Question

Two Indian tourists in the US cycled towards each other, one from point A and the other from point B. The first tourist left point A 6 hrs later than the second left point B, and it turned out on their meeting that he had traveled 12 km less than the second tourist. After their meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hours later and the second arrived at A 9 hours later. Find the speed of the faster tourist.

Answer 1

Let the total distance to D and speed of A & B be $S_A\&S_B$ respectively.
Let the time when both A&B meet be t
$\therefore s_n(t-6)+S_{B}t=D$
As distance covered by A is 12 km less than B
$\therefore S_{B}t-S_A(t-6)=12$
Now remaining distance covered by A in B hours
$\frac{S_{B}t}{8}=S_A\to \frac{S_B}{S_A}=\frac{B}{t}$
Remaining distance covered by B in 9 hours
$\frac{S_A(t-6)}{9}=S_B\Rightarrow \frac{S_B}{S_A}=\frac{t-6}{9}$
Equating $\frac{S_B}{S_A}=\frac{8}{t}=\frac{t-6}{9}$
$\Rightarrow t^2-6t-72=0$
$\Rightarrow t=12$
$\therefore S_A=6km/hr$
$S_B=6km/hr$