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Download Boats and Stream Questions for IBPS PO Prelims<\/a><\/p>\n Download IBPS PO Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>A boat can cover 336 km distance downstream in 4.2 hours. The same boat can cover the same distance in still water in 6.72 hours. Find out the speed of the stream.<\/p>\n a)\u00a035 km\/h<\/p>\n b)\u00a025 km\/h<\/p>\n c)\u00a040 km\/h<\/p>\n d)\u00a030 km\/h<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The same boat can cover the same distance in still water in 6.72 hours. Question 2:\u00a0<\/b>A boat covers a 20 km distance towards downstream in 4 hours. If the speed of the boat is 1.5 times the speed of the current then find the time taken by the boat to cover the same distance in upstream.<\/p>\n a)\u00a022 hours<\/p>\n b)\u00a020 hours<\/p>\n c)\u00a025 hours<\/p>\n d)\u00a0Can\u2019t be determined<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat and speed of the stream be x km\/hr and y km\/hr respectively. Time taken to cover 20 km in upstream, let it be \u2018T\u2019 Question 3:\u00a0<\/b>If the ratio between the time taken by the boat to cover 280 km distance in upstream to the same distance in downstream is 5:2 respectively. The speed of the stream is 42.85% of the speed of a boat. Find out the speed of the boat.<\/p>\n a)\u00a07 km\/h<\/p>\n b)\u00a014 km\/h<\/p>\n c)\u00a06 km\/h<\/p>\n d)\u00a0Cannot be determined.<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The speed of the stream is 42.85% of the speed of a boat. $\\frac{280}{4y} = 5t$ Downstream \u21d2 $\\frac{280}{7y+3y} = 2t$<\/p>\n $\\frac{280}{10y} = 2t$ Here Eq.(i) and Eq.(ii) are the same. So from the given information, we cannot determine the speed of the boat. Question 4:\u00a0<\/b>A boat can cover 720 km distance downstream in 40 hours. The speed of boat in still water is double the speed of the stream. Then find out the time taken by boat to cover 480 km distance in upstream.<\/p>\n a)\u00a0100 hours<\/p>\n b)\u00a060 hours<\/p>\n c)\u00a040 hours<\/p>\n d)\u00a080 hours<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the speed of boat in still water and the speed of stream is B and C respectively. B+C = 18 km\/h Eq.(i) = $\\frac{480}{6}$<\/p>\n = 80 hours Question 5:\u00a0<\/b>A boat can travel 72 km against the stream in 8 hours. If the speed of the stream is $\\dfrac{1}{3}$rd of the speed of the boat upstream, then the time taken by the boat to travel 285 km along the stream is:<\/p>\n a)\u00a017.25 hours<\/p>\n b)\u00a018.5 hours<\/p>\n c)\u00a021 hours<\/p>\n d)\u00a019 hours<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the boat upstream = $\\dfrac{72}{8} = 9$ km\/hr. Question 6:\u00a0<\/b>A boat can cover 390 km distance upstream in 78 hours. The speed of the stream is 44.44% of the speed of the boat. Then find out the time taken by boat to cover the same distance downstream.<\/p>\n a)\u00a030 hours<\/p>\n b)\u00a035 hours<\/p>\n c)\u00a025 hours<\/p>\n d)\u00a020 hours<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the speed of the boat in still water is B and the speed of the stream is C. $\\frac{5}{9}\\times B = 5$ = 30 hours Question 7:\u00a0<\/b>The speed of a boat downstream is five times the speed of the stream. If the boat can travel 120 km in 5 hours while in upstream, then find the time taken by the boat to travel 480 km downstream and 180 km upstream.<\/p>\n a)\u00a019.5 hours<\/p>\n b)\u00a016.8 hours<\/p>\n c)\u00a012.75 hours<\/p>\n d)\u00a018.5 hours<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the stream be \u2018x\u2019 km\/hr. Speed of the boat upstream = 3x = 24 km\/hr. Question 8:\u00a0<\/b>A boat travels 82 km downstream and returns. If speed of the boat is 25 km\/hr and speed of the stream is 8km\/hr then the time taken by the boat to reach back its starting point is ________?<\/p>\n a)\u00a05.3 hr<\/p>\n b)\u00a06.3 hr<\/p>\n c)\u00a07.3 hr<\/p>\n d)\u00a08.3 hr<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Relative speed of boat in downstream = 25 + 8 = 33 km\/hr Question 9:\u00a0<\/b>A boat covers a 28 km distance towards upstream in 4 hours. If the speed of the boat is three times the speed of the current then find the time taken by the boat to cover the 56km distance downstream.<\/p>\n a)\u00a02 hour<\/p>\n b)\u00a04.5 hour<\/p>\n c)\u00a04 hour<\/p>\n d)\u00a03.5 hour<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat and speed of the stream be x km\/hr and y km\/hr respectively. Question 10:\u00a0<\/b>Total time taken by a boat to cover 136km in downstream and 60km in upstream is 32 hours. If the speed of the boat in still water is three times the speed of stream. Then what is the downstream speed of that boat?<\/p>\n a)\u00a06 km\/hr<\/p>\n b)\u00a012 km\/hr<\/p>\n c)\u00a08 km\/hr<\/p>\n d)\u00a010 km\/hr<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the Speed of boat in still water and speed of stream is B and C respectively. If the speed of boat is three times the speed of stream. $\\dfrac{136}{4C} + \\dfrac{60}{2C} = 32 $<\/p>\n $\\dfrac{34}{C} + \\dfrac{30}{C} = 32 $<\/p>\n $\\dfrac{64}{C} = 32 $<\/p>\n $\\dfrac{64}{32} = C $<\/p>\n C = 2km\/hr. Eq.(3) Take Free IBPS PO Mock Tests<\/a><\/p>\n Question 11:\u00a0<\/b>Speed of a boat while travelling along the stream and against the stream are in the ratio 5 : 3. If the speed of the stream is 8 km\/hr, then in how much time can the boat travel 32 km when the water is still?<\/p>\n a)\u00a01.5 hours<\/p>\n b)\u00a01 hour<\/p>\n c)\u00a02.25 hours<\/p>\n d)\u00a01.75 hours<\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat downstream = 5x km\/hr. Speed of the stream = Speed of the boat in Still water – Speed of the boat upstream = Speed of the boat downstream – Speed of the boat in still water = 4x-3x (or) 5x-4x = x km\/hr.<\/p>\n Given, x = 8 km\/hr. Question 12:\u00a0<\/b>Speed of boat in still water is 125% more than the speed of stream. Total time taken by boat to cover 260km in downstream and same distance in upstream is 72 hours. Then find out the speed of the boat in still water?<\/p>\n a)\u00a010 km\/hr<\/p>\n b)\u00a012 km\/hr<\/p>\n c)\u00a06 km\/hr<\/p>\n d)\u00a09 km\/hr<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the speed of the boat in still water is B and speed of stream is C. $B = \\frac{9}{4} \\times C$<\/p>\n $B = \\frac{9}{4} C$ Eq.(1)<\/p>\n Total time taken by boat to cover 260km in downstream and same distance in upstream is 72 hours. Put Eq.(1) in Eq.(2). $\\dfrac{260}{\\frac{(9+4)}{4}C} + \\dfrac{260}{\\frac{(9-4)}{4}C} = 72$<\/p>\n $\\dfrac{260}{\\frac{(13)}{4}C} + \\dfrac{260}{\\frac{(5)}{4}C} = 72$<\/p>\n $\\dfrac{1040}{13C} + \\dfrac{1040}{5C} = 72$<\/p>\n $\\dfrac{80}{1C} + \\dfrac{208}{1C} = 72$<\/p>\n $\\dfrac{288}{1C} = 72$<\/p>\n C = $\\dfrac{288}{72}$ = 4 km\/hr. Eq.(3)<\/p>\n Put Eq.(3) in Eq.(1). $B = \\frac{9}{4} \\times 4$<\/p>\n Speed of the boat in still water = B = 9 km\/hr. Question 13:\u00a0<\/b>A boat covers a 30 km distance towards downstream in 4 hours. If the speed of the boat is two times the speed of the current then find the time taken by the boat to cover the same distance in upstream?<\/p>\n a)\u00a012hour<\/p>\n b)\u00a014.5hour<\/p>\n c)\u00a010hour<\/p>\n d)\u00a0Can\u2019t be determined<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat and speed of the stream be x km\/hr and y km\/hr respectively. Question 14:\u00a0<\/b>If a boat’s speed is increased by 50% then it can cover 25% more distance than at its normal speed taking 1 hour less than its original time in still water. Find the time taken by the boat to cover its actual distance?<\/p>\n a)\u00a05 hr<\/p>\n b)\u00a06 hr<\/p>\n c)\u00a04 hr<\/p>\n d)\u00a0Can\u2019t be determined<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat be \u2018V\u2019 km\/hr and distance be \u2018D\u2019 km and time be \u2018T\u2019 hour Question 15:\u00a0<\/b>A boat covers 20 km downstream distance in 2 hours and the ratio of upstream speed to the speed of the current is 2:3. Find the speed of the boat in still water.<\/p>\n a)\u00a07.25km\/hr<\/p>\n b)\u00a03.75km\/hr<\/p>\n c)\u00a07km\/hr<\/p>\n d)\u00a06.25km\/hr<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat be x km\/hr and speed of current be y km\/hr Question 16:\u00a0<\/b>In a river, Speed of boat in still water is 36km\/h and speed of current is $\\frac{1}{9}$ of the speed of the boat. Distance covered by the boat in downstream during 4 hours is 32km more than the distance covered by boat in upstream during 4 hours. Then find out the distance covered by boat in upstream during 4 hours?<\/p>\n a)\u00a0128km<\/p>\n b)\u00a0160km<\/p>\n c)\u00a0150km<\/p>\n d)\u00a0148km<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of boat in still water = B = 36km\/h Eq.(1) Speed of current = C = $\\frac{1}{9}$ of 36km\/h = $\\frac{1}{9}\\times36$ = 4km\/h Eq.(2)<\/p>\n Speed of boat in downstream = (B+C) = (36+4) = 40 Eq.(3) $\\dfrac{d}{32} = \\dfrac{(d+32)}{40}$<\/p>\n $\\dfrac{d}{4} = \\dfrac{(d+32)}{5}$<\/p>\n 5d = 4(d+32) Question 17:\u00a0<\/b>In a river, the speed of a boat is six times the speed of stream. Time taken by boat to cover the same distance in upstream and downstream is 14 hours and T hours respectively. Then find out the value of T?<\/p>\n a)\u00a010 hours<\/p>\n b)\u00a012 hours<\/p>\n c)\u00a011 hours<\/p>\n d)\u00a09 hours<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the speed of a boat is B and speed of stream is C. Question 18:\u00a0<\/b>A boat can row 105km upstream in 6hr and same distance covered by that boat downstream in 4hr. Then what is the speed of that boat in still water?<\/p>\n a)\u00a03.575 km\/hr.<\/p>\n b)\u00a04.375 km\/hr.<\/p>\n c)\u00a05.305 km\/hr.<\/p>\n d)\u00a06.105 km\/hr.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let\u2019s assume the speed of the boat in still water is B. (B-C) = 17.5 Eq.(1) (B+C) = 26.25 Eq.(2)<\/p>\n Subtract Eq.(1) from Eq.(2). Speed of that boat in still water = C = 4.375 km\/hr. Question 19:\u00a0<\/b>A boat can travel 10 km in 12 minutes when the water is still. If the ratio between the speeds of the boat in still water and the speed of the current is 5 : 1, then find the time taken by the boat to travel 80 km against the stream.<\/p>\n a)\u00a03.5 hours<\/p>\n b)\u00a02.75 hours<\/p>\n c)\u00a02 hours<\/p>\n d)\u00a01.25 hours<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Speed of the boat in still water = $\\dfrac{10}{\\dfrac{12}{60}} = 50$ km\/hr. Speed of the boat against the stream = Speed of the boat in still water – Speed of the stream = 50-10 = 40 km\/hr. Question 20:\u00a0<\/b>A boat travels 30 km upstream in 10 hours and travels 52 km downstream in 4 hours. What is the time taken to cover 121 km downstream if the speeds of both stream and boat are decreased by 1 km\/hr ?<\/p>\n a)\u00a012 hrs<\/p>\n b)\u00a011 hrs<\/p>\n c)\u00a010 hrs<\/p>\n d)\u00a09 hrs<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the speed of the boat be \u2018b\u2019 and stream be \u2018s\u2019. Question 21:\u00a0<\/b>A boat travels 120 km upstream in 6 hours and travels 90 km downstream with double the speed of upstream. What is the time taken to cover the 90km downstream ?<\/p>\n a)\u00a02 hr 10 min<\/p>\n b)\u00a02 hr 15 min<\/p>\n c)\u00a02 hr 20 min<\/p>\n d)\u00a02 hr 30 min<\/p>\n 21)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the speed of the boat be \u2018b\u2019 and stream be \u2018s\u2019. Instructions<\/b><\/p>\n Question 22:\u00a0<\/b>A boat takes 4 hours to travel same distance \u2018x\u2019 upstream and downstream.If the speed of the boat is 4 km\/hr and stream is 2 km\/hr then find the value of x ?<\/p>\n a)\u00a06 km<\/p>\n b)\u00a08 km<\/p>\n c)\u00a04 km<\/p>\n d)\u00a010 km<\/p>\n 22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Upstream speed =4-2=2 km\/hr Instructions<\/b><\/p>\n Question 23:\u00a0<\/b>A boat takes 10 hours to travel same distance \u2018x\u2019 upstream and downstream.If the speed of the boat is 10 km\/hr and stream is 6 km\/hr then find the value of x ?<\/p>\n a)\u00a024 km<\/p>\n b)\u00a028 km<\/p>\n c)\u00a030 km<\/p>\n d)\u00a032 km<\/p>\n 23)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Upstream speed =10-6=4 km\/hr Question 24:\u00a0<\/b>A boat can travel with double the speed downstream than upstream. If the speed of current is 24 km\/hr, then find the speed of the boat in still water.<\/p>\n a)\u00a064 km\/hr<\/p>\n b)\u00a084 km\/hr<\/p>\n c)\u00a072 km\/hr<\/p>\n d)\u00a096 km\/hr<\/p>\n 24)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat upstream = 2x km\/hr Question 25:\u00a0<\/b>Speed of a boat downstream is 5 times of the speed of boat upstream. Then find the speed of the boat in still water if speed of current is 12 km\/hr.<\/p>\n a)\u00a018 km\/hr<\/p>\n b)\u00a024 km\/hr<\/p>\n c)\u00a027 km\/hr<\/p>\n d)\u00a021 km\/hr<\/p>\n 25)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the speed of the boat upstream be x km\/hr Then, Speed of boat in still water = 3x = 3*6 = 18 km\/hr<\/p>\n
\nSpeed of the boat in still water = $\\frac{336}{6.72}$ = 50 km\/h
\nA boat can cover 336 km distance downstream in 4.2 hours.
\n$\\frac{336}{50 + \\text{speed of the stream}} = 4.2$
\n80 = 50 + speed of the stream
\nspeed of the stream = 80-50 = 30 km\/h
\nHence, option d is the correct answer.<\/p>\n
\nATQ,
\nDistance = Speed x Time
\nUpstream speed = x-y
\nDownstream speed = x+y
\n20 = 4(x+y)
\nx+y = 5
\nGiven x=1.5y,
\nSo,
\n1.5y + y = 5
\n2.5y = 5
\ny = 2km\/hr
\nx = 3 km\/hr<\/p>\n
\n20 = T(x-y)
\n20 = T x 1
\nT = 20 hours<\/p>\n
\nLet\u2019s assume the speed of a boat is 7y.
\nSpeed of stream = 42.85% of 7y
\n= (3\/7) of 7y
\n= 3y
\nIf the ratio between the time taken by the boat to cover 280 km distance in upstream to the same distance in downstream is 5:2 respectively.
\nLet\u2019s assume the time taken by the boat in upstream to downstream is 5t and 2t respectively.
\nUpstream \u21d2 $\\frac{280}{7y-3y} = 5t$<\/p>\n
\n20yt = 280 Eq.(i)<\/p>\n
\n20yt = 280 Eq.(ii)<\/p>\n
\nHence, option d is the correct answer.<\/p>\n
\nA boat can cover 720 km distance downstream in 40 hours.
\nB+C = $\\frac{720}{40}$<\/p>\n
\nThe speed of boat in still water is double the speed of stream.
\nB = 2C Eq.(ii)
\nPut Eq.(ii) in Eq.(i).
\n2C+C = 18 km\/h
\n3C = 18 km\/h
\nC = 6 km\/h Eq.(iii)
\nPut Eq.(iii) in Eq.(ii).
\nB = $2\\times6$ = 12 km\/h
\nThe time taken by boat to cover 480 km distance in upstream = $\\frac{480}{B-C}$
\n= $\\frac{480}{12-6}$<\/p>\n
\nHence, option d is the correct answer.<\/p>\n
\nSpeed of the stream = $\\dfrac{1}{3}\\times9 = 3$ km\/hr.
\nSpeed of the boat in still water = 9+3 = 12 km\/hr.
\nSpeed of the boat downstream = 12+3 = 15 km\/hr.
\nHence, The time taken by the boat to travel 285 km downstream = $\\dfrac{285}{15} = 19$ hours.<\/p>\n
\nThe speed of the stream is 44.44% of the speed of the boat.
\nC = 44.44% of B
\n$C = \\frac{4}{9}\\times B$ Eq.(i)
\nA boat can cover 390 km distance upstream in 78 hours.
\n$\\frac{390}{78} = B – C$
\nB – C = 5 Eq.(ii)
\nPut Eq.(i) in Eq.(ii).
\n$B – \\frac{4}{9}\\times B = 5$<\/p>\n
\nB = 9 km\/h Eq.(iii)
\nPut Eq.(iii) in Eq.(ii).
\n9 – C = 5
\nC = 9 – 5 = 4 km\/h
\nTime taken by boat to cover the same distance downstream = $\\frac{390}{9+4}$
\n= $\\frac{390}{13}$<\/p>\n
\nHence, option a is the correct answer.<\/p>\n
\nSpeed of the boat downstream = 5x km\/hr.
\nSpeed of the boat in still water = 5x-x = 4x km\/hr
\nSpeed of the boat upstream = 4x-x = 3x km\/hr.
\nGiven, $\\dfrac{120}{5} = 3x$\u21d2 x = 8.<\/p>\n
\nSpeed of the boat downstream = 5x = 40 km\/hr.
\nHence, Time taken by the boat to travel 480 km downstream and 180 km upstream = $\\dfrac{480}{40}+\\dfrac{180}{24} = 12+7.5 = 19.5$ hours.<\/p>\n
\nTime taken by boat to cover 82 km in downstream = 82\/33 = 2.48 hr
\nBoat returns its starting point so,
\nRelative speed of boat in upstream = 25 – 8 = 17 km\/hr
\nTime taken by boat to cover 82 km in upstream = 82\/17 = 4.82 hr
\nTotal time = 2.48 + 4.82 = 7.3 hours<\/p>\n
\nATQ,
\nDistance = Speed x Time
\nUpstream speed = x-y
\nDownstream speed = x+y
\n28 = 4(x-y)
\nGiven x=3y,
\nSo,
\n28 = 4(3y-y)
\n28 = 8y
\ny = 3.5km\/hr
\nx = 10.5 km\/hr (as, x = 3y)
\nTime taken to cover 56 km in downstream, let it be \u2018T\u2019
\n56 = T(x+y)
\n56 = 4y x T
\nT = 56\/14
\nT = 4 hour.<\/p>\n
\nSo upstream speed will be (B-C) and downstream speed will be (B+C).
\nTotal time taken by a boat to cover 136km in downstream and 60km in upstream is 32 hours.
\n$\\dfrac{136}{B + C} + \\dfrac{60}{B – C} = 32 $ Eq.(1)<\/p>\n
\nB = 3C Eq.(2)
\nPut Eq.(2) in Eq.(1).
\n$\\dfrac{136}{3C + C} + \\dfrac{60}{3C – C} = 32 $<\/p>\n
\nPut Eq.(3) in Eq.(2).
\nB = $3 \\times 2$
\nB = 6km\/hr. Eq.(4)
\nDownstream speed of that boat = (B+C)
\nPut Eq.(4) and Eq.(3) here.
\n(6+2)km\/hr.
\n8 km\/hr.
\nHence, option c is the correct answer.<\/p>\n
\nSpeed of the boat upstream = 3x km\/hr.
\nThen, The speed of the boat in still water = $\\dfrac{5x+3x}{2} = 4x$ km\/hr.<\/p>\n
\nThen, Speed of the boat in Still water = 4x = 4*8 = 32 km\/hr.
\nTherefore, The boat can travel 32 km in 1 hour when the water is still.<\/p>\n
\nSpeed of the boat in still water is 125% more than the speed of stream.
\nB = (100 + 125)% of C
\nB = 225% of C
\n$B = \\frac{225}{100} \\times C$<\/p>\n
\n$\\dfrac{260}{B + C} + \\dfrac{260}{B – C} = 72$ Eq.(2)<\/p>\n
\n$\\dfrac{260}{\\frac{9}{4}C + C} + \\dfrac{260}{\\frac{9}{4}C – C} = 72$<\/p>\n
\n$B = \\frac{9}{4} C$<\/p>\n
\nHence, option d is the correct answer.<\/p>\n
\nATQ,
\nDistance = Speed x Time
\nUpstream speed = x-y
\nDownstream speed = x+y
\n30 = 4(x+y)
\nGiven x=2y,
\nSo,
\n30 = 4(2y+y)
\n30 = 12y
\ny = 2.5 km\/hr
\nx = 5 km\/hr (as, x = 2y)
\nTime taken to cover 30 km in upstream, let it be \u2018T\u2019
\n30 = T(x-y)
\n30 = y x T
\n$T = 30\/2.5$
\nT = 12 hour.<\/p>\n
\nATQ,
\nD = VT \u2014\u2014(i)
\n1.25D = 1.5V(T-1)
\n5D = 6V(T-1) \u2014\u2014\u2014\u2014(ii)
\nPutting D = VT from eq (i) in eq (ii), we get
\n5VT = 6V(T-1)
\n5T = 6(T-1)
\n5T = 6T – 6
\nT = 6 hour.<\/p>\n
\nATQ,
\n20 = 2(x+y)
\nx+y = 10 \u2014\u2014\u2014(i)
\n(x-y)\/y = 2:3
\n3x-3y = 2y
\n3x = 5y
\ny = 3x\/5 \u2014\u2014\u2014(ii)
\nPutting eq (ii) value in eq (i) we get,
\nx+ 3x\/5 = 10
\n8x = 50
\nx = 50\/8
\nx = 6.25 km\/hr<\/p>\n
\nSpeed of current is $\\frac{1}{9}$ of the speed of the boat.<\/p>\n
\nSpeed of boat in upstream = (B-C) = (36-4) = 32 Eq.(4)
\nDistance covered by the boat in downstream during 4 hours is 32km more than the distance covered by boat in upstream during 4 hours.
\nSo let\u2019s assume distance covered by boat in upstream = d Eq.(5)
\nThen distance covered by boat in downstream = (d+32) Eq.(6)
\nAs per the question time is equal.
\n$\\dfrac{Eq.(5)}{Eq.(4)} = \\dfrac{Eq.(6)}{Eq.(3)}$<\/p>\n
\n5d = 4d+128
\n5d-4d = 128
\nd = 128km
\nHence, option a is the correct answer.<\/p>\n
\nSo upstream speed = (B – C)
\nAnd downstream speed = (B + C)
\nIn a river, the speed of a boat is six times the speed of stream.
\nSo B = 6C Eq.(1)
\nTime taken by boat to cover the same distance in upstream and downstream is 14 hours and T hours respectively.
\n14(B – C) = T(B + C) Eq.(2)
\nPut Eq.(1) in Eq.(2).
\n14(6C – C) = T(6C + C)
\n$14 \\times 5C = T \\times 7C$
\n$14 \\times 5 = T \\times 7$
\n$70 = T \\times 7$
\n$T = \\frac{70}{7}$
\nT = 10 hours.
\nHence, option a is the correct answer.<\/p>\n
\nLet\u2019s assume the speed of the stream is C.
\nSo upstream and downstream speeds will be (B-C) and (B+C) respectively.
\nA boat can row 105km upstream in 6hr
\n(B-C) = $\\frac{105}{6}$<\/p>\n
\nSame distance covered by that boat downstream in 4hr.
\n(B+C) = $\\frac{105}{4}$<\/p>\n
\nB+C – (B-C) = 26.25 – 17.5
\nB+C – B+C = 8.75
\n2C = 8.75
\nC = $\\frac{8.75}{2}$<\/p>\n
\nHence, option b is the correct answer.<\/p>\n
\nSpeed of the boat in still water : Speed of the stream = 5 : 1.
\nLet the speed of the boat in still water = 5x km\/hr
\n5x = 50 \u21d2 x = 10.
\nThen, Speed of the stream = x = 10 km\/hr.<\/p>\n
\nTherefore, The boat can travel 80 km upstream in $\\dfrac{80}{40} = 2$ hours.<\/p>\n
\n30\/(b-s) =10
\nb-s=3
\n52\/(b+s) =4
\nb+s=13
\nb-s=3
\n2b=16
\nb=8 km\/hr
\ns=5 km\/hr
\nEach is decreased by 1 km\/hr so
\nb=7 km\/hr
\ns=4 km\/hr
\nTime taken for 121 km downstream=121\/11
\n=11 hrs<\/p>\n
\n120\/(b-s) =6
\nb-s=20
\nGiven b+s=2(b-s)
\nb+s=2b-2s
\n3s=b
\n3s-s=20
\n2s=20
\ns=10
\nb=30
\nTime taken for 90 km downstream=90\/40
\n=2.25 hrs<\/p>\n
\nDownstream speed=4+2=6 km\/hr
\nTherefore we have $\\frac{x}{6}+\\frac{x}{2}$=6
\n4x\/6 =4
\nx=6 km<\/p>\n
\nDownstream speed=10+6=16 km\/hr
\nTherefore we have $\\frac{x}{16}+\\frac{x}{4}$=10
\n5x\/16 =10
\nx=32<\/p>\n
\nThen the speed of the boat downstream = 4x km\/hr
\nThen, Speed of boat in still water = (2x+4x)\/2 = 6x\/2 = 3x km\/hr
\nSpeed of current = Speed of boat in still water – Speed of boat upstream = 4x-3x = x km\/hr
\nGiven, x = 24 km\/hr
\nThen, Speed of boat in still water = 3x = 3*24 = 72 km\/hr<\/p>\n
\nThen, the speed of the boat downstream = 5x km\/hr
\nSpeed of the boat in still water = (x+5x)\/2 = 6x\/2 = 3x km\/hr
\nSpeed of current = Speed of boat in still water – Speed of boat upstream = 3x – x = 2x km\/hr
\nGiven, 2x = 12 km\/hr
\n\u21d2 x = 6 km\/hr<\/p>\n