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.


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 $$\frac {t-5}5+\frac {t-7}2=1\\2t-10+5t-35=10\\7t=55\\t=7\frac67\approx 7:51:25.714$$

  • $\begingroup$ Thanks for the reply.Please, Can you also state whats the error in my working ? Because I am getting 18/7 hours which is 7:34 am. Please correct me. $\endgroup$ – Harish Kayarohanam Aug 15 '13 at 21:25
  • $\begingroup$ in your line "so the cars meet at", the part $\frac {2D}5$ should be $\frac {3D}5$ because that is how much is left. $\endgroup$ – Ross Millikan Aug 15 '13 at 22:06
  • $\begingroup$ thanks for pointing out the problem . $\endgroup$ – Harish Kayarohanam Aug 15 '13 at 22:57

Let's say that cars velocity is constant.

Then car B travells $\frac{1}{2}$ of the path in one hour and car A travells $\frac{1}{5}$ of the path. We know that they move each to each other so when they meet the sum of the distances they've passed will be equals as the distance between place A and place B.

Then we'll have:

$$\frac{x}{2} + \frac{x+2}{5} = 1$$

$$\frac{5x + 4 + 2x}{10} = 1$$

$$7x + 4 = 10$$

$$7x = 6$$

$$x = \frac{6}{7}$$

It means that they'll meat $\frac{6}{7}$ hours after the second car have started travelling.

  • $\begingroup$ Ya I have edited it , It is waiting his approval $\endgroup$ – Harish Kayarohanam Aug 15 '13 at 21:34
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    $\begingroup$ Thanks for editing, it's just a typo and i continued later using $3x$, instead of $2x$ $\endgroup$ – Stefan4024 Aug 15 '13 at 21:38

Normalize the distance to $1$. Let $x$ be the location they meet at time $t$. The distance traveled by P is $x=v_Pt, $ similarly for Q the distance traveled is $1-x=v_Q(t-2).$ $v_P$ and $v_Q$ will be $\frac15,$ and $\frac12$ respectively based on the total travel time. Solve the two equations for $t$ and $x$.


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