Math, asked by 95eshmsumitp6aok4, 1 year ago

A plane Left 30 min late than its scheduled time and in order to reach the destinations 1500 km away in time, it had to increase it speed by 100 km/h from the usual speed.find its usual speed.

Answers

Answered by dk361646
1
0.5+1500/(x+100)=t
And 1500/x=t
Equating both eq.
1500(1/x +1/(x+100) )=0.5
x=500km/h
Answered by Gitali7101
2




A plane left 30 min later than the scheduled time and in order to reach the destination 1500km away in time it has to increase the speed by 250km/h from the usual speed. find its usual speed

A plane left 30 min later than the scheduled time and in order to reach the destination 1500km away in time it has to increase the speed by 250km/h from the usual speed. Find its usual speed

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/hrs.

Answer 2

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.


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