Question

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Answer 1

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

Answer 2

$\underline {\bf{refer \: to \: the \: attachment}}$

Continuation from pic :-


$\sf \frac{1500(x + 100)-1500x}{x(x + 100)} = \frac{1}{2}$


$\sf\frac{1500x - 1500x + 150000}{x {}^{2} + 100x } = \frac{1}{2}$



$\sf\frac{ 150000}{x {}^{2} + 100x } = \frac{1}{2}$



$\sf \: 150000 \times 2 = {x}^{2} + 100x$



$\sf \: 300000 = {x}^{2} + 100x$



$\sf \: 0 = {x}^{2} + 100x -300000$


$\sf \: \sqrt{ D }\: = \sqrt{ {b}^{2} - 4ac}$



$= \sqrt{ (100) {}^{2} - 4 \times 1 \times - 300000}$



$= \sqrt{ 10000 + 1200000}$



$= \sqrt{1210000}$


$D \: = 1100$


So, the two roots will be :-


-b-√D/2a and -b +√D/2a


$\frac{ - 100 + 1100}{2} \: and \: \frac{ - 100 - 1100}{2}$


$\frac{1000}{2} \: and \: \frac{ - 1200}{2}$



$500 \: and \: - 600$


Since speed can't be negative, so x=500


So, original speed of aeroplane=500 km/hr