Question

A plane left 30 minutes later than the scheduled time and in order to arrive its destination 1500 km away in time, it has to increase its speed by 250 km/h from its usual speed. Find its usual speed.

Answer 1

Answer:

Let the usual time taken by the aeroplane = x km/hr

Distance to the destination = 1500 km

--------Speed = Distance / Time = (1500 / x) Hrs

 

------------Time taken by the aeroplane = (x - 1/2) Hrs

            Distance to the destination = 1500 km

Speed = Distance / Time = 1500 / (x - 1/2) Hrs

 

Increased speed = 100 km/hr

 

[1500 / (x - 1/2)] - [1500 / x] = 100

 150((10/x-1/2)-10/x0=100

 15 /x -1/2- 15/x =1 

30x -30x +15=2x^2-x

15=2x^2-x

2x^2-x-15=0

2x^2 - 6x+5x-15=0

2x(x-3)+5(x-3)

2x+5=0  or x-3 =0

x=-5/2   or x=3

Since, the time can not be negative,

The usual time taken by the aeroplane = 3hrs

and the usual speed = (1500 / 3) = 500km/hr

Answer 2

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