Question

A car starting from station A at 8:00am reached station B at 12:00 noon. If another car starting from station B at 9:00 am reaches sattion A at 11:30 am.Then the two cars meet each at what time

Answer 1

Let the distance from A to B be 'x' km.

Let C be the point at which the 1st car arrives after driving for 1 hour (i.e., at 9 am, the time at which the 2nd car starts).

Let D be the point at which the 2 cars meet.

Let the distance from C to D be 'y' km.

Now, total time travelled by 1st car = 12-8=4 hours
And, total time travelled by 2nd car = 11.5-9=2.5 hours

So, speed of 1st car = $\frac{x}{4}$ km/h
And, speed of 2nd car = $\frac{x}{2.5}=\frac{2x}{5}$ km/h

Now, distance from A to C = $$\begin{gathered} \frac{x}{4}km/hour\times 1hour \\ =\frac{x}{4}km \end{gathered}$$

Therefore, distance from B to D $$\begin{gathered} =x-\frac{x}{4}-y \\ =\frac{3x}{4}-y \end{gathered}$$

Now, since the 2 cars meet at D, therefore:-

Time required by the 1st car to move from C to D = Time required by the 2nd car to move from B to D.
$\Rightarrow \frac{y}{{\displaystyle \frac{x}{4}}}=\frac{{\displaystyle \frac{3x}{4}}-y}{{\displaystyle \frac{2x}{5}}}$
$$\begin{gathered} \Rightarrow \frac{4y}{x}=\left(\frac{3}{4}-\frac{y}{x}\right)\frac{5}{2} \\ \Rightarrow \frac{4y}{x}+\frac{5y}{2x}=\frac{15}{8} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{13y}{2x}=\frac{15}{8} \\ \Rightarrow \frac{y}{x}=\frac{15}{52} \end{gathered}$$

Now, time required by the 1st car to move from C to D
= $\frac{y}{{\displaystyle \frac{x}{4}}}=\frac{4y}{x}$
$=4\times \frac{15}{52}hours=\frac{15}{13}hours$
$$\begin{gathered} =1hourand\left(\frac{2}{13}\times 60\right)mins \\ =1hour9minutes13.8seconds \end{gathered}$$

So, the time at which the 2 cars will meet = $$\begin{gathered} 9am+1hour9minutes13.8seconds \\ i.e.,10:09:13.8am \end{gathered}$$