1cyclist bikes at 6 mph and the 2nd starts 3 hrs later at 10 mph. When will the 2nd catch up in hours?

Answer 1

The distance cyclist 1 travels before cyclist 2 starts is :
Distance = Speed x time taken.
Distance = 6 mph x 3 hr = 18 miles.

Now if cyclist 2 were to start traveling at 6 miles per hour, then a distance of 18 miles would remain between the two cyclists. This is not the case. Cyclist 2 is traveling 4 mph (10 - 6) quicker then cyclists 1. So it's this 4 miles per hours that closes the gap between them. How long will this take ?

10mph - 6mph = 4mph extra.
6mph x 3h = 18 miles.
The extra distance is 18 miles between them at first.
Gap Distance / Extra speed = Time to close the gap.
18m / 4mph = 4∙5hr.

To check answer:
6mph x 4∙5h + 18 miles = 27 + 18 = 45 miles.
10mph x 4∙5h = 45 miles.
Answer confirmed.

Answer 2

Suppose t is the time when they met (starting measuring since the first cyclist started cycling). Then the first cyclist made 6t miles. The second cyclist made 10(t - 3) miles. Since they've met, the distance is the same for both of them, so we can write:

6t = 10(t - 3) --> 4t = 30 --> t = 7,5

The second catch up with the first 7 and a half hours after the first one started cycling

Answer 3

Jack forgot his "t" was the time for the first cyclist, that the solution to the equation was not the answer to the question. If you want it to be, write the equation as:

6(t+3) = 10t
6t + 18 = 10t
18 = 4t
4.5 = t

Answer 4

6(t+3)=10t
6t+18=10t
4t=18
t=9/2 hours

that is
t=4 hours 30 minutes

Answer 5

d1=6t
d2=10(t-3)
6t=10t-30
4t=30
t=7.5

Answer 6

4.5hours
4hrs 30mins