Question 37 (OR 1st question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at May 29, 2023 by Teachoo

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.

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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 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑇𝑖𝑚𝑒
x = 360/𝑇𝑖𝑚𝑒
Time = 360/𝑥
Speed 5 km/h more
Distance = 360 km
Speed = (x + 5) km/hr
Time = (360/𝑥 " – " 48/60) hours
Speed = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑇𝑖𝑚𝑒
x + 5 = 360/((360/𝑥 " – " 48/60) )
(x + 5) (360/𝑥 " – " 48/60) = 360
From (1)
(x + 5) (360/𝑥 " – " 48/60) = 360
(x + 5) (360/𝑥 " – " 4/5) = 360
(x + 5) ((5 × 360 − 4𝑥)/5𝑥) = 360
(x + 5) ((1800 − 4𝑥)/5𝑥) = 360
(x + 5) (1800 – 4x) = 360 × 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 = (− 𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎
Putting values
x = (−5 ± √(5^2 − 4 × 1 × (−2250) ))/(2 × 1)
x = (−5 ± √(25 + 4 × 2250 ))/2
x = (−5 ± √(25 + 9000))/2
x = (−5 ± √9025)/2
x = (−5 ± √(5^2×〖19〗^2 ))/2
x = (−5 ± 5×19)/2
x = (−5 ± 95)/2
x = (−5 + 95)/2
x = 90/2
x = 45
x = (−5 − 95)/2
x = (−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

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.

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