Question

A plane left 30min later than the schedule time and in order to reach its destinatin 1500km away in time ,it has to increase its speed by 250km/hr its usual speed. Find its usual speed

Answer 1

Let the usual speed of plane be x kmph
then speed=distance / time
X=1500/time
time = 1500/X
hence usual time taken by the plane is 1500/X hours
Now new speed = X + 250
Then new time taken = 1500/X + 250
now equating it we get
$\frac{1500}{x} - \frac{1}{2} = \frac{1500}{x + 250}$

(1/2 because 30 minutes are converted into hours to match the equation)
now solve this equation further
$\frac{3000 - x}{2x} = \frac{1500}{x + 250}$
$(3000 - x)(x + 250) = 1500 \times 2x$
$3000x + 750000 - x^{2} - 250x = 3000x$
$x^{2}+250x - 750000=0$
Now solve this equation for roots..the root which is positive is the answer...
Thank you

Answer 2

Solution :-

Let the original speed of train be x km/hr
New speed = (x + 250) 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 + 250) = 1/2
=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2
=> 2(37500) = x(x + 250)
=> 75000 = x² + 250x
=> x² + 250x - 75000 = 0
=> x² + 1000x - 750x - 75000 = 0
=> x(x + 1000) - 750(x + 1000) = 0
=> (x - 750) (x + 1000) = 0
=> x = 750 or x = - 1000

∴ x ≠ - 1000 (Because speed can't be negative)


Hence,
Its usual speed = 750 km/hr