Flight Speed and Delay Calculation

Let initial speed be $F\text{ km/hr}$ and initial planned time be $t\text{ hours}$. $F \cdot t = 11200 \implies t = \frac{11200}{F}$.

Delayed departure by 3 hours, overall delay reduced to 1 hour $\implies$ travel time reduced by $3 - 1 = 2\text{ hours}$. So new travel time is $t - 2$. $(F + 100)(t - 2) = 11200 \implies (F + 100)\left(\frac{11200}{F} - 2\right) = 11200$ $11200 - 2F + \frac{1120000}{F} - 200 = 11200$ $-2F^2 - 200F + 1120000 = 0 \implies F^2 + 100F - 560000 = 0$. Solving: $(F - 700)(F + 800) = 0 \implies F = 700\text{ km/hr}$.

If speed is increased by $350\text{ km/hr}$, new speed $= 700 + 350 = 1050\text{ km/hr}$. New travel time $= \frac{11200}{1050} = \frac{32}{3}\text{ hours} = 10\text{ hours } 40\text{ minutes}$.

Actual departure time $= 6:30\text{ AM} + 3\text{ hours} = 9:30\text{ AM}$. Arrival time $= 9:30\text{ AM} + 10\text{ hours } 40\text{ minutes} = 8:10\text{ PM}$.