Question

A jet plane covers 4500 km in some time. If the regular speed is decreased by 150 km/hr it take one and a half hour more to complete the journey. find original speed of jet plane.​

Answer 1

Answer: 750 km/hr.

Step by step explanation:

We've to find the original speed of the jet plane.

Let's assume the original speed to be "x" km/hr.

Distance covered: 4500 km

Given that: The speed decreases by 150km/hr.

Thus, new speed = (x - 150) km/hr.

Thus, time taken = (4500/x) hrs ( with original speed )

Time taken ( with decreased speed ) = (4500/x - 150) km/hr.

As given, To complete the journey, it takes 3/2 hrs (1.5 hrs)

Thus, ATQ

[(4500/x - 150) - (4500/x)] = 3/2

=> [4500 (x - x + 150) / x(x - 150) ] = 3/2

=> [4500 (150) × 2/3] = x(x - 150)

=> 1500 × 150 × 2 = x² - 150x

=> 450000 = x² - 150x

=> x² - 150x - 450000 = 0

=> x² - 750x + 600x - 450000 = 0

=> x(x - 750) + 600(x - 750) = 0

=> (x + 600) (x - 750) = 0

=> x = -600 or x = 750

We know that, Speed cannot be negative.

So, x = 750 km/hr.

So, the original speed of the jet is 750 km/hr.

Answer 2

Given :

• A jet plane covers 4500 km in some time.
• If the speed is decreased by 150 km/hr it takes one and half hour more to complete the journey.

To Find :

• The original speed of the jet plane.

Solution :

Let the original speed of the jet plane be x km/hr.

Case 1 :

Distance covered is 4500 km.

Speed = x km/hr.

Equation :

We know,

$\large{\boxed{\sf{Time\:=\:\dfrac{Distance}{Speed}}}}$

Block in the data,

$\sf{\longrightarrow{Time\:taken\:=\:\dfrac{4500}{x}\:hrs\:\:\:\:(1)}}$

Case 2 :

If the original speed is reduced by 150 km/hr the time taken increases by 1 and a half hour.

Speed = (x - 150) km/hr.

Equation :

Using the same formula for time,

$\sf{\longrightarrow{Time\:=\:\dfrac{4500}{x-150}\:hrs\:\:\:(2)}}$

From the two given conditions,

$\sf{\longrightarrow{\dfrac{4500}{x-150}-\dfrac{4500}{x}=1\:+\:\dfrac{1}{2}}}$

$\sf{\longrightarrow{\dfrac{4500}{x-150}-\dfrac{4500}{x}=\dfrac{2+1}{2}}}$

$\sf{\longrightarrow{4500\Big(\dfrac{1}{x-50}-\dfrac{1}{x}\Big)\:=\:\dfrac{3}{2}}}$

$\sf{\longrightarrow{4500\Big(\dfrac{x-x+150}{x(x-150)}=\dfrac{3}{2}}}$

$\sf{\longrightarrow{4500\Big(\dfrac{150}{x^2-150x}\Big)=\dfrac{3}{2}}}$

$\sf{\longrightarrow{\dfrac{675000}{x^2-3x}=\dfrac{3}{2}}}$

$\sf{\longrightarrow{675000\:\times\:2\:=\:3\:(x^2-150x)}}$

$\sf{\longrightarrow{1350000=3x^2-450x}}$

$\sf{\longrightarrow{450000=x^2-150x}}$

$\bold{\big[Dividing\:throughout\:by\:3\:\big]}$

$\sf{\longrightarrow{x^2-150x=450000}}$

$\sf{\longrightarrow{x^2-150x-450000=0}}$

Now, we can find value of x using factorization method.

$\sf{\longrightarrow{x^2-750x+600x-450000=0}}$

$\sf{\longrightarrow{x(x-750)+600(x-750)=0}}$

$\sf{\longrightarrow{(x-750)(x+600)=0}}$

$\sf{\longrightarrow{x-750=0\:\:or\:\:x+600=0}}$

$\sf{\longrightarrow{x=750\:\:\:or\:\:\:x=-600}}$

Since speed cannot be negative.

•°• x = - 600 is neglected.

$\large{\boxed{\bold{The\:original\:speed\:of\:jet\:plane\:=\:x\:=\:750\:km\:/h}}}$