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in Quadratic Equations by (44.9k points)
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An aeroplane takes off 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time , it has to increse its speed by 250km/h from its usual speed . Find its usual speed .

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Let the usual speed be x km/hr.
Actual speed `=(x+250)km/hr.`
Time taken at usual speed `=((1500)/(x))hr.`
Time taken at actual speed `=((1500)/(x+250))hr.`
Difference between the two times taken `=(1)/(2)hr.`
`:." "(1500)/(x)-(1500)/(x+250)=(1)/(2)`
`implies" "(1)/(x)-(1)/((x+250))=(1)/(3000)implies((x+250)-x)/(x(x+250))=(1)/(3000)`
`implies" "(250)/((x^(2)+250x))=(1)/(3000)impliesx^(2)+250x-750000=0`
`implies" "x^(2)+1000x-750x-750000=0`
`implies" "x(x+1000)-750(x+1000)=0implies(x+1000)=0implies(x+1000)(x-750)=0`
`implies" "x+1000=0" or "x-750=0impliesx=-1000" or "x=750`
`implies" "x=750" "[because" speed cannot be negative"]`
Hence, the usual speed of the aeroplane was 750 km/hr.

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