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Question
A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Solution
Let the usual speed of the plane = x km/hr
Increased speed of the plane = (x + 100) km/hr
Time take to reach the destination at the usual speed, `t_1 = 1500/x` hr
Time take to reach the destination at the increased speed, `t_2 = 1500/(x + 100) hr`
Difference of both the times = t1 - t2 = 30 mins = 1/2 hr
`1500/x - 1500/(x + 100) = 1/2`
`=> 1500 (x + 100) - 1500x = (x(x + 100))/2`
`=> 1500x + 150000 - 1500x = (x(x+100))/2`
`=> 300000 = x^2 + 100x`
`=> x^2 + 100x - 300000 = 0`
`=> x^2 + 600x - 500x - 300000 = 0`
`=> (x + 600) (x - 500) = 0`
x = 500 or x = -600
As speed cannot be negative so x = 500 km/hr.
Hence, the usual speed of the plane is 500 km/hr
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