An aeroplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.

Given:

An aeroplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed.

To do:

We have to find the usual speed of the aeroplane.

 Solution:

Let the usual 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 1200 km at usual speed$=\frac{1200}{x}$ hours

Time taken by the plane to travel 1200 km at new speed$=\frac{1200}{x+100}$ hours

According to the question,

$\frac{1200}{x}-\frac{1200}{x+100}=1$

$\frac{1200(x+100)-1200(x)}{(x)(x+100)}=1$

$\frac{1200(x+100-x)}{x^2+100x}=1$

$(1200)(100)=1(x^2+100x)$   (On cross multiplication)

$120000=x^2+100x$

$x^2+100x-120000=0$

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

$x^2+400x-300x-120000=0$

$x(x+400)-300(x+400)=0$

$(x+400)(x-300)=0$

$x+400=0$ or $x-300=0$

$x=-400$ or $x=300$

Speed cannot be negative. Therefore, the value of $x$ is $300$ km/hr.

The usual speed of the aeroplane is $300$ km/hr. 

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Updated on: 10-Oct-2022

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