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 Nov. 1, 2019 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 = π·ππ π‘ππππ/ππππ
x = 360/ππππ
Time = 360/π₯
Speed 5 km/h more
Distance = 360 km
Speed = (x + 5) km/hr
Time = (360/π₯ " β " 48/60) hours
Speed = π·ππ π‘ππππ/ππππ
x + 5 = 360/((360/π₯ " β " 48/60) )
(x + 5) (360/π₯ " β " 48/60) = 360
From (1)
(x + 5) (360/π₯ " β " 48/60) = 360
(x + 5) (360/π₯ " β " 4/5) = 360
(x + 5) ((5 Γ 360 β 4π₯)/5π₯) = 360
(x + 5) ((1800 β 4π₯)/5π₯) = 360
(x + 5) (1800 β 4x) = 360 Γ 5x
x(1800 β 4x) + 5(1800 β 4x) = 1800x
1800x β 4x2 + 5(1800) β 20x = 1800x
1800x β 4x2 + 9000 β 20x = 1800x
1800x β 4x2 + 9000 β 20x β 1800x = 0
β 4x2 β 20x + 9000 = 0
4x2 + 20x β 9000 = 0
4(x2 + 5x β 2250) = 0
x2 + 5x β 2250 = 0
Comparing with ax2 + bx + c = 0
a = 1, b = 5, c = β2250
Roots of the equation are given by
x = (β π Β± β(π^2 β 4ππ))/2π
Putting values
x = (β5 Β± β(5^2 β 4 Γ 1 Γ (β2250) ))/(2 Γ 1)
x = (β5 Β± β(25 + 4 Γ 2250 ))/2
x = (β5 Β± β(25 + 9000))/2
x = (β5 Β± β9025)/2
x = (β5 Β± β(5^2Γγ19γ^2 ))/2
x = (β5 Β± 5Γ19)/2
x = (β5 Β± 95)/2
x = (β5 + 95)/2
x = 90/2
x = 45
x = (β5 β 95)/2
x = (β100)/2
x = β50
Hence x = 45, x = β50 are the roots of the equation
We know that Speed of train = x
So, x cannot be negative
β΄ x = 45 is the solution
So, Speed of train = x = 45 km/hr

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.