Question

When there is a decrease of 5 km/hr in the usual uniform speed of a goods train, due to track repair work going on it takes 4 hours more than the usual time for travelling the distance of 400 km. Find the usual speed of the train.

Answer 1

Dear student,

Answer: Speed of train is 25 km/h

Solution: Let the speed of train be x km/h

As we know that Time taken in a journey = Distance / speed

when speed decreased by 5 km/h

than speed of train( x-5) km/h

Time taken by usual speed = 400/x

Time taken by decreased speed 400 /(x-5),extra time 4 hours

So,

$\frac{400}{x} + 4 = \frac{400}{x-5} \\ \\ \frac{400+4x}{x} =\frac{400}{x-5}\\ \\ ( 400+4x)(x-5) = 400x\\ \\ 400x+4x^{2} -2000-20x = 400x\\ \\ 4x^{2} -20x - 2000 =0$

solve the quadratic equation

$x^{2} -5x-500 =0\\ \\ x^{2} -25x+20x-500 =0\\ \\ x(x-25)+20(x-25) =0 \\ \\ (x-25)(x+20) =0\\ \\ x-25 = 0\\ \\x = 25 \\ \\ x+ 20 =0\\ \\ x= -20$

Discard the negative value since speed can't be negative.

So, speed of train is 25 km/h

Hope it helps you.

Answer 2

Hi ,

i ) Let uniform speed of the train = x km/h

distance ( d ) = 400 km

time = t1 hour

t1 = 400/x ----( 1 )

ii ) If the speed decreased ,

new speed of the train = ( x - 5 ) km/hr

distance = d = 400 km

time ( t2 ) = 400/( x - 5 ) km/hr ---( 2 )

According to the problem given ,

t2 - t1 = 4

400/( x - 5 ) - 400/x = 4

( 400/4 )[ 1/( x - 5 ) - 1/x ] = 1

100[ ( x - x + 5 )/(x² - 5x )] = 1

100 × 5 = x² - 5x

x² - 5x - 500 = 0

x² -25x + 20x - 500 = 0

x( x - 25 ) + 20( x - 25 ) = 0

( x - 25 )( x + 20 ) = 0

x - 25 = 0 or x + 20 = 0

x = 25 or x = -20

speed should be negative .

Therefore ,

Usual speed of the train = x = 25km/hr

I hope this helps you.

: )