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At 12:00 hours Jake starts to walk from his house at 6 Km an hour. At 13:30 hours, Paul follows him from Jake’s house on his bicycle at 8 Km per hour. When will Jake be 3 Km behind Paul?
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- ans is 15:30
- 9 years agoHelpfull: Yes(2) No(0)
- From the question, Where will they behind 3 km-------> so the distance should be 3 km plus for Jake....
equate both of them distance by formula, distance = speed * time
distance of Jake + 3 km = distance of Paul ( as he should behind 3 km)
3 + (6 km)(x + 3/2hr) = (x hr) (8)
2x = 12
x =6
13:30 + 6 hr = 19:30 hr - 9 years agoHelpfull: Yes(0) No(0)
- At 19:30 jack will 3 km behind paul
- 9 years agoHelpfull: Yes(0) No(0)
- Answer: 19:00
Jake is at 9 kms when Paul starts so at 14:00 hours they will be at 12kms and 4kms respectively and if we keeps on adding 6 and 8 to both their distances at 18:00 hours Paul would be 1km ahead of Jake and since the difference between their speed is 2 at the next hour i.e 19:00 hours Paul would be 3kms ahead of Jake - 9 years agoHelpfull: Yes(0) No(1)
- jake travelled in 4 hour and 30 minute or 4:30 hours=27km
paul should travel 3 hours to maintain a distance 3 km
so answer is 4:30 hours - 9 years agoHelpfull: Yes(0) No(0)
- for every half an hour jake goes 3kmph
at 1.00 jake goes 6km and at 1.30 jake goes 9 km
next at 2.30 jake goes 9+6=15km and paul goes 8km
3.30 - Jake=15+6=21km and Paul=8+8=16km
5.30-jake=33km and paul=32km
7.30-jake=45km and paul=48km
so at 7.30 jake will be 3km behind paul
- 9 years agoHelpfull: Yes(0) No(0)
- here we have to observe one thing.. that if they travel together paul will be 1km ahead of jake for every .5hrs if he has to overtake jake he has to first cover 1.5*6(becasue he has started late)=9km=9*.5hrs=4.5hrs
so by 6 pm he will be reaching jake...
now he has to be ahead of him and they have started together at 6pm..3 more .5 hrs so by 7:30pm he will be ahead of jake.
- 9 years agoHelpfull: Yes(0) No(0)
- ans:19:30
upto 13:30 jake walks 9 km (.". he travels for 1:30 hrs)
W.K.T Time taken to cover a distance when objects moving in same direction = distance/difference of their speeds
now the distance between jake and pual is 9 kms. paul reach jake after T= 9/(8-6)= 4.5 hrs
to make the distance between them 3km time taken = 3/(8-6) = 1.5 hr
total time taken = 4.5+1.5 = 6
after 6 hrs the time becomes 13:30+6 = 19:30
- 9 years agoHelpfull: Yes(0) No(0)
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