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A train covered a certain distance at uniform speed. If train 10 km

Ex 3.7, 3 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.7, 3 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3 Ex 3.7, 3 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4 Ex 3.7, 3 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5 Ex 3.7, 3 (Optional) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6

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Question 3 A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. Let Speed of train = x km/h & Time taken = y hours. We know that, Speed = π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’/π‘‡π‘–π‘šπ‘’ Distance = Speed Γ— Time Distance = xy If the train would have been 10 km/h faster I.e. Speed = x + 10 It would have taken 2 hours less i.e. Time = y βˆ’ 2 Now, Distance = Speed Γ— time Distance = (x + 10) (y βˆ’ 2) Putting Distance = xy from Equation (1) xy = (x + 10) (y βˆ’ 2) xy = x (y βˆ’ 2) + 10 (y βˆ’ 2) xy = xy βˆ’ 2x + 10y βˆ’ 20 2x βˆ’ 10y + 20 = xy βˆ’ xy 2x βˆ’ 10y + 20 = 0 Also, If the train were slower by 10 km/h Speed = x βˆ’ 10, it would have taken 3 hours more Time = y + 3. Now Distance = Speed Γ— time Distance = (x βˆ’ 10) (y + 3) Putting Distance = xy from equation (1) xy = (x βˆ’ 10) (y + 3) xy = x (y + 3) βˆ’ 10y βˆ’ 30 xy = xy + 3x βˆ’ 10y βˆ’ 30 xy βˆ’ xy = 3x βˆ’ 10y βˆ’ 30 3x βˆ’ 10y βˆ’ 30 = 0 Hence, the equations are 2x βˆ’ 10y + 20 = 0 …(2) 3x βˆ’ 10y βˆ’ 30 = 0 …(3) From equation (2) 2x βˆ’ 10y + 20 = 0 2 (x βˆ’ 5y + 10) = 0 x βˆ’ 5y + 10 = 0 x = 5y βˆ’ 10 Putting (4) in equation (3) 3x βˆ’ 10y βˆ’ 30 = 0 3 (5y βˆ’ 10) βˆ’ 10y βˆ’ 30 = 0 15y βˆ’ 30 βˆ’ 10y βˆ’ 30 = 0 5y βˆ’ 60 = 0 5y = 60 y = 60/5 y = 12 Putting y = 12 in equation (4) x = 5y βˆ’ 10 x = (5 Γ— 12) βˆ’ 10 x = 60 βˆ’ 10 x = 50 Thus, Speed of train = x = 50 km/ h & Time taken by the train = y = 12 hours Now, Distance = Speed Γ— time Distance = 50 Γ— 12 Distance = 600 km

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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.