Quantitative Aptitude :: Time and Distance - Discussion
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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?
A24 km
B72 km
C18 km
DData Insufficient
Show Explanation
Let x km be the distance between his house and office.
While travelling at 24kmph, he would take x/24 hours.
While travelling at 25% faster speed,
i.e. 24 25% of 24=24×(1/4)=30 kmph, he would take x/30 hr
Now as per the problem, time difference = 5 min late 4 min early = 9 min
x/24-x/30=9 min
Solving this we get ⇒ x= 18 km
Asked In ::
Let x km be the distance between his house and office.
While travelling at 24kmph, he would take x/24 hours.
While travelling at 25% faster speed,
i.e. 24 25% of 24=24×(1/4)=30 kmph, he would take x/30 hr
Now as per the problem, time difference = 5 min late 4 min early = 9 min
x/24-x/30=9 min
Solving this we get ⇒ x= 18 km
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