An airplane took off from the starting point 45 minutes later than the scheduled time. The destination was 2100 km away from the starting point. To reach on time, the pilot had to increase the speed by 40% of its usual speed. What was the increased speed (in km/h)?

Let's assume the usual speed of an airplane is 'y' and the usual time is taken to cover the given distance is 't'.

$$speed\times\ time\ =\ distance$$

yt = 2100    Eq.(i)

An airplane took off from the starting point 45 minutes later than the scheduled time. The destination was 2100 km away from the starting point. To reach on time, the pilot had to increase the speed by 40% of its usual speed.

60 minutes = 1 hour

1 minutes = $$\frac{1}{60}$$ hour

45 minutes = $$\frac{1}{60}\times45$$ hour = 0.75 hours

(100+40)% of y $$\times$$ (t-0.75) = 2100

140% of y $$\times$$ (t-0.75) = 2100     Eq.(ii)

Equating Eq.(i) and Eq.(ii).

yt = 140% of y $$\times$$ (t-0.75)

t = 1.4 $$\times$$ (t-0.75)

t = 1.4t-1.05

1.4t-t = 1.05

0.4t = 1.05

t = 2.625

Put the value of 't' in Eq.(i).

$$2.625\times y = 2100$$

y = 800

Increased speed (in km/h) = 140% of y

= 140% of 800

= $$\frac{140}{100}\times\ 800$$

= 1120 km/h