Suppose that, after #h" hours"# past #10:00" AM,"# Lucy and John
meet each other.
Lucy started running at #10:00" AM"# at an average rate of #3.5#
miles/hour, so, from the starting point (s.p.) to the meeting point
(m.p.), Lucy ran #3.5h# miles...................#(1).#
Now, let us calculate John's distance from the s.p. to the m.p.,
keeping in mind that, he started running at #10:30" AM,"# so, he ran
for #1/2# hour less time, i.e., for #(h-1/2)" hours"# at an average
rate of #5# miles/hour; hence, John's distance from s.p. to m.p. is
#5(h-1/2)..................................(2)#
Both the Distances #(1) and (2)# are the same.
#:. 3.5h=5(h-1/2)=5h-5/2=5h-2.5#
#:. 3.5h-5h=-1.5h=-2.5#
#rArr h=2.5/1.5=5/3#
Therefore, after #5/3=1 2/3 #hours, i.e., after #1" hour & "40#
#(=2/3xx60)" minutes"# past #10:00" AM"# they meet.
Thus, at #11:40" AM"# John catch up with Lucy.
Enjoy maths.!
