A plane left 30 min late than its scheduled time and in order to reach the destination 1500 km away in time it had to increase its speed by 100 km hr from its usual sped find its usual speed
Answers
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
Let the usual time that the plane takes be x hr.
Distance = 1500 km
Time = x hr
usual Speed = 1500 / x ........................eqn (1)
to increase the speed we know that the time will decrease so
Time = x hr - 30 min
now we 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 ) ................................eqn (2)
now we have to find the value of the x
it is given that
Original speed has to be increased by 100 km/hr
So
Final speed - usual speed = 100 km / hr
From eqns (1) and (2)
1500 / ( x - 0.5 ) - 1500 / x = 100
= > 1500 [ 1 / ( x - 0.5 ) - 1 / x ] = 100
= > 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
=> 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
now
x - 3 = 0
= > x = 3
Or
x + 2.5 = 0
= > x = -2.5
now we know that time is a scalar quantity and it cannot be negative so the time is of the first equation
Time = 3 hr
hence
usual speed = distance / time
= 1500 km / 3 hr
= 500 km /hr
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