Last updated at June 22, 2017 by Teachoo

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Example 19 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 current(stream) be y km/hr Speed downstream = Speed of boat in still water + Speed of stream Speed downstream = x + y Speed upstream = Speed of boat in still water – Speed of stream 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 (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 30 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚) + (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 44 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚) = 10 30/(𝑥 − 𝑦) + 44/(𝑥 + 𝑦) = 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 (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 40 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚) + (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 55 𝑘𝑚)/(𝑆𝑝𝑒𝑒𝑑 𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚) = 13 40/(𝑥 − 𝑦) + 55/(𝑥 + 𝑦) = 13 Our equations are 30 (1/(𝑥 − 𝑦))+44(1/(𝑥 + 𝑦))=10 …(1) 40 (1/(𝑥 − 𝑦))+55(1/(𝑥 + 𝑦))=13 …(2) Solving 30u + 44v = 10 …(3) 40u + 55v = 13 …(4) From (3) 30u + 44v = 10 30u = 10 – 44v u = (10 −44𝑣)/30 Putting value of u in (4) 40u + 55v = 13 40((10−44𝑣)/30)+55𝑣=13 4((10−44𝑣)/3)+55𝑣=13 Multiplying both sides by 3 3 × 4((10−44𝑣)/3)+"3 ×" 55𝑣="3 ×" 13 4(10 – 44v) + 165𝑣= 39 40 – 176v + 165v = 39 – 176v + 165v = 39 – 40 – 11v = -1 v = (−1)/(−11) 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 = 1/5 So, u = 1/5 & v = 1/11 So, u = 1/5 & v = 1/11 But we need to find x & y We know that So, our equations become x – y = 5 …(6) x + y = 11 …(7) From (6) x – y = 5 x = 5 + y Putting value of x in (7) x + y = 11 (5 + y) + y = 11 5 + 2y = 11 2y = 11 – 5 2y = 6 y =6/2 y = 3 Put y = 3 in (6) x – y = 5 x – 3 = 5 x = 5 + 3 x = 8 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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.