Question #50fa1

Question

1465 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-29T18:38:54

At $11 : 40 AM$ $11:40" AM"$ John catch up with Lucy.

Suppose that, after $h hours$ $h" hours"$ past $10 : 00 AM,$ $10:00" AM,"$ Lucy and John

meet each other.

Lucy started running at $10 : 00 AM$ $10:00" AM"$ at an average rate of $3.5$ $3.5$

miles/hour, so, from the starting point (s.p.) to the meeting point

(m.p.), Lucy ran $3.5 h$ $3.5h$ miles................... $(1) .$ $(1).$

Now, let us calculate John's distance from the s.p. to the m.p.,

keeping in mind that, he started running at $10 : 30 AM,$ $10:30" AM,"$ so, he ran

for $\frac{1}{2}$ $1/2$ hour less time, i.e., for $(h - \frac{1}{2}) hours$ $(h-1/2)" hours"$ at an average

rate of $5$ $5$ 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.!