Starting from 12:01 pm, every minute a car leaves point A and moves towards point B. All cars are travelling at the same speed. Forty- five such cars reach B by 13:00 pm. A man leaves point A at 12:00 pm and reaches B at 13:00 pm. If the man doubled his speed, how many cars would have reached point B by the time the cars reached point B?
Answers
Step-by-step explanation:
Time taken for car to travel from B to A is:
T=
v
s
=
60
60
=1 hr=60 min
Distance between any two consecutive cars leaving from A is constant and is given by:
s=10/60×60=10 km
Relative speed between any car leaving from A and car leaving from B is v
r
=60−(−60)=120 km/hr
Hence, on crossing one car, next car is met in time,
t=
v
s
=
120
10
hr=5 min
Hence, total number of cars met are:
n=
t
T
=12
This excludes the first car met at B. Hence, car starting from B meets a total of 13 cars.
Answer:
15 cars would have reached point B by the time the cars reached point B.
Step-by-step explanation:
Let speed of car be m and that of man be c.
Since there are 45 cars reaching in 1 hour, so the latest car must start by 12:15 from A.
Now, man doubles its speed, so man reaches B at 10:30 because if speed gets doubled then time gets halved, so maximum cars that can start are from 10:01 to 10:15 is 15.
So, 15 is the correct answer.