Question 547829
This is just another way of saying
" At what time will the busses meet? "
The key is:
" How much of a head start over B does A have?
In 15 min, bus A goes
{{{ d[1] = 20*(15/60) }}} mi
{{{ d[1] = 5 }}} mi
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Start a stopwatch when bus B leaves
and call the time when they meet {{{t}}}
Let {{{d}}} = distance B will now travel
equation for bus A:
(1) {{{ d - 5 = 20t }}}
equation for bus B:
(2) {{{ d = 24t }}}
substitute (2) into (1)
(1) {{{ 24t - 5 = 20t }}}
(1) {{{ 4t = 5 }}}
(1) {{{ t= 5/4 }}} hrs
They will have traveled the same distance 
in 1 hr and 15 min, which will be
9:15 + 1.25 = 10:30
check answer:
(2) {{{ d = 24*1.25 }}}
(2) {{{ d = 30 }}}
and
(1) {{{ d - 5 = 20*1.25 }}}
(1) {{{ d - 5 = 25 }}}
(1) {{{ d = 30 }}}
OK