Question

plz someone answer...

A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And ,if the train was slower by 6 km/h it would have taken 6 hours more than the scheduled time. Find the length of the journey.​

Answer 1

Let the actual speed of the train be x km/hr and the actual time taken be y hours. Then,

Distance covered =(xy)km ..(i) [∴ Distance = Speed × Time]

If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x+6)km/hr, time of journey is (y−4) hours.

∴ Distance covered =(x+6)(y−4)

⇒xy=(x+6)(y−4) [Using (i)]

⇒−4x+6y−24=0

⇒−2x+3y−12=0 ..(ii)

When the speed is reduced by 6 km/hr, then the time of journey is increased by 6 hours i.e., when speed is (x−6) km/hr, time of journey is (y−6) hours.

∴ Distance covered =(x−6)(y+6)

⇒xy=(x−6)(y+6) [Using (i)]

⇒6x−6y−36=0

⇒x−y−6=0 (iii)

Thus, we obtain the following system of equations:

−2x+3y−12=0

x−y−6=0

By using cross-multiplication, we have,

3×−6−(−1)×−12

x

=

−2×−6−1×−12

−y

=

−2×−1−1×3

1

⇒

−30

x

=

24

−y

=

−1

1

⇒x=30 and y=24

Putting the values of x and y in equation (i), we obtain

Distance =(30×24)km =720km.

Hence, the length of the journey is 720km.