PS Princeton CAT1

i believe 3+10t=4+15t
t is negative is negative, which is not possible so the correct answer is E

Hi pinkbox! sorry, i couldnt understand your reasoning. can you explain for me?

.. alternative approach (i appreciate a more structure-algebraic approach):

3+10=13
+10=23
+10=33
+10=43, always finish with 3...

4+15=19
+15=34
+15=49
+15=64
+15=79, always finish with 9 or 4...

so, letter E

i will gladly try again
basically, bus B is starting at 3:00 and then running every hr so its schedule will be following a function that looks like this: 3+10t
ie it will run again at 1:00 AM, then at 11:00 AM, etc
likewise, bus C will be starting at 4:00 and running every 15 hrs so its schedule will be following a function that looks like this: 4+15t
now, we are looking for a time (t) for which these functions are equal.
we set 3+10t=4+15t
if u do the algebra you get t=-1/5
time can only be positive!
hence, "So long as they continue operating in this manner, they will never leave at the same time"
hope that's clear now