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A boat goes 30 km upstream and 44 km downstream in 10 hours [Video]

Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3 Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4 Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5 Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6 Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7 Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 8 Example 19 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 9

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Question 9 A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water. Let the speed of boat in still water be x km/hr & let the speed of stream be y km/hr Now, Speed downstream = x + y Speed upstream = x – y Given that A boat goes 30 km upstream and 44 km downstream in 10 hours Time taken to go 30 km upstream + Time taken to go 44 km downstream (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 30 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘’π‘π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) + (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 44 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘‘π‘œπ‘€π‘›π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) = 10 πŸ‘πŸŽ/(𝒙 βˆ’ π’š) + πŸ’πŸ’/(𝒙 + π’š) = 10 Similarly, A boat goes 40 km upstream and 55 km downstream in 13 hours Time taken to go 40 km upstream + Time taken to go 55 km downstream (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 40 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘’π‘π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) + (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 55 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘‘π‘œπ‘€π‘›π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) = 13 πŸ’πŸŽ/(𝒙 βˆ’ π’š) + πŸ“πŸ“/(𝒙 + π’š) = 13 Our equations are 30 (1/(π‘₯ βˆ’ 𝑦))+44(1/(π‘₯ + 𝑦))=10 …(1) 40 (1/(π‘₯ βˆ’ 𝑦))+55(1/(π‘₯ + 𝑦))=13 …(2) So, our equations become 30u + 44v = 10 40u + 55v = 13 Solving 30u + 44v = 10 …(3) 40u + 55v = 13 …(4) From (3) 30u + 44v = 10 30u = 10 – 44v u = (10 βˆ’ 44𝑣)/30 Putting value of u in (4) 40u + 55v = 13 40((10 βˆ’ 44𝑣)/30)+55𝑣=13 4((10 βˆ’ 44𝑣)/3)+55𝑣=13 Multiplying both sides by 3 3 Γ— 4((10 βˆ’ 44𝑣)/3)+"3 Γ—" 55𝑣="3 Γ—" 13 4(10 – 44v) + 165𝑣= 39 40 – 176v + 165v = 39 – 176v + 165v = 39 – 40 – 11v = –1 v = 𝟏/𝟏𝟏 Putting v = 1/11 in equation (3) 30u + 44v = 10 30u + 44(1/11) = 10 30u + 4 = 10 30u = 10 – 4 30u = 6 u = 6/30 u = 𝟏/πŸ“ So, u = 1/5 & v = 1/11 But we need to find x & y We know that u = 𝟏/(𝒙 βˆ’ π’š) 1/5 = 1/(π‘₯ βˆ’ 𝑦) x – y = 5 v = 𝟏/(𝒙 + π’š) 1/11 = 1/(π‘₯ + 𝑦) x + y = 11 So, our equations become x – y = 5 …(6) x + y = 11 …(7) Adding (6) and (7) (x – y) + (x + y) = 5 + 11 2x = 16 x = 16/2 x = 8 Putting x = 8 in (7) x + y = 11 8 + y = 11 y = 11 – 8 y = 3 So, x = 8, y = 3 is the solution of the given equation Hence Speed of boat in still water = x = 8 km/hr Speed of stream = y = 3 km/hr

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.