Question

if the speed of the car is increased by 15 kmph, It takes 2 hours less to cover a distance of 600km. Find usual time for the journey

Answer 1

suppose that the speed the car is = x km/ h

we considered that the car takes y hours to cover 600 km

$v = s \div t$

x= 600/ y

xy=600

now the speed is increased so the speed is = x +15 km/ h

and the car takes (y-2) hours to cover the 600 km.

$v = s \div t$

(x+15)=600/(y-2)

(x+15)(y-2)= 600

xy-2x +15y- 30 = 600

600 - 2x + 15y -30 =600

2x -15y + 30 = 0

2 (600/y) - 15y +30 = 0

1200/ y - 15y + 30 =0

1200/y = 15y - 30

1200 =15yy -30 y

15yy -30y-1200=0

yy - 2y -80=0

so y = 10 and y= - 8

here y=-8 is not exist

so y = 10 h