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A motorboat covers a distance of 16km upstream and 24km downstream in 6 hours. In the same time it covers a distance of 12 km upstream and 36km downstream. Find the speed of the boat in still water and that of the stream.

A motorboat covers a distance of 16km upstream and 24km downstream inA

Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 2
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 3
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 4
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 5
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 6
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 7
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 8
Question 36 - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 9

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Note : This is similar to Example 19 ,Β  Chapter 3 Class 10 Pair of Linear Equations in 2 Variables (NCERT Book)

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Question 36 A motorboat covers a distance of 16km upstream and 24km downstream in 6 hours. In the same time it covers a distance of 12 km upstream and 36km downstream. Find the speed of the boat in still water and that of the stream.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 16 km upstream and 24 km downstream in 6 hours Time taken to go 16 km upstream + Time taken to go 24 km downstream (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 16 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘’π‘π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) + (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 24 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘‘π‘œπ‘€π‘›π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) = 6 πŸπŸ”/(𝒙 βˆ’ π’š) + πŸπŸ’/(𝒙 + π’š) = 6 We know that, Speed = (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ )/π‘‡π‘–π‘šπ‘’ Time = (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ )/𝑆𝑝𝑒𝑒𝑑 Similarly, A boat goes 12 km upstream and 36 km downstream in 6 hours Time taken to go 12 km upstream + Time taken to go 36 km downstream (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 12 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘’π‘π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) + (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘œπ‘“ 36 π‘˜π‘š)/(𝑆𝑝𝑒𝑒𝑑 π‘‘π‘œπ‘€π‘›π‘ π‘‘π‘Ÿπ‘’π‘Žπ‘š) = 6 𝟏𝟐/(𝒙 βˆ’ π’š) + πŸ‘πŸ”/(𝒙 + π’š) = 6 We know that, Speed = (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ )/π‘‡π‘–π‘šπ‘’ Time = (π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ )/𝑆𝑝𝑒𝑒𝑑 Our equations are 16(1/(π‘₯ βˆ’ 𝑦))+24(1/(π‘₯ + 𝑦))=6 …(1) 12(1/(π‘₯ βˆ’ 𝑦))+36(1/(π‘₯ + 𝑦))=6 …(2) So, our equations become Solving 16u + 24v = 6 …(3) 12u + 36v = 6 …(4) Let 1/(π‘₯ βˆ’ 𝑦) = u 1/(π‘₯ + 𝑦) = v 12u + 36v = 6 From (3) 16u + 24v = 6 16u = 6 – 24v u = (6 βˆ’ 24𝑣)/16 Putting value of u in (4) 12u + 36v = 6 12((6 βˆ’ 24𝑣)/16)+36𝑣=6 3((6 βˆ’ 24𝑣)/4)+36𝑣=6 Multiplying both sides by 4 4 Γ— 3((6 βˆ’ 24𝑣)/4)+"4 Γ—" 36𝑣="4 Γ—" 6 3(6 – 24v) + 144𝑣= 24 18 – 72v + 144v = 24 – 72v + 144v = 24 – 18 72v = 6 v = πŸ”/πŸ•πŸ v = 𝟏/𝟏𝟐 Putting v = 1/12 in equation (3) 12u + 36v = 6 12u + 36(1/12) = 6 12u + 3 = 6 12u = 6 βˆ’ 3 12u = 3 u = 3/12 u = 𝟏/πŸ’ So, u = 1/4 & v = 1/12 But we need to find x & y We know that u = 𝟏/(𝒙 βˆ’ π’š) 1/4 = 1/(π‘₯ βˆ’ 𝑦) x – y = 4 v = 𝟏/(𝒙 + π’š) 1/12 = 1/(π‘₯ + 𝑦) x + y = 12 So, our equations become x – y = 4 …(6) x + y = 12 …(7) Adding (6) and (7) (x – y) + (x + y) = 4 + 12 2x = 16 x = 16/2 x = 8 Putting x = 8 in (7) x + y = 12 8 + y = 12 y = 12 – 8 y = 4 So, x = 8, y = 4 is the solution of the given equation Hence Speed of boat in still water = x = 8 km/hr Speed of stream = y = 4 km/hr y = 12 – 8 y = 4 So, x = 8, y = 4 is the solution of the given equation Hence Speed of boat in still water = x = 8 km/hr Speed of stream = y = 4 km/hr

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.