Question

an aeroplane left 50 minutes later than its scheduled time and in order to reach the destination 1250 km away in time had to increase its speed by 250km/h from its usual time. find its usual time

Answer 1

Solution:-
Let the usual speed of the aeroplane = 'x' km/hr
Increased speed of the aeroplane = (x+250) km/hr
Time taken to reach the destination at usual speed = T1 = (1250/x) hr
Time taken to reach the destination at increased speed = T2 = (1250/x+250) hr.
Given : T1 - T2 = 50 minutes.
Therefore,
(1250/x) hr - (1250/x+250) hr = 50/60
After taking the L.C.M. we get
(1250x +312500 - 1250x)/x(x+250) = 5/6
Cross multiplying, we get
312500*6 = 5(x²+250x)
1875000 = 5x²+1250x
Dividing it by 5, we get
375000 = x²+250x
x²+250x-375000
x² + 750x - 500x - 375000
x(x+750) - 500(x+750)
(x+750) (x-500)
x + 750 = 0 
x = - 750 
x -500 = 0
x = 500
The value of x is 500 because the speed cannot be in negative.
so, the usual speed of the plane is 500 km/hr.
Answer.

Answer 2

the answer is 500kmph
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