटा है , तो x का मान ज्ञात कीजिए।A. 10B. 20C. 25D. 35E. 40Option “E” is correct.Stream speed = 4 kmphStream speed = 1/7 * downstream speed4 = 1/7 * downstream speedDownstream speed = 28 km /hrUpstream speed =28 – 4 -4 = 20 kmph(x+100)/28 + (x)/20 =7(5x + 500 + 7x) /140 =75x+ 500+7x =98012x = 980 -50012x = 480x =40 km