# Time and Distance Part 7

Problem 1: If a man walks 5km/hr, he misses a train by 7 min. However, if he walked at the rate of 6km/hr, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station?

Solution:

5km/hr ————–>+7

6km/hr—————>-5

7-(-5)=12

Distance=$\frac{S_{1}\times&space;S_{2}}{S_{1}\sim&space;S_{2}}\times&space;\Delta&space;t$

$\frac{5\times&space;6}{1}\times&space;\frac{12}{60}=6km$

Problem 2: If a train runs at the rate of 40km/hr it reaches its destination late by 11min but if it reaches 50km/hr it is late by 5 minutes early. the correct time for the train to complete its journey is?

Solution:

40km/hr———–>+11

50km/hr————->+5

11-5=6

Distance=$\frac{S_{1}\times&space;S_{2}}{S_{1}\sim&space;S_{2}}\times&space;\Delta&space;t$

=$\frac{40\times&space;50}{10}\times&space;\frac{6}{60}=20$

$Speed=\frac{Distance}{time}$

$40=\frac{20}{t}$

$t=\frac{1}{2}=30min$

Original time=30-11=19min

Problem 3: A and B ate 285 km apart. A train starts from A at 9:30 am and travels towards B at 60km/hr another train starts from B at 10:30 AM  and travels towards A at 40km/hr. At what time they will meet?

Solution: Time gap between Train A and B is 1 hour

285 – 60=225

Time = $\frac{Distance}{Speed}$

=$\frac{225}{40+60}=\frac{225}{100}=2.25&space;hrs$

At 12:55 both trains will meet