Question

a plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its usual speed and find its usual speed

Answer 1

Question is Incomplete :
A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time it has to increase the speed by 100 km/h from the usual speed.find its usual speed.

SOLUTION :
Let x km/h be the usual speed of an aeroplane .
Total Distance to reach the destination = 1500 km
ATQ..
Case 1.
Original Time = Distance /speed = (1500 / x) h

Case 2.
Let( x +100 )km/h be the new speed of an aeroplane

New Time = Distance / Speed = 1500 / (x +100) Hrs

Original time - new time = 30 min [1500 / x] - [1500 / x+100] = 30 min [1500 / x] - [1500 / x+100] = 30/60
[ 1 min = 1/60 h]
[1500 / x] - [1500 / x+100] = 1/2 h
1500 [ 1/x - 1 /(x+100)] = ½
1500 × 2 [ x +100 - x / (x(x+100)] = 1
3000 [ 100/(x²+100x) ]= 1
300000= x² +100x

x² +100x -300000= 0
x² +600x -500x -300000= 0
x(x + 600) - 500(x +600)= 0
(x +600 )(x -500) = 0
x = -600 or x= 500
Since, the speed can not be negative.

Hence, the usual speed of an aeroplane = 500 km/h.

HOPE THIS WILL HELP YOU..

Answer 2

Given:

Late = 30 min

Distance = 1500 km

Speed increase = 100 km / hr

To find:

The usual speed.

Solution:

By formula,

Speed = Distance / Time  

Since the time is unknown,  

Consider x for time,

Speed original = 1500 / x  

Time taken = ( x - 1 / 2 )  

Therefore,

Substituting the value of x,

Speed new = 1500 / (x - 1/2)  

Speed New - Speed original = 100

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

150 ( ( 10 / x - 1 / 2 ) - 10 / x = 100  

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

15 = 2x^2 - x

2x^2 - x -15 = 0

Solving,

x = -5/2

x = 3

As time cannot be negative,

The usual time taken by the plane is 3 hr

Speed original = ( 1500 / 3 ) = 500 km / hr

Hence, the usual speed of the plane is 500 km / hr

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