A passenger train takes one hour less for a journey of 150km if its speed is increased by 5km/hr from its usual speed. How do you find the usual speed of the train?

2 Answers
Narad T.
Apr 4, 2018

The usual speed of the train is #=25kmh^-1#

Explanation:

Let the time for the normal journey be #=t "h"#

The distance is #d=150km#

The speed is #v=(150/t) km h^-1#

#t=150/v#...............#(1)#

The new speed is #v_1=(v+5) kmh^-1#

The new time is #t_1=(t-1)h#

#v_1=v+5=150/(t-1)#.................#(2)#

#(v+5)(t-1)=150#

Solving for #v# in equations #(1)# and #(2)#

#(v+5)(150/v-1)=150#

#(v+5)(150-v)=150v#

#150v-v^2+750-5v=150v#

#v^2+5v-750=0#

#v=(-5+-sqrt(5^2+4xx750))/(2)#

#=(-5+-55)/(2)#

#=25kmh^-1#

Junaid Mirza
Apr 4, 2018

#"25 kmph"#

Explanation:

Let usual speed be #"u kmph"#. Then

#"u" = "S"/"t" = "150"/"t"#

#"t" = "150"/"u" color(white)(.)……[1]#

When #"u"# is increased by #"5 kmph"#

#"u + 5" = "S"/"t - 1" = "150"/"t - 1"#

#"u + 5" = (150)/(150/"u" - 1)color(white)(...)[∵ "t" = "150"/"u"]#

#"u + 5" = "150u"/"150 - u"#

#("u" + 5)(150 - "u") = "150u"#

#"150u - u"^2 + 750 - "5u" = "150u"#

#"u"^2 + "5u" - 750 = 0#

#"u"^2 + "30u" - "25u" - 750 = 0#

#"u(u + 30) - 25(u + 30) = 0"#

#"(u - 25)(u + 30) = 0"#

#"u = 25 (or) u = -30"#

Speed can’t be negative. Therefore, usual speed of train is #"25 kmph"#

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