Two pigs entered a race around a track. Piggly takes 6 minutes to run one lap. Wiggly takes 5 minutes to run one lap. If both pigs begin the race at the same time, what is the shortest amount of minutes it will take for them to be back at the starting line at the same time?

36 minutes.

Step-by-step explanation:

Piggly takes 6 minutes and Wiggly takes 5 minutes. Therefore Wiggly is travelling one minute slower.

If we just think about the first lap to start, and break it up into minute sections (like in the diagram attached), we can think about each pigs placement on the diagram for each lap.

At the end of the first lap, Piggly will have reached the end of the 6th minute section and will be back at the start/finish. Wiggly will only be at the end of the 5th minute section.

For the second lap, Piggly will again end up back at the start/finish, but Wiggly, having started his lap one section behind (at the end of the 5th section) will now end up even further behind (at the end of the 4th section).

Go ahead and count it out. Every lap he only covers 5 section, so every lap he falls one section further behind.

After six laps he will finish back at the start, at the same time as Piggly (but Piggly will be a lap ahead of him. Our sections have 6 minutes in each, and 6 laps are completed.

6 x 6 = 36 minutes

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)