While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

Given:

The flight is delayed by 30 minutes and the destination is 1500 km away.

Increase in the speed of the flight$=100$ km/hr.

To do:

We have to find the original speed of the plane.

 Solution:

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

This implies,

New speed of the plane$=x+100$ km/hr

Time taken by the plane to travel 1500 km at original speed$=\frac{1500}{x}$ hours

Time taken by the plane to travel 1500 km at new speed$=\frac{1500}{x+100}$ hours

$30$ minutes in hours$=\frac{30}{60}$ hours.    (since 1 hour = 60 minutes)

According to the question,

$\frac{1500}{x}-\frac{1500}{x+100}=\frac{30}{60}$

$\frac{1500(x+100)-1500(x)}{(x)(x+100)}=\frac{30\times1}{30\times2}$

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

$2(1500)(100)=1(x^2+100x)$   (On cross multiplication)

$300000=x^2+100x$

$x^2+100x-300000=0$

Solving for $x$ by factorization method, we get,

$x^2+600x-500x-300000=0$

$x(x+600)-500(x+600)=0$

$(x+600)(x-500)=0$

$x+600=0$ or $x-500=0$

$x=-600$ or $x=500$

Speed cannot be negative. Therefore, the value of $x$ is $500$ km/hr.

The original speed of the plane is $500$ km/hr. 

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Updated on: 10-Oct-2022

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