A TRAIN COVERD A CERTAIN DISTANCE AT A UNIFORM SPEED. IF THE TRAIN WOULD HAVE BEEN 6KM/H. FASTER,IT WOULD HAVE TAKEN 4 HOUR LESS THAN THE SCHEDULE.A TIME AND IF THE TRAIN WOULD HAVE SLOWED ASUME BY 6KM/H IT WOULD HAVE TAKEN 6 HOURS MORE THAN THE SCHEDULED TIME. FIND THE LENGTH OF THE JOUNERY.
Answers
Answer:
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.
