A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/hour more, it would have taken 48 minutes less for the journey. Find the original speed of the train.

A train covers a distance of 360 km at a uniform speed. Had the speed

Question 37 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 2
Question 37 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 3 Question 37 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 4 Question 37 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 5

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Note : This is similar to Ex 4.3, 8 of NCERT – Chapter 4 Class 10

Check the answer here https://www.teachoo.com/1561/508/Ex-4.3--8---A-train-travels-360-km-at-a-uniform-speed./category/Ex-4.3/

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Transcript

Question 37 (OR 1st question) A train covers a distance of 360 km at a uniform speed. Had the speed been 5 km/hour more, it would have taken 48 minutes less for the journey. Find the original speed of the train. Let the speed of train be x km/hr Normal speed Distance = 360 km Speed = x km/hr Speed = π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’/π‘‡π‘–π‘šπ‘’ x = 360/π‘‡π‘–π‘šπ‘’ Time = 360/π‘₯ Speed 5 km/h more Distance = 360 km Speed = (x + 5) km/hr Time = (360/π‘₯ " – " 48/60) hours Speed = π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’/π‘‡π‘–π‘šπ‘’ x + 5 = 360/((360/π‘₯ " – " 48/60) ) (x + 5) (360/π‘₯ " – " 48/60) = 360 From (1) (x + 5) (360/π‘₯ " – " 48/60) = 360 (x + 5) (360/π‘₯ " – " 4/5) = 360 (x + 5) ((5 Γ— 360 βˆ’ 4π‘₯)/5π‘₯) = 360 (x + 5) ((1800 βˆ’ 4π‘₯)/5π‘₯) = 360 (x + 5) (1800 – 4x) = 360 Γ— 5x x(1800 – 4x) + 5(1800 – 4x) = 1800x 1800x – 4x2 + 5(1800) – 20x = 1800x 1800x – 4x2 + 9000 – 20x = 1800x 1800x – 4x2 + 9000 – 20x – 1800x = 0 – 4x2 – 20x + 9000 = 0 4x2 + 20x – 9000 = 0 4(x2 + 5x – 2250) = 0 x2 + 5x – 2250 = 0 Comparing with ax2 + bx + c = 0 a = 1, b = 5, c = –2250 Roots of the equation are given by x = (βˆ’ 𝑏 Β± √(𝑏^2 βˆ’ 4π‘Žπ‘))/2π‘Ž Putting values x = (βˆ’5 Β± √(5^2 βˆ’ 4 Γ— 1 Γ— (βˆ’2250) ))/(2 Γ— 1) x = (βˆ’5 Β± √(25 + 4 Γ— 2250 ))/2 x = (βˆ’5 Β± √(25 + 9000))/2 x = (βˆ’5 Β± √9025)/2 x = (βˆ’5 Β± √(5^2Γ—γ€–19γ€—^2 ))/2 x = (βˆ’5 Β± 5Γ—19)/2 x = (βˆ’5 Β± 95)/2 x = (βˆ’5 + 95)/2 x = 90/2 x = 45 x = (βˆ’5 βˆ’ 95)/2 x = (βˆ’100)/2 x = –50 Hence x = 45, x = –50 are the roots of the equation We know that Speed of train = x So, x cannot be negative ∴ x = 45 is the solution So, Speed of train = x = 45 km/hr

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