Question

By increasing the speed of car by 10 km/hr, the time
of journey for a distance of 72 km is reduced by 36
min. Find the original speed of the car. (ICSE 2005)​

Answer 1

here

d= s×t

72= s×t

s= 72/t (1)

Now after increasing the speed which is

= s+10

the time is reduced by 36 min, which is 36/60 hr or

3/5 hr,

= t-3/5

The car is covering the same distance,

Thus,

D= s×t

72= (s+10)(t-3/5)

72= (72/t+10)(5t-3)/5

72×5= (72+10t)/t × (5t-3)

360t= 360t-216+50t²-30t

0= 50t²-30t-216

2(25t²-15t-108)=0

25t²-15t-108=0

25t²-60t+45t-108=0

5t(5t-12)+9(5t-12)=0

(5t+9) (5t-12)=0

either t= -9/5

or t= 12/5

since time cannot be negative, t= 12/5 hr

Putting the value of t in equation (1)

s= 72km/t= 72km/12/5 hr

= 6×5km/hr

= 30km/hr