Question

A man driving his bike at 24 kmph reaches his office 5 minutes late. had he driven 25% faster on an average he would have reached 4 minutes earlier than the scheduled time. how far is his office?

Answer 1

Let x km be the distance between his house and office.

While travelling at  24 kmph,he would take x/24hours.

While travelling at 25 % faster speed,

i.e. 24+[25%of24] = 24×(1/4)=30kmph,he would take x/30 hours.

Now as per the problem ,time  difference=5 min late + 4min early =9 min

⇒[  x /24]− [x/30] = 9 min

⇒ x= 18 km ANS

Answer 2

Step-by-step explanation:

initial speed 24kmph

+25% speed = 30kmph

speed ratio 24:30 = 4: 5

time 5:4 (as time is inversly propertional to speed)

so 5 - 4 = 1hr = 60 mins late when dist is lcm(24,30)=120 km

(4+5)=9mins late when dist is = (120*90)/60 = 18 km