Question

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?

Answer 1

By the time P starts J is 1.5 hr x 6 = 9 km away from his house.

J is 3 km behind when P is 3 km ahead of him.

ie.,

P has to cover 12 km.

So,

He takes 12 / (8 - 6) = 6 hrs after 13.30.

So,

The required time is 19.30Hrs

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OR
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First ,let us calculate the time at which Paul catches Jake...

To catch Jake,both has to cover equal distance...

6*x=8*(x-3/2),where x is the time taken from 12:00 hrs at which paul overtakes jake

x=6 hrs from 12 i.e 18:00 hrs

At 18:00 hrs both are at same point...

Next,we have to take the difference of distance b/w the two as 3km

therefore,8(x)-6(x)=3,where x is the time taken from 18:00 hrs

Hence x=1hr 30 minutes

Therefore total time taken=18+1hr 30min=19:30

Answer 2

Answer: 19 hours and 30 minutes

Step-by-step explanation:

1. data given in the question:

   let jake be x

    let paul be y

2. x starts to walk from his house:

                                  time= 12:00

                                   speed= 6 km/hour

3. y starts following x

                                   time = 13:30

                                   speed= 8km/ hour

4. to find : when will x be 3km behind y

5. First ,let us calculate the time at which y catches x

6. $6*l=8*(l-3/2)$

here l is the time taken from 12:00 hrs at which y overtakes

7. we see that at 18:00 hrs both are at same point.

the difference of distance between the two is  3km

therefore,

$8(l)-6(l)=3$

by solving the equation we get the answer as 19hours and 30 mins

#SPJ3

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