Question

1. In a flight of 6000 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 400 km/hour and time increased by 30 minutes. Find the original duration of flight.

2. A train travels a distance of 300 km at a constant speed. If the speed of the train is increased by 5 km an hour, the journey would have taken 2 hours less. Find the original speed of the train.

Answer 1

1.Sol:
Let the original speed of the aircraft be 'x' km/hr
Decreased speed of the aircraft = (x - 400) km/hr

Time = Distance/Speed

Original duration of the flight = (6000 / x) hrs
Duration of the flight with decreased speed = (6000 / x-400) hrs

Difference of time = 30 min = 1/2 hrs
⇒ (6000 / x-400) - (6000 / x) = 1/2
⇒ (6000x - 6000x + 2400000) / x(x - 400) = 1/2
⇒ x2 - 400x = 4800000
⇒ x2 - 400x - 4800000 = 0
⇒ (x - 2400)(x + 2000) = 0
⇒ x = 2400 or -2000

Positive value of x is 2400.

Therefore original speed of the aircraft is 2400 km/hr.

Original duration of the aircraft = 6000/2400 = 2.5 hrs.
2.If 'v' is speed, and 't' is time taken for first trip, then 300 = v * t

For next trip, speed is 'v+5' and time taken is 't-2'. So 300 = ( v+5)*(t-2)

300 = vt -2v + 5t - 10

Since t = 300/v,

300 = 300 -2v + 5*300/v - 10

10 = -2v + 1500/v

Multiply by v,

2v2+10v−1500=0

v2+5v−750=0

(v−25)(v+30)=0

Hence v = 25 km/hr ( v cannot be -30 since it is a positive quantity ).