Question

Aeroplane left 30 minutes later than its scheduled time and in order to reach destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed. determine its usual speed.

Answer 1

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

Answer 2

ANSWER:-

Let the usual speed of the aeroplane = x km/h

Then it's increased speed = (x+250)km/h

Time difference in the two cases = 30 min. = 1/2 hr.

$\sf\red{\frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} }$

$\longrightarrow$$\sf\red{2 \times 1500((x + 250) - x) = x(x + 250)}$

$\longrightarrow$$\sf\red{{x}^{2} + 250x - 750000 = 0}$

$\longrightarrow$$\sf\red{(x - 750)(x + 1000) = 0}$

$\longrightarrow$$\sf\red{x = 750 \: or \: x = - 1000}$

As the speed cannot be negative, x ≠ -1000 ,so x = 750

Hence, the usual speed of the aeroplane = 750km/h