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Ex 3.6

Ex 3.6, 1 (i) and (ii)
Important
Deleted for CBSE Board 2023 Exams

Ex 3.6, 1 (iii) and (iv) Deleted for CBSE Board 2023 Exams

Ex 3.6, 1 (v) and (vi) Important Deleted for CBSE Board 2023 Exams

Ex 3.6, 1 (vii) and (viii) Important Deleted for CBSE Board 2023 Exams

Ex 3.6, 2 (i) Important You are here

Ex 3.6, 2 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 3.6, 2 (iii) Important Deleted for CBSE Board 2023 Exams

Last updated at July 24, 2021 by Teachoo

Ex 3.6, 2 Formulate the following problems as a pair of equations, and hence find their solutions: (i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water & speed of the current. Let the speed of boat in still water be x km/hr & let the speed of current be y km/hr Now, Speed downstream = x + y Speed upstream = x β y Ritu can row 20 km downstream in 2 hours For downstream Distance = 20 km Time = 2 hours Speed = x + y We know that Speed = (π·ππ π‘ππππ )/ππππ x + y = 20/2 x + y = 10 Ritu can row 4 km upstream in 2 hours For upstream Distance = 4 km Time = 2 hours Speed = x β y We know that Speed = (π·ππ π‘ππππ )/ππππ x β y = 4/2 x β y = 2 Hence, our equations are x + y = 10 β¦(1) x β y = 2 β¦(2) From (1) x + y = 10 x = 10 β y Putting x = 10 β y in (2) x β y = 2 (10 β y) β y = 2 β 2y = 2 β 10 β 2y = β8 y = (β8)/(β2) y = 4 Putting y = 4 in (1) x + y = 10 x + 4 = 10 x = 10 β 4 x = 6 Thus, x = 6, y = 4 is the solution Hence Speed of boat in still water = x = 6 km/hr Speed of stream = y = 4 km/hr