Question

A train covers a distance of 90 km at a uniform speed. If the speed had
been 15 km/hr more, it would have taken 30 minutes less for the same
journey. Find the original speed of the train
ICBSE 2006 C]​

Answer 1

AnswEr :

• Fixed Distance is 90 km
• Let the Original Speed of Train be x km/h
• New Speed of Train be (x + 15) km/h

$\leadsto\sf{New \: Time - Original \: Time = \dfrac{1}{2}hr }$

$\leadsto\sf{ \dfrac{Distance}{New \: Speed } - \dfrac{Distance}{Old\: Speed } = \dfrac{1}{2} }$

$\leadsto\sf{ \dfrac{90}{(x + 15)} - \dfrac{90}{x} = \dfrac{1}{2} }$

$\leadsto \sf{ \dfrac{90x - 90(x + 15)}{(x + 15)x}= \dfrac{1}{2} }$

$\leadsto\sf{ \dfrac{ \cancel{90x} - \cancel{90x} + 1350}{ {x}^{2} + 15x}= \dfrac{1}{2} }$

$\leadsto\sf{2700 = {x}^{2} + 15x }$

$\leadsto\sf{{x}^{2} + 15x - 2700 = 0}$

$\leadsto\sf{{x}^{2} + 60x - 45x - 2700 = 0}$

$\leadsto\sf{x(x + 60) - 45(x + 60) = 0}$

$\leadsto\sf{(x + 60)(x - 45) = 0}$

$\leadsto \sf{x = - 60 \: \: \: or \: \: x = 45}$

⋆ we will Ignore x = - 60 as Speed can't be Negative.

⠀

$\therefore$ Original Speed of Train is 45 km/h.