Date : 1
An aeroplane left 30 min later than its
Shedualed time and in order to such reach
it destination 300 km in time at had
to increase its speed byy 250 km/hr from it usual speed
Answers
Answer:
Step-by-step explanation:
Sol:
Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
Case (i)
Speed = Distance / Time = (1500 / x) Hrs
Case (iI)
Time taken by the aeroplane = (x - 1/2) Hrs
Distance to the destination = 1500 km
Speed = Distance / Time = 1500 / (x - 1/2) Hrs
Increased speed = 250 km/hr
⇒ [1500 / (x - 1/2)] - [1500 / x] = 250
⇒ 1/(2x2 - x) = 1/6
⇒ 2x2 - x = 6
⇒ (x - 2)(2x + 3) = 0
⇒ x = 2 or -3/2
Since, the time can not be negative,
The usual time taken by the aeroplane = 2 hrs
and the usual speed = (1500 / 2) = 750 km/hr.
Recommend(9) Comment (0)
person
Syeda , SubjectMatterExpert
Member since Jan 25 2017
Answer,
Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
Case (i)
Speed = Distance / Time = (1500 / x) Hrs
Case (iI)
Time taken by the aeroplane = (x - 1/2) Hrs
Distance to the destination = 1500 km
Speed = Distance / Time = 1500 / (x - 1/2) Hrs
Increased speed = 250 km/hr
⇒ [1500 / (x - 1/2)] - [1500 / x] = 250
⇒ 1/(2x2 - x) = 1/6
⇒ 2x2 - x = 6
⇒ (x - 2)(2x + 3) = 0
⇒ x = 2 or -3/2
Since, the time can not be negative,
The usual time taken by the aeroplane = 2 hrs
and the usual speed = (1500 / 2) = 750 km/hr.