Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Last updated at Aug. 1, 2017 by Teachoo

Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Transcript

Ex 3.7, 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 = ๐ท๐๐ ๐ก๐๐๐๐/๐๐๐๐ Distance = Speed ร 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 โ 2 Now, Distance = Speed ร time Distance = (x + 10) (y โ 2) Putting Distance = xy from equation (1) xy = (x + 10) (y โ 2) xy = x (y โ 2) + 10 (y โ 2) xy = xy โ 2x + 10y โ 20 2x โ 10y + 20 = xy โ xy 2x โ 10y + 20 = 0 Also, If the train were slower by 10 km/h Speed = x โ 10, it would have taken 3 hours more Time = y + 3. Now Distance = Speed ร time Distance = (x โ 10) (y + 3) Putting Distance = xy from equation (1) xy = (x โ 10) (y + 3) xy = x (y + 3) โ 10y โ 30 xy = xy + 3x โ 10y โ 30 xy โ xy = 3x โ 10y โ 30 3x โ 10y โ 30 = 0 Hence, the equations are 2x โ 10y + 20 = 0 โฆ(2) 3x โ 10y โ 30 = 0 โฆ(3) From equation (2) 2x โ 10y + 20 = 0 2 (x โ 5y + 10) = 0 x โ 5y + 10 = 0 x = 5y โ 10 Putting (4) in equation (3) 3x โ 10y โ 30 = 0 3 (5y โ 10) โ 10y โ 30 = 0 15y โ 30 โ 10y โ 30 = 0 5y โ 60 = 0 5y = 60 y = 60/5 y = 12 Putting y = 12 in equation (4) x = 5y โ 10 x = (5 ร 12) โ 10 x = 60 โ 10 x = 50 Therefore, speed of the train = 50 km/ h and time taken by the train = 12 hours Now, the distance covered by the train = Speed ร time Distance = 50 ร 12 Distance = 600 km Thus, Speed of train = x = 50 km/ h & Time taken by the train = y = 12 hours Now, Distance = Speed ร time Distance = 50 ร 12 Distance = 600 km

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.