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.!
