# Question #50fa1

Sep 29, 2016

At $11 : 40 \text{ AM}$ John catch up with Lucy.

#### Explanation:

Suppose that, after $h \text{ hours}$ past $10 : 00 \text{ AM,}$ Lucy and John

meet each other.

Lucy started running at $10 : 00 \text{ AM}$ at an average rate of $3.5$

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

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

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

keeping in mind that, he started running at $10 : 30 \text{ AM,}$ so, he ran

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

rate of $5$ miles/hour; hence, John's distance from s.p. to m.p. is

$5 \left(h - \frac{1}{2}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$

Both the Distances $\left(1\right) \mathmr{and} \left(2\right)$ are the same.

$\therefore 3.5 h = 5 \left(h - \frac{1}{2}\right) = 5 h - \frac{5}{2} = 5 h - 2.5$

$\therefore 3.5 h - 5 h = - 1.5 h = - 2.5$

$\Rightarrow h = \frac{2.5}{1.5} = \frac{5}{3}$

Therefore, after $\frac{5}{3} = 1 \frac{2}{3}$hours, i.e., after $1 \text{ hour & } 40$

$\left(= \frac{2}{3} \times 60\right) \text{ minutes}$ past $10 : 00 \text{ AM}$ they meet.

Thus, at $11 : 40 \text{ AM}$ John catch up with Lucy.

Enjoy maths.!