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Question :

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Answer 1

Q: A plane left 30 min later than the scheduled time and in order to reach the destination 1500 km away in time it has to increase the speed by 250km/h from the usual speed. Find its usual speed?

Solutions :-


We know that,
Speed = Distance / Time

Given : Distance = 1500 km

Let the original time be x hours

Speed = Distance / Time
= 1500/x km/hr

Time taken = x - 30 minutes = x - 0.5 hrs


Now,
Substitute the value of x

So, New speed = Distance / Time
= 1500/x - 0.5 km/hr


Find the usual time of plane :-

New Speed - Original Speed = 100

=> (1500/x - 0.5) - (1500/x) = 100
=> 150 { (10/x - 0.5) - 10/x} = 100
=> 30x - 30x + 15 = 2x² - x
=> 15 = 2x² - x
=> 2x² - x - 15 = 0

Solving quadratic equation,
=> 2x² - x - 15 = 0
=> 2x² - 6x + 5x - 15 = 0
=> 2x (x - 3) + 5 (x - 3) = 0
=> (2x + 5) (x - 3) = 0
=> x = - 5/2 or 3

Time is always taken positively.
So, time = 3 hrs


Find the usual speed of plane :-

Usual speed = 1500/3 = 500 km/hr


Hence,
Usual speed of plane = 500 km/hr

Answer 2

Question :

A plane left 30 min later than the scheduled time and in order to reach the destination 1500 km away in time it has to increase the speed by 100 km/hr from the usual speed. Find its usual speed .  

Answer:

500 km/hr

Step-by-step explanation:

I think you know the formula :

Speed = distance / time

Let the usual time that the plane takes be x hr.

Usual speed

Distance = 1500 km

Time = x hr

Speed = 1500  / x  ........................(1)

Speed to be increased

Time = x hr - 30 min

You should know that 1 hr = 60 min

                              ==> 1 / 2 hr = 60 / 2 min

                              ==> 0.5 hr = 30 min

Time = x hr - 0.5 hr

        == >  ( x - 0.5 ) hr

Speed = distance / time

           == >  ( 1500  ) / ( x - 0.5 )

           == >  1500  /  ( x - 0.5 )  ................................(2)

First find the value of x

Given :

Original speed has to be increased by 100 km/hr

So : Final speed - usual speed = 100 km / hr

From (1) and (2) you can calculate easily :-

1500 / ( x - 0.5 ) - 1500 / x = 100

Take 1500 common to get this :

== > 1500 [  1 / ( x - 0.5 ) - 1 / x ] = 100

Dividing by 100 both sides to get this :

== > 15 [ 1 / ( x - 0.5 )  - 1 / x ] = 1

== > 15 [ ( x - x + 0.5 ) / x ( x - 0.5 ) ] = 1

== > 15 [ 0.5 / ( x² - 0.5 x ) ] = 1

== > 0.5 / ( x² - 0.5 x ) = 1 / 15

Cross multiply to get  :-

==> x² - 0.5 x = 15 × 0.5

== > x² - 0.5 x = 7.5

== > x² - 0.5 x - 7.5 = 0

== > x² - 3 x + 2.5 x - 7.5 = 0

== > x ( x - 3 ) + 2.5 ( x - 3 ) = 0

== > ( x - 3 )( x + 2.5 ) = 0

Either :-

x - 3 = 0

== > x = 3

Or :-

x + 2.5 = 0

== > x = -2.5

How can time be negative ?

Nope that's not possible in this real world !

Time = 3 hr

So : usual speed = distance / time

                            = 1500 km / 3 hr

                            = 500 km /hr

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

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