Question

A plan left 30 min late then it's scheduled time and in order to reach the distance 1500km away in time it had to increase its speed by100km/h from the usual speed find usual speed

Answer 1

distance = 1500km
let the usual speed be x
usual time (T1) = dist./ speed = 1500/x
new time (T2) = 1500/x+100 (as speed increases by 100km/h)

we know that the new time is less than the usual time by 30min or 1/2 an hour

it means

T2 = T1 - 1/2
1500/(x+100) = 1500/x - 1/2
1/2 = 1500/x - 1500/(x+100)

(by taking LCM)

1/2 = 1500x + 150,000 - 1500x / x^2 + 100
1/2 = 150,000/x^2 + 100
x^2 + 100 = 300,000
x^2 + 100 - 300,000
so by solving this we get
x = -1783 , x = 1683

speed cannot be in negative
so
x = 1683 km/h
so the usual speed of plane is 1683 km/hr

If there us any mistake or any point u dont understand please let me know.

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