John and Jacob set out together on bicycle traveling at 15 and 12 mile

distance travelled by john in 40 minutes = 15(40/60) = 10 km
distance travelled by jacob in 40 minutes = 12(40/60) =8 km
after 40 minutes john leads jacob by 2 km.
after 40 minutes jacob continues to ride for 1 hour, while john stops to fix a flat tire. during this time jacob covers 12km. but since initially john was leading the jacob by 2km thus, distance now between john and jacob =12-2 =10km

thus time taken by john to catch up with jacob = 10/(15-12)
=3 1/3

hence B

I agree with Paresh. Just wanted to make a slight modification since it looks like there is a typo in the last line:

Total catch up time taken \(= \frac{10+40}{15} = 3\frac{1}{3}\) Hrs

John and Jacob set out together on bicycle traveling at 15 and 12 miles per hour, respectively. After 40 minutes, John stops to fix a flat tire. If it takes John one hour to fix the flat tire and Jacob continues to ride during this time, how many hours will it take John to catch up to Jacob assuming he resumes his ride at 15 miles per hour? (consider John's deceleration/acceleration before/after the flat to be negligible)?

At 1 hour and 40 minutes (time at which John resumes after his 1 hour halt) = John's distance covered is 15*40/60 = 10 miles and distance covered by Jacob is 20 (12*40/60+12).

Difference in their speeds = 3
Time that will be taken to catch up = difference in distances / differences in speeds = (20-10)/3 = 10/3 (B)

10 miles and distance covered by Jacob is 20 (12*40/60+12). Why do you add +12 here?

Johns Travel:
40 minutes 15 mph + 1 hour time to fix tire + Catch up time 15 mph (Let us assume it as T)
Distance travelled before catch up = Distance traveled in 40 min + 0(when fixing the tire) + Distance traveled during catch up time T which is equal to:
15mph x 2/3 hr( 2/3 hr = 40 mins) + 0 + 15 mph x T

Jacobs Travel:
40 minutes 12 mph + 1 hour time 12 mph + T minutes 12 mph
ie. Distance travelled by Jacob = 12 x 2/3 + 12 x 1 + 12 x T.

When John catches up Jacob, both would have traveled the same distance.

So the above calculations for John and Jacob can be converted to an equation.

15 x 2/3 + 0 + 15T = 12 x 2/3 + 12 x 1 + 12T
10 + 15T = 20 + 12T
3T = 10

T = 10/3 which 3 1/3 hours.

So time taken to catch up is 3 1/3 hours.

How is it that the time taken by Jacob would be the same as time taken by John?

Hi Bunuel, after 40 mins John has covered 10 miles and Jacob has covered 8. After 1 hour, John is still on 10 and Jacob is on 20 but is still moving - so why should that not be considered while calculating the answer? John should cover the distance of 10 miles in 40 minutes and Jacob would cover 8 more miles in this time - so shouldn't John be covering 18 miles in total to catch up with Jacob?

Hi, trying to answer your query,
Your part reasoning is correct, but not in later part.
After 1 hour, John is still on 10 and Jacob is on 20 but is still moving - : Correct
so shouldn't John be covering 18 miles in total to catch up with Jacob?[/quote] : incorrect

For catch and chase questions ,important thing is relative speed. Hoping you know that if both move in same direction speeds are deducted :
So relative speed is : 15-12 = 3.
Now in these questions, its eas when you think the object in lead as stationery and the former catching it . If you keep calculating the 1st ran how much in time required to catch. it will make things difficult. So, moving on,
Speed : 3 kmph
Distance travelled by john in 1.40 hours : 10
Distance travelled by jacub in 1.40 hours : (100/60 )*12 =20

distance left to be covered : 10 : Now if you see i am not considering Jakub is still travelling and keeping him stationery at 20 will make this easier.
Time : D/T : 10/3 or 3 1/3

Hope you understood this catch.


Give Kudos if you like answer

@Bunnel
Why cant this Q be solved by equation: 15(t-1) = 12t

John travelled for (t-1) hr and Jacob travelled for t hrs.
From Q its not clear we have to calculate time after 40 minutes.

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