Question

a plane left 30 minutes late than its scheduled time and in order to reach the destination 1500km away in time , it had to increase its speed by 100km,/h from the usual speed.find its actual speed

Answer 1

$thirty \: minutes \: = \frac{30}{60 } = \frac{1}{2} \\ \\ \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2} \\ = \frac{1500x - 150000 - 1500x}{ {(x }^{2} - 100x)} = \frac{1}{2} \\ by \: cancelling \: 1500x \: and - 1500x \\ by \: cross \: multiplication \\ {x }^{2} - 100x - 300000 = 0 \\ by \: splitting \: the \: middle \: terms \\ {x}^{2} + 500x - 600x - 300000 = 0 \\ x(x + 500) - 600(x + 500) = 0\\ (x + 500)(x - 600) = 0 \\ x = - 500 \\ x = 600$
As x can't be negative.
x= 600.

THIS IS THE QUESTION IN 10TH CBSE MATHEMATICS EXAMINATION 2018.

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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