Question

a plane started late by 30 minutes. I to reach a destination 1500 km away in time the piolet increased the speed by 100km/hr. the find the original speed of the plane

Answer 1

okay so, let's take the time taken by the plane to travel 1500km normally be 't', the normal speed be 's'. the equation here becomes t=1500/s. now since the plane is 30 min late but it still reaches in time by increasing the speed by 100, the equation 2 can be shown as, (t-30)=1500/(s+100)
dividing equation 1 by 2,
t/t-30=s+100/s
st= (s+100) (t-30)
solving that we get,
3s-10t=30.
now from eq 1, t=1500/s
s=1500/t
substitute that value in the now equation, we get,
4500/t-10t=30
450-t^2=3t
t^2+3t-450=0
solving that would give you the time 't'
put that in equation 1, t=1500/s and you'll find the original speed of the plane.

Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 100) km/hr

We know that,

Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

=> 1500/x - 1500/(x + 100) = 1/2

=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2

=> 2(15000) = x(x + 100)

=> 30000 = x² + 100x

=> x² + 100x - 30000 = 0

=> x² + 600x - 500x - 30000 = 0

=> x(x + 600) - 500(x + 600) = 0

=> (x - 500) (x + 600) = 0

=> x = 500 or x = - 600

∴ x ≠ - 600 (Because speed can't be negative)

Hence,

Its usual speed = 500 km/hr