A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.

Given:

A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed.

To do:

We have to find the usual speed of the plane.

 Solution:

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

This implies,

New speed of the plane$=x+400$ km/hr

Time taken by the plane to travel 1600 km at usual speed$=\frac{1600}{x}$ hours

Time taken by the plane to travel 1600 km at new speed$=\frac{1600}{x+400}$ hours

$40$ minutes in hours$=\frac{40}{60}$ hours.    (since 1 hour = 60 minutes)

According to the question,

$\frac{1600}{x}-\frac{1600}{x+400}=\frac{40}{60}$

$\frac{1600(x+400)-1600(x)}{(x)(x+400)}=\frac{20\times2}{20\times3}$

$\frac{1600(x+400-x)}{x^2+400x}=\frac{2}{3}$

$3(1600)(400)=2(x^2+400x)$   (On cross multiplication)

$4800(200)=x^2+400x$

$x^2+400x-960000=0$

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

$x^2+1200x-800x-960000=0$

$x(x+1200)-800(x+1200)=0$

$(x+1200)(x-800)=0$

$x+1200=0$ or $x-800=0$

$x=-1200$ or $x=800$

Speed cannot be negative. Therefore, the value of $x$ is $800$ km/hr.

The usual speed of the plane is $800$ km/hr. 

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Updated on: 10-Oct-2022

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