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 $9$ hrs later. Find the speed of faster tourist.
$\bf\text{options}$ a.)$4$ km/h $\quad$ b.) $6$ km/h $\quad$ c.) $9$ km/h $\quad$ d.) $2$ km/h
let they meet at point X as shown in the diagram.
From the question i concluded that first cyclist is fastest .
let the time and distance taken by first cyclist be $(t-6)$ hrs and $(m)$ km to travel upto point X
let the time and distance taken by second cyclist be $(t)$ hrs $(m+12)$ km to travel upto point X
i used the formula that
if two persons starting from point x and y with speed $s_1$ ans $s_2$ coming towards
each other from opposite direction after meeting each other take time $t_1$ and $t_2 $
to reach points y and x respectively then
$\large \frac{s_1}{s_2}=\sqrt{\frac{t_2}{t_1}}$
and i concluded.
$\large \dfrac{\frac{m}{t-6}}{\frac{m+12}{t}}=\sqrt{\frac{9}{8}}$
now stucked for two unknowns.
Best Answer
Your $t+6$ should be $t-6$ as A left after B. We can then write an equation for each that says they traveled at constant speed $$\frac m{t-6}=\frac {m+12}8 \\\frac {m+12}t=\frac m9\\\frac m{m+12}=\frac {t-6}8\\\frac m{m+12}=\frac 9t\\\frac {t-6}8=\frac 9t\\t^2-6t=72$$ and you have a quadratic to solve