Β

Β

Get live Maths 1-on-1 Classs - Class 6 to 12

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

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

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

Last updated at March 29, 2023 by Teachoo

Ex 3.6, 2 Formulate the following problems as a pair of equations, and hence find their solutions: (iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. Let speed of train be x km/hr & speed of bus be y km/hr Given that , Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train & the remaining by bus. Train Distance = 60 km Speed = x Time = ? We know that Time = (π·ππ π‘ππππ )/πππππ Time = 60/π₯ Bus Distance = 240 km Speed = y Time = ? We know that Time = (π·ππ π‘ππππ )/πππππ Time = 240/π¦ Now, Total time taken = 4 hours 60/π₯ + 240/π¦ = 4 Also, She travels 100 km by train & remaining by bus, she takes 10 min longer. So, she now takes 4 hours, 10 minutes Train Distance = 100 km Speed = x Time = ? We know that Time = (π·ππ π‘ππππ )/πππππ Time = 100/π₯ Bus Distance = 200 km Speed = y Time = ? We know that Time = (π·ππ π‘ππππ )/πππππ Time = 200/π¦ Now, Total time taken = 4 hours 10 minutes 100/π₯ + 200/π¦ = 4 + 10/60 hours 100/π₯ + 200/π¦ = 4 + 1/6 hours 100/π₯ + 200/π¦ = (4 (6) + 1)/6 100/π₯ + 200/π¦ = (24 + 1)/6 100/π₯ + 200/π¦ = 25/6 So, our two equations are 60/π₯ + 240/π¦ = 4 β¦(1) 100/π₯ + 200/π¦ = 25/6 β¦(2) So, our equations become 60u + 240v = 4 100u + 200v = 25/6 6/25 (100u + 200v) = 1 24u + 48v = 1 Now, we have to solve 60u + 240v = 4 β¦(3) 24u + 48v = 1 β¦(4) From (3) 60u + 240v = 4 60u = 4 β 240v u = (4 β 240π£)/60 Putting value of u in (4) 24u + 48v = 1 24 ((4 β 240π£)/60) + 48v = 1 2 ((4 β 240π£)/5) + 48v = 1 Multiplying both sides by 5 5 Γ 2 ((4 β 240π£)/5) + 5 Γ 48v = 5 Γ 1 2(4 β 240v) + 240v = 5 8 β 480v + 240v = 5 β 480v + 240v = 5 β 8 β240v = β3 v = (β3)/(β240) v = 1/80 Putting v = 1/80 in (3) 60u + 240v = 4 60u + 240 (1/80) = 4 60u + 3 = 4 60u = 4 β 3 60u = 1 u = 1/60 Hence, u = π/ππ , v = π/ππ But we have to find x & y We know that u = π/π 1/60 = 1/π₯ x = 60 v = π/π 1/80 = 1/π¦ y = 80 So, x = 60, y = 80 is the solution of the given equation Therefore Speed of train = x = 60 km/hr & Speed of bus = y = 80 km/hr