Question

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Two persons A and B start simultaneously from
two places c km apart and walk in the same
direction. If A travels at the rate of p km/h and B
travels at the rate of q km/h, then A has
travelled before he overtakes B a distance of
(a) ac/p+q km
(b) pc/p-q km
(c) qc/p-q km
(d) pc/p+q km​

Answer 1

A has  traveled before he overtakes B a distance of  $\frac{pc}{p-q} \text{ km}$.

Step-by-step explanation:

We are given that two persons A and B start simultaneously from two places c km apart and walk in the same direction. If A travels at the rate of p km/h and B  travels at the rate of q km/h.

The distance traveled by A before he overtakes B is given by;

       Distance =  $({\text{Speed of A}) \times (\text{Time taken by A to overtake B}})$

We know about the speed of A which is p km/hr but we have to find the time taken by A to overtake B.

Now, the time taken by A to overtake B =  $\frac{\text{Distance between A and B }}{\text{Relative speed of A and B}}$

Here, the distance between A and B = c km

The Relative speed of A and B when they both traveling in the same direction is = Speed of A - Speed of B

                   = p km/hr - q km/hr

So, the time taken by A to overtake B =  $\frac{c}{p-q}$

Now, using our first formula ;

Distance traveled by A to overtake B = $({\text{Speed of A}) \times (\text{Time taken by A to overtake B}})$

          =  $p \times \frac{c}{p-q}$  =  $\frac{pc}{p-q} \text{ km}$.