A jet plane covers 4500 km is some time. If the regular speed is decreasesd by 150 km/h, it take one and a half hours more to complete the journey. Find the original speed of the jet plane.

Explanation

Let the original speed of the jet plane be `x` km/h.
Time taken to covers `4500 km=("distance")/("speed")=4500/x` hours
The new speed is `(x-150)km//h`
Time taken to cover 4500 km `=("distance")/("speed")=4500/(x-150)` hours
From the given condition.
`4500/(x-150)-4500/x=3/2` (one and a half hours `=3/2`)
`:. 4500(1/(x-150)-1/x)=3/2`
`:.4500[(x-x+150)/(x(x-150))]=3/2`
`:. 4500(150)=3/2x(x-150)`
`:.4500xx150xx2/3=3/2x (x-150)xx2/3` ..........[ Multiplying both the sides by `2/3`]
`:.4500xx100=x^(3)-150x`
`:.x^(2)-150x-45000=0` brgt `:.x^(2)-750x+600x-450000=0`

`:.x(x-750)+600(x-750)=0`
`:.(x-750)(x+600)=0`
`:.x-750=0` or `x+600=0`
`:.x=750` or `x=-600`
But the speed cannot be negative `:.x=-600` is unacceptable.
`:.x=750`
Ans. The original speed of the jet plane is `750km//h`