A plane left 30 minutes late than its scheduled time and in order to reach the destination $1500\ km$ away in time, it had to increase its speed by $100\ km/h$ from the usual speed. Find its usual speed.

Given: Time delayed by the plane to leave$=30\ minutes$, Distance traveled by the plan$=1500\ km$, increment of speed$=100\ km/h$.

To do: To find its usual speed.

Solution:

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

Time taken to cover $1500\ km$ with usual speed $x\ km/hr =\frac{1500}{x}$

Time taken to cover $1500\ km$ with speed of  $(x+100) km/hr=\frac{1500}{x+100}$

According to the given condition,

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

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

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

$\Rightarrow x^{2}+100x=300000$

$\Rightarrow x^{2}+100x-300000=0$

$\Rightarrow x^{2}+600x-500x-300000=0$

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

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

$\Rightarrow x=-600\ or\ x=500$

$\because$ Speed can't be negative .

$\therefore$ Usual speed of the plane is $500\ km/hr$.
 

Updated on: 10-Oct-2022

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