Question

A plane left 30 minutes later than its schedule time and in order to reach its destination 1500km away in time it had to increase its speed by 100 km

Answer 1

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

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

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

= [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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