An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Given:

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 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$=x+250$ km/hr

Time taken by the plane to travel 1250 km at usual speed$=\frac{1250}{x}$ hours

Time taken by the plane to travel 1250 km at new speed$=\frac{1250}{x+250}$ hours

$50$ minutes in hours$=\frac{50}{60}$ hours.    (since 1 hour = 60 minutes)

According to the question,

$\frac{1250}{x}-\frac{1250}{x+250}=\frac{50}{60}$

$\frac{1250(x+250)-1250(x)}{(x)(x+250)}=\frac{10\times5}{10\times6}$

$\frac{1250(x+250-x)}{x^2+250x}=\frac{5}{6}$

$6(1250)(250)=5(x^2+250x)$   (On cross multiplication)

$7500(50)=x^2+250x$

$x^2+250x-375000=0$

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

$x^2+750x-500x-375000=0$

$x(x+750)-500(x+750)=0$

$(x+750)(x-500)=0$

$x+750=0$ or $x-500=0$

$x=-750$ or $x=500$

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

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

Tutorialspoint

Updated on: 10-Oct-2022

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