There is a roadway between city A and B . A car P starts at 5:00 am from A and reaches B at 10:00 am. Another car Q starts from B at 7:00 am and reaches A at 9:00 am. Find the time when car P meets car Q ?
I did as follows
for car P , travelling from 5:00 am to 7:00 am cannot definitely meet car Q as Q has not even started .
Lets say at 7:00 am car P reached a point R, and lets assume total road way distance be D
so $D_{AR} = \frac{2\times D}{5} $
so lets see from 7:00 am to 9:00 am when they will meet .
Let the cars P and Q are said to meet at a distance $D_{1}$ from R and $D_{1}$ from B.
The velocity ratio is inversely proportional to time ratio so it is $2/5$
$$ \frac{D_{1}}{D_{2}} = \frac{2}{5}$$
As distance travelled by car P between 7:00 to 9:00 (that is 2 hrs ) is $\frac{2\times D}{5} $
so the cars meet at $D_{AR} + \frac{2}{7}\times \frac{2\times D}{5} $ = $ \frac{18 \times D }{35} $ from city A
SO Time is calculated as like
To travel D distance if it takes 5 hours , then to travel $ \frac{18 \times D }{35} $ it should take
$\displaystyle\frac{5}{D} \times \frac{18D}{35}$
So $\frac{18}{7}$ hours . Am I right ? If wrong please correct .
Even if my answer is correct please suggest if there are easier and logical ways to solve this sort of problems without equations . In the above also I have tried avoiding equations by using simple logic Distance proportional to velocity when speed is constant.
Best Answer
Let the meeting time be $t$, a real number with the whole portion being hours. At time $t$, $P$ has covered $\frac {t-5}5$ of the distance. $Q$ has covered $\frac {t-7}2$ of the distance. These have to add to the whole distance, so $$\frac {t-5}5+\frac {t-7}2=1\\2t-10+5t-35=10\\7t=55\\t=7\frac67\approx 7:51:25.714$$